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✓ 3 results · all verified · 0 also independently AI-judged

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Fourier Multipliers and Sobolev Characterisations — Examples

1 · Prerequisites

2 · Summary

These examples and recorded leaves exercise the multiplier and Sobolev conventions of the companion page, all in the negative-sign $2 π$ normalization with complex scalars and Countable Choice as declared on the individual items.

The heat semigroup multiplies the Fourier transform by $e^{- 4 π^{2} t ∣ ξ ∣^{2}}$ and the Poisson semigroup by $e^{- 2 π t ∣ ξ ∣}$ ; both are bounded $L^{2}$ multiplier extensions with norm one, compose as semigroups, and solve their distributional evolution equations for $t > 0$ . Translation has the unimodular symbol $e^{- 2 π i a \cdot ξ}$ and so extends to an $L^{2}$ isometry, whereas the distributional derivative $\partial_{j}$ has the unbounded symbol $2 π i ξ_{j}$ ; explicit frequency-localized bumps show that no bounded $L^{2}$ extension exists, which is the sharp contrast between bounded and unbounded multiplier symbols.

Two recorded leaves are orientation only and are not proved here. The ball indicator is an $L^{2}$ multiplier of norm one, but Fefferman's theorem records that for $n \geq 2$ it is an $L^{p}$ multiplier only at $p = 2$ ; and the shifted signum symbol $sgn (ξ - 1)$ is a bounded jump multiplier on $R$ outside the punctured-domain Mihlin class, its $L^{p}$ bound for $1 < p < \infty$ being the Hilbert-transform theorem deferred to the later singular-integral page. Neither leaf is used as a supplier anywhere.

The final example evaluates the weighted Fourier criterion at a Dirac mass: since $F δ_{0} = 1$ , the membership $δ_{0} \in H^{s} (R^{n})$ is equivalent to the finiteness of $\int_{R^{n}} ⟨ ξ ⟩^{2 s} d ξ$ , which polar coordinates reduce to a real $p$ -series block comparison. The outcome is the strict criterion $δ_{0} \in H^{s} ⟺ s < - n / 2$ , with the borderline $s = - n / 2$ failing through the logarithmic divergence of $\int_{1}^{R} r^{- 1} d r$ .

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-09-30Open item page →

Heat and Poisson semigroups as Fourier multipliers

Example

Assume Countable Choice, let $n \geq 1$ , and use the complex $L^{2}$ conventions of Complex Lp classes and Euclidean test-function conventions. For $t \geq 0$ define the bounded continuous frequency symbols $m_{t} (ξ) = e^{- 4 π^{2} t ∣ ξ ∣^{2}}, p_{t} (ξ) = e^{- 2 π t ∣ ξ ∣}, ξ \in R^{n},$ and let $H_{t} : = T_{m_{t}}, P_{t} : = T_{p_{t}}$ be the unique bounded extensions to $L^{2} (R^{n}; C)$ that Exact L2 Fourier multiplier norm supplies for the Schwartz-core multiplier of Translation-invariant Fourier multiplier on the Schwartz core; explicitly $H_{t} = F_{2}^{- 1} M_{m_{t}} F_{2}, P_{t} = F_{2}^{- 1} M_{p_{t}} F_{2} .$ These are the operators customarily written $e^{t Δ}$ and $e^{- t \sqrt{- Δ}}$ . They are defined here only through the bounded Fourier symbols: no spectral theorem, generator, or functional calculus is assumed. Then:

Both families are $L^{2}$ contractions, with operator norm exactly one: $∥ H_{t} f ∥_{2} \leq ∥ f ∥_{2}$ and $∥ P_{t} f ∥_{2} \leq ∥ f ∥_{2}$ for all $f \in L^{2} (R^{n}; C)$ and all $t \geq 0$ .
$H_{0} = P_{0} = {i d}_{L^{2} (R^{n}; C)}$ .
$H_{t} H_{r} = H_{t + r}$ and $P_{t} P_{r} = P_{t + r}$ for all $t, r \geq 0$ .
For every $f \in L^{2} (R^{n}; C)$ and every $t > 0$ the map $s \mapsto H_{s} f$ is differentiable from $(0, \infty)$ into $L^{2}$ with derivative $u^{'} (t) = F_{2}^{- 1} (- 4 π^{2} ∣ ξ ∣^{2} m_{t} (ξ) F_{2} f)$ , and the distributional Laplacian in $x$ of $u (t) = H_{t} f$ is the regular distribution of the $L^{2}$ class $u^{'} (t)$ ; that is, $u$ solves the heat equation $\partial_{t} u = Δ u$ for $t > 0$ . Likewise $s \mapsto P_{s} f$ is twice differentiable on $(0, \infty)$ and $w (t) = P_{t} f$ satisfies the upper-half-space Laplace equation $\partial_{t}^{2} w + Δ_{x} w = 0$ in $S^{'} (R^{n})$ for every $t > 0$ .

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , an $L^{2}$ class $f \in L^{2} (R^{n}; C)$ , and real parameters $t, r \geq 0$ and $h$ with $0 < ∣ h ∣ \leq t / 2$ wherever these appear.

[A1]

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

[F1]

On $S (R^{n})$ the multiplier with symbol $m$ has domain $D_{m} = {f : m \hat{f} locally integrable with tempered regular distribution}$ and acts by $T_{m} f = F^{- 1} (u_{m \hat{f}})$ (Translation-invariant Fourier multiplier on the Schwartz core).

[F2]

If $m$ is measurable with $M = ∥ m ∥_{\infty} = ess sup ∣ m ∣ < \infty$ , then $S \subseteq D_{m}$ , $T_{m} f = F_{2}^{- 1} (m F_{2} f)$ as $L^{2}$ classes for Schwartz $f$ , and $T_{m}$ has a unique bounded $L^{2}$ extension $T_{m} = F_{2}^{- 1} M_{m} F_{2}$ whose operator norm is exactly $M$ (Exact L2 Fourier multiplier norm).

[F3]

Plancherel $F_{2}$ is a surjective complex-linear isometry of $L^{2} (R^{n}; C)$ , so $F_{2}^{- 1}$ exists, is complex-linear, and preserves norms (Plancherel theorem).

[F4]

[F5]

For an $L^{2}$ class $h$ with regular distribution $u_{h}$ , $F u_{h} = u_{F_{2} h}$ in $S^{'} (R^{n})$ (Fourier transform agrees with l one and plancherel transforms).

[F6]

$F$ is a topological automorphism of $S^{'} (R^{n})$ , in particular injective with inverse $F^{- 1}$ (Fourier transform is a topological automorphism of tempered distributions).

