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✓ 5 results · all verified · 1 also independently AI-judged

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Hilbert and Riesz Transforms — Examples

1 · Prerequisites

2 · Summary

These examples compute the transforms and mark the endpoint obstructions that the companion page's strict-range theorems leave open. All of them assume Countable Choice.

The interval indicator $1_{(0, 1)}$ has symmetric principal value $π^{- 1} \log ∣ x / (x - 1) ∣$ away from the two endpoints, obtained by the exact logarithmic antiderivative of $1 / (x - y)$ on the two sides of the interval and identified with the $L^{2}$ multiplier extension through smooth approximations. That single computation powers both endpoint counterexamples: the transform is not integrable, because its $1 / ∣ x ∣$ tail at infinity has divergent integral, so there is no bounded strong-type $(1, 1)$ extension compatible with the $L^{2}$ transform; and it is essentially unbounded near $0$ and $1$ , so there is no bounded action on $L^{\infty}$ agreeing with the $L^{2}$ transform on the intersection. Neither argument refutes a weak $(1, 1)$ estimate or a BMO bound.

The positive examples compute the line Poisson kernel: the transform of $P_{a} (x) = a / [π (a^{2} + x^{2})]$ is the conjugate Poisson kernel $Q_{a} (x) = x / [π (a^{2} + x^{2})]$ , which is established pointwise and in $L^{2}$ from the locally proved Fourier transform $\hat{P_{a}} (ξ) = e^{- 2 π a ∣ ξ ∣}$ . Finally, the finite sum of Riesz squares is evaluated on $L^{2} (R^{n})$ , where $\sum_{j = 1}^{n} R_{j}^{2} = - I$ reduces to the elementary trigonometric identity $\sum_{j} ξ_{j}^{2} / ∣ ξ ∣^{2} = 1$ for $ξ \neq 0$ ; the assigned value at $ξ = 0$ has no effect on an $L^{2}$ statement.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Hilbert transform of an interval indicator

Statement

Assume Countable Choice and let $f : = 1_{(0, 1)}$ be the indicator of the open unit interval, with the $L^{p}$ conventions of Complex Lp classes and Euclidean test-function conventions. Write

$q (x) : = \frac{1}{π} \log \frac{∣ x ∣}{∣ x - 1 ∣} (x \notin {0, 1}) .$

Then:

for every $x \notin {0, 1}$ the symmetric principal value $\lim_{ε ↓ 0} H_{ε} f (x)$ exists and equals $q (x)$ , the logarithm being taken at the positive argument $∣ x ∣ / ∣ x - 1 ∣$ ;
the function $q$ represents the $L^{2}$ Hilbert transform of $f$ almost everywhere, that is, $q = H f$ in $L^{2} (R; C)$ .

The values at the two endpoints are immaterial: every assertion is about the complement of the Lebesgue-null set ${0, 1}$ , and no claim is made about $H_{ε} f$ at $x \in {0, 1}$ .

Facts & Assumptions

Given: Countable Choice, the indicator $f = 1_{(0, 1)} \in L^{1} (R) \cap L^{2} (R)$ with $0 \leq f \leq 1$ , and the truncated Hilbert transform of Truncated Hilbert transform and principal value.

[F1]

For $ε > 0$ and $x \in R$ , $H_{ε} f (x) = \frac{1}{π} \int_{∣ t ∣ > ε} \frac{f (x - t)}{t} d t$ , absolutely convergent for $f \in L^{p}$ , $1 \leq p < \infty$ ; $H_{p v} f (x)$ is the $ε ↓ 0$ limit where it exists. Truncated Hilbert transform and principal value

[F2]

For Schwartz $g$ the principal value exists at every $x$ and equals $(W * g) (x)$ for the tempered convolution with $pv \frac{1}{π x}$ , and the $L^{2}$ extension $H$ has symbol $m (ξ) = - i sgn (ξ)$ and satisfies $∥ H g ∥_{2} = ∥ g ∥_{2}$ . The Hilbert transform is the tempered convolution with pv(1/(pi x)) and has signum Fourier multiplier The Hilbert transform is an L2 isometry and squares to minus the identity

[F3]

There is $χ \in C_{c}^{\infty} (R)$ with $0 \leq χ \leq 1$ , $χ = 1$ on $[- 1, 1]$ and $χ = 0$ off $(- 2, 2)$ . Explicit compactly supported smooth cutoffs

[F4]

For real $a \leq b$ the interval $[a, b]$ is Lebesgue measurable with $λ_{1} ([a, b]) = b - a$ ; and if $0 \leq u \leq v$ are measurable then $\int u \leq \int v$ . A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included Monotonicity and nonnegative homogeneity of the nonnegative integral

[F5]

A function $φ \in C_{c}^{\infty} (R)$ with $\int φ = 1$ generates the mollifier family $φ_{ε} (x) = ε^{- 1} φ (x / ε)$ , and $(φ_{ε})_{ε > 0}$ is an $L^{1}$ approximate identity. The mollifier family generated by a unit-mass smooth bump A unit-mass smooth bump generates an $L^{1}$ approximate identity

[F6]

If $1 \leq p < \infty$ and $g \in L^{p} (R)$ , then $∥ g * φ_{ε} - g ∥_{p} \to 0$ ; in particular $g * φ_{ε} \to g$ in $L^{p}$ . Every $L^{1}$ approximate identity converges to the identity in $L^{p}$ for $1 \leq p < \infty$

[F7]

For locally integrable $g$ the convolution $g * φ_{ε}$ is smooth; and $supp (g * φ_{ε}) \subseteq \overline{supp (g) + supp (φ_{ε})}$ . Convolution with a mollifier is smooth, and derivatives pass under the integral sign The support of a convolution lies in the closure of the support sumset

[F8]

On $(0, \infty)$ : $\log$ is differentiable with $\log^{'} = 1 / x$ , $\log x = \int_{1}^{x} d t / t$ , and $\log (x / y) = \log x - \log y$ ; $\log$ is strictly increasing. With the chain rule this gives $\frac{d}{d t} \log ∣ t ∣ = \frac{1}{t}$ for $t \in R ∖ {0}$ . The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm 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 $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$

[F9]

Oriented additivity over subintervals and the second fundamental theorem: on a compact interval on which the integrand is continuous with the displayed antiderivative, the integral is the antiderivative difference, and $\int_{u}^{v} f + \int_{v}^{w} f = \int_{u}^{w} f$ . 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 $a, b, c$ The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$

[F10]

Norm-convergent sequences in $L^{2}$ have subsequences converging almost everywhere to a representative of the limit. Complex Lp completeness and almost-everywhere subsequences

Proof

technique · direct

1.1F1

Let $x \notin {0, 1}$ and $ε > 0$ . Substituting $t = x - y$ in [F1] and using $1_{(0, 1)} (x - t) = 1$ exactly for $t \in (x - 1, x)$ gives $H_{ε} f (x) = \frac{1}{π} \int_{(x - 1, x) \cap {∣ t ∣ > ε}} \frac{d t}{t}$ , the integrand being continuous on each piece because $t = 0$ is either excluded by the truncation or avoided.

1.2F3F4F5F7

Construction of approximants. Put $φ : = χ / \int χ$ with $χ$ as in [F3]. The bounds $0 \leq χ \leq 1$ and [F4] give $2 = λ_{1} ([- 1, 1]) \leq \int χ \leq λ_{1} ([- 2, 2]) = 4$ , so $0 < \int χ < \infty$ and $φ \in C_{c}^{\infty} (R)$ is nonnegative with $\int φ = 1$ . Let $(φ_{ε})$ be its mollifier family and put $f_{j} : = f * φ_{1 / j}$ for $j \geq 1$ . By [F7] each $f_{j}$ is smooth, and since $supp (f) \subseteq [0, 1]$ and $supp (φ_{1 / j}) \subseteq [- 2 / j, 2 / j]$ , the support inclusion gives $supp (f_{j}) \subseteq [- 2 / j, 1 + 2 / j]$ . Also, if $dist (y, {0, 1}) > 2 / j$ , the bump samples only where $f$ is constant, so $f_{j} (y) = f (y)$ ; hence $g_{j} = f_{j} - f$ is supported within distance $2 / j$ of the endpoints, and $f_{j} \in C_{c}^{\infty} (R)$ . Since $0 \leq φ$ and $\int φ_{1 / j} = 1$ , moreover $0 \leq f_{j} \leq 1$ pointwise: $f_{j} (x) = \int f (x - y) φ_{1 / j} (y) d y \in [0, 1]$ .

