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✓ 16 results · all verified · 13 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 3 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

The Heat Kernel and the Cauchy Problem

1 · Prerequisites

2 · Summary

This page defines the heat operator $\partial_{t} - Δ_{x}$ , its classical solutions and the Cauchy problem with data on the time-zero slice, and then constructs the Gaussian heat kernel $Γ (x, t) = (4 π t)^{- n / 2} e^{- ∣ x ∣^{2} / (4 t)}$ together with its causal extension. The kernel is normalised to unit mass, obeys the parabolic scaling law and the semigroup identity $Γ_{t} * Γ_{s} = Γ_{t + s}$ , satisfies the heat equation with explicit Gaussian derivative bounds, and forms an $L^{1}$ approximate identity as $t ↓ 0^{+}$ ; its causal extension is proved to be the fundamental solution of $\partial_{t} - Δ_{x}$ , with weak convergence to the Dirac mass at the origin. The heat evolution $H_{t}$ of initial data is then defined on $L^{p}$ by convolution and developed: derivatives and the heat operator pass through the convolution for positive time, bounded uniformly continuous data and $L^{p}$ data ( $1 \leq p < \infty$ ) give classical solutions with the expected initial behaviour, the semigroup is the unique solution of the mild relation in $C ([0, T]; L^{p})$ , and the generator at zero is computed on compactly supported smooth data.

The contractive, order-preserving and mass-conserving properties of the flow are recorded, together with the $L^{p}$ to $L^{q}$ smoothing estimate with its exact Gaussian constant and the derivative estimates that follow from it by parabolic scaling. At positive time the flow of any $L^{p}$ datum is spatially real analytic with factorial derivative bounds, and nonnegative nonzero data propagate with infinite speed. A closing remark compares the diffusivity normalisation with the ball and half-space Poisson kernels. Countable Choice is assumed throughout because the convolution, Fubini–Tonelli, change of variables and approximate-identity interfaces carry it, and each item states its assumption explicitly; the spatial analyticity proof records the exact choice cost of its $L^{p}$ supplier steps.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

The heat operator, the heat equation, and the Cauchy problem

Definition

Let $n \geq 1$ and let $Ω \subseteq R^{n + 1}$ be open in the space-time variable $(x, t) \in R^{n} \times R$ ; points of $Ω$ are thus written with a spatial slot $x \in R^{n}$ and a time slot $t \in R$ . Partial derivatives are those of Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ , multi-indices and the classes $C^{k}$ are those of $C^{k}$ maps and multi-index derivative notation in Euclidean space, and $Δ_{x}$ denotes the Laplacian of The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field applied in the spatial variables with $t$ held fixed. The vocabulary of differential operators, their order, and of classical solutions is that of Scalar partial differential equations, order, and classical solutions.

The heat operator is $\partial_{t} - Δ_{x}$ , a linear second-order operator in the sense of Linear, semilinear, quasilinear, and fully nonlinear partial differential equations. A classical solution of the heat equation on $Ω$ is a function $u \in C^{2} (Ω)$ with

$\partial_{t} u - Δ_{x} u = 0 on Ω,$

and the inhomogeneous heat equation is the equation $\partial_{t} u - Δ_{x} u = f$ for a prescribed source $f$ . A complex-valued $u$ is a classical solution when its real and imaginary parts are. Since the spatial quadratic form of $\partial_{t} - Δ_{x}$ is positive definite while the time direction enters only through a first derivative, the operator is parabolic at every point in the classification of Elliptic, hyperbolic, and parabolic principal symbols.

A Cauchy problem for the heat equation on a space-time domain consists of the equation together with initial data $u (\cdot, 0) = u_{0}$ prescribed on the time-zero slice; values prescribed on the lateral (spatial) boundary of the domain are boundary data. Initial and spatial boundary data are different sets of constraints and are not interchanged.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

The heat kernel on $R^{n}$ and its causal extension

Definition

Let $n \geq 1$ and $t > 0$ . The heat kernel on $R^{n}$ is the function

$Γ (x, t) : = (4 π t)^{- n / 2} \exp ⁣ (- \frac{∣ x ∣^{2}}{4 t}), x \in R^{n},$

where $∣ x ∣^{2} = ⟨ x, x ⟩$ is the Euclidean square of The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ and $\exp$ is the real exponential function of The real exponential function and the number $e$ by a power series. For fixed $t > 0$ the map $x \mapsto Γ (x, t)$ is a composite of the quadratic form with the exponential and a scalar multiple, so it is $C^{\infty}$ in the sense of $C^{k}$ maps and multi-index derivative notation in Euclidean space by The exponential function is smooth and $(\exp)^{'} = \exp$ and $C^{k}$ Euclidean maps are closed under componentwise algebra and composition, and it is strictly positive by The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ . The spatial Laplacian entering the heat equation is that of The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field.

The causal extension of the heat kernel is $Γ (x, t)$ for $t > 0$ and $Γ (x, t) : = 0$ for $t \leq 0$ ; the displayed formula is not evaluated at $t = 0$ as a function value. The normalisation is fixed by the unit-mass identity proved for the kernel on this page, and the causal extension is used only as a locally integrable function or, after embedding, as a distribution.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Normalisation, parabolic scaling, heat equation and derivative bounds for the heat kernel

Statement

Assume Countable Choice and let $n \geq 1$ . For every $t > 0$ : (i) $Γ (\cdot, t) > 0$ and $\int_{R^{n}} Γ (x, t) d x = 1$ ; (ii) parabolic scaling: $Γ (λ x, λ^{2} t) = λ^{- n} Γ (x, t)$ for every $λ > 0$ and $x \in R^{n}$ ; (iii) $Γ$ is $C^{\infty}$ on $R^{n} \times (0, \infty)$ and solves the heat equation there, $\partial_{t} Γ (x, t) = Δ_{x} Γ (x, t)$ ; (iv) for every multi-index $α$ there is $C_{n, α} < \infty$ with $∣ D_{x}^{α} Γ (x, t) ∣ \leq C_{n, α} t^{- (∣ α ∣ + n) / 2} e^{- ∣ x ∣^{2} / (8 t)}$ for all $x \in R^{n}$ , $t > 0$ ; in particular $∥ Γ (\cdot, t) ∥_{1} = 1$ and $(Γ (\cdot, t))_{t > 0}$ is an $L^{1}$ approximate identity on $R^{n}$ .

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , together with $t, λ > 0$ , a multi-index $α$ and $δ > 0$ wherever these appear.

[A1]

Countable Choice is the hypothesis carried by the measure-theoretic and change-of-variables suppliers below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

For $n \geq 1$ and $t > 0$ the heat kernel is $Γ (x, t) = (4 π t)^{- n / 2} \exp (- ∣ x ∣^{2} / (4 t))$ , it is strictly positive, and its causal extension vanishes for $t \leq 0$ (The heat kernel on $R^{n}$ and its causal extension).

[F2]

$\int_{- \infty}^{\infty} e^{- x^{2}} d x = \sqrt{π}$ (The Gaussian integral $\int_{- \infty}^{\infty} e^{- x^{2}} d x = \sqrt{π}$ ).

[F3]

Under $R^{m + n} = R^{m} \times R^{n}$ , the Lebesgue measure $λ_{m + n}$ is the completion of the product measure $λ_{m} \times λ_{n}$ (The Euclidean Lebesgue measure is the completion of the product of the factor Lebesgue measures).

[F4]

On completed sigma-finite product measure spaces, a nonnegative completed-product-measurable $f$ has measurable sections outside measurable null sets. Set the inner integrals to zero on those exceptional sets; the resulting measurable functions have integrals equal to $\int f d \overline{μ \times ν}$ (Tonelli and Fubini for the completed product, with only almost-everywhere section measurability). For the continuous Euclidean Gaussian integrands used here, every section is measurable, so the ordinary iterated integrals give the same value.

[F5]

An invertible linear $T : R^{n} \to R^{n}$ carries Lebesgue measurable sets to Lebesgue measurable sets and satisfies $λ_{n} (T [E]) = ∣ \det A ∣ λ_{n} (E)$ for every Lebesgue measurable $E$ (A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not).

[F6]

For a $C^{1}$ diffeomorphism $T : U \to V$ of open sets and every nonnegative Lebesgue measurable $f : V \to [0, \infty]$ , $\int_{V} f (y) d y = \int_{U} f (T (x)) ∣ \det D T (x) ∣ d x$ (A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions).

[F7]

If $f_{n} \to f$ almost everywhere and $∣ f_{n} ∣ \leq g$ almost everywhere for a single integrable nonnegative $g$ , then $\int f_{n} \to \int f$ (Dominated convergence).

[F8]

For every $m \in N$ and real $a > 0$ , $x^{m} / \exp (a x) \to 0$ as $x \to + \infty$ (The exponential dominates every fixed nonnegative integer power at $+ \infty$ ).

[F9]

If $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ (The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ ).

[F10]

Derivatives of sums, scalar multiples, products and quotients obey the sum, scalar, product and quotient rules (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

[F11]

Finite componentwise sums, products and scalar multiples of $C^{k}$ Euclidean maps are $C^{k}$ , and composites of composable $C^{k}$ Euclidean maps are $C^{k}$ , for every $k \in N$ ( $C^{k}$ Euclidean maps are closed under componentwise algebra and composition).

[F12]

The real exponential function is $C^{\infty}$ and $\exp^{(m)} = \exp$ for every $m \in N$ (The exponential function is smooth and $(\exp)^{'} = \exp$ ).

[F13]

An $L^{1}$ approximate identity on $R^{n}$ is a family $(K_{ε})_{ε > 0} \subseteq L^{1} (R^{n})$ with $\int K_{ε} = 1$ , with $∥ K_{ε} ∥_{1}$ bounded independently of $ε$ , and with $\int_{∣ x ∣ > δ} ∣ K_{ε} ∣ \to 0$ as $ε \to 0^{+}$ for every $δ > 0$ (An $L^{1}$ approximate identity on $R^{n}$ ).

Proof

technique · direct

1.1A1F1F2F3F4F5F6givenalgebra

Work under [A1] and fix $t > 0$ . By [F1], $Γ (x, t) = (4 π t)^{- n / 2} \exp (- ∣ x ∣^{2} / (4 t))$ . Let $T (u) = (4 t)^{1 / 2} u$ , an invertible linear self-map of $R^{n}$ with $\det D T (u) = (4 t)^{n / 2} > 0$ in the sense of [F5], and put $φ (x) = \exp (- ∣ x ∣^{2} / (4 t)) \geq 0$ ; since $T$ is a $C^{1}$ diffeomorphism and $φ$ is nonnegative and measurable, [F6] applied with $U = V = R^{n}$ gives $\int_{R^{n}} Γ (x, t) d x = (4 π t)^{- n / 2} \int_{R^{n}} \exp (- ∣ u ∣^{2}) (4 t)^{n / 2} d u = π^{- n / 2} \int_{R^{n}} e^{- ∣ u ∣^{2}} d u$ . By [F3] the Lebesgue integral over $R^{n}$ is the completed product integral, so Tonelli's theorem [F4] factorises $\int_{R^{n}} e^{- ∣ u ∣^{2}} d u = \prod_{i < n} \int_{R} e^{- u_{i}^{2}} d u_{i} = (\sqrt{π})^{n}$ by the one-dimensional Gaussian integral [F2]; hence $\int_{R^{n}} Γ (x, t) d x = 1$ , absolutely and as a nonnegative integral, and with the strict positivity recorded in [F1] this proves (i).

1.2F1givenalgebra

Scaling: for $λ > 0$ , [F1] gives $Γ (λ x, λ^{2} t) = (4 π λ^{2} t)^{- n / 2} \exp (- ∣ λ x ∣^{2} / (4 λ^{2} t)) = λ^{- n} (4 π t)^{- n / 2} \exp (- ∣ x ∣^{2} / (4 t)) = λ^{- n} Γ (x, t)$ , using $∣ λ x ∣^{2} = λ^{2} ∣ x ∣^{2}$ and $(4 π λ^{2} t)^{- n / 2} = λ^{- n} (4 π t)^{- n / 2}$ , which is (ii).

1.3F1F11F12givenalgebra

Smoothness: the map $(x, t) \mapsto - ∣ x ∣^{2} / (4 t)$ is a quotient of polynomials defined and smooth on the open set $R^{n} \times (0, \infty)$ , the map $\exp$ is $C^{\infty}$ by [F12], and $t \mapsto (4 π t)^{- n / 2}$ is a nonzero scalar multiple of $t^{- n / 2}$ , smooth for $t > 0$ ; closure under products, scalar multiples and composition [F11] makes $(x, t) \mapsto Γ (x, t)$ $C^{\infty}$ on $R^{n} \times (0, \infty)$ .

1.4F1F8F12givenalgebra

Derivative bound at $t = 1$ : by induction on $∣ α ∣$ we show that $D^{α} Γ (x, 1) = P_{α} (x) e^{- ∣ x ∣^{2} / 4}$ for a polynomial $P_{α}$ with $K_{α} : = \sup_{x \in R^{n}} ∣ P_{α} (x) ∣ e^{- ∣ x ∣^{2} / 8} < \infty$ . For $α = 0$ , [F1] gives $P_{0} = (4 π)^{- n / 2}$ , and $K_{0} < \infty$ because $e^{- ∣ x ∣^{2} / 8}$ is bounded and tends to $0$ at infinity. If the claim holds for $α$ , then differentiating once more in a coordinate multiplies by a linear polynomial (the derivative of $P_{α}$ plus $- (x_{i} / 2) P_{α}$ ) and keeps the Gaussian factor, so the polynomial form is preserved; for the finiteness, each monomial $x^{β}$ of $∣ P_{α} ∣$ satisfies $∣ x^{β} ∣ e^{- ∣ x ∣^{2} / 8} \leq ∣ x ∣^{∣ β ∣} e^{- ∣ x ∣^{2} / 8} \to 0$ as $∣ x ∣ \to \infty$ by [F8] applied to the radial variable $∣ x ∣$ , and a continuous function on $R^{n}$ that tends to $0$ at infinity is bounded, so the supremum is finite. Hence $∣ D^{α} Γ (x, 1) ∣ \leq K_{α} e^{- ∣ x ∣^{2} / 8}$ for every $x$ .