[F7]

A smooth function $a$ whose derivatives are all polynomially bounded multiplies $S^{'}$ by $⟨ a u, φ ⟩ = ⟨ u, a φ ⟩$ (Smooth polynomially bounded multipliers on schwartz space); if $g, a g \in L^{2}$ , their regular distributions are tempered by [F5], and for every Schwartz test $φ$ the integrals $⟨ a u_{g}, φ ⟩ = \int g (a φ) = \int (a g) φ = ⟨ u_{a g}, φ ⟩$ converge by Cauchy–Schwarz, since $a φ \in S \subseteq L^{2}$ . Thus $a u_{g} = u_{a g}$ in the case used below. The underlying compact-test regular functional is that of Regular distribution from a locally integrable function.

[F8]

Dominated convergence: if measurable $F_{j}$ satisfy $F_{j} \to F$ almost everywhere and $∣ F_{j} ∣ \leq G$ almost everywhere for one nonnegative integrable $G$ , then $\int ∣ F_{j} - F ∣ \to 0$ (Dominated convergence).

[F9]

$\exp (x + y) = \exp (x) \exp (y)$ for real $x, y$ , and the real exponential is the power series of The real exponential function and the number $e$ by a power series, so $\exp (0) = 1$ (The exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$ ).

[F10]

$1 + x \leq \exp (x)$ for every real $x$ ( $1 + x \leq \exp (x)$ for every real $x$ , hence $(1 - p)^{m} \leq \exp (- m p)$ ).

[F11]

If $g$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , there is $c \in (a, b)$ with $g (b) - g (a) = g^{'} (c) (b - a)$ (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 $f (b) - f (a) = f^{'} (c) (b - a)$ ).

[F12]

$(\exp)^{'} = \exp$ , and the chain rule computes $(f \circ g)^{'} (s) = f^{'} (g (s)) g^{'} (s)$ (The exponential function is smooth and $(\exp)^{'} = \exp$ , 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 $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ ).

[F13]

The essential supremum of a measurable function is the least essential bound: if $∣ a ∣ \leq L$ almost everywhere then $∥ a ∥_{\infty} \leq L$ (The essential supremum is attained as the least essential bound).

[F14]

Every nonempty Euclidean ball has positive finite Lebesgue measure (Euclidean balls have positive finite Lebesgue measure).

Verification

Given: The data and conventions of the Example above and of [F1]-[F14].

1.1F9F10F13F14algebra

Each map $ξ \mapsto m_{t} (ξ) = e^{- 4 π^{2} t ∣ ξ ∣^{2}}$ is continuous, and $∣ m_{t} ∣ \leq 1$ everywhere because $- 4 π^{2} t ∣ ξ ∣^{2} \leq 0$ ; the same holds for $p_{t}$ with $∣ p_{t} ∣ \leq 1$ . At $ξ = 0$ both symbols equal $1$ by [F9]. For $ε > 0$ , continuity at $0$ gives a ball on which $∣ m_{t} ∣ > 1 - ε$ and $∣ p_{t} ∣ > 1 - ε$ , these balls have positive measure by [F14], so no number below $1$ is an essential bound; with [F13] this gives $∥ m_{t} ∥_{\infty} = ∥ p_{t} ∥_{\infty} = 1$ for every $t \geq 0$ .

1.2F10algebra

For all $t > 0$ and $r \geq 0$ : (i) $4 π^{2} r^{2} e^{- 4 π^{2} t r^{2}} \leq \frac{1}{e t}$ , (ii) $2 π r e^{- 2 π t r} \leq \frac{1}{e t}$ , (iii) $4 π^{2} r^{2} e^{- 2 π t r} \leq \frac{4}{e^{2} t^{2}}$ . Indeed [F10] with $x = s - 1$ gives $0 < s e^{- s} \leq e^{- 1}$ for every real $s > 0$ ; substituting $s = 4 π^{2} t r^{2}$ proves (i), substituting $s = 2 π t r$ proves (ii), and writing $s^{2} e^{- s} = (2 \cdot \frac{s}{2} e^{- s / 2})^{2} \leq (2 e^{- 1})^{2}$ with the same substitution proves (iii).

2.1F1F2F3F9step 1.1

Step 1.1 bounds the symbols, so the operators under discussion are the domain-qualified Schwartz multipliers $T_{m_{t}}, T_{p_{t}}$ of [F1], and [F2] gives $S \subseteq D_{m_{t}} \cap D_{p_{t}}$ together with, for every $t \geq 0$ , the unique bounded extensions $H_{t} = T_{m_{t}}$ and $P_{t} = T_{p_{t}}$ on $L^{2} (R^{n}; C)$ with $H_{t} = F_{2}^{- 1} M_{m_{t}} F_{2}$ , $P_{t} = F_{2}^{- 1} M_{p_{t}} F_{2}$ and operator norm $∥ m_{t} ∥_{\infty} = ∥ p_{t} ∥_{\infty} = 1$ ; in particular $∥ H_{t} f ∥_{2} \leq ∥ f ∥_{2}$ and $∥ P_{t} f ∥_{2} \leq ∥ f ∥_{2}$ for every $f \in L^{2}$ . Since $m_{0} = p_{0} = 1$ by [F9], the same formulas and [F3] give $H_{0} = P_{0} = F_{2}^{- 1} F_{2} = {i d}_{L^{2}}$ .

2.2F3step 1.2algebra

Fix $f \in L^{2}$ and $t > 0$ and put $g_{t} = - 4 π^{2} ∣ ξ ∣^{2} m_{t} F_{2} f$ , $q_{t} = - 2 π ∣ ξ ∣ p_{t} F_{2} f$ and ${\tilde{g}}_{t} = 4 π^{2} ∣ ξ ∣^{2} p_{t} F_{2} f$ . Each is a measurable function of $ξ$ and step 1.2 bounds its modulus by $\frac{1}{e t} ∣ F_{2} f ∣$ , $\frac{1}{e t} ∣ F_{2} f ∣$ and $\frac{4}{e^{2} t^{2}} ∣ F_{2} f ∣$ respectively; hence by [F3], $g_{t}, q_{t}, {\tilde{g}}_{t} \in L^{2} (R^{n}; C)$ with $∥ g_{t} ∥_{2} \leq \frac{1}{e t} ∥ f ∥_{2}$ , $∥ q_{t} ∥_{2} \leq \frac{1}{e t} ∥ f ∥_{2}$ and $∥ {\tilde{g}}_{t} ∥_{2} \leq \frac{4}{e^{2} t^{2}} ∥ f ∥_{2}$ .

3.1F3F9step 2.1

For all $t, r \geq 0$ the addition law [F9] gives the pointwise symbol identities $m_{t} m_{r} = m_{t + r}$ and $p_{t} p_{r} = p_{t + r}$ . Substituting the explicit formulas of step 2.1 and using that composition of multiplication operators multiplies symbols, $H_{t} H_{r} = F_{2}^{- 1} M_{m_{t}} F_{2} F_{2}^{- 1} M_{m_{r}} F_{2} = F_{2}^{- 1} M_{m_{t} m_{r}} F_{2} = H_{t + r},$ and identically $P_{t} P_{r} = P_{t + r}$ ; the case $t = r = 0$ reduces to $H_{0}^{2} = H_{0}$ , consistent with step 2.1.