2.1step 1.1F8F9

Case $x > 1$ . For $0 < ε < x - 1$ one has $(x - 1, x) \subseteq (ε, \infty)$ , so $H_{ε} f (x) = \frac{1}{π} \int_{x - 1}^{x} \frac{d t}{t} = \frac{1}{π} (\log x - \log (x - 1)) = \frac{1}{π} \log \frac{x}{x - 1}$ by [F8] and [F9].

2.2step 1.1F8F9

Case $x < 0$ . For $0 < ε < - x$ one has $(x - 1, x) \subseteq (- \infty, - ε)$ , so $H_{ε} f (x) = \frac{1}{π} \int_{x - 1}^{x} \frac{d t}{t} = \frac{1}{π} (\log ∣ x ∣ - \log ∣ x - 1 ∣) = \frac{1}{π} \log \frac{∣ x ∣}{∣ x - 1 ∣}$ by [F8], the antiderivative of $1 / t$ on the negative axis being $\log ∣ t ∣$ .

2.3step 1.1F8F9

Case $0 < x < 1$ . For $0 < ε < \min (x, 1 - x)$ the set $(x - 1, x) \cap {∣ t ∣ > ε}$ is $(x - 1, - ε) \cup (ε, x)$ , so by [F9] $H_{ε} f (x) = \frac{1}{π} [\log ∣ - ε ∣ - \log ∣ x - 1 ∣ + \log x - \log ε] = \frac{1}{π} [\log x - \log (1 - x)] = \frac{1}{π} \log \frac{x}{1 - x}$ , the two $\log ε$ terms cancelling exactly because $\log ∣ - ε ∣ = \log ε$ ; since $∣ x - 1 ∣ = 1 - x > 0$ this is $\frac{1}{π} \log \frac{∣ x ∣}{∣ x - 1 ∣}$ .

2.4step 1.2F2F6

By [F6] applied with $p = 1$ and $p = 2$ , the sequence of 1.2 satisfies $∥ f_{j} - f ∥_{1} \to 0$ and $∥ f_{j} - f ∥_{2} \to 0$ ; consequently $f_{j} \to f$ in $L^{2}$ , and the $L^{2}$ boundedness of [F2] gives $∥ H f_{j} - H f ∥_{2} \to 0$ , where $H f_{j}$ is both the $L^{2}$ transform of $f_{j}$ and the pointwise principal value of [F2].

2.5step 1.2F2algebra

Fix $x \notin {0, 1}$ and put $δ : = \frac{1}{2} dist (x, {0, 1}) > 0$ ; let $c \in {0, 1}$ be the constant value of $f$ on $(x - δ, x + δ)$ and set $r : = δ / 4$ . The mollifier is supported in $[- 2 / j, 2 / j]$ , so for $j > 4 / δ$ its convolution samples only points of $(x - δ, x + δ)$ when the argument lies in $(x - δ / 2, x + δ / 2)$ ; hence 1.2 gives $f_{j} = c$ there. Thus for $0 < η < r$ the part of $H_{η} f_{j} (x)$ over $η < ∣ x - y ∣ < r$ is the integral of $c / (π (x - y))$ over a symmetric annulus, hence is zero. The remaining integral is absolutely convergent because $f_{j}$ has compact support and $∣ x - y ∣ \geq r$ there. Letting $η ↓ 0$ in [F2] gives $H f_{j} (x) = \frac{1}{π} \int_{∣ x - y ∣ > r} \frac{f_{j} (y)}{x - y} d y .$

3.1step 2.1step 2.2step 2.3

By 2.1, 2.2 and 2.3, for every $x \notin {0, 1}$ and every $0 < ε < r (x)$ , where $r (x) : = x - 1$ for $x > 1$ , $r (x) : = - x$ for $x < 0$ and $r (x) : = \min (x, 1 - x)$ for $0 < x < 1$ , one has $H_{ε} f (x) = q (x) = \frac{1}{π} \log \frac{∣ x ∣}{∣ x - 1 ∣}$ . Since $r (x) > 0$ , the symmetric principal value exists at every $x \notin {0, 1}$ and equals $q (x)$ ; this proves assertion 1.

3.2step 1.2step 2.4step 2.5

The function $g_{j} = f_{j} - f$ is supported in ${y : dist (y, {0, 1}) \leq 2 / j}$ by 1.2, so for $j > 4 / δ$ and $y \in supp (g_{j})$ one has $∣ x - y ∣ \geq 2 δ - \frac{2}{j} \geq \frac{δ}{2}$ ; hence $∣ \frac{1}{π} \int_{R} \frac{g_{j} (y)}{x - y} d y ∣ \leq \frac{∥ g_{j} ∥_{1}}{π} \cdot \frac{2}{δ} \to 0$ as $j \to \infty$ by 2.4.

4.1step 3.1algebra

Since $f = c$ on $∣ x - y ∣ < r$ , the same symmetric cancellation shows that for every $0 < η < r$ , $H_{η} f (x) = \frac{1}{π} \int_{∣ x - y ∣ > r} \frac{f (y)}{x - y} d y$ . This outer integral is absolutely convergent because $f$ has compact support and its denominator is bounded away from zero. By 3.1 its value is $q (x)$ .

5.1step 2.5step 3.2step 4.1

Combining 2.5, 3.2 and 4.1, $H f_{j} (x) \to q (x)$ for every fixed $x \notin {0, 1}$ .

6.1step 2.4step 5.1F10∎

By 2.4, $H f_{j} \to H f$ in $L^{2}$ ; by [F10] a subsequence converges almost everywhere to a representative of the class $H f$ , while 5.1 makes that same subsequence converge to $q$ at every point of the full-measure set $R ∖ {0, 1}$ . Therefore $q = H f$ almost everywhere: $q$ represents the $L^{2}$ multiplier extension of $f$ , which is assertion 2.

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-10-02Open item page →

Hilbert transform is not strong type (1,1)

Statement refuted

The claim that the Schwartz-core Hilbert transform has a bounded $C$ -linear extension $T : L^{1} (R; C) \to L^{1} (R; C)$ agreeing with the $L^{2}$ Hilbert transform on $L^{1} (R) \cap L^{2} (R)$ — equivalently, that the Hilbert transform is of strong type $(1, 1)$ — is false. The interval indicator supplies the witness: it lies in $L^{1}$ , while its $L^{2}$ transform $q (x) = \frac{1}{π} \log \frac{∣ x ∣}{∣ x - 1 ∣}$ has a nonintegrable $\frac{1}{∣ x ∣}$ tail and therefore is not an $L^{1}$ class.

This refutes only strong type $(1, 1)$ . No weak-type $(1, 1)$ estimate is refuted or asserted here.

Facts & Assumptions

Given: Countable Choice, the indicator $f = 1_{(0, 1)}$ , and the function $q (x) = \frac{1}{π} \log \frac{∣ x ∣}{∣ x - 1 ∣}$ for $x \notin {0, 1}$ , with the $L^{p}$ conventions of Complex Lp classes and Euclidean test-function conventions.