2.1step 1.3F1F9F10F12givenalgebra

Derivatives: differentiating the formula of [F1] in the coordinate $x_{i}$ with the one-variable chain and product rules [F9, F10] and $\exp^{'} = \exp$ [F12] gives $\partial_{x_{i}} Γ = - (x_{i} / (2 t)) Γ$ and, differentiating once more, $\partial_{x_{i}} \partial_{x_{i}} Γ = - (1 / (2 t)) Γ + (x_{i}^{2} / (4 t^{2})) Γ$ ; differentiating in $t$ gives $\partial_{t} Γ = (- n / (2 t) + ∣ x ∣^{2} / (4 t^{2})) Γ$ . Summing the spatial identities over $i$ yields $Δ_{x} Γ = \sum_{i < n} \partial_{x_{i}} \partial_{x_{i}} Γ = (- n / (2 t) + ∣ x ∣^{2} / (4 t^{2})) Γ = \partial_{t} Γ$ on $R^{n} \times (0, \infty)$ , which together with step 1.3 is (iii).

2.2step 1.2step 1.4F9givenalgebra

Derivative bound at general $t$ : by step 1.2 applied with $λ = \sqrt{t}$ and with $x$ replaced by $x / \sqrt{t}$ , $Γ (x, t) = t^{- n / 2} Γ (x / \sqrt{t}, 1)$ for every $x$ and $t > 0$ ; differentiating this identity $α$ times in $x$ , the chain rule [F9] contributes one factor $t^{- 1 / 2}$ for each spatial derivative, so $D^{α} Γ (x, t) = t^{- (n + ∣ α ∣) / 2} (D^{α} Γ) (x / \sqrt{t}, 1)$ , and step 1.4 yields $∣ D^{α} Γ (x, t) ∣ \leq K_{α} t^{- (n + ∣ α ∣) / 2} e^{- ∣ x ∣^{2} / (8 t)}$ , which is (iv) with $C_{n, α} = K_{α}$ .

2.3step 1.1step 1.2F1F6F7F13given

Tail estimate: by step 1.2, $Γ (x, t) = t^{- n / 2} Γ (x / \sqrt{t}, 1)$ ; applying the diffeomorphism substitution [F6] to $x = \sqrt{t} z$ gives $\int_{∣ x ∣ > δ} Γ (x, t) d x = \int_{∣ z ∣ > δ / \sqrt{t}} Γ (z, 1) d z$ for every $δ > 0$ . As $t ↓ 0^{+}$ the integrands $1_{{∣ z ∣ > δ / \sqrt{t}}} Γ (z, 1)$ are dominated by the fixed integrable function $Γ (\cdot, 1)$ from step 1.1 and converge pointwise to $0$ at every $z$ , including $z = 0$ , so dominated convergence [F7] gives $\int_{∣ z ∣ > δ / \sqrt{t}} Γ (z, 1) d z \to 0$ ; with unit mass and positivity from step 1.1 and [F1], the three defining clauses of [F13] hold for the family $K_{ε} : = Γ (\cdot, ε)$ , so $(Γ (\cdot, t))_{t > 0}$ is an $L^{1}$ approximate identity.

3.1step 1.1step 1.2step 1.3step 2.1step 1.4step 2.2step 2.3F13∎

Steps 1.1, 1.2, 1.3, 2.1, 1.4, 2.2 and 2.3 prove (i) unit mass and positivity, (ii) parabolic scaling, (iii) smoothness and the heat equation, (iv) the derivative bounds with finite constants $C_{n, α} = K_{α}$ , and the unit $L^{1}$ norm together with the approximate-identity property of [F13]; this is the whole statement.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passOpen item page →

First and second Gaussian heat-kernel moments

Statement

Assume Countable Choice. For $n \geq 1$ and $t > 0$ , all first and second moments are absolutely integrable and $\int_{R^{n}} x_{i} Γ (x, t) d x = 0, \int_{R^{n}} x_{i} x_{j} Γ (x, t) d x = 2 t δ_{i j} .$ In particular $\int_{R^{n}} ∣ x ∣^{2} Γ (x, t) d x = 2 n t$ .

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $t > 0$ , and coordinate indices $0 \leq i, j < n$ wherever they appear.

[A1]

Countable Choice is the hypothesis carried by the integration and change-of-variables suppliers below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

For $t > 0$ the heat kernel is $Γ (x, t) = (4 π t)^{- n / 2} \exp (- ∣ x ∣^{2} / (4 t)) > 0$ on $R^{n}$ (The heat kernel on $R^{n}$ and its causal extension).

[F2]

$\int_{- \infty}^{\infty} e^{- u^{2}} d u = \sqrt{π}$ (The Gaussian integral $\int_{- \infty}^{\infty} e^{- x^{2}} d x = \sqrt{π}$ ).

[F3]

Under $R^{m + n} = R^{m} \times R^{n}$ the Lebesgue measure $λ_{m + n}$ is the completion of the product measure $λ_{m} \times λ_{n}$ (The Euclidean Lebesgue measure is the completion of the product of the factor Lebesgue measures).

[F4]

On completed sigma-finite product measure spaces, Tonelli applies to nonnegative completed-product-measurable functions and Fubini to $L^{1}$ functions. Sections are measurable (and in the Fubini case integrable) outside measurable null sets; define their inner integrals to be zero on those exceptional sets before taking the outer integral (Tonelli and Fubini for the completed product, with only almost-everywhere section measurability). The Gaussian products and moment integrands used here have measurable sections everywhere; their one-dimensional factors are integrable, so their ordinary iterated integrals agree with these modified integrals.

[F5]

For a $C^{1}$ diffeomorphism $T : U \to V$ of open sets and every $f \in L^{1} (V)$ , $\int_{V} f (y) d y = \int_{U} f (T (x)) ∣ \det D T (x) ∣ d x$ ; in particular for $n = 1$ both $u \mapsto - u$ and $u \mapsto c u$ with $c > 0$ qualify (A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions, A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not).

[F6]

For every $m \in N$ and real $a > 0$ , $s^{m} / \exp (a s) \to 0$ as $s \to + \infty$ (The exponential dominates every fixed nonnegative integer power at $+ \infty$ ).

[F7]

If $F, G$ are continuous on $[a, b]$ and differentiable on $(a, b)$ with $F^{'} = f$ , $G^{'} = g$ Riemann integrable there, then $\int_{a}^{b} F g + \int_{a}^{b} f G = F (b) G (b) - F (a) G (a)$ (Integration by parts for continuous factors with Riemann-integrable extensions of their interior derivatives).

[F8]

If $T$ preserves a measure $μ$ and $f$ is integrable, then $\int f \circ T d μ = \int f d μ$ (Integral invariance under measure-preserving maps); the reflection $s \mapsto - s$ preserves Lebesgue measure on $R$ .

Proof

technique · direct

1.1A1F1F2F3F4F5F6givenalgebra

Work under [A1] and fix $t > 0$ . For $m \in {1, 2}$ set $K_{m} (t) : = \sup_{s \in R} ∣ s ∣^{m} e^{- s^{2} / (8 t)} < \infty$ : the function is continuous and tends to $0$ at infinity by [F6] applied to the radial variable, so the supremum is finite; consequently, by [F1], $∣ x_{i} ∣^{m} Γ (x, t) \leq (4 π t)^{- n / 2} K_{m} (t) \prod_{k < n} e^{- x_{k}^{2} / (8 t)}$ . By the identification [F3] and Tonelli's theorem [F4], $\int_{R^{n}} \prod_{k} e^{- x_{k}^{2} / (8 t)} d x = \prod_{k} \int_{R} e^{- x_{k}^{2} / (8 t)} d x_{k} = (\sqrt{8 π t})^{n} < \infty$ , where each one-dimensional factor is computed by the substitution $x_{k} = 2 \sqrt{2 t} u$ of [F5] and the Gaussian integral [F2]; hence $∣ x_{i} ∣^{m} Γ (\cdot, t) \in L^{1} (R^{n})$ for $m = 1, 2$ and every $i$ , and $∣ x_{i} x_{j} ∣ \leq (x_{i}^{2} + x_{j}^{2}) / 2$ gives absolute integrability of the mixed moments as well, so Fubini's clause of [F4] applies to them.

2.1F4F5F8step 1.1givenalgebra

First moments: by Fubini's theorem [F4] applied to the integrable function $x \mapsto x_{i} Γ (x, t)$ , the integral is the iterated integral in which the $i$ -th factor is $\int_{R} s e^{- s^{2} / (4 t)} d s$ ; the function $s \mapsto s e^{- s^{2} / (4 t)}$ is odd and integrable, so by the change of variables $s = - u$ of [F5] its integral equals its own negative and is therefore $0$ , while all other factors are finite by step 1.1; hence $\int x_{i} Γ (x, t) d x = 0$ .

2.2F4F5F8step 1.1givenalgebra

Off-diagonal second moments: for $i \neq j$ , Fubini [F4] applied to the integrable $x \mapsto x_{i} x_{j} Γ (x, t)$ factors the integral into the product of the one-dimensional integrals $\int_{R} s e^{- s^{2} / (4 t)} d s$ in the $i$ -th and $j$ -th coordinates and the finite Gaussian factors in the remaining coordinates; each of the two odd factors vanishes by the change of variables $s = - u$ of [F5], so $\int x_{i} x_{j} Γ (x, t) d x = 0$ for $i \neq j$ .

2.3step 1.1F2F4F5F6F7givenalgebra

Diagonal second moments: fix $i$ and put $F (s) = s$ , $G (s) = e^{- s^{2} / (4 t)}$ on $[- R, R]$ ; integration by parts [F7] gives $\int_{- R}^{R} s^{2} e^{- s^{2} / (4 t)} d s = 2 t \int_{- R}^{R} e^{- s^{2} / (4 t)} d s - 4 t R e^{- R^{2} / (4 t)}$ . Letting $R \to \infty$ , the boundary term tends to $0$ by [F6] and the remaining integral equals $\sqrt{4 π t}$ by the substitution $s = 2 \sqrt{t} u$ of [F5] and [F2], so $\int_{R} s^{2} e^{- s^{2} / (4 t)} d s = 2 t \sqrt{4 π t}$ ; Fubini [F4] applied to the integrable function $x \mapsto x_{i}^{2} Γ (x, t)$ now gives $\int_{R^{n}} x_{i}^{2} Γ (x, t) d x = (4 π t)^{- n / 2} \cdot 2 t \sqrt{4 π t} \cdot (\sqrt{4 π t})^{n - 1} = 2 t$ .

3.1step 1.1step 2.1step 2.2step 2.3algebra∎

Steps 1.1, 2.1, 2.2 and 2.3 give absolute integrability, vanishing first moments, the covariance identity $\int x_{i} x_{j} Γ = 2 t δ_{i j}$ for every pair $i, j$ , and, summing the $n$ diagonal identities by linearity of the integral, $\int ∣ x ∣^{2} Γ (x, t) d x = \sum_{i < n} 2 t = 2 n t$ .

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

Gaussian kernels form an approximate identity

Statement

Assume Countable Choice. For $n \geq 1$ , the kernels $Γ_{t}$ are positive with unit mass and $∥ Γ_{t} ∥_{1} = 1$ . For every $δ > 0$ , $\int_{∣ x ∣ > δ} Γ (x, t) d x \to 0$ as $t ↓ 0$ . Thus they form an $L^{1}$ approximate identity.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , and $δ > 0$ wherever it appears.

[A1]

Countable Choice is the hypothesis carried by the cited integration interface (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

For every $t > 0$ the kernel satisfies $Γ (\cdot, t) > 0$ , $\int_{R^{n}} Γ (x, t) d x = 1$ , and the parabolic scaling identity $Γ (λ x, λ^{2} t) = λ^{- n} Γ (x, t)$ for every $λ > 0$ and $x \in R^{n}$ (Normalisation, parabolic scaling, heat equation and derivative bounds for the heat kernel).

[F2]

[F3]

If $f_{n} \to f$ almost everywhere and $∣ f_{n} ∣ \leq g$ almost everywhere for a single integrable nonnegative $g$ , then $\int f_{n} \to \int f$ (Dominated convergence).

[F4]

For a $C^{1}$ diffeomorphism $T : U \to V$ of open sets and every nonnegative Lebesgue measurable $f : V \to [0, \infty]$ , $\int_{V} f (y) d y = \int_{U} f (T (x)) ∣ \det D T (x) ∣ d x$ ; the scaling $x = \sqrt{t} z$ is such a diffeomorphism with $∣ \det D T ∣ = t^{n / 2}$ (A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions, A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not).

Proof

technique · direct

1.1A1F1given

Work under [A1] and fix $t > 0$ . By [F1] the kernel is strictly positive with $\int_{R^{n}} Γ (x, t) d x = 1$ , so $Γ_{t} \in L^{1} (R^{n})$ , $Γ_{t} > 0$ and $∥ Γ_{t} ∥_{1} = 1$ .

2.1step 1.1F1F3F4given

Tail estimate: by the scaling clause of [F1] with $λ = \sqrt{t}$ and with $x$ replaced by $x / \sqrt{t}$ , $Γ (x, t) = t^{- n / 2} Γ (x / \sqrt{t}, 1)$ ; the diffeomorphism substitution $x = \sqrt{t} z$ of [F4] therefore gives $\int_{∣ x ∣ > δ} Γ (x, t) d x = \int_{∣ z ∣ > δ / \sqrt{t}} Γ (z, 1) d z$ for every $δ > 0$ . As $t ↓ 0^{+}$ the integrands $1_{{∣ z ∣ > δ / \sqrt{t}}} Γ (z, 1)$ are dominated by the fixed integrable function $Γ (\cdot, 1)$ from step 1.1 and converge at every $z \neq 0$ to $0$ , so dominated convergence [F3] gives $\int_{∣ z ∣ > δ / \sqrt{t}} Γ (z, 1) d z \to 0$ .

3.1step 1.1step 2.1F2given∎

Steps 1.1 and 2.1 verify the three clauses of [F2] for the family $K_{ε} : = Γ (\cdot, ε)$ : unit integral, the uniform $L^{1}$ bound $∥ K_{ε} ∥_{1} = 1$ , and the vanishing of the tails; hence $(Γ (\cdot, t))_{t > 0}$ is an $L^{1}$ approximate identity.

LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passOpen item page →

The heat kernel semigroup identity $Γ_{t} * Γ_{s} = Γ_{t + s}$

Statement

Assume Countable Choice and let $n \geq 1$ . For all $s, t > 0$ the convolution $Γ (\cdot, t) * Γ (\cdot, s)$ converges absolutely at every $x \in R^{n}$ and equals $Γ (x, t + s)$ ; that is, $Γ_{t} * Γ_{s} = Γ_{t + s}$ as functions on $R^{n}$ .

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $s, t > 0$ , and $x \in R^{n}$ .

[A1]

Countable Choice is the hypothesis carried by the integration and change-of-variables suppliers below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

For $t > 0$ the heat kernel is $Γ (y, t) = (4 π t)^{- n / 2} \exp (- ∣ y ∣^{2} / (4 t))$ , positive and integrable (The heat kernel on $R^{n}$ and its causal extension); the convolution $f * g$ of Convolution of two functions on $R^{n}$ is defined at $x$ when $y \mapsto f (x - y) g (y)$ is measurable and integrable.