3.2F3F8F11F12step 1.2step 2.1step 2.2

(Heat, first time derivative.) Fix $f \in L^{2}$ and $t > 0$ . For $0 < ∣ h ∣ \leq t / 2$ , step 2.1 and linearity of $F_{2}^{- 1}$ write the difference quotient as $\frac{H_{t + h} f - H_{t} f}{h} = F_{2}^{- 1} (\frac{m_{t + h} - m_{t}}{h} F_{2} f)$ . For fixed $ξ$ the function $s \mapsto m_{s} (ξ) = e^{- 4 π^{2} s ∣ ξ ∣^{2}}$ is differentiable on the interval with endpoints $t + h$ and $t$ (both positive), with derivative $s \mapsto - 4 π^{2} ∣ ξ ∣^{2} e^{- 4 π^{2} s ∣ ξ ∣^{2}}$ by [F12]; [F11] therefore gives a point $s_{h} (ξ)$ between $t + h$ and $t$ with $\frac{m_{t + h} (ξ) - m_{t} (ξ)}{h} = - 4 π^{2} ∣ ξ ∣^{2} e^{- 4 π^{2} s_{h} (ξ) ∣ ξ ∣^{2}}$ . As $h \to 0$ one has $s_{h} (ξ) \to t$ , so the quotient tends to $- 4 π^{2} ∣ ξ ∣^{2} m_{t} (ξ)$ pointwise. Since $s_{h} (ξ) \geq t / 2$ , step 1.2(i) with $t / 2$ bounds the quotient by $\frac{2}{e t}$ , while step 1.2(i) also gives $∣ 4 π^{2} ∣ ξ ∣^{2} m_{t} (ξ) ∣ \leq \frac{1}{e t}$ . Hence $F_{h} = ∣ \frac{m_{t + h} - m_{t}}{h} + 4 π^{2} ∣ ξ ∣^{2} m_{t} ∣^{2} ∣ F_{2} f ∣^{2}$ tends to $0$ pointwise and is dominated by $\frac{9}{e^{2} t^{2}} ∣ F_{2} f ∣^{2}$ , which is integrable; for every sequence $h_{j} \to 0$ with $∣ h_{j} ∣ \leq t / 2$ , [F8] gives $\int F_{h_{j}} \to 0$ , and the isometry [F3] converts this into $∥ \frac{H_{t + h_{j}} f - H_{t} f}{h_{j}} - F_{2}^{- 1} (g_{t}) ∥_{2} \to 0$ with $g_{t}$ from step 2.2, which is the two-sided limit statement. Thus $s \mapsto H_{s} f$ is differentiable on $(0, \infty)$ with $u^{'} (t) = F_{2}^{- 1} (g_{t}) = F_{2}^{- 1} (- 4 π^{2} ∣ ξ ∣^{2} m_{t} F_{2} f)$ .

3.3F3F8F11F12step 1.2step 2.1step 2.2

(Poisson, first time derivative.) The same computation with $p_{t}$ in place of $m_{t}$ : for fixed $ξ$ the function $s \mapsto p_{s} (ξ) = e^{- 2 π s ∣ ξ ∣}$ has derivative $- 2 π ∣ ξ ∣ e^{- 2 π s ∣ ξ ∣}$ by [F12], so [F11] gives points with quotient tending pointwise to $- 2 π ∣ ξ ∣ p_{t} (ξ)$ ; step 1.2(ii) bounds the quotient by $\frac{2}{e t}$ and the limit symbol by $\frac{1}{e t}$ , so [F8] and [F3] give that $s \mapsto P_{s} f$ is differentiable on $(0, \infty)$ with $w^{'} (t) = F_{2}^{- 1} (q_{t})$ , $q_{t}$ as in step 2.2.

4.1F4F5F6F7step 2.1step 2.2step 3.2

(Heat equation.) Let $u (t) = H_{t} f$ . By step 2.1, $F_{2} u (t) = m_{t} F_{2} f$ , so [F5] gives $F (u_{u (t)}) = u_{m_{t} F_{2} f}$ . Summing the coordinate identities of [F4] with $∣ α ∣ = 2$ gives $F (Δ u_{u (t)}) = \sum_{j} (2 π i ξ_{j})^{2} F (u_{u (t)}) = - 4 π^{2} ∣ ξ ∣^{2} u_{m_{t} F_{2} f}$ , and the polynomial $- 4 π^{2} ∣ ξ ∣^{2}$ together with [F7] identifies this as the regular distribution $u_{g_{t}}$ of the $L^{2}$ class $g_{t}$ of step 2.2. Since [F5] applied to $h = F_{2}^{- 1} g_{t}$ gives $F (u_{F_{2}^{- 1} g_{t}}) = u_{g_{t}}$ , injectivity of $F$ on $S^{'}$ [F6] yields $Δ u_{u (t)} = u_{F_{2}^{- 1} g_{t}}$ . By step 3.2 the class $u^{'} (t)$ is exactly $F_{2}^{- 1} g_{t}$ , so for every $t > 0$ the distributional Laplacian of $u (t) = H_{t} f$ is the regular distribution of the strong $L^{2}$ derivative $u^{'} (t)$ : the solution satisfies $\partial_{t} u = Δ u$ for $t > 0$ .

4.2F3F8F11F12step 1.2step 2.2step 3.3

(Poisson, second time derivative.) Apply the argument of step 3.2 to the family the scalar symbols $b_{s} (ξ) = - 2 π ∣ ξ ∣ p_{s} (ξ)$ , so that $q_{s} = b_{s} F_{2} f$ is the $L^{2}$ class of step 2.2: for fixed $ξ$ , the map $s \mapsto - 2 π ∣ ξ ∣ e^{- 2 π s ∣ ξ ∣}$ has derivative $4 π^{2} ∣ ξ ∣^{2} p_{s} (ξ)$ by [F12], so [F11] gives points $σ_{h} (ξ) \geq t / 2$ with $\frac{b_{t + h} (ξ) - b_{t} (ξ)}{h} = 4 π^{2} ∣ ξ ∣^{2} e^{- 2 π σ_{h} (ξ) ∣ ξ ∣}$ tending to $4 π^{2} ∣ ξ ∣^{2} p_{t} (ξ)$ pointwise, and step 1.2(iii) with $t / 2$ in place of $t$ bounds these quotients by $\frac{16}{e^{2} t^{2}}$ while step 1.2(iii) bounds $∣ 4 π^{2} ∣ ξ ∣^{2} p_{t} (ξ) ∣$ by $\frac{4}{e^{2} t^{2}}$ . Hence $F_{h} = ∣ \frac{b_{t + h} - b_{t}}{h} - 4 π^{2} ∣ ξ ∣^{2} p_{t} ∣^{2} ∣ F_{2} f ∣^{2} \to 0$ pointwise, dominated by $(\frac{20}{e^{2} t^{2}})^{2} ∣ F_{2} f ∣^{2}$ ; [F8] and [F3] give $∥ \frac{w^{'} (t + h) - w^{'} (t)}{h} - F_{2}^{- 1} ({\tilde{g}}_{t}) ∥_{2} \to 0$ with ${\tilde{g}}_{t}$ of step 2.2, so $s \mapsto P_{s} f$ is twice differentiable on $(0, \infty)$ with $w^{''} (t) = F_{2}^{- 1} ({\tilde{g}}_{t})$ .