[F1]

$f \in L^{1} (R) \cap L^{2} (R)$ with $0 \leq f \leq 1$ ; the symmetric principal value of $\frac{1}{π} \int \frac{f (y)}{x - y} d y$ exists at every $x \notin {0, 1}$ and equals $q (x)$ ; and $q = H f$ in $L^{2} (R; C)$ for the $L^{2}$ Hilbert transform. Hilbert transform of an interval indicator

[F2]

$H$ is complex-linear on $L^{2} (R; C)$ and $∥ H g ∥_{2} = ∥ g ∥_{2}$ for every $g \in L^{2}$ . The Hilbert transform is an L2 isometry and squares to minus the identity

[F3]

For $0 < a < b$ one has $\int_{a}^{b} \frac{d t}{t} = \log b - \log a$ , and $\log (1 + u) = \int_{1}^{1 + u} \frac{d t}{t}$ for $u \geq 0$ ; $\log$ is strictly increasing on $(0, \infty)$ . The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm

[F4]

The nonnegative integral is monotone and positively homogeneous, and monotone convergence passes to the limit of an increasing sequence of truncations. Monotonicity and nonnegative homogeneity of the nonnegative integral Monotone convergence for the integral

[F5]

There is $χ \in C_{c}^{\infty} (R)$ with $0 \leq χ \leq 1$ , $χ = 1$ on $[- 1, 1]$ and $χ = 0$ off $(- 2, 2)$ ; $φ : = χ / \int χ$ is a nonnegative $C_{c}^{\infty}$ function of integral one; and its mollifiers $φ_{ε} (x) = ε^{- 1} φ (x / ε)$ satisfy: $g * φ_{ε}$ is smooth, $supp (g * φ_{ε}) \subseteq \overline{supp g + supp φ_{ε}}$ , and $∥ g * φ_{ε} - g ∥_{p} \to 0$ for $g \in L^{p} (R)$ and $1 \leq p < \infty$ . Every $C_{c}^{\infty} (R)$ function is a Schwartz function. Explicit compactly supported smooth cutoffs The mollifier family generated by a unit-mass smooth bump A unit-mass smooth bump generates an $L^{1}$ approximate identity Every $L^{1}$ approximate identity converges to the identity in $L^{p}$ for $1 \leq p < \infty$ Convolution with a mollifier is smooth, and derivatives pass under the integral sign The support of a convolution lies in the closure of the support sumset Schwartz space and its seminorms

[F6]

Holder: $∣ \int u v d x ∣ \leq ∥ u ∥_{1} ∥ v ∥_{\infty}$ and $∣ \int u v d x ∣ \leq ∥ u ∥_{2} ∥ v ∥_{2}$ . Complex Holder, Minkowski, and the quotient norm

[F7]

Dominated convergence: if $∣ h_{R} ∣ \leq G$ with $G \in L^{1}$ and $h_{R} \to h$ almost everywhere, then $\int h_{R} \to \int h$ . Dominated convergence

[F8]

On a compact interval a bounded Riemann integrable function is Lebesgue integrable with the same integral, and the interval $(0, 1)$ has Lebesgue measure one. A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included

Counterexample

technique · direct

1.1F1F8givenalgebra

The indicator $f$ is measurable with $0 \leq f \leq 1$ and ${f \neq 0} = (0, 1)$ of measure one, so $∣ f ∣^{p} \leq 1$ on a set of measure one and vanishes elsewhere; hence $f \in L^{p} (R; C)$ for every $1 \leq p \leq \infty$ , with $∥ f ∥_{p} \leq 1$ .

1.2F1F3givenalgebra

On $(1, \infty)$ one has $q (x) = \frac{1}{π} \log \frac{x}{x - 1} = \frac{1}{π} \log (1 + \frac{1}{x - 1}) > 0$ because $\log$ is strictly increasing and $\log 1 = 0$ by [F3].

1.3F3F4algebra

For $0 \leq u \leq 1$ one has $\log (1 + u) = \int_{1}^{1 + u} \frac{d t}{t} \geq \int_{1}^{1 + u} \frac{d t}{1 + u} = \frac{u}{1 + u} \geq \frac{u}{2}$ , using monotonicity of the integral and $1 + u \leq 2$ .

2.1F3F4F8step 1.2step 1.3algebra

$q \notin L^{1} (R)$ : for $x \geq 2$ put $u = \frac{1}{x - 1} \in (0, 1]$ ; steps 1.2 and 1.3 give $q (x) \geq \frac{1}{2 π} \cdot \frac{1}{x - 1} \geq \frac{1}{2 π x} \geq 0$ . Hence for every $R > 2$ , using additivity over the interval and [F8] with the antiderivative of [F3], $\int_{(2, R)} q \geq \frac{1}{2 π} \int_{(2, R)} \frac{d x}{x} = \frac{1}{2 π} (\log R - \log 2)$ , which tends to $+ \infty$ ; monotone convergence [F4] gives $\int_{(2, \infty)} q = + \infty$ , and monotonicity in the domain gives $\int_{R} ∣ q ∣ \geq \int_{(2, \infty)} q = + \infty$ . So $q$ is not an $L^{1}$ class.

2.2F5step 1.1algebra

Let $φ$ be the unit-mass bump of [F5] and for $j \in N$ put $f_{j} : = f * φ_{1 / (j + 1)}$ . Each $f_{j}$ is smooth with $supp f_{j} \subseteq [- 2 / (j + 1), 1 + 2 / (j + 1)]$ , so $f_{j} \in C_{c}^{\infty} (R) \subseteq S (R)$ ; and $0 \leq f_{j} \leq 1$ because $0 \leq f \leq 1$ and $φ_{1 / (j + 1)} \geq 0$ has integral one. Since $∥ f_{j} - f ∥_{p} \to 0$ for $p = 1$ and $p = 2$ , each $f_{j}$ lies in $L^{1} \cap L^{2}$ and the sequence converges to $f$ in both norms.

3.1F1F2step 2.2algebra

$∥ H f_{j} - q ∥_{2} \to 0$ : by [F1] $q = H f$ , and by [F2] $H$ is a linear isometry, so $∥ H f_{j} - q ∥_{2} = ∥ H (f_{j} - f) ∥_{2} = ∥ f_{j} - f ∥_{2} \to 0$ by step 2.2.

4.1step 2.2step 3.1F6given

Suppose, for contradiction, that $T : L^{1} (R; C) \to L^{1} (R; C)$ is bounded and linear with $T g = H g$ for every $g \in S (R)$ . Since $f_{j} \in S$ , one has $T f_{j} = H f_{j}$ as $L^{1}$ classes, and $∥ T f_{j} - T f ∥_{1} \leq ∥ T ∥ ∥ f_{j} - f ∥_{1} \to 0$ by step 2.2. Fix $ψ \in C_{c}^{\infty} (R)$ . Then $∣ \int (T f_{j} - T f) ψ ∣ \leq ∥ T f_{j} - T f ∥_{1} ∥ ψ ∥_{\infty} \to 0$ and, by step 3.1, $∣ \int (H f_{j} - q) ψ ∣ \leq ∥ H f_{j} - q ∥_{2} ∥ ψ ∥_{2} \to 0$ ; since the $j$ -th integrals of $T f_{j}$ and $H f_{j}$ agree, it follows that $\int (T f) ψ = \int q ψ$ , i.e. $\int (T f - q) ψ = 0$ for every $ψ \in C_{c}^{\infty} (R)$ .

5.1step 2.1step 4.1F5F7algebra

For $R > 8$ set $ψ_{R} : = 1_{[4, R]} * φ_{1}$ with $φ_{1}$ as in [F5]. Then $ψ_{R} \in C_{c}^{\infty} (R)$ , its support is contained in $[2, R + 2]$ , where $q \geq 0$ by step 1.2, $0 \leq ψ_{R} \leq 1$ , and $ψ_{R} = 1$ on $[6, R - 2]$ : indeed $ψ_{R} (x) = \int_{x - R}^{x - 4} φ_{1} (y) d y$ , and for $x \in [6, R - 2]$ the interval $[x - R, x - 4]$ contains the support of $φ_{1}$ , which is contained in $[- 2, 2]$ . Step 4.1 gives $\int (T f) ψ_{R} = \int q ψ_{R}$ for every $R > 8$ . As $R \to \infty$ the functions $ψ_{R}$ converge pointwise to the bounded function $x \mapsto \int_{- \infty}^{x - 4} φ_{1}$ , so dominated convergence [F7] with majorant $∣ T f ∣ \in L^{1}$ shows that the left-hand sides converge to a finite limit; but step 2.1 and $ψ_{R} = 1$ on $[6, R - 2]$ give $\int q ψ_{R} \geq \int_{[6, R - 2]} q \geq \frac{1}{2 π} (\log (R - 2) - \log 6) \to \infty$ . A sequence cannot converge to a finite limit while equalling terms that tend to $+ \infty$ , so no such $T$ exists.