[F2]

For real $u, v$ the exponential satisfies $\exp (u) \exp (v) = \exp (u + v)$ (The exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$ ).

[F3]

$\int_{- \infty}^{\infty} e^{- u^{2}} d u = \sqrt{π}$ (The Gaussian integral $\int_{- \infty}^{\infty} e^{- x^{2}} d x = \sqrt{π}$ ).

[F4]

[F5]

[F6]

[F7]

For every $t > 0$ the kernel satisfies $Γ (λ z, λ^{2} t) = λ^{- n} Γ (z, t)$ for every $λ > 0$ and $z \in R^{n}$ (Normalisation, parabolic scaling, heat equation and derivative bounds for the heat kernel).

Proof

technique · direct

1.1A1F1F2F7given

Work under [A1] and fix $s, t > 0$ and $x \in R^{n}$ . By the product form [F1] and the addition formula [F2], the convolution integrand is $Γ (x - y, t) Γ (y, s) = (4 π t)^{- n / 2} (4 π s)^{- n / 2} \exp (- ∣ x - y ∣^{2} / (4 t) - ∣ y ∣^{2} / (4 s))$ , a measurable function of $y$ that is strictly positive everywhere.

1.2F3F4F5F6givenalgebra

Gaussian evaluation: for $A > 0$ , $\int_{R^{n}} e^{- A ∣ z ∣^{2}} d z = (π / A)^{n / 2}$ . Indeed, by the identification [F4] and Tonelli's theorem [F5] the integral factorises over the coordinates, each one-dimensional factor is $\int_{R} e^{- A u^{2}} d u = A^{- 1 / 2} \sqrt{π}$ by the substitution $u = A^{- 1 / 2} v$ of [F6] and the Gaussian integral [F3], and the product is $(π / A)^{n / 2}$ .

2.1step 1.1givenalgebra

Completing the square in the exponent: with $A : = (t + s) / (4 t s)$ and $b : = x s / (t + s)$ , the algebraic identity $- ∣ x - y ∣^{2} / (4 t) - ∣ y ∣^{2} / (4 s) = - A ∣ y - b ∣^{2} - ∣ x ∣^{2} / (4 (t + s))$ holds, since the quadratic terms give $- A ∣ y ∣^{2}$ , the linear terms give $2 A b \cdot y = x \cdot y / (2 t)$ , and the constant term gives $A ∣ b ∣^{2} - ∣ x ∣^{2} / (4 t) = - ∣ x ∣^{2} / (4 (t + s))$ .

3.1step 2.1step 1.2F1F6givenalgebra

Therefore the absolutely convergent (indeed nonnegative) integral defining the convolution equals $(4 π t)^{- n / 2} (4 π s)^{- n / 2} e^{- ∣ x ∣^{2} / (4 (t + s))} \int_{R^{n}} e^{- A ∣ y - b ∣^{2}} d y$ , where the last integral is $\int_{R^{n}} e^{- A ∣ z ∣^{2}} d z = (π / A)^{n / 2} = (4 π t s / (t + s))^{n / 2}$ by the translation case of [F6] and step 1.2; the prefactor simplifies to $(4 π t)^{- n / 2} (4 π s)^{- n / 2} (4 π t s / (t + s))^{n / 2} = (4 π (t + s))^{- n / 2}$ , so $Γ (x - y, t) Γ (y, s)$ is integrable in $y$ and $Γ_{t} * Γ_{s} (x) = (4 π (t + s))^{- n / 2} e^{- ∣ x ∣^{2} / (4 (t + s))} = Γ (x, t + s)$ by [F1].

4.1F1step 3.1given∎

Steps 1.1, 2.1, 1.2 and 3.1 show that for every fixed $x$ the convolution integral converges absolutely and equals $Γ (x, t + s)$ ; as $s, t > 0$ and $x$ were arbitrary, $Γ_{t} * Γ_{s} = Γ_{t + s}$ as functions on $R^{n}$ .

TheoremStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-6.1-sol)Open item page →

The causal heat kernel is the fundamental solution of the heat operator

Statement

Assume Countable Choice and let $n \geq 1$ , and let $Γ$ be the causal extension of the heat kernel. Then $Γ \in L_{l o c}^{1} (R^{n + 1})$ , and its regular distribution $u_{Γ} \in D^{'} (R^{n + 1})$ satisfies $(\partial_{t} - Δ) u_{Γ} = δ_{(0, 0)}$ , that is, $⟨ (\partial_{t} - Δ) u_{Γ}, φ ⟩ = φ (0, 0) for every φ \in C_{c}^{\infty} (R^{n + 1}),$ so the causal extension is a fundamental solution of $\partial_{t} - Δ$ . Moreover $Γ (\cdot, t) \to δ_{0}$ in $D^{'} (R^{n})$ as $t ↓ 0^{+}$ , that is, $\int_{R^{n}} Γ (x, t) φ (x) d x \to φ (0)$ for every $φ \in C_{c}^{\infty} (R^{n})$ . The derivative $\partial_{t}$ is the distributional derivative in the last space-time coordinate.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , a test function $φ \in C_{c}^{\infty} (R^{n + 1})$ , a real $R > 0$ with $supp φ$ contained in the open box $B_{R} \times (- R, R) \subseteq R^{n + 1}$ , and $0 < ε < R$ .

[A1]

Countable Choice is the hypothesis carried by the integration and embedding suppliers below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

The causal extension of the heat kernel is $Γ (x, t)$ for $t > 0$ and $Γ (x, t) = 0$ for $t \leq 0$ , with $Γ (x, t) = (4 π t)^{- n / 2} \exp (- ∣ x ∣^{2} / (4 t))$ positive and $C^{\infty}$ on $R^{n} \times (0, \infty)$ (The heat kernel on $R^{n}$ and its causal extension).

[F2]

For every $t > 0$ , $\int_{R^{n}} Γ (x, t) d x = 1$ , $\partial_{t} Γ = Δ_{x} Γ$ on $R^{n} \times (0, \infty)$ , and $(Γ (\cdot, t))_{t > 0}$ is an $L^{1}$ approximate identity on $R^{n}$ (Normalisation, parabolic scaling, heat equation and derivative bounds for the heat kernel).

[F3]

For $f \in L_{l o c}^{1} (Ω)$ the regular functional $⟨ u_{f}, φ ⟩ = \int_{Ω} f φ$ is well-defined and depends only on the almost-everywhere class of $f$ (Regular distribution from a locally integrable function).

[F4]

The distributional derivative is $⟨ \partial^{α} u, φ ⟩ = (- 1)^{∣ α ∣} ⟨ u, \partial^{α} φ ⟩$ (Distributional derivative); thus a first time derivative contributes a sign $- 1$ and a second spatial derivative a sign $+ 1$ .

[F5]

Assuming Countable Choice, $f \mapsto u_{f}$ is an injection from $L_{l o c}^{1} (Ω)$ modulo almost-everywhere equality into $D^{'} (Ω)$ , and local $L^{1}$ convergence implies strong distribution convergence (Locally integrable functions embed in distributions).

[F6]

The Dirac distribution satisfies $δ_{a} (φ) = φ (a)$ and $⟨ \partial^{α} δ_{a}, φ ⟩ = (- 1)^{∣ α ∣} \partial^{α} φ (a)$ (Dirac delta and its derivatives).

[F7]

[F8]

On completed sigma-finite product measure spaces, Tonelli's theorem holds for nonnegative measurable functions and Fubini's theorem for $L^{1}$ functions (Tonelli and Fubini for the completed product, with only almost-everywhere section measurability); under $R^{m + n} = R^{m} \times R^{n}$ , $λ_{m + n}$ is that completed product measure (The Euclidean Lebesgue measure is the completion of the product of the factor Lebesgue measures).

[F9]

If $(K_{ε})$ is an $L^{1}$ approximate identity and $f$ is bounded and continuous, then $(f * K_{ε}) (x) \to f (x)$ uniformly for $x$ in every compact set ( $L^{1}$ approximate identities converge uniformly on compacta for bounded continuous functions).

[F10]

For continuous $f$ on $[a, b]$ , differentiable on $(a, b)$ , there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ (The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ ).

Proof

technique · direct

1.1A1F1F2F3F5F8given

Local integrability: let $K \subseteq R^{n + 1}$ be compact and choose $R > 0$ with $K \subseteq B_{R} \times [- R, R]$ . By Tonelli's theorem [F8] over the completed product measure, $\int_{K} ∣ Γ ∣ \leq \int_{0}^{R} \int_{R^{n}} Γ (x, t) d x d t = \int_{0}^{R} 1 d t = R < \infty$ using unit mass from [F2] and $Γ = 0$ for $t \leq 0$ from [F1]; hence $Γ \in L_{l o c}^{1} (R^{n + 1})$ and its regular functional $u_{Γ}$ of [F3] is a distribution by [F5].

1.2F1F2F6F9givenalgebra

Dirac limit at time zero: fix a spatial test function $ψ \in C_{c}^{\infty} (R^{n})$ . For every $t > 0$ , $\int_{R^{n}} Γ (x, t) ψ (x) d x = (ψ * Γ_{t}) (0)$ because $Γ$ is spatially even, and by [F9] applied on a compact set containing $0$ and $supp ψ$ this converges to $ψ (0)$ as $t ↓ 0^{+}$ , so $Γ (\cdot, t) \to δ_{0}$ in $D^{'} (R^{n})$ by the definition of $δ_{0}$ in [F6].

2.1step 1.1F3F4given

Pairing: fix $φ$ with support in $B_{R} \times (- R, R)$ . The definitions of the regular functional and of the distributional derivative give $⟨ (\partial_{t} - Δ_{x}) u_{Γ}, φ ⟩ = - \int_{R^{n + 1}} Γ \partial_{t} φ - \int_{R^{n + 1}} Γ Δ_{x} φ = - \int_{R^{n + 1}} Γ (\partial_{t} φ + Δ_{x} φ)$ , the two integrals being absolutely convergent by step 1.1 and the compact support of $φ$ , with the signs as in [F4].

2.2step 1.1F1F2F7F8given

Truncated integration by parts: for $0 < ε < R$ , choose $T > R$ with $supp φ \subseteq R^{n} \times (- T, T)$ . Fubini [F8] on the strip $ε \leq t \leq T$ , the scalar integration by parts [F7] in the time variable at each fixed $x$ , and [F7] twice in each spatial coordinate at each fixed $t$ (the boundary terms vanish because $φ$ and all its derivatives are supported in the open box $B_{R} \times (- R, R)$ ) give $\int_{ε}^{\infty} \int_{R^{n}} Γ \partial_{t} φ d x d t = - \int_{R^{n}} Γ (x, ε) φ (x, ε) d x - \int_{ε}^{\infty} \int_{R^{n}} \partial_{t} Γ φ d x d t$ and $\int_{ε}^{\infty} \int_{R^{n}} Γ Δ_{x} φ d x d t = \int_{ε}^{\infty} \int_{R^{n}} Δ_{x} Γ φ d x d t$ ; adding the two identities and substituting the heat equation $\partial_{t} Γ = Δ_{x} Γ$ of [F2] cancels the interior terms and yields $\int_{ε}^{\infty} \int_{R^{n}} Γ (\partial_{t} φ + Δ_{x} φ) d x d t = - \int_{R^{n}} Γ (x, ε) φ (x, ε) d x$ .

3.1step 2.1step 2.2F1F2F9F10givenalgebra

Limit as $ε ↓ 0$ : by step 2.2 and step 2.1, $⟨ (\partial_{t} - Δ_{x}) u_{Γ}, φ ⟩ = \lim_{ε ↓ 0} \int_{R^{n}} Γ (x, ε) φ (x, ε) d x$ . Write the last integral as $\int_{R^{n}} Γ (x, ε) φ (x, 0) d x + \int_{R^{n}} Γ (x, ε) (φ (x, ε) - φ (x, 0)) d x$ . The first term equals $(φ (\cdot, 0) * Γ_{ε}) (0)$ because $Γ$ is even in its spatial variable, and it tends to $φ (0, 0)$ by the approximate-identity corollary [F9] applied to the bounded continuous compactly supported function $x \mapsto φ (x, 0)$ on a compact set containing $0$ ; the second term is bounded in modulus by $\sup_{x} ∣ φ (x, ε) - φ (x, 0) ∣ \cdot ∥ Γ (\cdot, ε) ∥_{1} \leq ε \sup ∣ \partial_{t} φ ∣$ by the mean value theorem [F10] in the time variable and unit mass from [F2], hence tends to $0$ .

4.1step 1.1step 3.1step 1.2F6given∎

Step 3.1 gives $⟨ (\partial_{t} - Δ_{x}) u_{Γ}, φ ⟩ = φ (0, 0) = ⟨ δ_{(0, 0)}, φ ⟩$ for every $φ \in C_{c}^{\infty} (R^{n + 1})$ by [F6], that is, $(\partial_{t} - Δ_{x}) u_{Γ} = δ_{(0, 0)}$ ; step 1.2 gives the weak Dirac limit at time zero; step 1.1 gives local integrability, so the causal extension is a fundamental solution of $\partial_{t} - Δ_{x}$ .

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

The heat evolution $H_{t}$ of initial data

Definition

Assume Countable Choice, let $n \geq 1$ and $1 \leq p \leq \infty$ , and let $Γ$ be the heat kernel of The heat kernel on $R^{n}$ and its causal extension, with $Γ_{t} : = Γ (\cdot, t)$ the $L^{1}$ function of unit norm supplied by Normalisation, parabolic scaling, heat equation and derivative bounds for the heat kernel. Convolution is that of Convolution of two functions on $R^{n}$ , and $L^{p}$ means the class space of The space $L^{p} (μ)$ as the quotient by null functions.

For $t > 0$ and $f \in L^{p} (R^{n})$ define $H_{t} f$ to be the $L^{p}$ class of the function

$x ⟼ \int_{R^{n}} Γ (x - y, t) f (y) d y .$

By Young's convolution inequality Young's convolution inequality under Countable Choice applied with the exponent triple $(p, 1, p)$ , which satisfies $1 / p = 1 / p + 1 / 1 - 1$ , this convolution is defined for almost every $x$ and belongs to $L^{p}$ with $∥ H_{t} f ∥_{p} \leq ∥ Γ_{t} ∥_{1} ∥ f ∥_{p} = ∥ f ∥_{p}$ ; hence $H_{t} f$ is a well-defined element of $L^{p}$ satisfying the contraction bound. For $p = \infty$ the integral converges absolutely for every $x$ because $∣ Γ (x - y, t) f (y) ∣ \leq ∥ f ∥_{\infty} Γ (x - y, t)$ almost everywhere in $y$ , with $\int Γ (x - y, t) d y = 1$ , so it defines a bounded representative with $∥ H_{t} f ∥_{\infty} \leq ∥ f ∥_{\infty}$ .