5.1F4F5F6F7step 2.1step 4.2

(Upper-half-space Laplace equation.) Let $w (t) = P_{t} f$ . By step 2.1, $F_{2} w (t) = p_{t} F_{2} f$ , so [F5] and [F4] with $∣ α ∣ = 2$ give $F (Δ_{x} w (t)) = - 4 π^{2} ∣ ξ ∣^{2} u_{p_{t} F_{2} f} = u_{- {\tilde{g}}_{t}}$ by [F7]. Step 4.2 gives $F (u_{w^{''} (t)}) = u_{{\tilde{g}}_{t}}$ by [F5], so by linearity of $F$ the distribution $u_{w^{''} (t)} + Δ_{x} w (t)$ has Fourier transform $u_{{\tilde{g}}_{t}} + u_{- {\tilde{g}}_{t}} = 0$ ; injectivity of $F$ [F6] gives $w^{''} (t) + Δ_{x} w (t) = 0$ in $S^{'} (R^{n})$ for every $t > 0$ , the upper-half-space Laplace equation with boundary control left entirely to the symbol $e^{- 2 π t ∣ ξ ∣}$ .

6.1A1step 2.1step 3.1step 4.1step 5.1∎

The operators $H_{t}, P_{t}$ are defined only through the bounded symbols $m_{t}, p_{t}$ by [F2] (step 2.1): step 2.1 gives the contraction and identity claims, step 3.1 the semigroup laws, step 4.1 the heat equation, and step 5.1 the upper-half-space Laplace equation, which are exactly the four asserted properties; Countable Choice enters only through the cited published interfaces of [A1], and no spectral theorem is used.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-09-30Open item page →

Translation and differentiation symbols

Example

Assume Countable Choice and let $n \geq 1$ with the complex $L^{2}$ conventions of Complex Lp classes and Euclidean test-function conventions. For $a \in R^{n}$ let $τ_{a} f (x) = f (x - a)$ be the translate of Translation of a function on $R^{n}$ , acting on Schwarz space, and let $\partial_{j}$ be the $j$ -th coordinate derivative. Then:

The Fourier multiplier with symbol $m (ξ) = e^{- 2 π i a \cdot ξ}$ acts on $S (R^{n})$ exactly as $τ_{a}$ : $T_{m} = τ_{a}$ on the Schwartz core, and $τ_{a}$ extends to an isometry of $L^{2} (R^{n}; C)$ .
The distributional derivative $\partial_{j}$ has the Fourier symbol $σ (ξ) = 2 π i ξ_{j}$ : for every $u \in S^{'} (R^{n})$ , $F (\partial_{j} u) = σ F u$ , and on Schwartz functions the multiplier with symbol $σ$ is the classical quotient derivative.
Although $σ$ is smooth and polynomially bounded, it is unbounded, and $\partial_{j}$ admits no bounded $L^{2}$ extension from the Schwartz core: no bounded linear $P : L^{2} (R^{n}; C) \to L^{2} (R^{n}; C)$ satisfies $P f = \partial_{j} f$ for every Schwartz $f$ .
No Mihlin-type criterion is invoked here; the obstruction in part 3 is the elementary frequency growth of $∣ σ (ξ) ∣ = 2 π ∣ ξ_{j} ∣$ .

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $a \in R^{n}$ , $j \in {1, \dots, n}$ , the Schwartz core conventions of Translation-invariant Fourier multiplier on the Schwartz core, and an $L^{2}$ class $g$ where a norm estimate is stated.

[A1]

Countable Choice is inherited from every cited interface below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

On $S (R^{n})$ the transform is the integral transform; $F$ is a topological automorphism of $S$ with $F^{- 1} = R F$ , $R f = f (- \cdot)$ , and for $f \in L^{1}$ the distributional transform of the regular distribution is $F u_{f} = u_{\hat{f}}$ (Fourier transform is a topological automorphism of Schwartz space, Fourier transform agrees with l one and plancherel transforms).

[F2]

Translation is continuous on $S$ , is the convention $τ_{a} f (x) = f (x - a)$ , and satisfies the $L^{1}$ translation law $\hat{τ_{a} f} (ξ) = e^{- 2 π i a \cdot ξ} \hat{f} (ξ)$ (Basic operations are continuous on Schwartz space, Translation of a function on $R^{n}$ , Translation, modulation, linear dilation and reflection laws).

[F3]

Every Schwartz function is integrable (Schwartz derivatives are integrable), and for $f \in D_{m}$ , the multiplier is $T_{m} f = F^{- 1} (u_{m \hat{f}})$ (Translation-invariant Fourier multiplier on the Schwartz core).

[F4]

If $m$ is measurable with finite essential supremum, then $S \subseteq D_{m}$ , $T_{m} f = F_{2}^{- 1} (m F_{2} f)$ as $L^{2}$ classes for Schwartz $f$ , and $T_{m}$ has a unique bounded $L^{2}$ extension with operator norm $∥ m ∥_{\infty}$ (Exact L2 Fourier multiplier norm).

[F5]

Plancherel $F_{2}$ is a surjective complex-linear isometry of $L^{2} (R^{n}; C)$ extending the Schwartz transform, and the Schwartz classes are dense in $L^{2}$ (Plancherel theorem, Schwartz space is dense in L2).

[F6]

For $u \in S^{'} (R^{n})$ and every multi-index $α$ , $F (\partial^{α} u) = (2 π i ξ)^{α} F u$ in $S^{'} (R^{n})$ (Fourier differentiation and multiplication identities on tempered distributions), and for $L^{2}$ classes $f, h$ with $u_{f} = u$ , $u_{h} = \partial^{α} u$ one has $F_{2} h = (2 π i ξ)^{α} F_{2} f$ almost everywhere (Distributional derivatives are polynomial Fourier multipliers).

[F7]

On Schwartz space, $\partial_{j}$ maps $S$ continuously into $S$ and $F (\partial_{j} f) (ξ) = 2 π i ξ_{j} \hat{f} (ξ)$ pointwise for every $f \in S$ (Basic operations are continuous on Schwartz space, Fourier transform acts continuously on Schwartz space).