6.1step 5.1∎

The compatibility hypothesis in step 4.1 was imposed only on Schwartz functions, which lie in $L^{1} \cap L^{2}$ ; hence there is no bounded linear $L^{1} \to L^{1}$ operator agreeing with the $L^{2}$ Hilbert transform on $L^{1} \cap L^{2}$ either. The witness $f \in L^{1}$ with transform $q \notin L^{1}$ therefore refutes strong type $(1, 1)$ . Nothing here addresses weak type $(1, 1)$ , which is a different assertion.

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-10-02Open item page →

Hilbert transform does not map L-infinity to L-infinity

Statement refuted

The claim that the Schwartz-core Hilbert transform extends to a bounded $C$ -linear operator $T : L^{\infty} (R; C) \to L^{\infty} (R; C)$ agreeing with the $L^{2}$ transform on the intersection $L^{\infty} (R) \cap L^{2} (R)$ is false. The bounded interval indicator lies in that intersection, but its $L^{2}$ transform $q (x) = \frac{1}{π} \log \frac{∣ x ∣}{∣ x - 1 ∣}$ is essentially unbounded near $0$ and $1$ ; a bounded action would have to keep the approximating transforms essentially bounded, and an almost-everywhere subsequence would then force $q$ itself to be essentially bounded.

This refutes a bounded $L^{\infty}$ action only. No $B M O$ -valued endpoint estimate is refuted or asserted here.

Facts & Assumptions

Given: Countable Choice, the indicator $f = 1_{(0, 1)}$ , the function $q (x) = \frac{1}{π} \log \frac{∣ x ∣}{∣ x - 1 ∣}$ for $x \notin {0, 1}$ , and the $L^{p}$ conventions of Complex Lp classes and Euclidean test-function conventions.

[F1]

The symmetric principal value of the indicator exists at every $x \notin {0, 1}$ and equals $q (x)$ , and $q = H f$ in $L^{2} (R; C)$ for the $L^{2}$ Hilbert transform; in particular $q (x) = \frac{1}{π} \log \frac{x}{1 - x}$ for $x \in (0, 1)$ . Hilbert transform of an interval indicator

[F2]

$H$ is complex-linear on $L^{2} (R; C)$ and satisfies $∥ H g ∥_{2} = ∥ g ∥_{2}$ . The Hilbert transform is an L2 isometry and squares to minus the identity

[F3]

$\log : (0, \infty) \to R$ is continuous, strictly increasing and onto, $\log 1 = 0$ , and $\exp : R \to (0, \infty)$ is its inverse; hence for real $M$ and $y > 0$ , $\log y > M$ holds exactly when $y > e^{M}$ . Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm The exponential is a continuous bijection from $R$ onto $(0, \infty)$

[F4]

For $a < b$ the interval $(a, b)$ is Lebesgue measurable with $λ_{1} ((a, b)) = b - a$ . A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included

[F5]

There is $χ \in C_{c}^{\infty} (R)$ with $0 \leq χ \leq 1$ , $χ = 1$ on $[- 1, 1]$ and $χ = 0$ off $(- 2, 2)$ ; $φ : = χ / \int χ$ is a nonnegative $C_{c}^{\infty}$ function of integral one; for $g \in L^{1} (R)$ , $g * φ_{ε}$ is smooth with support in $\overline{supp g + supp φ_{ε}}$ ; $∥ g * φ_{ε} - g ∥_{2} \to 0$ for $g \in L^{2} (R)$ ; and $C_{c}^{\infty} (R) \subseteq S (R)$ . Explicit compactly supported smooth cutoffs The mollifier family generated by a unit-mass smooth bump A unit-mass smooth bump generates an $L^{1}$ approximate identity Every $L^{1}$ approximate identity converges to the identity in $L^{p}$ for $1 \leq p < \infty$ Convolution with a mollifier is smooth, and derivatives pass under the integral sign The support of a convolution lies in the closure of the support sumset Schwartz space and its seminorms

[F6]

Every norm-convergent sequence in $L^{2}$ has a subsequence of measurable representatives converging almost everywhere to a representative of the limit, and countable unions of Lebesgue-null sets are Lebesgue null. Complex Lp completeness and almost-everywhere subsequences Finite and countable subadditivity of measures

[F7]

For a bounded linear $T$ on a normed space, $∥ T g ∥ \leq ∥ T ∥ ∥ g ∥$ ; in particular $∥ T g ∥_{\infty} \leq ∥ T ∥ ∥ g ∥_{\infty}$ . The operator norm as the least bound and as the unit-sphere or unit-ball supremum

Counterexample

technique · direct

1.1F1F4givenalgebra

The indicator $f$ is measurable with $0 \leq f \leq 1$ , so $∥ f ∥_{\infty} \leq 1$ , and $\int_{R} ∣ f ∣^{2} = λ_{1} ((0, 1)) = 1$ by [F4]; hence $f \in L^{\infty} (R) \cap L^{2} (R)$ .

1.2F1F3F4algebra

$q$ is not essentially bounded. Indeed, fix $M > 0$ ; by [F1] and [F3], for $x \in (0, 1)$ one has $q (x) > M$ exactly when $\log \frac{x}{1 - x} > π M$ , i.e. $\frac{x}{1 - x} > e^{π M}$ , i.e. $x > \frac{1}{1 + e^{- π M}}$ . Hence the set $E_{M} : = {x \in (0, 1) : q (x) > M}$ is the interval $(\frac{1}{1 + e^{- π M}}, 1)$ , which by [F4] has measure $\frac{e^{- π M}}{1 + e^{- π M}} > 0$ . Since $M$ was arbitrary, no real number bounds $q$ from above almost everywhere, so $q \notin L^{\infty} (R)$ .

2.1F5step 1.1algebra

Let $φ$ be the unit-mass bump of [F5] and for $j \in N$ put $f_{j} : = f * φ_{1 / (j + 1)}$ . Then $f_{j}$ is smooth with support in $[- 2 / (j + 1), 1 + 2 / (j + 1)]$ , hence $f_{j} \in C_{c}^{\infty} (R) \subseteq S (R)$ ; and $0 \leq f_{j} \leq 1$ because $0 \leq f \leq 1$ and $φ_{1 / (j + 1)} \geq 0$ has integral one. By [F5], $∥ f_{j} - f ∥_{2} \to 0$ .

3.1F1F2step 2.1algebra

$∥ H f_{j} - q ∥_{2} \to 0$ : by [F1] $q = H f$ and by [F2] $H$ is a linear isometry, so $∥ H f_{j} - q ∥_{2} = ∥ f_{j} - f ∥_{2} \to 0$ by step 2.1.

3.2F7step 2.1

Suppose, for contradiction, that $T : L^{\infty} (R; C) \to L^{\infty} (R; C)$ is bounded and linear with $T g = H g$ almost everywhere for every $g \in L^{\infty} (R) \cap L^{2} (R)$ . Each $f_{j}$ of step 2.1 lies in this intersection, so $T f_{j} = H f_{j}$ almost everywhere; by [F7] and $∥ f_{j} ∥_{\infty} \leq 1$ , $∥ H f_{j} ∥_{\infty} = ∥ T f_{j} ∥_{\infty} \leq ∥ T ∥ ∥ f_{j} ∥_{\infty} \leq ∥ T ∥$ .

