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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]

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).

[F5]

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.

[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); a translation of $R^{n}$ has $∣ \det D T ∣ = 1$ (Integral invariance under measure-preserving maps).

[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 $R^{n}$ .

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Sources

John K. Hunter, Notes on Partial Differential Equations (revised 18 June 2014, UC Davis) (standard reference, not scraped)
Gerald Teschl, Partial Differential Equations: From Classical to Modern (2025 archived author manuscript, AMS Graduate Studies in Mathematics) (standard reference, not scraped)
Jared Speck, MIT 18.152 Introduction to Partial Differential Equations, Class Meeting #5: The Fundamental Solution for the Heat Equation (Fall 2011) (standard reference, not scraped)

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