[F8]

$F$ is a topological automorphism of $S^{'} (R^{n})$ , in particular injective with inverse $F^{- 1}$ , so $F^{- 1} F u = u$ for every tempered distribution (Fourier transform is a topological automorphism of tempered distributions).

[F9]

If $a$ is smooth with every derivative polynomially bounded, it maps Schwartz functions to Schwartz functions and multiplies tempered distributions by $⟨ a u, φ ⟩ = ⟨ u, a φ ⟩$ (Smooth polynomially bounded multipliers on schwartz space). For the Schwartz function $h = \hat{f}$ used below, both $h$ and $a h$ are Schwartz. Their regular pairings are absolutely convergent and $⟨ a u_{h}, φ ⟩ = \int h (a φ) = \int (a h) φ = ⟨ u_{a h}, φ ⟩$ , so $a u_{h} = u_{a h}$ (Regular distribution from a locally integrable function, [F3]).

[F10]

$∣ \exp (x + i y) ∣ = e^{x}$ for real $x, y$ , so $∣ e^{i θ} ∣ = 1$ for every real $θ$ ( $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ).

[F11]

For every compact $K$ inside an open $U \subseteq R^{n}$ there is a smooth $ρ : R^{n} \to [0, 1]$ with $ρ = 1$ on $K$ and $supp ρ \subseteq U$ (A Euclidean bump for a compact set inside an open set).

[F12]

The natural inclusion $D (R^{n}) = C_{c}^{\infty} (R^{n}) ↪ S (R^{n})$ is continuous with dense image, so each compactly supported smooth function is Schwartz (Test function inclusion in schwartz space is continuous).

[F13]

A closed bounded subset of Euclidean space is compact (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

Verification

Given: The data, symbols and conventions of the Example and of [F1]-[F13].

1.1F10algebra

The symbol $m (ξ) = e^{- 2 π i a \cdot ξ}$ is continuous, and [F10] with $x = 0$ and $y = - 2 π a \cdot ξ$ gives $∣ m (ξ) ∣ = 1$ for every $ξ$ ; hence $∥ m ∥_{\infty} = 1$ . The symbol $σ (ξ) = 2 π i ξ_{j}$ is smooth with $∣ σ (ξ) ∣ = 2 π ∣ ξ_{j} ∣ \leq 2 π (1 + ∣ ξ ∣)$ , so all its derivatives are polynomially bounded, and $∣ σ (t e_{j}) ∣ = 2 π ∣ t ∣ \to \infty$ as $t \to \infty$ shows it is unbounded.

1.2F1F2F3F7

For $f \in S$ : [F2] gives $τ_{a} f \in S$ with $\hat{τ_{a} f} (ξ) = m (ξ) \hat{f} (ξ)$ at every $ξ$ , and $f \in L^{1}$ by [F3], so the $L^{1}$ case of [F1] gives the distributional identity $F u_{τ_{a} f} = u_{m \hat{f}}$ . Likewise [F7] gives $\partial_{j} f \in S$ with $F (\partial_{j} f) = σ \hat{f}$ and hence $F u_{\partial_{j} f} = u_{σ \hat{f}}$ .

1.3F1F11F12F13algebra

Fix $N > 0$ and put $η = (N + 2) e_{j}$ . The singleton $K = {η}$ is compact, $U = B (η, 1) \subseteq {ξ \in R^{n} : ξ_{j} > N}$ is bounded and open, and $K \subseteq U$ ; [F11] gives a smooth $ψ_{N}$ with $0 \leq ψ_{N} \leq 1$ , $ψ_{N} (η) = 1$ and $supp ψ_{N} \subseteq U$ . Its support is closed and bounded, hence compact by [F13], so $ψ_{N} \in C_{c}^{\infty}$ . Then $ψ_{N} \neq 0$ , and $ψ_{N} \in S$ by [F12]; by [F1] the class $f_{N} : = F^{- 1} ψ_{N}$ lies in $S$ with $F f_{N} = ψ_{N}$ .

2.1F4F8step 1.1step 1.2

Since $m$ is bounded and measurable by step 1.1, [F4] gives $S \subseteq D_{m}$ and $T_{m} f = F^{- 1} (u_{m \hat{f}})$ for $f \in S$ ; substituting step 1.2, $T_{m} f = F^{- 1} (F u_{τ_{a} f}) = u_{τ_{a} f}$ by [F8]. Thus the multiplier with symbol $m$ equals the translate $τ_{a}$ on the Schwartz core, so $m$ is the multiplier symbol of $τ_{a}$ .

2.2F1F6F8F9step 1.1step 1.2

Since $σ$ is smooth and polynomially bounded by step 1.1, $σ \hat{f} \in S$ for $f \in S$ ; the distributional derivative identity [F6] gives $F (\partial_{j} u_{f}) = σ F u_{f} = σ u_{\hat{f}} = u_{σ \hat{f}}$ , using [F1] and [F9] for the middle identifications, while step 1.2 gives $F u_{\partial_{j} f} = u_{σ \hat{f}}$ . Injectivity of $F$ on $S^{'}$ [F8] yields $\partial_{j} u_{f} = u_{\partial_{j} f}$ : on Schwartz classes the distributional derivative is the classical derivative. Consequently $T_{σ} f = F^{- 1} (u_{σ \hat{f}}) = F^{- 1} (F u_{\partial_{j} f}) = u_{\partial_{j} f}$ by [F8], so the Schwarz-core multiplier with symbol $σ = 2 π i ξ_{j}$ is exactly $\partial_{j}$ .

2.3F5F7step 1.3

For the $f_{N}$ of step 1.3, Plancherel [F5], the derivative identity of [F7] and $F f_{N} = ψ_{N}$ give $∥ \partial_{j} f_{N} ∥_{2} = ∥ F_{2} (\partial_{j} f_{N}) ∥_{2} = ∥ 2 π i ξ_{j} ψ_{N} ∥_{2} = 2 π ∥ ξ_{j} ψ_{N} ∥_{2}$ . On $supp ψ_{N}$ one has $ξ_{j} > N$ , so $∥ ξ_{j} ψ_{N} ∥_{2} \geq N ∥ ψ_{N} ∥_{2} = N ∥ f_{N} ∥_{2}$ by [F5]; hence $∥ \partial_{j} f_{N} ∥_{2} \geq 2 π N ∥ f_{N} ∥_{2}$ , and $f_{N} \neq 0$ since $ψ_{N} (η) = 1$ .