4.1step 3.1step 3.2F6

By step 3.1 and [F6] there is a subsequence $(H f_{j_{k}})_{k}$ converging almost everywhere to $q$ . The sets where $T f_{j_{k}} \neq H f_{j_{k}}$ are null, the sets where $∣ H f_{j_{k}} ∣ > ∥ T ∥$ are null by step 3.2, and the set where the subsequence fails to converge to $q$ is null; their countable union is null by [F6]. Off that union one has $∣ H f_{j_{k}} ∣ \leq ∥ T ∥$ for every $k$ by step 3.2 and $H f_{j_{k}} \to q$ , so $∣ q ∣ \leq ∥ T ∥$ almost everywhere. Hence $q \in L^{\infty} (R)$ with $∥ q ∥_{\infty} \leq ∥ T ∥$ .

5.1step 1.2step 4.1∎

Step 4.1 contradicts step 1.2, so no such bounded linear operator $T$ exists. The compatibility required of $T$ was only on $L^{\infty} \cap L^{2}$ , hence also holds for every Schwartz function; therefore no bounded $L^{\infty}$ action agreeing with the $L^{2}$ Hilbert transform on the intersection exists. A $B M O$ -valued endpoint is a different assertion and is not addressed.

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Hilbert transform of the line Poisson kernel

Statement

Assume Countable Choice and fix $a > 0$ , with the Fourier convention $e^{- 2 π i x ξ}$ of Fourier transform on complex L1 classes. Put

$P_{a} (x) : = \frac{a}{π (a^{2} + x^{2})}, Q_{a} (x) : = \frac{x}{π (a^{2} + x^{2})} .$

Then:

for every $ξ \in R$ , $\hat{P_{a}} (ξ) = e^{- 2 π a ∣ ξ ∣}$ ;
for every $x \in R$ the symmetric principal value $\lim_{ε ↓ 0} H_{ε} P_{a} (x)$ of Truncated Hilbert transform and principal value exists and equals $Q_{a} (x)$ , the conjugate Poisson kernel;
$Q_{a} \in L^{2} (R; C)$ , and $Q_{a} = H P_{a}$ in $L^{2} (R; C)$ for the $L^{2}$ Hilbert transform $H$ with symbol $m (ξ) = - i sgn (ξ)$ of The Hilbert transform is an L2 isometry and squares to minus the identity.

This is the line Poisson kernel, not the periodic Poisson kernel on the circle; no statement is made about $L^{p}$ mapping for $p \neq 2$ .

Facts & Assumptions

Given: $a > 0$ , Countable Choice, the $L^{p}$ conventions of Complex Lp classes and Euclidean test-function conventions, the Fourier convention of Fourier transform on complex L1 classes, and the truncated Hilbert transform, $H_{ε} f (x) = \frac{1}{π} \int_{∣ t ∣ > ε} f (x - t) / t d t = \frac{1}{π} \int_{∣ x - y ∣ > ε} f (y) / (x - y) d y$ , absolutely convergent for $f \in L^{p}$ , $1 \leq p < \infty$ , whose principal value is the $ε ↓ 0$ limit wherever it exists.

[F1]

For $ε > 0$ and $x \in R$ , $H_{ε} f (x) = \frac{1}{π} \int_{∣ t ∣ > ε} \frac{f (x - t)}{t} d t$ is the absolutely convergent truncation of Truncated Hilbert transform and principal value for $f \in L^{p}$ , $1 \leq p < \infty$ ; $H_{p v} f (x)$ is its $ε ↓ 0$ limit where that exists, and no almost-everywhere existence and no $L^{p}$ bound is asserted by the definition.

[F2]

For Schwartz $g$ the principal value exists at every $x$ and equals $(W * g) (x)$ for the tempered convolution with $W = pv \frac{1}{π x}$ , whose pairing with a Schwartz test function is the two-piece formula $\frac{1}{π} \int_{∣ x ∣ > 1} \frac{φ (x)}{x} d x + \frac{1}{π} \int_{∣ x ∣ < 1} \frac{φ (x) - φ (0)}{x} d x$ ; the $L^{2}$ extension $H$ has symbol $m (ξ) = - i sgn (ξ)$ , extends the Schwartz-core action uniquely and satisfies $∥ H g ∥_{2} = ∥ g ∥_{2}$ . The Hilbert transform is the tempered convolution with pv(1/(pi x)) and has signum Fourier multiplier The Hilbert transform is an L2 isometry and squares to minus the identity

[F3]

There is $χ \in C_{c}^{\infty} (R)$ with $0 \leq χ \leq 1$ , $χ = 1$ on $[- 1, 1]$ and $χ = 0$ off $(- 2, 2)$ . Explicit compactly supported smooth cutoffs

[F4]

For $f \in L^{1} (R; C)$ the transform is the absolutely convergent integral $\hat{f} (ξ) = \int_{R} f (x) e^{- 2 π i x ξ} d x$ of the Fourier-transform definition, which defines a function at every frequency; $F$ is complex-linear on $L^{1}$ and maps it into the bounded uniformly continuous functions, with $\sup_{ξ} ∣ \hat{f} (ξ) ∣ \leq ∥ f ∥_{1}$ ; and if $\hat{f} \in L^{1}$ , then $g (x) = \int_{R} \hat{f} (ξ) e^{2 π i x ξ} d ξ$ is bounded and continuous, equals $f$ almost everywhere, and equals the value of $f$ at every Lebesgue point of $f$ . Fourier transform on complex L1 classes The L1 transform is bounded and uniformly continuous L1 Fourier inversion with an integrable transform

[F5]

A $C_{c}^{\infty} (R)$ function is a Schwartz function: all seminorms $p_{α β} (f) = \sup_{x} ∣ x^{α} \partial^{β} f (x) ∣$ are finite because they are suprema of continuous functions of compact support. Schwartz space and its seminorms

[F6]

On a compact interval a continuous function is Riemann integrable and hence Lebesgue integrable with the same integral; a nonnegative function Riemann integrable on every $[a, R]$ whose improper integral $\int_{a}^{\infty}$ converges is Lebesgue integrable on $[a, \infty)$ with the same integral; oriented additivity over subintervals holds, and the second fundamental theorem gives $\int_{u}^{v} G^{'} = G (v) - G (u)$ for a differentiable $G$ with integrable derivative. A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral A nonnegative improper Riemann integral on a half-line agrees with the Lebesgue integral 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 $a, b, c$ The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$

[F7]

Chain rule, the principal arctangent, and the natural logarithm: $(\arctan)^{'} = 1 / (1 + x^{2})$ and $\arctan x = \int_{0}^{x} d t / (1 + t^{2})$ ; $\arctan$ is the continuous, strictly increasing inverse of $\tan$ on $(- π / 2, π / 2)$ , so its image is $(- π / 2, π / 2)$ and its supremum is $π / 2$ ; $\log$ is continuous on $(0, \infty)$ , $\log^{'} = 1 / x$ , $\log x = \int_{1}^{x} d t / t$ , $\log 1 = 0$ and $\log (x / y) = \log x - \log y$ ; and for differentiable $φ$ the mean value theorem bounds a difference quotient by $∥ φ^{'} ∥_{\infty}$ . 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 $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ Principal arctangent: derivative, integral, power series, and the Gregory–Leibniz series The principal inverse tangent $\arctan : R \to (- π / 2, π / 2)$ The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm 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 $f (b) - f (a) = f^{'} (c) (b - a)$

[F8]

Dominated convergence for complex-valued functions, and the a.e.-subsequence property of $L^{2}$ -convergent sequences. Dominated convergence Complex Lp completeness and almost-everywhere subsequences

[F9]

A quotient of polynomials is continuous wherever its denominator does not vanish, so $y \mapsto (x + y) / (a^{2} + y^{2})$ is continuous on $R$ . Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function

[F10]