The value depends only on the class of $f$ : if $f = f^{'}$ almost everywhere then the null set where they differ is carried by translation to a null set (Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation), so at every $x$ the two $y$ -integrands agree almost everywhere. Their absolute convergence holds at the same points, and wherever they converge the integrals agree by Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree. Each $H_{t}$ is complex-linear on $L^{p}$ , by linearity of the integral of each representative.

Set $H_{0} f : = f$ , the identity operator on $L^{p}$ ; the singular kernel formula is never evaluated at $t = 0$ .

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

Spatial and time derivatives pass through heat convolution for positive time

Statement

Assume Countable Choice. Let $n \geq 1$ , $1 \leq p \leq \infty$ , and $f \in L^{p} (R^{n})$ (bounded measurable data are included). The absolutely convergent representative $u (x, t) = \int_{R^{n}} Γ (x - y, t) f (y) d y$ is $C^{\infty}$ for $t > 0$ , and for every multi-index $α$ and $k \geq 0$ $D_{x}^{α} \partial_{t}^{k} u = (D_{x}^{α} \partial_{t}^{k} Γ_{t}) * f$ . Moreover $∣ D_{x}^{α} \partial_{t}^{k} Γ (x, t) ∣ \leq C_{n, α, k} t^{- (n + ∣ α ∣ + 2 k) / 2} e^{- ∣ x ∣^{2} / (8 t)}$ . Domination is uniform on compact subsets $x \in K$ , $τ \leq t \leq T$ with $0 < τ < T < \infty$ ; $u_{t} = Δ u$ .

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $1 \leq p \leq \infty$ , $f \in L^{p} (R^{n})$ , a multi-index $α$ , an integer $k \geq 0$ , and $0 < τ < T$ with $K \subseteq R^{n}$ compact.

[A1]

Countable Choice is the hypothesis carried by the differentiation and integration suppliers below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

For $1 \leq p \leq \infty$ and $f \in L^{p} (R^{n})$ the heat evolution $H_{t} f$ is the $L^{p}$ class of $x \mapsto \int_{R^{n}} Γ (x - y, t) f (y) d y$ , defined almost everywhere with the contraction bound $∥ H_{t} f ∥_{p} \leq ∥ f ∥_{p}$ (The heat evolution $H_{t}$ of initial data).

[F2]

For every $t > 0$ the kernel is $C^{\infty}$ on $R^{n} \times (0, \infty)$ , satisfies $\partial_{t} Γ = Δ_{x} Γ$ , and for every multi-index $β$ there is $C_{n, β} < \infty$ with $∣ D^{β} Γ (z, t) ∣ \leq C_{n, β} t^{- (n + ∣ β ∣) / 2} e^{- ∣ z ∣^{2} / (8 t)}$ ; also $\int_{R^{n}} Γ (z, t) d z = 1$ (Normalisation, parabolic scaling, heat equation and derivative bounds for the heat kernel).

[F3]

A $C^{2}$ function on an open set has equal mixed partials, $\partial_{i} \partial_{j} ϕ = \partial_{j} \partial_{i} ϕ$ (Clairaut--Schwarz theorem for continuous second partial derivatives).

[F4]

If $f_{k} \to f$ almost everywhere and $∣ f_{k} ∣ \leq g$ almost everywhere with $g$ integrable, then $\int f_{k} \to \int f$ (Dominated convergence).

[F5]

For conjugate exponents $p, q \in [1, \infty]$ and measurable $F, G$ with $F \in L^{p}$ , $G \in L^{q}$ , the product is integrable and $\int ∣ F G ∣ \leq ∥ F ∥_{p} ∥ G ∥_{q}$ (Holder's inequality for integrals, including the endpoint cases).

[F6]

Let $I$ be an open interval and $F : X \times I \to C$ satisfy: for each $t \in I$ , $x \mapsto F (x, t)$ is integrable; for almost every $x$ , $t \mapsto F (x, t)$ is differentiable; the $t$ -derivative is measurable in $x$ ; and the derivative is dominated by one integrable function of $x$ uniformly in $t$ . Then $t \mapsto \int F (x, t) d x$ is differentiable on $I$ with derivative $\int \partial_{t} F (x, t) d x$ (Differentiation under the integral sign).

Proof

technique · direct

1.1A1F1F2F3givenalgebra

Work under [A1]; let $u (x, t) = \int_{R^{n}} Γ (x - y, t) f (y) d y$ be the everywhere-defined representative of the evolution $H_{t} f$ of [F1]. Kernel derivatives and their $L^{p^{'}}$ bounds: by [F2] the identity $\partial_{t} Γ = Δ_{x} Γ$ holds on $R^{n} \times (0, \infty)$ , and $Γ$ is $C^{\infty}$ , so Clairaut–Schwarz [F3] lets the time and space derivatives be interchanged; iterating the heat equation gives $\partial_{t}^{k} Γ = Δ_{x}^{k} Γ$ and therefore $D_{x}^{α} \partial_{t}^{k} Γ = Δ_{x}^{k} D_{x}^{α} Γ$ , a finite sum of terms $D^{β} Γ$ with $∣ β ∣ = ∣ α ∣ + 2 k$ , so [F2] yields $∣ D_{x}^{α} \partial_{t}^{k} Γ (x, t) ∣ \leq C_{n, α, k} t^{- (n + ∣ α ∣ + 2 k) / 2} e^{- ∣ x ∣^{2} / (8 t)}$ for a finite constant. Writing $p^{'}$ for the conjugate exponent and $m : = ∣ α ∣ + 2 k$ , for $1 < p \leq \infty$ (so $p^{'} < \infty$ ) the function $z \mapsto e^{- p^{'} ∣ z ∣^{2} / (8 t)}$ is $e^{- ∣ z ∣^{2} / (4 t^{'})} = (4 π t^{'})^{n / 2} Γ (z, t^{'})$ with $t^{'} = 2 t / p^{'}$ , so its integral is $(4 π t^{'})^{n / 2} = (8 π t / p^{'})^{n / 2}$ by unit mass in [F2], while for $p = 1$ the function $z \mapsto e^{- ∣ z ∣^{2} / (8 t)}$ is bounded by $1$ ; in every case $D_{x}^{α} \partial_{t}^{k} Γ (\cdot, t) \in L^{p^{'}}$ with norm at most $C_{n, α, k} t^{- (n + m) / 2} (8 π t / p^{'})^{n / (2 p^{'})}$ when $p > 1$ , and at most $C_{n, α, k} t^{- (n + m) / 2}$ when $p = 1$ .

2.1step 1.1F5givenalgebra

Absolute convergence and compact domination: fix $x \in R^{n}$ and $t > 0$ ; by Hölder [F5] and step 1.1, $\int ∣ D_{x}^{α} \partial_{t}^{k} Γ (x - y, t) f (y) ∣ d y \leq ∥ D_{x}^{α} \partial_{t}^{k} Γ (\cdot, t) ∥_{p^{'}} ∥ f ∥_{p} < \infty$ , so every derivative integral converges absolutely and defines $u_{α}^{k} (x, t) : = \int D_{x}^{α} \partial_{t}^{k} Γ (x - y, t) f (y) d y$ . If $K \subseteq B_{R}$ and $0 < τ \leq t \leq T$ , then $∣ x - y ∣^{2} \geq ∣ y ∣^{2} / 2 - R^{2}$ gives $∣ D_{x}^{α} \partial_{t}^{k} Γ (x - y, t) ∣ \leq C_{n, α, k} τ^{- (n + m) / 2} e^{R^{2} / (8 τ)} e^{- ∣ y ∣^{2} / (16 T)}$ for every $x \in K$ and $t \in [τ, T]$ , an $x$ -independent, $t$ -independent bound whose product with $∣ f ∣$ is integrable by [F5] because the Gaussian $e^{- ∣ y ∣^{2} / (16 T)}$ is in every $L^{q}$ .

3.1step 1.1step 2.1F4F6given

Differentiation under the integral sign: for fixed $x$ apply [F6] on any open parameter interval $I = (τ, T)$ with $0 < τ < T$ to $F (y, t) = Γ (x - y, t) f (y)$ , whose $t$ -derivative is dominated uniformly on $I$ by the integrable function from step 2.1 with $α = 0, k = 1$ ; this gives $\partial_{t} u (x, t) = \int \partial_{t} Γ (x - y, t) f (y) d y$ , and the same argument with $\partial_{x_{i}}$ in place of $\partial_{t}$ gives $\partial_{x_{i}} u (x, t) = \int \partial_{x_{i}} Γ (x - y, t) f (y) d y$ . Reapplying [F6] to the resulting integral representations, whose integrands are dominated on each compact set exactly as in step 2.1 for the next multi-index, produces every mixed derivative $D_{x}^{α} \partial_{t}^{k} u$ as the corresponding convolution integral; the domination of step 2.1 is uniform in $(x, t)$ on $K \times [τ, T]$ , and dominated convergence [F4] makes each such integral a continuous function of $(x, t)$ there, so $u$ is $C^{\infty}$ on $R^{n} \times (0, \infty)$ and $D_{x}^{α} \partial_{t}^{k} u = (D_{x}^{α} \partial_{t}^{k} Γ_{t}) * f$ .

4.1step 3.1F2givenalgebra

Heat equation: by step 3.1 with $α = 0$ , $k = 1$ and with the spatial derivatives, $\partial_{t} u = (\partial_{t} Γ_{t}) * f$ and $Δ_{x} u = (Δ_{x} Γ_{t}) * f$ ; the kernel identity $\partial_{t} Γ = Δ_{x} Γ$ of [F2] makes the two convolution integrands equal, so $\partial_{t} u = Δ_{x} u$ on $R^{n} \times (0, \infty)$ .

5.1step 1.1step 2.1step 3.1step 4.1given∎

Steps 1.1 and 2.1 give the derivative bounds of the kernel, absolute convergence of the representative and the uniform compact domination; steps 3.1 and 4.1 give the $C^{\infty}$ property, the derivative identity $D_{x}^{α} \partial_{t}^{k} u = (D_{x}^{α} \partial_{t}^{k} Γ_{t}) * f$ and the heat equation $u_{t} = Δ u$ , which is the whole statement.

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The heat Cauchy problem for bounded uniformly continuous data

Statement

Assume Countable Choice and let $n \geq 1$ , and let $u_{0} : R^{n} \to C$ be bounded and uniformly continuous. Define $u (x, t) : = \int_{R^{n}} Γ (x - y, t) u_{0} (y) d y$ for $t > 0$ and $u (x, 0) : = u_{0} (x)$ . Then $u$ is $C^{\infty}$ on $R^{n} \times (0, \infty)$ , satisfies $\partial_{t} u = Δ_{x} u$ there, is bounded with $∥ u (\cdot, t) ∥_{\infty} \leq ∥ u_{0} ∥_{\infty}$ , and converges locally uniformly at the initial time: $\sup_{x \in K} ∣ u (x, t) - u_{0} (x) ∣ \to 0$ as $t ↓ 0^{+}$ for every compact $K \subseteq R^{n}$ .

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , a bounded uniformly continuous $u_{0} : R^{n} \to C$ , a multi-index $α$ , an integer $k \geq 0$ , and $0 < τ < T$ with $K \subseteq R^{n}$ compact.

[A1]

Countable Choice is the hypothesis carried by the differentiation and integration suppliers below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

For $p = \infty$ the heat evolution $H_{t}$ of The heat evolution $H_{t}$ of initial data is defined on bounded measurable data $g$ by the everywhere absolutely convergent integral $(H_{t} g) (x) = \int_{R^{n}} Γ (x - y, t) g (y) d y$ , with $∥ H_{t} g ∥_{\infty} \leq ∥ g ∥_{\infty}$ , and $H_{0} g = g$ .

[F2]

For every $t > 0$ the kernel is $C^{\infty}$ with $\partial_{t} Γ = Δ_{x} Γ$ , for every multi-index $β$ there is $C_{n, β} < \infty$ with $∣ D^{β} Γ (z, t) ∣ \leq C_{n, β} t^{- (n + ∣ β ∣) / 2} e^{- ∣ z ∣^{2} / (8 t)}$ , the unit-mass identity $\int Γ (z, t) d z = 1$ holds, and for every $δ > 0$ $\int_{∣ z ∣ > δ} Γ (z, t) d z \to 0$ as $t ↓ 0^{+}$ (Normalisation, parabolic scaling, heat equation and derivative bounds for the heat kernel).

[F3]

For $f \in L^{p}$ , $1 \leq p \leq \infty$ , the absolutely convergent heat convolution is $C^{\infty}$ in space and time for $t > 0$ , every derivative passes through the integral, and $u_{t} = Δ_{x} u$ (Spatial and time derivatives pass through heat convolution for positive time).

[F4]

If $(K_{ε})$ is an $L^{1}$ approximate identity and $g$ is bounded and continuous, then $(g * K_{ε}) (x) \to g (x)$ uniformly on compact sets ( $L^{1}$ approximate identities converge uniformly on compacta for bounded continuous functions).

Proof

technique · direct

1.1A1F1F2given

Work under [A1] and put $M : = ∥ u_{0} ∥_{\infty} < \infty$ . For every $x$ and $t > 0$ the integral defining $u (x, t)$ is absolutely convergent with $∣ u (x, t) ∣ \leq M \int Γ (x - y, t) d y = M$ by unit mass [F2], so $u$ is a bounded function with $∥ u (\cdot, t) ∥_{\infty} \leq M = ∥ u_{0} ∥_{\infty}$ and $u (\cdot, 0) = u_{0}$ by the definition of $H_{0}$ in [F1].

2.1step 1.1F3given

Since $u_{0}$ is bounded and measurable, it represents an $L^{\infty}$ class. Applying [F3] with $p = \infty$ gives the $C^{\infty}$ representative $u$ on $R^{n} \times (0, \infty)$ and the heat equation $\partial_{t} u = Δ_{x} u$ .

3.1step 2.1F2F4given

Initial convergence: by the evenness of $Γ (\cdot, t)$ in its first argument, $u (x, t) = \int_{R^{n}} Γ (x - y, t) u_{0} (y) d y = \int_{R^{n}} Γ (y - x, t) u_{0} (y) d y = (u_{0} * Γ_{t}) (x)$ for every $t > 0$ ; the datum $u_{0}$ is bounded and continuous and the family $(Γ_{t})_{t > 0}$ is an $L^{1}$ approximate identity by [F2], so the published approximate-identity corollary [F4] gives $\sup_{x \in K} ∣ u (x, t) - u_{0} (x) ∣ \to 0$ as $t ↓ 0^{+}$ for every compact $K \subseteq R^{n}$ , which is the asserted local uniform convergence.