3.1F4F5step 1.1step 2.1

Since $∣ m ∣ = 1$ everywhere by step 1.1, [F4] and [F5] give, for every $g \in L^{2}$ , $T_{m} g = F_{2}^{- 1} (m F_{2} g)$ and $∥ T_{m} g ∥_{2} = ∥ m F_{2} g ∥_{2} = ∥ F_{2} g ∥_{2} = ∥ g ∥_{2}$ ; thus $T_{m}$ is an isometry of $L^{2}$ (indeed unitary, with inverse $T_{\overline{m}}$ ). By step 2.1 the operator $T_{m}$ agrees with $τ_{a}$ on the dense Schwartz subspace [F5]; the $L^{2}$ extension of $τ_{a}$ is unique by [F4], so that extension is this isometry and $τ_{a}$ is an $L^{2}$ isometry.

4.1A1step 2.1step 2.2step 3.1step 2.3∎

Suppose a bounded linear $P : L^{2} \to L^{2}$ extended $\partial_{j}$ from the Schwartz core, say $∥ P g ∥_{2} \leq C ∥ g ∥_{2}$ and $P f = \partial_{j} f$ for every $f \in S$ . Applying this to $f_{N} \in S$ of step 1.3 and using step 2.3 gives $2 π N ∥ f_{N} ∥_{2} \leq ∥ \partial_{j} f_{N} ∥_{2} = ∥ P f_{N} ∥_{2} \leq C ∥ f_{N} ∥_{2}$ , hence $2 π N \leq C$ for every $N > 0$ with $f_{N} \neq 0$ , which is impossible. Therefore no bounded $L^{2}$ extension exists, while steps 2.1 and 2.2 identify the symbols $m$ and $σ$ and step 3.1 establishes the $L^{2}$ isometry of $τ_{a}$ ; no Mihlin or other smoothness criterion was used.

RemarkRemark: Literature-sourcedProof: Not applicable‡ sources checked 2026-09-30‡ not proved hereOpen item page →

‡ Recorded, not proved here. This statement is included so the library can refer to it honestly, with a citation to the literature. It is not proved anywhere in this library: the track that would prove it has not been developed here yet.

Fefferman ball multiplier obstruction

Remark

Assume Countable Choice for the library Fourier and $L^{p}$ conventions, and let $n \geq 2$ . Fix a Euclidean ball $B \subseteq R^{n}$ and let $1_{B}$ be its indicator, a bounded measurable symbol. Recorded boundary result, not proved here: in the usual range $1 \leq p < \infty$ the ball indicator is an $L^{p} (R^{n})$ Fourier multiplier in the sense of Lp Fourier multiplier and its norm only at $p = 2$ .

The $p = 2$ half is elementary on this page. Since $∣ 1_{B} ∣ \leq 1$ everywhere, Exact L2 Fourier multiplier norm gives the unique bounded $L^{2}$ extension of the Schwartz-core multiplier with symbol $1_{B}$ and its operator norm is exactly $1$ . The case $p \neq 2$ is the deep part: Fefferman proved in Theorem 1 of The multiplier problem for the ball that for $n > 1$ the ball multiplier is bounded on $L^{p}$ if and only if $p = 2$ . Williams records the same statement in Remark 3.12 and refers to Grafakos §10.1 for the argument.

This remark is a bibliographic leaf: it records the exact limitation of the $L^{2}$ criterion of Exact L2 Fourier multiplier norm and supplies no proof or dependency for any later item. No Kakeya or geometric measure-theoretic argument is reproduced here, and no $L^{p}$ multiplier claim other than the recorded quotation is asserted. Countable Choice is inherited from the cited $L^{2}$ multiplier interface (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

RemarkRemark: Literature-sourcedProof: Not applicable‡ sources checked 2026-09-30‡ not proved hereOpen item page →

Jump multipliers may lie outside the Mihlin criterion

Remark

Assume Countable Choice for the library Fourier and $L^{p}$ conventions. In one dimension the Mihlin convention of Mihlin smoothness convention above half the dimension uses $q = ⌊ 1 / 2 ⌋ + 1 = 1$ : a symbol must agree almost everywhere with a $C^{1}$ function $m_{0}$ on $R ∖ {0}$ satisfying $∣ \partial^{α} m_{0} (ξ) ∣ \leq C_{α} ∣ ξ ∣^{- ∣ α ∣}$ for $∣ α ∣ \leq 1$ and $ξ \neq 0$ . In this convention:

Unshifted signum satisfies the criterion. The function $m_{0} (ξ) = sgn ξ$ is constant on each of the two components $(- \infty, 0)$ and $(0, \infty)$ of the punctured line, hence $C^{1}$ there with all derivatives zero; the bounds hold with $C_{0} = 1$ and $C_{1} = 0$ . Its only jump is at the excluded frequency $0$ . (This corrects a claim that the unshifted signum symbol fails the punctured-domain condition.)

Shifted signum fails the criterion. The symbol $m (ξ) = sgn (ξ - 1)$ equals $- 1$ on $(0, 1)$ and $+ 1$ on $(1, \infty)$ . A function $m_{0}$ continuous on $R ∖ {0}$ that agreed with $m$ almost everywhere would have to equal $- 1$ everywhere on $(0, 1)$ and $+ 1$ everywhere on $(1, \infty)$ , since it is continuous and the two open intervals have full measure in themselves; continuity at the nonzero point $ξ = 1$ would then fail. So the jump at the nonzero frequency $1$ is not excused by the punctured domain, and $m$ is not a Mihlin symbol here.

Recorded $L^{p}$ boundedness. Write $h (ξ) = - i sgn ξ$ ; the Hilbert transform on $R$ is the Fourier multiplier with symbol $h$ (Williams §3.1). Then $m (ξ) = sgn (ξ - 1) = i h (ξ - 1) .$ The shifted symbol is a unimodular constant times the frequency translate of the Hilbert multiplier, and frequency translation preserves the $L^{p}$ multiplier norm (Grafakos Proposition 2.5.14; modulation of the operator is an $L^{p}$ isometry). So the $L^{p} (R)$ boundedness of $sgn (ξ - 1)$ for $1 < p < \infty$ , hence its membership in this library's $M^{p}$ , is exactly the $L^{p}$ boundedness of the Hilbert transform, a recorded result assigned to the later singular-integral material and not proved or used on this page.

Thus the Mihlin condition is sufficient but not necessary: a multiplier with a jump at a nonzero frequency can still be bounded on every $L^{p} (R)$ , $1 < p < \infty$ . This remark is a recorded orientation leaf; the only locally checked content is the elementary smoothness comparison and the constant modulation identity above. Countable Choice is inherited from the cited conventions (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-09-30Open item page →

A Dirac mass has precisely sufficiently negative Sobolev order

Statement

Assume Countable Choice and let $n \geq 1$ . Let $δ_{0} \in S^{'} (R^{n})$ be the Dirac mass at the origin and let $H^{s} = H^{s} (R^{n})$ be the real-order Bessel-potential completion, identified with its canonical image in $S^{'} (R^{n})$ under $E_{s}$ (Real-order H^s as weighted Fourier distributions). Then $δ_{0} \in H^{s} ⟺ s < - \frac{n}{2} .$ Equivalently the threshold is $2 s + n < 0$ ; at the strict endpoint $s = - n / 2$ the membership fails, and the radial integrand decays like $1 / r$ , so the failure is a logarithmic divergence. Membership is read through the weighted Fourier characterization: $δ_{0} \in H^{s}$ means that $⟨ ξ ⟩^{s} F δ_{0}$ is the regular distribution of an $L^{2}$ class $g$ , the class is unique, and then $∥ δ_{0} ∥_{H^{s}} = ∥ g ∥_{2}$ . No pointwise-function assumption is made on $δ_{0}$ or on any representative of $g$ .