Complex and real calculus on intervals: for complex $C^{1}$ functions $u, v$ on $[a, b]$ , $\int_{a}^{b} u^{'} = u (b) - u (a)$ ; $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ and $∣ \exp (x + i y) ∣ = e^{x}$ ; $(\sin t)^{'} = \cos t$ and $(\cos t)^{'} = - \sin t$ ; the real exponential is smooth with $(\exp)^{'} = \exp$ ; and the sum, scalar-multiple and product rules and the chain rule for real derivatives hold. Complex integration by parts on intervals and decaying lines $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ The derivatives of sine and cosine are cosine and minus sine The exponential function is smooth and $(\exp)^{'} = \exp$ 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 $g (c) \neq 0$ 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 $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$

[F11]

Balls, averages and Lebesgue points: every Euclidean ball $B (x, r)$ is Lebesgue measurable with $0 < λ (B (x, r)) < \infty$ , so the ball average $A_{r} f (x) = λ (B (x, r))^{- 1} \int_{B (x, r)} f d λ$ is defined for $f \in L_{l o c}^{1} (R^{n})$ ; a point $x$ is a Lebesgue point of $f$ exactly when $A_{r} (∣ f - f (x) ∣) (x) \to 0$ as $r \to 0^{+}$ ; and $\int_{E} f d λ : = \int f χ_{E} d λ$ for integrable real or complex $f$ , this indefinite integral being countably additive on pairwise disjoint measurable families. Every continuous function is Borel measurable. Euclidean balls have positive finite Lebesgue measure The average of a locally integrable function over a Euclidean ball Lebesgue points and the Lebesgue set of an $L_{l o c}^{1}$ class A locally integrable function on $R^{n}$ Integral over a measurable subset The indefinite integral of an integrable function is countably additive on measurable sets Continuous functions on Euclidean spaces are Borel measurable

[F12]

Reflection and order rules: the reflection $T (x) = - x$ of $R^{n}$ is a $C^{1}$ diffeomorphism with $∣ \det D T ∣ = 1$ , so $\int_{R^{n}} F (T (x)) d λ (x) = \int_{R^{n}} F (y) d λ (y)$ for every integrable $F$ ; if $0 \leq f \leq g$ are measurable then $\int f d μ \leq \int g d μ$ , and $\int c f d μ = c \int f d μ$ for $c \geq 0$ ; the nonnegative integral agrees with the simple integral, and the simple integral of a constant multiple of an indicator is $\int_{s i m p l e} c χ_{E} d μ = c μ (E)$ . A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions Monotonicity and nonnegative homogeneity of the nonnegative integral The nonnegative integral agrees with the simple integral on simple functions The integral of a nonnegative simple function

[F13]

Continuity and decay: sums, scalar multiples and products of continuous real functions, and the absolute value, are continuous, and composites of continuous functions are continuous; the real exponential is $C^{\infty}$ and hence continuous; and $1 + s \leq e^{s}$ for every real $s$ , so $e^{- s} \leq (1 + s)^{- 1} \to 0$ as $s \to + \infty$ . Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function A composite of continuous functions is continuous, with no side hypothesis of the kind the composition of limits needs The exponential function is smooth and $(\exp)^{'} = \exp$ A function differentiable at $c$ is continuous at $c$ $1 + x \leq \exp (x)$ for every real $x$ , hence $(1 - p)^{m} \leq \exp (- m p)$

Proof

technique · direct

Steps 1.1, 2.1, 3.1 and 4.1 settle assertion 1; the remaining steps settle assertions 2 and 3. Nothing in the principal-value computation uses assertion 1.

1.1F6F11F12F13

Let $q (u) : = e^{- 2 π a ∣ u ∣}$ for $u \in R$ . Then $q$ is continuous and real-valued: $u \mapsto ∣ u ∣$ is continuous, so is $u \mapsto - 2 π a ∣ u ∣$ , and the composite with the continuous exponential is continuous [F13]; in particular $q$ is Borel measurable [F11]. For $R > 0$ the second fundamental theorem [F6] applied on $[0, R]$ to the antiderivative $u \mapsto - e^{- 2 π a u} / (2 π a)$ gives $\int_{0}^{R} e^{- 2 π a u} d u = \frac{1 - e^{- 2 π a R}}{2 π a}$ , and $e^{- 2 π a R} \leq \frac{1}{1 + 2 π a R} \to 0$ as $R \to \infty$ by [F13]; hence the improper Riemann integral of the nonnegative continuous function $q$ over $[0, \infty)$ converges to $\frac{1}{2 π a}$ , and [F6] makes $q$ Lebesgue integrable on $(0, \infty)$ with $\int_{(0, \infty)} q d λ = \frac{1}{2 π a}$ . The function $q χ_{[0, \infty)}$ is already integrable by [F6]. Apply [F12] to this function and the reflection $T (u) = - u$ , whose Jacobian has absolute value one: its pullback $q χ_{(- \infty, 0]}$ is integrable and has the same integral $1 / (2 π a)$ (the singleton ${0}$ has measure zero). Thus $q$ is the sum of two known integrable functions $q χ_{(- \infty, 0]}$ and $q χ_{(0, \infty)}$ , so $q \in L^{1} (R)$ before applying additivity [F11], which gives $\int_{R} q d λ = \frac{1}{π a} < \infty$ , that is, $q \in L^{1} (R)$ . For every ball $B$ monotonicity [F12] gives $\int_{B} q d λ \leq \int_{R} q d λ < \infty$ , so $q \in L_{l o c}^{1} (R)$ as well.

1.2F6F12algebra

For the integrability of $P_{a}$ used repeatedly below, note that $0 < P_{a} (y) \leq 1 / (π a)$ for all $y$ , while $0 < P_{a} (y) \leq a / (π y^{2})$ for $y \neq 0$ , because $y^{2} \leq a^{2} + y^{2}$ ; hence $P_{a} \leq \min (1 / (π a), a / (π y^{2}))$ pointwise. By [F6] the continuous bounded function $P_{a}$ is integrable over $[- a, a]$ , and the improper integrals $\int_{a}^{\infty} a / (π y^{2}) d y$ and $\int_{a}^{\infty} a^{2} / (π^{2} y^{4}) d y$ converge by the second fundamental theorem applied to the antiderivatives $- a / (π y)$ and $- a^{2} / (3 π^{2} y^{3})$ with vanishing limits at infinity. Reflecting the already integrable positive-tail majorants by [F12] gives the corresponding negative-tail bounds. Together with integrability on $[- a, a]$ , these bounds give $P_{a} \in L^{1} (R) \cap L^{2} (R)$ , with $\int_{R} ∣ P_{a} ∣ \leq 4 / π$ and $\int_{R} P_{a}^{2} \leq 8 / (3 π^{2} a)$ .

1.3F1F6algebra

Substituting $y = x - t$ in the displayed truncation of [F1] shows that for every $x$ , every $0 < ε < R$ and every $f \in L^{1} (R)$ , $\frac{1}{π} \int_{ε < ∣ t ∣ < R} \frac{f (x - t)}{t} d t = \frac{1}{π} \int_{ε < ∣ x - y ∣ < R} \frac{f (y)}{x - y} d y$ . Subtracting the constant $f (x)$ , whose integral against $1 / (x - y)$ vanishes over the symmetric domain $ε < ∣ x - y ∣ < R$ (the substitution $u = y - x$ makes the integrand odd), gives the identity $\frac{1}{π} \int_{ε < ∣ x - y ∣ < R} \frac{f (y)}{x - y} d y = \frac{1}{π} \int_{ε < ∣ x - y ∣ < R} \frac{f (y) - f (x)}{x - y} d y$ , valid when $f$ is bounded near $x$ ; the subtraction changes no value.