4.1step 1.1step 2.1step 3.1given∎

Steps 1.1, 2.1 and 3.1 prove boundedness with the stated sup-norm bound, $C^{\infty}$ smoothness and the heat equation at positive time, and locally uniform recovery of the initial datum, which is the whole statement.

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The heat Cauchy problem for $L^{p}$ data

Statement

Assume Countable Choice, let $n \geq 1$ and $1 \leq p < \infty$ . For every $f \in L^{p} (R^{n})$ the heat evolution of The heat evolution $H_{t}$ of initial data satisfies: (i) $∥ H_{t} f - f ∥_{p} \to 0$ as $t ↓ 0^{+}$ ; (ii) $∥ H_{t} f ∥_{p} \leq ∥ f ∥_{p}$ for every $t > 0$ ; (iii) the semigroup law $H_{t + s} = H_{t} H_{s}$ on $L^{p}$ for all $s, t \geq 0$ , with $H_{0}$ the identity.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $1 \leq p < \infty$ , $f \in L^{p} (R^{n})$ , and $s, t > 0$ .

[A1]

Countable Choice is the hypothesis carried by the integration and approximate-identity suppliers below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

For $1 \leq p \leq \infty$ and $g \in L^{p} (R^{n})$ , $H_{t} g$ is the $L^{p}$ class of $x \mapsto \int Γ (x - y, t) g (y) d y$ , defined almost everywhere, each $H_{t}$ is complex-linear, and $H_{0}$ is the identity on $L^{p}$ (The heat evolution $H_{t}$ of initial data).

[F2]

For every $t > 0$ the kernel satisfies $∥ Γ_{t} ∥_{1} = 1$ and the family $(Γ_{t})_{t > 0}$ is an $L^{1}$ approximate identity (Normalisation, parabolic scaling, heat equation and derivative bounds for the heat kernel).

[F3]

For all $s, t > 0$ , $Γ_{t} * Γ_{s} = Γ_{t + s}$ as functions on $R^{n}$ , the convolution converging absolutely everywhere (The heat kernel semigroup identity $Γ_{t} * Γ_{s} = Γ_{t + s}$ ).

[F4]

Assume countable choice; if $(K_{ε})$ is an $L^{1}$ approximate identity and $f \in L^{p}$ with $1 \leq p < \infty$ , then $∥ f * K_{ε} - f ∥_{p} \to 0$ as $ε \to 0^{+}$ (Every $L^{1}$ approximate identity converges to the identity in $L^{p}$ for $1 \leq p < \infty$ ).

[F5]

Assume Countable Choice; for $1 \leq p, q, r \leq \infty$ with $1 / r = 1 / p + 1 / q - 1$ , convolution gives $∥ f * g ∥_{r} \leq ∥ f ∥_{p} ∥ g ∥_{q}$ (Young's convolution inequality under Countable Choice).

[F6]

On sigma-finite product spaces, Tonelli's theorem holds for nonnegative product-measurable functions and Fubini's theorem for $L^{1}$ functions (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product, Fubini's theorem for L^1 functions on a sigma-finite product); the Lebesgue measure on $R^{2 n}$ is the completed product of two copies of $λ_{n}$ (The Euclidean Lebesgue measure is the completion of the product of the factor Lebesgue measures).

Proof

technique · direct

1.1A1F1F2F5givenalgebra

Contraction: by [F1] the class $H_{t} f$ is represented by the convolution $Γ_{t} * f$ , so Young's inequality [F5] with the exponent triple $(p, 1, p)$ , which satisfies $1 / p = 1 / p + 1 / 1 - 1$ , gives $∥ H_{t} f ∥_{p} \leq ∥ Γ_{t} ∥_{1} ∥ f ∥_{p} = ∥ f ∥_{p}$ by [F2]; at $t = 0$ , $H_{0} f = f$ has the same norm.

2.1step 1.1F1F2F4given

Strong convergence at time zero: by [F2] the family $(Γ_{t})_{t > 0}$ is an $L^{1}$ approximate identity, so the published $L^{p}$ approximate-identity theorem [F4] gives $∥ f * Γ_{t} - f ∥_{p} \to 0$ as $t ↓ 0^{+}$ ; by [F1] the class $H_{t} f$ is exactly the class of $f * Γ_{t}$ , so $∥ H_{t} f - f ∥_{p} \to 0$ , which is (i).

2.2step 1.1F1F3F5F6givenalgebra

Semigroup law: assume first $s, t > 0$ and put $g (z) : = \int Γ_{s} (z - y) f (y) d y$ , a representative of $H_{s} f$ defined for almost every $z$ by [F1]. For fixed $x$ consider the nonnegative function $(y, z) \mapsto ∣ Γ_{t} (x - z) ∣ Γ_{s} (z - y) ∣ f (y) ∣$ on $R^{n} \times R^{n}$ ; by the identification of [F6] and Tonelli's theorem, its iterated integral equals $\int_{R^{n}} ∣ f (y) ∣ (\int_{R^{n}} Γ_{t} (x - z) Γ_{s} (z - y) d z) d y$ , and the inner integral is $(Γ_{t} * Γ_{s}) (x - y) = Γ_{t + s} (x - y)$ by [F3], so the whole integral is $(Γ_{t + s} * ∣ f ∣) (x)$ , finite for almost every $x$ because $Γ_{t + s} \in L^{1}$ and Young's inequality [F5] applies. Hence the signed double integral is absolutely convergent, Fubini [F6] applies, and $\int Γ_{t} (x - z) g (z) d z = \int f (y) Γ_{t + s} (x - y) d y$ for almost every $x$ ; that is, $H_{t} (H_{s} f) = H_{t + s} f$ . If $s = 0$ or $t = 0$ the identity is the operator identity $H_{0} = i d$ of [F1], so the semigroup law holds for all $s, t \geq 0$ , which is (iii).

3.1step 1.1step 2.1step 2.2given∎

Steps 1.1, 2.1 and 2.2 prove the contraction bound (ii), the strong convergence (i) and the semigroup law (iii) for every $f \in L^{p} (R^{n})$ , $1 \leq p < \infty$ , which is the whole statement.

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Uniqueness of strongly continuous mild heat solutions

Statement

Assume Countable Choice. Let $n \geq 1$ , $1 \leq p < \infty$ , $T > 0$ , $u \in C ([0, T]; L^{p} (R^{n}))$ , $u (0) = f$ , and $u (t) = H_{t - s} u (s)$ for every $0 < s < t \leq T$ . Then $u (t) = H_{t} f$ for all $0 \leq t \leq T$ . Hence the heat evolution is the unique solution in this class. This is uniqueness for the semigroup relation, not an unrestricted classical uniqueness assertion or a claim of strong continuity on all $L^{\infty}$ .

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $1 \leq p < \infty$ , $T > 0$ , $u \in C ([0, T]; L^{p} (R^{n}))$ with $u (0) = f$ and $u (t) = H_{t - s} u (s)$ for all $0 < s < t \leq T$ , and $t \in (0, T]$ with $s \in (0, t)$ .

[A1]

Countable Choice is the hypothesis carried by the evolution suppliers below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

For $1 \leq p < \infty$ the heat evolution satisfies the semigroup law $H_{t + s} = H_{t} H_{s}$ on $L^{p}$ for all $s, t \geq 0$ with $H_{0}$ the identity, the contraction bound $∥ H_{r} f ∥_{p} \leq ∥ f ∥_{p}$ , and strong continuity at zero, $∥ H_{r} f - f ∥_{p} \to 0$ as $r ↓ 0^{+}$ , for every $f \in L^{p}$ (The heat Cauchy problem for $L^{p}$ data).

[F2]

If $(K_{ε})$ is an $L^{1}$ approximate identity and $f \in L^{p}$ , $1 \leq p < \infty$ , then $∥ f * K_{ε} - f ∥_{p} \to 0$ (Every $L^{1}$ approximate identity converges to the identity in $L^{p}$ for $1 \leq p < \infty$ ); this is the mechanism behind the strong continuity recorded in [F1].

Proof

technique · direct

1.1A1F1given

Fix $t \in (0, T]$ and $0 < s < t$ . By the assumed semigroup relation at time $t$ and at time $s$ , and by linearity of $H_{t - s}$ and the semigroup law of [F1], $u (t) - H_{t} f = H_{t - s} u (s) - H_{t - s} H_{s} f = H_{t - s} (u (s) - H_{s} f)$ .

2.1step 1.1F1given

Norm bound: applying the contraction clause of [F1] to the last expression and then the triangle inequality gives $∥ u (t) - H_{t} f ∥_{p} \leq ∥ u (s) - H_{s} f ∥_{p} \leq ∥ u (s) - f ∥_{p} + ∥ f - H_{s} f ∥_{p}$ .

3.1F1F2step 2.1givenalgebra

Limit: the continuity of $u$ at $0$ in the $L^{p}$ norm gives $∥ u (s) - f ∥_{p} = ∥ u (s) - u (0) ∥_{p} \to 0$ as $s ↓ 0^{+}$ , and the strong continuity clause of [F1] gives $∥ H_{s} f - f ∥_{p} \to 0$ ; hence the right-hand side of step 2.1 tends to $0$ , so the nonnegative number $∥ u (t) - H_{t} f ∥_{p}$ is $0$ and $u (t) = H_{t} f$ in $L^{p}$ for every $t \in (0, T]$ . At $t = 0$ the identity $u (0) = f = H_{0} f$ is the hypothesis and the identity case of [F1].

4.1step 3.1F1given

Existence in the same class: the map $t \mapsto H_{t} f$ lies in $C ([0, T]; L^{p})$ : at $t = 0$ this is the strong continuity of [F1], and for $t > 0$ and $h > 0$ the semigroup law and contraction give $∥ H_{t + h} f - H_{t} f ∥_{p} \leq ∥ H_{h} f - f ∥_{p}$ , while for $h < 0$ with $t + h \geq 0$ they give $∥ H_{t + h} f - H_{t} f ∥_{p} \leq ∥ H_{- h} f - f ∥_{p}$ , and both bounds tend to $0$ as $∣ h ∣ \to 0$ .

5.1step 3.1step 4.1given∎

Steps 1.1, 2.1 and 3.1 show that any $u$ in the stated class coincides with $t \mapsto H_{t} f$ on $[0, T]$ , and step 4.1 shows that $t \mapsto H_{t} f$ itself lies in that class, so the heat evolution is the unique solution of the semigroup relation with the given initial datum in $C ([0, T]; L^{p})$ .

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Heat generator at zero on compactly supported smooth data

Statement

Assume Countable Choice. For $n \geq 1$ , $φ \in C_{c}^{\infty} (R^{n})$ and $1 \leq p < \infty$ , $∥ (H_{t} φ - φ) / t - Δ φ ∥_{p} \to 0$ as $t ↓ 0$ .

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $φ \in C_{c}^{\infty} (R^{n})$ , $1 \leq p < \infty$ , and $0 < ε < t$ .

[A1]

Countable Choice is the hypothesis carried by the differentiation, integration and evolution suppliers below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

$C_{c}^{\infty} (R^{n})$ is the test-function space of Test function space d of an open set; $φ$ and every derivative of it is smooth with compact support, so $Δ φ \in C_{c}^{\infty} (R^{n})$ and in particular $Δ φ \in L^{p} (R^{n})$ . In this item $H_{s} ψ$ denotes the evolution of the class $ψ$ as in The heat evolution $H_{t}$ of initial data.

[F2]

For $s > 0$ the function $H_{s} φ$ is $C^{\infty}$ on $R^{n}$ and every spatial and time derivative passes through the convolution, $D^{α} \partial_{s}^{k} H_{s} φ = (D^{α} \partial_{s}^{k} Γ_{s}) * φ$ (Spatial and time derivatives pass through heat convolution for positive time).

[F3]

The kernel satisfies $\partial_{s} Γ = Δ_{x} Γ$ on $R^{n} \times (0, \infty)$ (Normalisation, parabolic scaling, heat equation and derivative bounds for the heat kernel).

[F4]

For continuous $F, G$ on $[a, b]$ differentiable on $(a, b)$ with $F^{'} = f$ , $G^{'} = g$ Riemann integrable there, $\int_{a}^{b} F g + \int_{a}^{b} f G = F (b) G (b) - F (a) G (a)$ (Integration by parts for continuous factors with Riemann-integrable extensions of their interior derivatives).

[F5]

If $G$ is continuous on $[a, b]$ , differentiable on $(a, b)$ and $f = G^{'}$ is Riemann integrable, then $\int_{a}^{b} f = G (b) - G (a)$ (Newton–Leibniz needs only continuity on $[a, b]$ , differentiability on $(a, b)$ , and a Riemann-integrable extension of the interior derivative).

[F6]

For $1 \leq p < \infty$ and $ψ \in L^{p} (R^{n})$ , $∥ H_{s} ψ - ψ ∥_{p} \to 0$ as $s ↓ 0^{+}$ , and $∥ H_{s} ψ ∥_{p} \leq ∥ ψ ∥_{p}$ (The heat Cauchy problem for $L^{p}$ data).

[F7]

Let $1 \leq p < \infty$ and let $F (x, s)$ be measurable with $\int_{0}^{t} ∥ F (\cdot, s) ∥_{L^{p}} d s < \infty$ ; then $∥ \int_{0}^{t} ∣ F (x, s) ∣ d s ∥_{L^{p}} \leq \int_{0}^{t} ∥ F (\cdot, s) ∥_{L^{p}} d s$ (Minkowski's integral inequality).

[F8]

Bounded uniformly continuous data are recovered locally uniformly at $t = 0$ by their heat convolution (The heat Cauchy problem for bounded uniformly continuous data). This applies to both $φ$ and $Δ φ$ , which are smooth with compact support.

Proof

technique · direct

1.1A1F1F2F3F4given

Derivative identity: fix $x \in R^{n}$ and $s > 0$ . By [F2] with $α = 0$ , $k = 1$ and $f = φ$ the function $s \mapsto H_{s} φ (x)$ is differentiable with $\partial_{s} H_{s} φ (x) = \int \partial_{s} Γ (x - y, s) φ (y) d y$ ; by [F3] and $\partial_{x_{i}} \partial_{x_{i}} Γ (x - y, s) = \partial_{y_{i}} \partial_{y_{i}} Γ (x - y, s)$ , applying the scalar integration by parts [F4] twice in each coordinate (the boundary terms vanish because $φ$ has compact support, and this works for every $n \geq 1$ including $n = 1$ ) turns the last integral into $\int Γ (x - y, s) Δ φ (y) d y = H_{s} Δ φ (x)$ ; hence $\partial_{s} H_{s} φ (x) = H_{s} Δ φ (x)$ for every $x$ and $s > 0$ .