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , the Dirac mass $δ_{0}$ , and the Japanese bracket $⟨ ξ ⟩ = (1 + ∣ ξ ∣^{2})^{1 / 2}$ .

[A1]

Countable Choice permits one selection from each nonempty set in a countable family (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

In the negative-sign $2 π$ normalization, $F δ_{0} = u_{1}$ , the regular distribution of the constant function $1$ ; the Dirac mass is a tempered distribution and constants are regular tempered distributions (Fourier transform of delta constants plane waves and polynomials).

[F2]

For every $s \in R$ , a tempered distribution $u$ lies in $E_{s} [H^{s}]$ if and only if there is a unique $g \in L^{2} (R^{n})$ with $⟨ ξ ⟩^{s} F u = u_{g}$ in $S^{'} (R^{n})$ , and then $∥ u ∥_{H^{s}} = ∥ g ∥_{2}$ (Real-order H^s as weighted Fourier distributions).

[F3]

A smooth symbol whose derivatives are all polynomially bounded acts on tempered distributions by $⟨ a u, φ ⟩ = ⟨ u, a φ ⟩$ (Smooth polynomially bounded multipliers on schwartz space). The bracket weight $a = ⟨ ξ ⟩^{s}$ preserves Schwartz space (Real powers of the Japanese bracket act on Schwartz space). In the case $u = u_{1}$ used below, $⟨ a u_{1}, φ ⟩ = \int a φ$ by [F1], so $a u_{1} = u_{a}$ is a regular tempered distribution. On compact tests this is exactly the regular functional of Regular distribution from a locally integrable function.

[F4]

The regular-distribution map $h \mapsto u_{h}$ is injective on $L_{l o c}^{1} (R^{n})$ modulo almost-everywhere equality (Locally integrable functions embed in distributions).

[F5]

Every $L^{2}$ class is locally integrable: for compact $K$ , $\int_{K} ∣ h ∣ \leq ∣ K ∣^{1 / 2} ∥ h ∥_{2} < \infty$ by Cauchy–Schwarz and finiteness of the measure of bounded sets (Complex completeness, density, and inner product: the consumer interface, Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure).

[F6]

Polar coordinates: for Borel measurable $f : R^{n} \to [0, \infty]$ , $\int_{R^{n}} f d λ_{n} = \int_{0}^{\infty} ⁣ ⁣ \int_{S^{n - 1}} f (r ω) r^{n - 1} d σ (ω) d r,$ where $σ$ is the finite Borel measure on $S^{n - 1}$ with $σ (E) = n λ_{n} ({r ω : ω \in E, 0 < r \leq 1})$ ; its total mass satisfies $0 < σ (S^{n - 1}) = n λ_{n} ({0 < ∣ x ∣ \leq 1}) < \infty$ , because the set contains $B (e_{1} / 2, 1 / 4)$ (where $e_{1}$ is the first coordinate unit vector) and is contained in $B (0, 2)$ ; these balls have positive finite measure (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma, The polar surface set function on the unit sphere, Euclidean balls have positive finite Lebesgue measure).

[F7]

For a nonnegative locally Riemann-integrable $φ$ , the tail integral $\int_{1}^{\infty} φ : = \sup_{R > 1} \int_{1}^{R} φ$ is finite exactly when the truncations are bounded, and changing a finite lower endpoint does not affect finiteness (A nonnegative improper integral converges iff its truncated integrals are bounded, Improper convergence is independent of finite truncations and split points).

[F8]

For a nonnegative series, convergence is equivalent to boundedness of its partial sums (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum); and for real $p$ , $\sum_{k \geq 1} 1 / k^{p}$ converges exactly when $p > 1$ (The p-series for a real exponent p converges exactly when p is greater than one).

[F9]

A continuous function on a compact interval is Riemann integrable; the integral is additive over adjacent intervals and monotone in the integrand (A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ , If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ ).

[F10]

For $R > 0$ , $\int_{1}^{R} \frac{d t}{t} = \log R$ , and $\log R \to + \infty$ as $R \to + \infty$ , so this integral diverges logarithmically (The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t).

[F11]

For $b > 0$ and real $u$ , $b^{u} = \exp (u \log b)$ (Real powers for positive bases, with the zero-base positive-exponent convention); the logarithm satisfies $\log^{'} (x) = 1 / x > 0$ on $(0, \infty)$ and is therefore strictly increasing (The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t), and $\exp$ is strictly increasing (The exponential function is strictly increasing), so $b \mapsto b^{u}$ is strictly increasing for $u > 0$ and strictly decreasing for $u < 0$ ; and $b^{u + v} = b^{u} b^{v}$ , $(b_{1} b_{2})^{u} = b_{1}^{u} b_{2}^{u}$ (The exponent, product, quotient, and iterated-power laws for positive real bases and real exponents).

[F12]

Every real number is exceeded by a natural number (Every complete ordered field is Archimedean).

[F13]

For a measurable $g$ , $g \in L^{2} (R^{n})$ exactly when $\int_{R^{n}} ∣ g ∣^{2} < \infty$ , and then $∥ g ∥_{2}^{2} = \int ∣ g ∣^{2}$ (Complex Lp classes and Euclidean test-function conventions).

[F14]

Bounded Riemann-integrable functions on compact intervals have equal Lebesgue and Riemann integrals (A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral). For nonnegative measurable functions, monotone convergence identifies the integral with the increasing limit of its truncations (Monotone convergence for the integral).

Proof

technique · reduce the membership to the finiteness of a radial integral, bracket that integral between blocks of a real $p$-series, and read off the threshold and its strict endpoint

1.1F1

The Fourier transform of the Dirac mass. By [F1], $F δ_{0} = u_{1}$ is the regular distribution of the constant function $1$ .