2.1step 1.1F6F8F10F13

For $ξ \in R$ put $z : = 2 π (a + i ξ)$ , so that $Re z = 2 π a > 0$ and $∣ z ∣ \geq 2 π a$ , and let $u (t) : = - z^{- 1} e^{- z t}$ for $t \in R$ . By [F10], $e^{- z t} = e^{- 2 π a t} (\cos (2 π ξ t) - i \sin (2 π ξ t))$ , and differentiating the two real components with the product, chain, trigonometric and exponential derivative rules of [F10] gives $\frac{d}{d t} e^{- z t} = - z e^{- z t}$ , so $u$ is complex $C^{1}$ on $R$ with $u^{'} (t) = e^{- z t}$ ; the complex fundamental theorem of calculus [F10] on $[0, R]$ then gives $\int_{0}^{R} e^{- z t} d t = u (R) - u (0) = \frac{1 - e^{- z R}}{z}$ , while $∣ e^{- z R} ∣ = e^{- 2 π a R} \leq \frac{1}{1 + 2 π a R} \to 0$ by [F10] and [F13], so the truncated integrals converge to $1 / z$ . Moreover $∣ e^{- z t} ∣ = q (t)$ for $t \geq 0$ and $q$ is Lebesgue integrable on $(0, \infty)$ with $\int_{(0, \infty)} q d λ = \frac{1}{2 π a}$ by step 1.1, so dominated convergence [F8] applied to the functions $1_{[0, R]} e^{- z \cdot}$ , which converge pointwise to $e^{- z \cdot}$ and are dominated by $q$ , gives

$\int_{(0, \infty)} e^{- z t} d λ (t) = \lim_{R \to \infty} \int_{[0, R]} e^{- z t} d λ (t) = \lim_{R \to \infty} \int_{0}^{R} e^{- z t} d t = \frac{1}{z} = \frac{1}{2 π (a + i ξ)},$

the middle equality because on the compact interval $[0, R]$ the continuous integrand has the same Riemann and Lebesgue integrals [F6]. Replacing $ξ$ by $- ξ$ throughout gives the companion identity $\int_{(0, \infty)} e^{- 2 π (a - i ξ) t} d λ (t) = \frac{1}{2 π (a - i ξ)}$ .

2.2step 1.1F11F12F13

Fix $x \in R$ . Since $q$ is continuous at $x$ [F13], for every $ε > 0$ there is $δ > 0$ such that $∣ q (y) - q (x) ∣ < ε$ whenever $∣ y - x ∣ < δ$ ; for $0 < r < δ$ the pointwise bound $∣ q - q (x) ∣ χ_{B (x, r)} \leq ε χ_{B (x, r)}$ , the monotonicity and homogeneity of the nonnegative integral, and the value $\int_{s i m p l e} ε χ_{B (x, r)} d λ = ε λ (B (x, r))$ of the simple integral [F12] give, since $λ (B (x, r))$ is positive and finite [F11] and $q \in L_{l o c}^{1} (R)$ by step 1.1, that the ball average $A_{r} (∣ q - q (x) ∣) (x)$ of [F11] satisfies $A_{r} (∣ q - q (x) ∣) (x) \leq ε$ for every $0 < r < δ$

$\int_{B (x, r)} ∣ q (y) - q (x) ∣ d λ (y) \leq ε λ (B (x, r)) .$

Since $ε > 0$ was arbitrary, the limit as $r \to 0^{+}$ of the average is $0$ , so every $x$ is a Lebesgue point of $q$ with value $q (x)$ [F11].

2.3step 1.3algebra

Applying step 1.3 to $f = P_{a}$ and using

$\frac{P_{a} (y) - P_{a} (x)}{x - y} = \frac{a}{π} \cdot \frac{x + y}{(a^{2} + x^{2}) (a^{2} + y^{2})},$

which is algebra from $P_{a} (y) - P_{a} (x) = \frac{a}{π} \cdot \frac{(a^{2} + x^{2}) - (a^{2} + y^{2})}{(a^{2} + x^{2}) (a^{2} + y^{2})}$ and $(a^{2} + x^{2}) - (a^{2} + y^{2}) = (x - y) (x + y)$ , gives for $0 < ε < R$

$\frac{1}{π} \int_{ε < ∣ t ∣ < R} \frac{P_{a} (x - t)}{t} d t = \frac{a}{π^{2}} \cdot \frac{1}{a^{2} + x^{2}} \int_{ε < ∣ x - y ∣ < R} \frac{x + y}{a^{2} + y^{2}} d y .$

2.4step 1.2F6algebra

For the $L^{2}$ assertion, note that $∣ Q_{a} (y) ∣ = \frac{∣ y ∣}{π (a^{2} + y^{2})} \leq \frac{1}{2 π a}$ for all $y$ and $∣ Q_{a} (y) ∣ \leq \frac{1}{π ∣ y ∣}$ for $∣ y ∣ \geq a$ ; the same elementary integration as in step 1.2, by [F6], gives $Q_{a} \in L^{2} (R)$ .

2.5step 1.2F3F5F8

For $j \geq 1$ define $ψ_{j} (y) : = P_{a} (y) χ (y / j)$ with $χ$ as in [F3]. Each $ψ_{j}$ lies in $C_{c}^{\infty} (R)$ and hence in $S (R)$ by [F5], with $0 \leq ψ_{j} \leq P_{a}$ ; and $ψ_{j} (y) = P_{a} (y)$ as soon as $j \geq ∣ y ∣$ , so $ψ_{j} \to P_{a}$ pointwise everywhere. Since $∣ ψ_{j} - P_{a} ∣ \leq 2 P_{a}$ with $P_{a} \in L^{1} \cap L^{2}$ by step 1.2, dominated convergence [F8] gives $∥ ψ_{j} - P_{a} ∥_{1} \to 0$ and $∥ ψ_{j} - P_{a} ∥_{2} \to 0$ .

3.1step 1.1step 2.1F4F11F12algebra

By the definition of the transform [F4], $\hat{q} (ξ) = \int_{R} q (t) e^{- 2 π i ξ t} d λ (t)$ for every $ξ$ ; the integrand $h_{ξ} : = q e^{- 2 π i ξ \cdot}$ satisfies $∣ h_{ξ} ∣ = q \in L^{1} (R)$ by step 1.1, so $h_{ξ}$ is integrable and its indefinite integral is countably additive on pairwise disjoint measurable families [F11]. Splitting over the disjoint measurable sets $(- \infty, 0]$ and $(0, \infty)$ , which cover $R$ , and applying the reflection change of variables [F12] to the integrable function $h_{ξ} χ_{(- \infty, 0]}$ , whose reflection is $q χ_{(0, \infty)} e^{2 π i ξ \cdot}$ because $q$ is even, gives, using step 2.1 on each half-line and step 2.1 again with $ξ$ replaced by $- ξ$ ,

$\int_{(- \infty, 0]} q (t) e^{- 2 π i ξ t} d λ (t) = \int_{(0, \infty)} q (s) e^{2 π i ξ s} d λ (s) = \int_{(0, \infty)} e^{- 2 π (a - i ξ) s} d λ (s) = \frac{1}{2 π (a - i ξ)},$

while the positive half contributes $\int_{(0, \infty)} q (t) e^{- 2 π i ξ t} d λ (t) = \int_{(0, \infty)} e^{- 2 π (a + i ξ) t} d λ (t) = \frac{1}{2 π (a + i ξ)}$ . Adding the two pieces and simplifying,

$\hat{q} (ξ) = \frac{1}{2 π} (\frac{1}{a + i ξ} + \frac{1}{a - i ξ}) = \frac{1}{2 π} \cdot \frac{2 a}{a^{2} + ξ^{2}} = \frac{a}{π (a^{2} + ξ^{2})} = P_{a} (ξ)$

for every $ξ \in R$ , since $(a + i ξ) (a - i ξ) = a^{2} + ξ^{2}$ .