2.1step 1.1F5given

Newton–Leibniz: by step 1.1 the map $s \mapsto H_{s} φ (x)$ is continuous on $[ε, t]$ with derivative $H_{s} Δ φ (x)$ on $(ε, t)$ , for its real and imaginary parts separately; the fundamental theorem [F5] applied on $[ε, t]$ gives $H_{t} φ (x) - H_{ε} φ (x) = \int_{ε}^{t} H_{s} Δ φ (x) d s$ for every $x$ .

3.1step 2.1F1F2F7F8given

Both $H_{ε} φ (x) \to φ (x)$ and $H_{s} Δ φ (x) \to Δ φ (x)$ hold at every $x$ by [F8]. Thus the integrand in step 2.1 extends continuously to $s = 0$ , and letting $ε ↓ 0$ gives $H_{t} φ (x) - φ (x) = \int_{0}^{t} H_{s} Δ φ (x) d s$ . Dividing by $t$ and subtracting $Δ φ (x)$ yields $\frac{H_{t} φ (x) - φ (x)}{t} - Δ φ (x) = \frac{1}{t} \int_{0}^{t} (H_{s} Δ φ (x) - Δ φ (x)) d s$ . The integrand is jointly continuous for $s > 0$ by [F2] applied to $Δ φ$ , hence measurable as required for [F7].

4.1step 3.1F6F7given

$L^{p}$ bound and limit: applying the Minkowski integral inequality [F7] to $F (x, s) : = H_{s} Δ φ (x) - Δ φ (x)$ on $R^{n} \times (0, t)$ — whose hypothesis holds because $∥ F (\cdot, s) ∥_{p} \leq 2 ∥ Δ φ ∥_{p}$ by the contraction clause of [F6] — gives $∥ \frac{H_{t} φ - φ}{t} - Δ φ ∥_{p} \leq \frac{1}{t} \int_{0}^{t} ∥ H_{s} Δ φ - Δ φ ∥_{p} d s$ . Given $η > 0$ , the strong convergence clause of [F6] applied to $Δ φ \in L^{p}$ supplies $δ > 0$ with $∥ H_{s} Δ φ - Δ φ ∥_{p} < η$ for every $0 < s < δ$ ; for every $0 < t < δ$ the right-hand side is then at most $η$ . Hence $∥ (H_{t} φ - φ) / t - Δ φ ∥_{p} \to 0$ as $t ↓ 0^{+}$ .

5.1step 1.1step 4.1given∎

Steps 1.1, 2.1, 3.1 and 4.1 prove the stated generator limit for compactly supported smooth data in every $L^{p}$ , $1 \leq p < \infty$ .

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Mass conservation and positivity of the heat flow

Statement

Assume Countable Choice, let $n \geq 1$ , and let $H_{t}$ be the heat evolution of The heat evolution $H_{t}$ of initial data. (i) If $f \in L^{1} (R^{n})$ then $\int_{R^{n}} H_{t} f (x) d x = \int_{R^{n}} f (x) d x$ for every $t > 0$ . (ii) If $f \in L^{p} (R^{n})$ , $1 \leq p \leq \infty$ , satisfies $f \geq 0$ almost everywhere, then $H_{t} f \geq 0$ almost everywhere for every $t > 0$ .

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $t > 0$ , and data $f$ in the class named in the respective clause.

[A1]

Countable Choice is the hypothesis carried by the integration suppliers below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

For $t > 0$ the heat kernel is positive and has $\int_{R^{n}} Γ (x, t) d x = 1$ (Normalisation, parabolic scaling, heat equation and derivative bounds for the heat kernel).

[F2]

For $1 \leq p \leq \infty$ , $t > 0$ and $f \in L^{p} (R^{n})$ , the heat evolution $H_{t} f$ is the $L^{p}$ class of the almost-everywhere defined function $x \mapsto \int_{R^{n}} Γ (x - y, t) f (y) d y$ (The heat evolution $H_{t}$ of initial data).

[F3]

On sigma-finite product spaces Tonelli's theorem gives $\int_{X \times Y} f d (μ \times ν) = \int_{X} \int_{Y} f_{x} d ν d μ$ for nonnegative product-measurable $f$ (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product).

[F4]

On sigma-finite product spaces Fubini's theorem gives the same iterated equality for $f \in L^{1} (μ \times ν)$ , the sections being integrable almost everywhere (Fubini's theorem for L^1 functions on a sigma-finite product).

[F5]

[F6]

If a measure-preserving $T$ and an integrable $f$ are given, then $\int f \circ T d μ = \int f d μ$ (Integral invariance under measure-preserving maps); for each fixed $y$ the translation $x \mapsto x + y$ preserves Lebesgue measure.

[F7]

$\int_{R^{n}} ∣ f ∣ d λ_{n}$ denotes the $L^{1}$ norm of The class $L^{1} (μ)$ of integrable functions.

Proof

technique · direct

1.1A1F1F2given

Work under [A1] and fix $t > 0$ . By [F2] the evolution $H_{t} f$ is the $L^{p}$ class of the representative $u (x) = \int_{R^{n}} Γ (x - y, t) f (y) d y$ , defined for almost every $x$ ; by [F1] the kernel is positive with unit mass.

2.1step 1.1F1F3F4F5F6F7given

Mass conservation: assume $f \in L^{1} (R^{n})$ , so that by [F2] and [F1] the function $(x, y) \mapsto ∣ f (y) ∣ Γ (x - y, t)$ is nonnegative and measurable. The identification [F5] makes the integral over $R^{2 n}$ the completed product integral, so Tonelli [F3] gives $\int_{R^{n}} \int_{R^{n}} ∣ f (y) ∣ Γ (x - y, t) d y d x = \int_{R^{n}} ∣ f (y) ∣ (\int_{R^{n}} Γ (x - y, t) d x) d y = \int_{R^{n}} ∣ f (y) ∣ d y < \infty$ , the inner integral being $1$ for every $y$ by the translation invariance [F6] and the unit mass of [F1]; hence $(x, y) \mapsto f (y) Γ (x - y, t)$ lies in $L^{1} (λ_{2 n})$ and Fubini [F4] gives $\int_{R^{n}} u (x) d x = \int_{R^{n}} f (y) (\int_{R^{n}} Γ (x - y, t) d x) d y = \int_{R^{n}} f (y) d y$ , which is (i), the integral of the class $H_{t} f$ being computed from its representative $u$ .

2.2step 1.1F1F2given

Positivity: assume $f \in L^{p} (R^{n})$ satisfies $f \geq 0$ almost everywhere and let $N$ be the null set where $f < 0$ . For every $x$ at which the defining integral converges, the function $y \mapsto Γ (x - y, t) f (y)$ is $\geq 0$ for every $y \notin N$ , because $Γ > 0$ by [F1]; a function that is nonnegative almost everywhere has nonnegative integral, so $u (x) \geq 0$ wherever $u$ is defined, and $u$ is defined almost everywhere by [F2]; hence the class $H_{t} f$ is $\geq 0$ almost everywhere, which is (ii).

3.1step 2.1step 2.2given∎

Steps 2.1 and 2.2 prove the mass-conservation clause (i) for $L^{1}$ data and the positivity clause (ii) for nonnegative $L^{p}$ data, so the corollary holds.

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Monotonicity and $L^{p}$ contractivity of the heat flow

Statement

Assume Countable Choice, let $n \geq 1$ and $1 \leq p \leq \infty$ . If $f, g \in L^{p} (R^{n})$ satisfy $f \leq g$ almost everywhere, then $H_{t} f \leq H_{t} g$ almost everywhere for every $t > 0$ ; and $∥ H_{t} f ∥_{p} \leq ∥ f ∥_{p}$ for every $f \in L^{p} (R^{n})$ and every $t \geq 0$ .

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $1 \leq p \leq \infty$ , $t > 0$ , and $f, g \in L^{p} (R^{n})$ .

[A1]

Countable Choice is the hypothesis carried by the evolution and convolution suppliers below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

Each $H_{t}$ is a complex-linear operator on $L^{p} (R^{n})$ , $H_{0}$ is the identity, and $H_{t} f$ is the class of $Γ_{t} * f$ (The heat evolution $H_{t}$ of initial data).

[F2]

If $f \in L^{p}$ satisfies $f \geq 0$ almost everywhere, then $H_{t} f \geq 0$ almost everywhere (Mass conservation and positivity of the heat flow).

[F3]

For $1 \leq p < \infty$ and $f \in L^{p}$ , $∥ H_{t} f ∥_{p} \leq ∥ f ∥_{p}$ (The heat Cauchy problem for $L^{p}$ data).

[F4]

For $1 \leq p, q, r \leq \infty$ with $1 / r = 1 / p + 1 / q - 1$ and $f \in L^{p}$ , $g \in L^{q}$ , $∥ f * g ∥_{r} \leq ∥ f ∥_{p} ∥ g ∥_{q}$ ; with exponent triple $(p, 1, p)$ this bounds convolution by an $L^{1}$ kernel (Young's convolution inequality under Countable Choice).

Proof

technique · direct

1.1A1F1F2given

Order preservation: assume $f \leq g$ almost everywhere and fix $t > 0$ . The difference $g - f$ is a class in $L^{p}$ with $g - f \geq 0$ almost everywhere, so $H_{t} (g - f) \geq 0$ almost everywhere by [F2]; by linearity of $H_{t}$ in [F1], $H_{t} g - H_{t} f = H_{t} (g - f)$ as classes, so any representatives satisfy $H_{t} f \leq H_{t} g$ almost everywhere, the comparison being independent of representatives because changing them on null sets does not affect an almost-everywhere inequality.

2.1step 1.1F1F3F4givenalgebra

Contractivity: for $1 \leq p < \infty$ the bound $∥ H_{t} f ∥_{p} \leq ∥ f ∥_{p}$ for every $t > 0$ is [F3], while at $t = 0$ it is the identity case of [F1]; for $p = \infty$ Young's inequality [F4] with the exponent triple $(\infty, 1, \infty)$ , which satisfies $1 / \infty = 1 / \infty + 1 - 1$ , gives $∥ H_{t} f ∥_{\infty} \leq ∥ Γ_{t} ∥_{1} ∥ f ∥_{\infty} = ∥ f ∥_{\infty}$ for every $t > 0$ , and again $∥ H_{0} f ∥_{\infty} = ∥ f ∥_{\infty}$ .

3.1step 1.1step 2.1given∎

Steps 1.1 and 2.1 prove the almost-everywhere monotonicity for $f \leq g$ and the $L^{p}$ contraction for all $t \geq 0$ and all $1 \leq p \leq \infty$ .

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$L^{p}$ to $L^{q}$ smoothing estimate for the heat flow

Statement

Assume Countable Choice, let $n \geq 1$ and $1 \leq p \leq q \leq \infty$ , and let $Q \in [1, \infty]$ be determined by $\frac{1}{Q} = 1 + \frac{1}{q} - \frac{1}{p}$ . Then for every $f \in L^{p} (R^{n})$ and every $t > 0$ , $∥ H_{t} f ∥_{q} \leq C_{n, p, q} t^{- \frac{n}{2} (\frac{1}{p} - \frac{1}{q})} ∥ f ∥_{p}, C_{n, p, q} : = (4 π)^{- \frac{n}{2} (\frac{1}{p} - \frac{1}{q})} Q^{- \frac{n}{2 Q}},$ with the endpoint $Q = \infty$ (which occurs exactly at $p = 1$ , $q = \infty$ ) read as $Q^{- n / (2 Q)} \to 1$ ; for $p = q$ the constant is $1$ and the estimate is the contraction clause.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $1 \leq p \leq q \leq \infty$ , the exponent $Q$ with $1 / Q = 1 + 1 / q - 1 / p$ , $f \in L^{p} (R^{n})$ and $t > 0$ .

[A1]

Countable Choice is the hypothesis carried by the convolution and integration suppliers below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

For $1 \leq p \leq \infty$ and $f \in L^{p} (R^{n})$ , $H_{t} f$ is the $L^{p}$ class of the convolution $Γ_{t} * f$ (The heat evolution $H_{t}$ of initial data).

[F2]

For every $s > 0$ the kernel satisfies $Γ (x, s) = (4 π s)^{- n / 2} e^{- ∣ x ∣^{2} / (4 s)}$ , the scaling identity $Γ (λ x, λ^{2} s) = λ^{- n} Γ (x, s)$ , and unit mass $\int_{R^{n}} Γ (x, s) d x = 1$ (Normalisation, parabolic scaling, heat equation and derivative bounds for the heat kernel).

[F3]

Assume Countable Choice; if $1 \leq p, q, r \leq \infty$ satisfy $1 / r = 1 / p + 1 / q - 1$ and $f \in L^{p}$ , $g \in L^{q}$ , then $∥ f * g ∥_{r} \leq ∥ f ∥_{p} ∥ g ∥_{q}$ (Young's convolution inequality under Countable Choice).

Proof

technique · direct

1.1A1F1F3given

Young estimate: the triple $(p, Q, q)$ is admissible because $1 / Q = 1 + 1 / q - 1 / p$ means exactly $1 / q = 1 / p + 1 / Q - 1$ , and $Q \in [1, \infty]$ lies in the Young range since $0 \leq 1 / p - 1 / q \leq 1$ ; hence, by [F3] applied to $f$ and $Γ_{t} \in L^{Q}$ , $∥ H_{t} f ∥_{q} \leq ∥ Γ_{t} ∥_{Q} ∥ f ∥_{p}$ for every $t > 0$ .

2.1step 1.1F2givenalgebra

Scaling of the kernel norm: the scaling identity of [F2] with $λ = \sqrt{t}$ gives $Γ_{t} (x) = t^{- n / 2} Γ_{1} (x / \sqrt{t})$ , so the substitution $x = \sqrt{t} z$ yields $∥ Γ_{t} ∥_{Q} = t^{- n / 2} t^{n / (2 Q)} ∥ Γ_{1} ∥_{Q} = t^{- \frac{n}{2} (1 - \frac{1}{Q})} ∥ Γ_{1} ∥_{Q}$ for $Q < \infty$ , and for $Q = \infty$ the same substitution gives $∥ Γ_{t} ∥_{\infty} = t^{- n / 2} ∥ Γ_{1} ∥_{\infty}$ .