1.2F6F7F9F11F14given

The polar reduction. Apply [F6] to $f (ξ) = ⟨ ξ ⟩^{2 s}$ ; since $f (r ω) = (1 + r^{2})^{s}$ is radial, the total mass of $σ$ factors out: $\int_{R^{n}} ⟨ ξ ⟩^{2 s} d ξ = σ (S^{n - 1}) \int_{0}^{\infty} φ (r) d r, φ (r) : = (1 + r^{2})^{s} r^{n - 1},$ with $σ (S^{n - 1}) \in (0, \infty)$ . The function $φ$ extends continuously to $r = 0$ , with value $1$ if $n = 1$ and $0$ if $n > 1$ . By [F14], its Lebesgue integral on each compact interval equals its Riemann integral, and monotone convergence of $φ 1_{[0, N]}$ identifies the full nonnegative Lebesgue integral with the supremum of these truncations, including when infinite. On $0 < r \leq 1$ one has $(1 + r^{2})^{s} \leq \max (1, 2^{s})$ and $r^{n - 1} \leq 1$ by [F11], so $\int_{0}^{1} φ \leq \max (1, 2^{s}) < \infty$ by [F9]. Since $\int_{0}^{R} φ = \int_{0}^{1} φ + \int_{1}^{R} φ$ for $R > 1$ by additivity [F9], the integral over $(0, \infty)$ is finite if and only if the tail truncations $\int_{1}^{R} φ$ are bounded, that is, if and only if the tail $\int_{1}^{\infty} φ$ is finite in the sense of [F7].

1.3F11algebra

Block bounds. Put $q : = 2 s + n - 1$ and write, by [F11], $φ (r) = r^{q} (1 + r^{- 2})^{s}$ . For an integer $k \geq 1$ and $r \in [k, k + 1]$ one has $1 + r^{- 2} \in [1, 2]$ , so $(1 + r^{- 2})^{s} \in [c_{1}, c_{2}]$ with $c_{1} = \min (1, 2^{s}) > 0$ and $c_{2} = \max (1, 2^{s})$ ; moreover $k + 1 \leq 2 k$ for $k \geq 1$ , so $2^{\min (q, 0)} \leq r^{q} / k^{q} \leq 2^{\max (q, 0)}$ , and with $c_{3} = c_{1} 2^{\min (q, 0)}$ , $c_{4} = c_{2} 2^{\max (q, 0)}$ , $c_{3} k^{q} \leq φ (r) \leq c_{4} k^{q}, r \in [k, k + 1], k \geq 1.$

2.1F2F3F4F5F13step 1.1

The membership criterion. By [F2], $δ_{0} \in H^{s}$ if and only if there is $g \in L^{2}$ with $⟨ ξ ⟩^{s} F δ_{0} = u_{g}$ . By step 1.1 and [F3] applied to the smooth symbol $a = ⟨ ξ ⟩^{s}$ and the locally integrable $h = 1$ , this product is $⟨ ξ ⟩^{s} u_{1} = u_{⟨ ξ ⟩^{s}}$ . The condition is therefore $u_{⟨ ξ ⟩^{s}} = u_{g}$ for some $g \in L^{2}$ . Both $⟨ ξ ⟩^{s}$ and $g$ are locally integrable by continuity and [F5], so [F4] forces $⟨ ξ ⟩^{s} = g$ almost everywhere; hence a qualifying $g$ exists exactly when $⟨ ξ ⟩^{s} \in L^{2} (R^{n})$ , that is, by [F13], exactly when $\int_{R^{n}} ⟨ ξ ⟩^{2 s} d ξ < \infty$ , and then $∥ δ_{0} ∥_{H^{s}} = ∥ g ∥_{2} = ∥ ⟨ ξ ⟩^{s} ∥_{2}$ .

2.2F7F8F9F12step 1.3

The block comparison. For every integer $N \geq 1$ , additivity and monotonicity of the integral [F9] applied to step 1.3 give $c_{3} \sum_{k = 1}^{N} k^{q} \leq \int_{1}^{N + 1} φ \leq c_{4} \sum_{k = 1}^{N} k^{q} .$ If $\sum_{k \geq 1} k^{q}$ converges with value $S$ , then for every $R > 1$ [F12] supplies an integer $N \geq R$ , and monotonicity [F9] together with step 1.3 gives $\int_{1}^{R} φ \leq \int_{1}^{N + 1} φ \leq c_{4} S$ , so the truncations are bounded and $\int_{1}^{\infty} φ \leq c_{4} S < \infty$ by [F7]. Conversely, if $\int_{1}^{\infty} φ < \infty$ , then $\sum_{k = 1}^{N} k^{q} \leq c_{3}^{- 1} \int_{1}^{N + 1} φ \leq c_{3}^{- 1} \int_{1}^{\infty} φ$ for every $N$ , so the partial sums are bounded and $\sum_{k \geq 1} k^{q}$ converges by [F8].

3.1F8step 2.1step 1.2step 2.2

The threshold. By step 2.2, $\int_{1}^{\infty} φ < \infty$ if and only if $\sum_{k \geq 1} k^{q} = \sum_{k \geq 1} 1 / k^{- q}$ converges, which by [F8] happens exactly when $- q > 1$ , that is, exactly when $2 s + n - 1 < - 1$ , i.e. $s < - n / 2$ . Combining with steps 2.1 and 1.2, this gives $δ_{0} \in H^{s} ⟺ s < - n / 2$ .

3.2F7F9F10F11step 2.1step 1.2

The strict endpoint. Let $s = - n / 2$ , so $q = - 1$ and $φ (r) = r^{- 1} (1 + r^{- 2})^{s} \geq c_{1} r^{- 1}$ for $r \geq 1$ by [F11]. Hence for every $R > 1$ , $\int_{1}^{R} φ \geq c_{1} \int_{1}^{R} \frac{d r}{r} = c_{1} \log R ⟶ + \infty (R \to \infty)$ by [F9] and [F10]; the truncations are unbounded, so $\int_{1}^{\infty} φ = \infty$ by [F7], and by steps 2.1 and 1.2, $δ_{0} \notin H^{- n / 2}$ . The divergence is logarithmic: the truncated integral grows like $\log R$ because the radial integrand behaves like $1 / r$ .

4.1A1F1F2F3F4F5F6F7F8F9F10F11F12F13step 1.1step 2.1step 1.2step 1.3step 2.2step 3.1step 3.2∎

Conclusion. Step 3.1 proves the equivalence $δ_{0} \in H^{s} ⟺ s < - n / 2$ and exhibits the strict threshold $2 s + n < 0$ , while step 3.2 proves that the borderline case $s = - n / 2$ fails by logarithmic divergence; step 2.1 identifies the exact norm with the $L^{2}$ norm of the unique weighted Fourier class, and no pointwise-function assumption is used anywhere. Countable Choice is used exactly through the cited characterization, regular-distribution and polar-coordinate interfaces, which carry it as their hypothesis.

Fourier Multipliers and Sobolev Characterisations — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Heat and Poisson semigroups as Fourier multipliers

Example

Facts & Assumptions

Verification

Translation and differentiation symbols

Example

Facts & Assumptions

Verification

Fefferman ball multiplier obstruction

Remark

Jump multipliers may lie outside the Mihlin criterion

Remark

A Dirac mass has precisely sufficiently negative Sobolev order

Statement

Facts & Assumptions

Proof

Sources