3.1step 2.1F2givenalgebra

Gaussian norm: by [F2] with $s = 1$ , $Γ_{1} (x) = (4 π)^{- n / 2} e^{- ∣ x ∣^{2} / 4}$ , and for $Q < \infty$ the identity $e^{- Q ∣ x ∣^{2} / 4} = (4 π / Q)^{n / 2} Γ (x, 1 / Q)$ holds by the explicit formula, so unit mass gives $∥ Γ_{1} ∥_{Q} = (4 π)^{- n / 2} (\int e^{- Q ∣ x ∣^{2} / 4} d x)^{1 / Q} = (4 π)^{- n / 2} (4 π / Q)^{n / (2 Q)}$ ; for $Q = \infty$ the same formula is read as $∥ Γ_{1} ∥_{\infty} = (4 π)^{- n / 2}$ .

4.1step 1.1step 2.1step 3.1givenalgebra

Assembling the estimate: since $1 - \frac{1}{Q} = \frac{1}{p} - \frac{1}{q}$ by the definition of $Q$ , steps 1.1, 2.1 and 3.1 give $∥ H_{t} f ∥_{q} \leq t^{- \frac{n}{2} (\frac{1}{p} - \frac{1}{q})} (4 π)^{- n / 2} (4 π / Q)^{n / (2 Q)} ∥ f ∥_{p}$ , and $(4 π)^{- n / 2} (4 π / Q)^{n / (2 Q)} = (4 π)^{- \frac{n}{2} (1 - \frac{1}{Q})} Q^{- n / (2 Q)} = (4 π)^{- \frac{n}{2} (\frac{1}{p} - \frac{1}{q})} Q^{- n / (2 Q)} = C_{n, p, q}$ , which is the displayed estimate.

5.1step 4.1givenalgebra

Endpoints: if $p = q$ then $1 / Q = 1$ and $Q = 1$ , so $C_{n, p, q} = 1$ and the estimate reads $∥ H_{t} f ∥_{p} \leq ∥ f ∥_{p}$ ; if $Q = \infty$ then $1 / p - 1 / q = 1$ , which forces $p = 1$ and $q = \infty$ , and the factor $Q^{- n / (2 Q)}$ tends to $1$ while the exponent is $t^{- n / 2}$ , so the estimate reads $∥ H_{t} f ∥_{\infty} \leq (4 π t)^{- n / 2} ∥ f ∥_{1}$ .

6.1step 4.1step 5.1given∎

Steps 1.1, 2.1, 3.1, 4.1 and 5.1 prove the displayed $L^{p}$ to $L^{q}$ estimate with the stated constant, including the endpoint conventions and the $p = q$ contraction case.

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Spatial derivative estimates for the heat flow

Statement

Assume Countable Choice, let $n \geq 1$ , $1 \leq p \leq q \leq \infty$ , and let $Q$ and the constants be as in $L^{p}$ to $L^{q}$ smoothing estimate for the heat flow. For every multi-index $α$ , every $f \in L^{p} (R^{n})$ and every $t > 0$ , the function $x \mapsto \int_{R^{n}} Γ (x - y, t) f (y) d y$ is $C^{\infty}$ on $R^{n}$ , with $D^{α} H_{t} f = (D^{α} Γ (\cdot, t)) * f$ as an absolutely convergent integral, and $∥ D^{α} H_{t} f ∥_{q} \leq C_{n, p, q, α} t^{- \frac{∣ α ∣}{2} - \frac{n}{2} (\frac{1}{p} - \frac{1}{q})} ∥ f ∥_{p}, C_{n, p, q, α} : = ∥ D^{α} Γ (\cdot, 1) ∥_{Q} < \infty .$

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $1 \leq p \leq q \leq \infty$ with exponent $Q$ determined by $1 / Q = 1 + 1 / q - 1 / p$ , a multi-index $α$ , $f \in L^{p} (R^{n})$ and $t > 0$ .

[A1]

Countable Choice is the hypothesis carried by the convolution and integration suppliers below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

For every $s > 0$ : $Γ (\cdot, s)$ is smooth, unit mass holds, $Γ (λ x, λ^{2} s) = λ^{- n} Γ (x, s)$ for every $λ > 0$ , and $D^{β} Γ (x, s) = s^{- (n + ∣ β ∣) / 2} (D^{β} Γ) (x / \sqrt{s}, 1)$ for every multi-index $β$ ; moreover $∣ D^{β} Γ (x, 1) ∣ \leq C_{n, β} e^{- ∣ x ∣^{2} / 8}$ (Normalisation, parabolic scaling, heat equation and derivative bounds for the heat kernel).

[F2]

$H_{t} f$ has the everywhere-defined absolutely convergent representative $u (x, t) = \int_{R^{n}} Γ (x - y, t) f (y) d y$ , which is $C^{\infty}$ in $x$ , and $D^{α} H_{t} f = (D^{α} Γ (\cdot, t)) * f$ for every multi-index $α$ (Spatial and time derivatives pass through heat convolution for positive time).

[F3]

For $1 \leq p, q, r \leq \infty$ with $1 / r = 1 / p + 1 / q - 1$ and $f \in L^{p}$ , $g \in L^{q}$ , $∥ f * g ∥_{r} \leq ∥ f ∥_{p} ∥ g ∥_{q}$ (Young's convolution inequality under Countable Choice).

[F4]

With $Q$ as in the statement, $1 / Q = 1 + 1 / q - 1 / p$ and the kernel estimate $∥ H_{t} f ∥_{q} \leq C_{n, p, q} t^{- \frac{n}{2} (\frac{1}{p} - \frac{1}{q})} ∥ f ∥_{p}$ holds ( $L^{p}$ to $L^{q}$ smoothing estimate for the heat flow); the exponent range $1 \leq p \leq q \leq \infty$ makes $Q \in [1, \infty]$ .

Proof

technique · direct

1.1A1F1F2given

Smoothness and derivative identity: by [F2] the representative $u$ is $C^{\infty}$ in $x$ for $t > 0$ and $D^{α} H_{t} f$ is the class of $(D^{α} Γ (\cdot, t)) * f$ , an absolutely convergent integral by the derivative bounds of [F1] together with Hölder and $f \in L^{p}$ .

2.1step 1.1F1givenalgebra

Scaling of the derivative kernel norms: by the derivative scaling identity of [F1], $D^{α} Γ (x, t) = t^{- (n + ∣ α ∣) / 2} (D^{α} Γ) (x / \sqrt{t}, 1)$ ; for $Q < \infty$ , substituting $x = \sqrt{t} z$ gives $∥ D^{α} Γ (\cdot, t) ∥_{Q}^{Q} = t^{- (n + ∣ α ∣) Q / 2} t^{n / 2} ∥ D^{α} Γ (\cdot, 1) ∥_{Q}^{Q}$ , hence $∥ D^{α} Γ (\cdot, t) ∥_{Q} = t^{- ∣ α ∣ / 2} t^{- \frac{n}{2} (1 - \frac{1}{Q})} ∥ D^{α} Γ (\cdot, 1) ∥_{Q}$ . For $Q = \infty$ , taking essential suprema in the same scaling identity gives $∥ D^{α} Γ (\cdot, t) ∥_{\infty} = t^{- (n + ∣ α ∣) / 2} ∥ D^{α} Γ (\cdot, 1) ∥_{\infty}$ , the same formula with $1 / Q = 0$ . The constant $C_{n, p, q, α} = ∥ D^{α} Γ (\cdot, 1) ∥_{Q}$ is finite because $∣ D^{α} Γ (x, 1) ∣ \leq C_{n, α} e^{- ∣ x ∣^{2} / 8}$ by [F1] and the Gaussian $e^{- ∣ \cdot ∣^{2} / 8}$ lies in every $L^{Q}$ , $1 \leq Q \leq \infty$ .

3.1step 1.1step 2.1F3F4givenalgebra

Young estimate for the derivative: applying Young's convolution inequality [F3] with the exponent triple $(p, Q, q)$ , admissible because $1 / Q = 1 + 1 / q - 1 / p$ , gives $∥ D^{α} H_{t} f ∥_{q} \leq ∥ D^{α} Γ (\cdot, t) ∥_{Q} ∥ f ∥_{p}$ , and step 2.1 together with the identity $1 - \frac{1}{Q} = \frac{1}{p} - \frac{1}{q}$ from [F4] turns this into $∥ D^{α} H_{t} f ∥_{q} \leq C_{n, p, q, α} t^{- \frac{∣ α ∣}{2} - \frac{n}{2} (\frac{1}{p} - \frac{1}{q})} ∥ f ∥_{p}$ , which is the stated estimate.

4.1step 1.1step 3.1given∎

Steps 1.1, 2.1 and 3.1 prove the smoothness of the representative, the absolutely convergent convolution formula for $D^{α} H_{t} f$ and the displayed derivative estimate with finite constant $C_{n, p, q, α} = ∥ D^{α} Γ (\cdot, 1) ∥_{Q}$ .

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Spatial analyticity of heat flow at positive time

Statement

Assume Countable Choice. Let $n \geq 1$ , $1 \leq p \leq \infty$ , $f \in L^{p} (R^{n})$ , and $t > 0$ . Then $H_{t} f$ is real analytic on $R^{n}$ (for complex data the real and imaginary parts are real analytic). For every $r > 0$ and multi-index $α$ , $∥ D^{α} H_{t} f ∥_{\infty} \leq α! r^{- ∣ α ∣} M_{n, p, t, r} ∥ f ∥_{p}$ , where $M_{n, p, t, r} = ∥ Γ (\cdot, t) \exp (r \sum_{i} ∣ x_{i} ∣ / (2 t) + n r^{2} / (4 t)) ∥_{p^{'}} < \infty$ , $1 / p + 1 / p^{'} = 1$ . The spatial Taylor series converges absolutely and equals $H_{t} f (x + h)$ for every real $h$ , uniformly for $h$ in each compact box. In particular taking $r = \sqrt{t}$ gives a factorial Gaussian derivative bound $C_{n, p} α! t^{- ∣ α ∣ / 2 - n / (2 p)} ∥ f ∥_{p}$ .

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $1 \leq p \leq \infty$ with conjugate $p^{'}$ , $f \in L^{p} (R^{n})$ , $t > 0$ , $r > 0$ , a centre $a \in R^{n}$ and a multi-index $α$ .

[A1]

Countable Choice is the hypothesis carried by the integration, differentiation and measure-theoretic suppliers below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

The heat kernel is $Γ (w, t) = (4 π t)^{- n / 2} e^{- ∣ w ∣^{2} / (4 t)}$ , where $∣ w ∣^{2} = \sum_{i} w_{i}^{2}$ (The heat kernel on $R^{n}$ and its causal extension).

[F2]

The real exponential is $\exp (u) = \sum_{k \geq 0} u^{k} / k!$ , with the factorial of The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ , and the series converges absolutely for every real $u$ (The real exponential function and the number $e$ by a power series, The exponential series converges absolutely for every real argument); for all real $u, v$ , $\exp (u) \exp (v) = \exp (u + v)$ (The exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$ ).

[F3]

The absolutely convergent representative $u (x, t) = \int_{R^{n}} Γ (x - y, t) f (y) d y$ is defined at every point and is $C^{\infty}$ in $x$ for $t > 0$ , with $D^{α} u (\cdot, t) = (D^{α} Γ_{t}) * f$ (Spatial and time derivatives pass through heat convolution for positive time); its class is $H_{t} f$ .

[F4]

For conjugate exponents and measurable $B, F$ with $B \in L^{p^{'}}$ and $F \in L^{p}$ , $\int ∣ B F ∣ \leq ∥ B ∥_{p^{'}} ∥ F ∥_{p}$ (Holder's inequality for integrals, including the endpoint cases).

[F5]

If $g_{N} \to g$ pointwise and $∣ g_{N} ∣ \leq G$ pointwise with $G$ integrable, then $\int g_{N} \to \int g$ (Dominated convergence).

[F6]

On sigma-finite product spaces, Tonelli's theorem equates the double integrals of a nonnegative measurable function with either iterated integral (Tonelli and Fubini for the completed product, with only almost-everywhere section measurability); the counting measure on $N^{n}$ is sigma-finite, so it may be used as one factor.

[F7]

$\int_{- \infty}^{\infty} e^{- s^{2}} d s = \sqrt{π}$ (The Gaussian integral $\int_{- \infty}^{\infty} e^{- x^{2}} d x = \sqrt{π}$ ).

[F8]

The multi-indexed power series of Multi-indexed power series in $C^{m}$ and their absolute convergence converge absolutely at a point exactly when the associated series of moduli does, and box partial sums converge to the sum of an absolutely convergent series.

[F9]

Fix $m \geq 1$ , $a \in C^{m}$ , a polyradius $r$ and coefficients with $∣ c_{α} ∣ \leq M \prod_{k < m} r_{k}^{- α_{k}}$ . Then $\sum_{α} c_{α} (z - a)^{α}$ converges absolutely and uniformly on each ${\overline{Δ}}_{θ r} (a)$ , $0 < θ < 1$ , its sum $g$ is holomorphic on $Δ_{r} (a)$ , and every iterated complex partial derivative exists there with $\partial_{z}^{β} g (a) = β! c_{β}$ (An absolutely convergent multi-indexed power series is holomorphic and differentiates termwise).

[F10]

A real analytic germ at $a$ is represented on a polydisc by $f (x) = \sum_{α} c_{α} (x - a)^{α}$ with real coefficients $c_{α} = D^{α} f (a) / α!$ , absolutely convergent there (Real analytic germs in several variables).

Proof

technique · direct

1.1A1F1F2F8givenalgebra

Expansion of the translated kernel: put $w = a - y$ and fix real $h$ with $∣ h_{i} ∣ \leq r$ for every $i$ . By [F1] and the addition formula [F2], $Γ (a + h - y, t) = Γ (w, t) \prod_{i < n} \exp (- w_{i} h_{i} / (2 t)) \exp (- h_{i}^{2} / (4 t))$ ; expanding each factor by the exponential series of [F2] and regrouping the finite products gives $Γ (a + h - y, t) = \sum_{α} b_{α} (w) h^{α}$ with $b_{α} (w) = Γ (w, t) \prod_{i < n} \sum_{m_{i} + 2 ℓ_{i} = α_{i}} \frac{(- w_{i} / (2 t))^{m_{i}}}{m_{i}!} \frac{(- 1 / (4 t))^{ℓ_{i}}}{ℓ_{i}!}$ , and the nonnegative series of moduli is bounded by $\sum_{α} ∣ b_{α} (w) ∣ r^{∣ α ∣} \leq Γ (w, t) \prod_{i < n} \exp (r ∣ w_{i} ∣ / (2 t)) \exp (r^{2} / (4 t)) = : B_{r} (w)$ , because the product of the two absolutely convergent exponential series is the absolutely convergent series for the product of the exponentials.

2.1step 1.1F1F6F7givenalgebra

The inequality $2 r ∣ w_{i} ∣ \leq w_{i}^{2} / 2 + 2 r^{2}$ gives $B_{r} (w) \leq (4 π t)^{- n / 2} \exp (3 n r^{2} / (4 t)) \exp (- ∣ w ∣^{2} / (8 t))$ . The last Gaussian is bounded and has integrable positive powers, by the Gaussian integral [F7], a scaling in each coordinate, and Tonelli [F6]. Thus $M_{n, p, t, r} : = ∥ B_{r} ∥_{p^{'}} < \infty$ for every $1 \leq p^{'} \leq \infty$ .

3.1step 1.1step 2.1F4F6given

Coefficient bounds: for every multi-index $α$ define $c_{α} (a) : = \int_{R^{n}} b_{α} (a - y) f (y) d y$ , absolutely convergent because $∣ b_{α} ∣ \leq r^{- ∣ α ∣} B_{r}$ and $B_{r} \in L^{p^{'}}$ with $f \in L^{p}$ by [F4]. Tonelli's theorem [F6] applied to the nonnegative summands over the counting index $α$ and $y \in R^{n}$ gives $\sum_{α} r^{∣ α ∣} ∣ c_{α} (a) ∣ \leq \sum_{α} r^{∣ α ∣} \int ∣ b_{α} (a - y) ∣ ∣ f (y) ∣ d y \leq \int B_{r} (a - y) ∣ f (y) ∣ d y \leq M_{n, p, t, r} ∥ f ∥_{p}$ by step 2.1 and [F4], the bound being uniform in the centre $a$ .

4.1step 1.1step 2.1step 3.1F3F5F8given

Power-series expansion of the representative: fix the centre $a$ and real $h$ with $∣ h_{i} ∣ \leq r$ . For the box partial sums $S_{N} (y) : = \sum_{α_{i} \leq N for all i} b_{α} (a - y) h^{α}$ of step 1.1 one has $S_{N} (y) \to Γ (a + h - y, t)$ pointwise in $y$ and $∣ S_{N} (y) ∣ \leq B_{r} (a - y)$ ; since $B_{r} (a - \cdot) ∣ f ∣ \in L^{1}$ by steps 2.1 and 3.1, dominated convergence [F5] gives $u (a + h, t) = \int Γ (a + h - y, t) f (y) d y = \sum_{α} c_{α} (a) h^{α}$ , where $u$ is the everywhere-defined representative of [F3]. The summable bound $\sum_{α} ∣ c_{α} (a) ∣ r^{∣ α ∣} < \infty$ from step 3.1 also gives uniform absolute convergence on this closed box, by [F8].

5.1step 3.1step 4.1F9given

Holomorphic extension and derivative formula: by step 3.1 the coefficients satisfy $∣ c_{α} (a) ∣ \leq r^{- ∣ α ∣} M_{n, p, t, r} ∥ f ∥_{p}$ for every $α$ , so [F9] with polyradius $(r, \dots, r)$ gives a holomorphic function $g$ on the polydisc $Δ_{r} (a) \subseteq C^{n}$ whose power series is $\sum_{α} c_{α} (a) (z - a)^{α}$ and whose iterated complex partial derivatives at $a$ are $\partial^{β} g (a) = β! c_{β} (a)$ ; by step 4.1 the restriction of $g$ to the real polydisc agrees with the representative $u (\cdot, t)$ , so $D^{α} u (a, t) = α! c_{α} (a)$ and $∣ D^{α} H_{t} f (a) ∣ = α! ∣ c_{α} (a) ∣ \leq α! r^{- ∣ α ∣} M_{n, p, t, r} ∥ f ∥_{p}$ , uniformly in $a$ .

6.1step 4.1step 5.1F3F10given

Real analyticity: if $f$ is real-valued, the coefficients $c_{α} (a)$ of step 4.1 are real and the absolute convergence of $\sum_{α} c_{α} (a) h^{α}$ for every real $h$ (step 4.1 with $r$ arbitrary) exhibits $u (\cdot, t)$ as a real analytic germ at every centre $a$ in the sense of [F10]; for complex $f$ apply the same conclusion to $Re f$ and $Im f$ , which lie in $L^{p}$ with $∥ \cdot ∥_{p} \leq ∥ f ∥_{p}$ , and add the two expansions using linearity of the integral, so the real and imaginary parts of $H_{t} f$ are real analytic.

6.2step 5.1givenalgebra

The choice $r = \sqrt{t}$ : the substitution $w = \sqrt{t} z$ gives $B_{\sqrt{t}} (\sqrt{t} z) = (4 π t)^{- n / 2} e^{- ∣ z ∣^{2} / 4} e^{\sum_{i} ∣ z_{i} ∣ / 2} e^{n / 4}$ and $d w = t^{n / 2} d z$ . For $p^{'} < \infty$ apply this to the $L^{p^{'}}$ integral; for $p^{'} = \infty$ take essential suprema. In both cases $M_{n, p, t, \sqrt{t}} = t^{- n / (2 p)} C_{n, p}$ with $C_{n, p} : = ∥ (4 π)^{- n / 2} e^{- ∣ \cdot ∣^{2} / 4} e^{\sum_{i} ∣ \cdot_{i} ∣ / 2} e^{n / 4} ∥_{p^{'}} < \infty$ ; step 5.1 with this $r$ gives the stated factorial Gaussian bound $C_{n, p} α! t^{- ∣ α ∣ / 2 - n / (2 p)} ∥ f ∥_{p}$ .

7.1step 5.1step 6.1step 6.2given∎

Steps 1.1, 2.1, 3.1, 4.1, 5.1, 6.1 and 6.2 establish the absolutely convergent spatial Taylor expansion of $H_{t} f$ at every centre with coefficients $c_{α} (a)$ , the uniform factorial bound $∥ D^{α} H_{t} f ∥_{\infty} \leq α! r^{- ∣ α ∣} M_{n, p, t, r} ∥ f ∥_{p}$ for every $r > 0$ , the real analyticity of $H_{t} f$ (real and imaginary parts for complex data), and the specialisation $r = \sqrt{t}$ ; the expansion converges for every real displacement because $r$ is arbitrary.

CorollaryStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-6.1-sol)Open item page →

Infinite propagation speed for nonnegative heat data

Statement

Assume Countable Choice and let $n \geq 1$ . Let $f \in L^{1} (R^{n})$ satisfy $f \geq 0$ almost everywhere and $f \neq 0$ . Then for every $t > 0$ and every $x \in R^{n}$ the everywhere-defined integral $H_{t} f (x) = \int_{R^{n}} Γ (x - y, t) f (y) d y$ is strictly positive. In particular, if $f$ is compactly supported, nonnegative and nonzero, then the support of the solution at time $t$ is all of $R^{n}$ for every $t > 0$ .

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $t > 0$ , $x \in R^{n}$ and a representative $f \in L^{1} (R^{n})$ with $f \geq 0$ almost everywhere and $f \neq 0$ .

[A1]

Countable Choice is the hypothesis carried by the kernel and integration suppliers below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

For every $t > 0$ the heat kernel is strictly positive, $Γ (y, t) > 0$ for all $y$ , and $∥ Γ_{t} ∥_{1} = 1$ (The heat kernel on $R^{n}$ and its causal extension, Normalisation, parabolic scaling, heat equation and derivative bounds for the heat kernel).

[F2]

For $f \in L^{1} (R^{n})$ , $H_{t} f$ is the $L^{1}$ class of the convolution $Γ_{t} * f$ , and the class is nonnegative almost everywhere when $f \geq 0$ almost everywhere (The heat evolution $H_{t}$ of initial data, Mass conservation and positivity of the heat flow).

[F3]

For a measurable $g \geq 0$ , $\int g = 0$ if and only if $g = 0$ almost everywhere (A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere).

Proof

technique · direct

1.1A1F3given

The set $A : = {y : f (y) > 0}$ is measurable with $λ_{n} (A) > 0$ : were $λ_{n} (A) = 0$ , then $f \leq 0$ almost everywhere together with the hypothesis $f \geq 0$ almost everywhere would make $f = 0$ almost everywhere, contradicting $f \neq 0$ in $L^{1} (R^{n})$ ; equivalently $\int_{R^{n}} f > 0$ by [F3] applied to the nonnegative function $f$ .

2.1step 1.1F1F3given

Fix $t > 0$ and $x \in R^{n}$ . The integrand $y \mapsto Γ (x - y, t) f (y)$ is measurable and nonnegative almost everywhere, by [F1] and the hypothesis on $f$ , and it is strictly positive for every $y \in A$ , since $Γ (x - y, t) > 0$ everywhere by [F1] and $f (y) > 0$ on $A$ ; as $A$ has positive measure by step 1.1, the nonnegative integrand is positive on a set of positive measure, so its integral is strictly positive by [F3].

3.1step 2.1F1F2given

The integral is finite for every $x$ , because $∣ Γ (x - y, t) f (y) ∣ \leq ∥ Γ_{t} ∥_{\infty} ∣ f (y) ∣$ with $∥ f ∥_{1} < \infty$ , so the integral defining $H_{t} f (x)$ converges absolutely at every point and defines the everywhere-positive representative $H_{t} f (x) > 0$ of the class of [F2].

4.1step 3.1given

Consequently, if in addition $supp f$ is compact, then the set where $H_{t} f$ is nonzero is all of $R^{n}$ by step 3.1, so $supp (H_{t} f) = R^{n}$ for every $t > 0$ ; that is, a compactly supported nonnegative nonzero datum has support spreading to the whole space at every positive time.

5.1step 3.1step 4.1given∎

Steps 1.1, 2.1, 3.1 and 4.1 prove strict positivity of the everywhere-defined integral for every nonnegative nonzero $L^{1}$ datum and the full-space support statement for compactly supported data.

RemarkRemark: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

Diffusivity, rescaling, and the heat kernel compared with the Poisson kernels

Remarks

Assume Countable Choice for the kernel identities cited below.

With diffusivity parameter $κ > 0$ , the equation $u_{t} - κ Δ u = 0$ is converted into the $κ = 1$ heat equation of The heat operator, the heat equation, and the Cauchy problem by $v (x, t) : = u (x, t / κ)$ , and its kernel is $Γ_{κ} (x, t) = (4 π κ t)^{- n / 2} e^{- ∣ x ∣^{2} / (4 κ t)}$ ; the kernel $Γ$ of The heat kernel on $R^{n}$ and its causal extension is the case $κ = 1$ . The rescaling is the chain rule: $\partial_{t} v (x, t) = κ^{- 1} \partial_{t} u (x, t / κ)$ while $Δ_{x} v (x, t) = (Δ u) (x, t / κ)$ , so $v_{t} = Δ v$ is equivalent to $u_{t} = κ Δ u$ ; the identity $Γ_{κ} (x, t) = Γ (x, κ t)$ gives the unit-mass normalisation of Normalisation, parabolic scaling, heat equation and derivative bounds for the heat kernel.

For $n \geq 3$ , the ball Poisson kernel $P_{R, a} (x, y) = (R^{2} - ∣ x - a ∣^{2}) / (R ω_{n - 1} ∣ x - y ∣^{n})$ of Poisson kernel of a Euclidean ball is a boundary-value kernel: its two slots are an interior point and a boundary point, it carries no time parameter, and its dependence on $x$ is not translation-invariant convolution, so it is not the heat kernel in other notation. For the half-space kernel of Poisson kernel and bounded Dirichlet problem on a half-space, a boundary mode $e^{2 π i ξ \cdot x^{'}}$ has bounded harmonic extension $e^{- 2 π ∣ ξ ∣ t} e^{2 π i ξ \cdot x^{'}}$ : direct differentiation shows it is harmonic and the bounded Dirichlet uniqueness identifies it with the Poisson integral. Its derivative at $t = 0$ is $- 2 π ∣ ξ ∣$ times that mode, whereas $Δ_{x^{'}}$ multiplies the mode by $- 4 π^{2} ∣ ξ ∣^{2}$ . This is the Fourier-symbol meaning of the Poisson generator $- \sqrt{- Δ}$ , and distinguishes its normal-time family from the heat semigroup. In three spatial dimensions the homogeneous wave representation uses spherical means at radius $c t$ , rather than a positive Gaussian convolution; this comparison is dimension-specific. A source's diffusivity normalisation must be matched before its kernel formula is quoted.

5 · Examples, counterexamples and false statements

None yet.

The Heat Kernel and the Cauchy Problem

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

The heat operator, the heat equation, and the Cauchy problem

Definition

The heat kernel on RnRn and its causal extension

Definition

Normalisation, parabolic scaling, heat equation and derivative bounds for the heat kernel

Statement

Facts & Assumptions

Proof

First and second Gaussian heat-kernel moments

Statement

Facts & Assumptions

Proof

Gaussian kernels form an approximate identity

Statement

Facts & Assumptions

Proof

The heat kernel semigroup identity Γt∗Γs=Γt+sΓt​∗Γs​=Γt+s​

Statement

Facts & Assumptions

Proof

The causal heat kernel is the fundamental solution of the heat operator

Statement

Facts & Assumptions

Proof

The heat evolution HtHt​ of initial data

Definition

Spatial and time derivatives pass through heat convolution for positive time

Statement

Facts & Assumptions

Proof

The heat Cauchy problem for bounded uniformly continuous data

Statement

Facts & Assumptions

Proof

The heat Cauchy problem for LpLp data

Statement

Facts & Assumptions

Proof

Uniqueness of strongly continuous mild heat solutions

Statement

Facts & Assumptions

Proof

Heat generator at zero on compactly supported smooth data

Statement

Facts & Assumptions

Proof

Mass conservation and positivity of the heat flow

Statement

Facts & Assumptions

Proof

Monotonicity and LpLp contractivity of the heat flow

Statement

Facts & Assumptions

Proof

LpLp to LqLq smoothing estimate for the heat flow

Statement

Facts & Assumptions

Proof

Spatial derivative estimates for the heat flow

Statement

Facts & Assumptions

Proof

Spatial analyticity of heat flow at positive time

Statement

Facts & Assumptions

Proof

Infinite propagation speed for nonnegative heat data

Statement

Facts & Assumptions

Proof

Diffusivity, rescaling, and the heat kernel compared with the Poisson kernels

Remarks

5 · Examples, counterexamples and false statements

Sources

The heat kernel on $R^{n}$ and its causal extension

The heat kernel semigroup identity $Γ_{t} * Γ_{s} = Γ_{t + s}$

The heat evolution $H_{t}$ of initial data

The heat Cauchy problem for $L^{p}$ data

Monotonicity and $L^{p}$ contractivity of the heat flow

$L^{p}$ to $L^{q}$ smoothing estimate for the heat flow