3.2step 2.3F6F7F9

Put $G (y) : = \frac{x}{a} \arctan \frac{y}{a} + \frac{1}{2} \log (a^{2} + y^{2})$ . By the chain rule, the arctangent and logarithm derivatives of [F7], and [F9], $G$ is differentiable on $R$ with $G^{'} (y) = \frac{x}{a^{2} + y^{2}} + \frac{y}{a^{2} + y^{2}} = \frac{x + y}{a^{2} + y^{2}}$ . Since the domain ${ε < ∣ x - y ∣ < R}$ is the disjoint union of the intervals $(x - R, x - ε)$ and $(x + ε, x + R)$ on which $y \mapsto (x + y) / (a^{2} + y^{2})$ is continuous, [F6] and the right-hand integral of step 2.3 give

$\int_{ε < ∣ x - y ∣ < R} \frac{x + y}{a^{2} + y^{2}} d y = G (x + R) - G (x - R) + G (x - ε) - G (x + ε),$

that is, with all logarithms of positive arguments,

$\frac{x}{a} [\arctan \frac{x + R}{a} - \arctan \frac{x - R}{a} + \arctan \frac{x - ε}{a} - \arctan \frac{x + ε}{a}] + \frac{1}{2} \log \frac{(a^{2} + (x - ε)^{2}) (a^{2} + (x + R)^{2})}{(a^{2} + (x + ε)^{2}) (a^{2} + (x - R)^{2})} .$

3.3step 2.5step 1.3F1F2

Fix $x \in R$ and $j > ∣ x ∣$ , and use the functions $ψ_{j}$ of step 2.5. For Schwartz $ψ_{j}$ , [F2] represents the principal value at $x$ by the two-piece pairing, and the oddness cancellation of step 1.3 identifies it with $(W * ψ_{j}) (x) = \frac{1}{π} \int_{∣ t ∣ > 1} \frac{ψ_{j} (x - t)}{t} d t + \frac{1}{π} \int_{∣ t ∣ < 1} \frac{ψ_{j} (x - t) - ψ_{j} (x)}{t} d t$ ; combining this with the same identity for $P_{a}$ in step 1.3, and abbreviating $δ_{j} : = ψ_{j} - P_{a}$ , gives for every $0 < ε < 1$

$H_{ε} ψ_{j} (x) - H_{ε} P_{a} (x) = \frac{1}{π} \int_{ε < ∣ t ∣ < 1} \frac{δ_{j} (x - t) - δ_{j} (x)}{t} d t + \frac{1}{π} \int_{∣ t ∣ > 1} \frac{δ_{j} (x - t)}{t} d t .$

3.4step 2.5F2F8

By [F2] the isometry $H$ is defined on $L^{2}$ and is linear, so $∥ H ψ_{j} - H P_{a} ∥_{2} = ∥ ψ_{j} - P_{a} ∥_{2} \to 0$ by step 2.5; that is, $H ψ_{j} \to H P_{a}$ in $L^{2} (R)$ .

4.1step 1.1step 1.2step 2.2step 3.1F4

Both $q \in L^{1} (R)$ (step 1.1) and $\hat{q} = P_{a} \in L^{1} (R)$ (step 1.2, step 3.1) are integrable, so the inversion theorem [F4] applied to $f : = q$ gives a bounded continuous function $g (x) = \int_{R} P_{a} (ξ) e^{2 π i x ξ} d λ (ξ)$ that agrees with $q$ almost everywhere and agrees with $q (x)$ at every Lebesgue point $x$ of $q$ ; every real $x$ is such a point by step 2.2, so $g (x) = e^{- 2 π a ∣ x ∣}$ everywhere, and writing the defining integral of $\hat{P_{a}}$ [F4] at the frequency $- x$ identifies $g (x) = \hat{P_{a}} (- x)$ , so $\hat{P_{a}} (- x) = e^{- 2 π a ∣ x ∣}$ for every $x$ , and replacing $x$ by $- ξ$ gives $\hat{P_{a}} (ξ) = e^{- 2 π a ∣ ξ ∣}$ for every $ξ$ ; this proves assertion 1.

4.2step 1.2step 3.2F6F7

Since $P_{a} \in L^{1}$ by step 1.2, for each fixed $ε > 0$ the full integral $H_{ε} P_{a} (x) = \frac{1}{π} \int_{∣ t ∣ > ε} P_{a} (x - t) / t d t$ converges absolutely and is the limit of its truncations at $R \to \infty$ ; hence passing to the limit $R \to \infty$ in step 3.2 is legitimate. As $R \to \infty$ , $\arctan \frac{x + R}{a} \to \frac{π}{2}$ and $\arctan \frac{x - R}{a} \to - \frac{π}{2}$ because $\arctan$ is increasing with supremum $π / 2$ and infimum $- π / 2$ on its range $(- π / 2, π / 2)$ ; the logarithmic argument tends to $\frac{a^{2} + (x - ε)^{2}}{a^{2} + (x + ε)^{2}} > 0$ , and log is continuous there by [F7]. Therefore

$H_{ε} P_{a} (x) = \frac{a}{π^{2} (a^{2} + x^{2})} [\frac{x}{a} (π + \arctan \frac{x - ε}{a} - \arctan \frac{x + ε}{a}) + \frac{1}{2} \log \frac{a^{2} + (x - ε)^{2}}{a^{2} + (x + ε)^{2}}] .$

5.1step 4.2F7

Letting $ε ↓ 0$ in step 4.2, continuity of $\arctan$ and $\log$ [F7] gives $\arctan \frac{x - ε}{a} - \arctan \frac{x + ε}{a} \to 0$ and $\log \frac{a^{2} + (x - ε)^{2}}{a^{2} + (x + ε)^{2}} \to \log 1 = 0$ ; hence

$\lim_{ε ↓ 0} H_{ε} P_{a} (x) = \frac{a}{π^{2} (a^{2} + x^{2})} \cdot \frac{x π}{a} = \frac{x}{π (a^{2} + x^{2})} = Q_{a} (x) .$

This holds for every $x \in R$ , including $x = 0$ , where both the display and the oddness of the truncated integrand give value $0$ . This proves assertion 2.

6.1step 5.1step 3.3F7F8

In the situation of step 3.3 one has $δ_{j} (x) = 0$ , and $\sup_{j} ∥ δ_{j}^{'} ∥_{\infty} < \infty$ : indeed $δ_{j}^{'} = P_{a}^{'} (χ (\cdot / j) - 1) + P_{a} χ^{'} (\cdot / j) / j$ with $∥ χ (\cdot / j) - 1 ∥_{\infty} \leq 1$ and $∣ χ^{'} (\cdot / j) / j ∣ \leq ∥ χ^{'} ∥_{\infty} / j$ , while $P_{a}$ and $P_{a}^{'}$ are bounded. Hence the mean value theorem [F7] bounds the difference quotient of $δ_{j}$ by a constant $C (a)$ uniformly in $j$ on $∣ t ∣ < 1$ , and the integrand of the first term of step 3.3 is dominated by the integrable constant $C (a)$ on $0 < ∣ t ∣ < 1$ ; letting $ε ↓ 0$ by dominated convergence [F8], and using that $H_{ε} ψ_{j} (x) \to H ψ_{j} (x)$ and $H_{ε} P_{a} (x) \to Q_{a} (x)$ by [F2] and step 5.1,

$H ψ_{j} (x) - Q_{a} (x) = \frac{1}{π} \int_{0 < ∣ t ∣ < 1} \frac{δ_{j} (x - t) - δ_{j} (x)}{t} d t + \frac{1}{π} \int_{∣ t ∣ > 1} \frac{δ_{j} (x - t)}{t} d t .$

7.1step 2.5step 6.1F8

The first term of step 6.1 tends to $0$ as $j \to \infty$ by dominated convergence [F8]: for each fixed $t \neq 0$ the integrand tends to $0$ because $δ_{j} \to 0$ pointwise and $δ_{j} (x) = 0$ for $j > ∣ x ∣$ , and it is dominated by $C (a)$ on the finite-measure set $0 < ∣ t ∣ < 1$ ; the second term tends to $0$ because $∣ \int_{∣ t ∣ > 1} δ_{j} (x - t) / t d t ∣ \leq ∥ δ_{j} ∥_{1} \to 0$ by step 2.5. Therefore $H ψ_{j} (x) \to Q_{a} (x)$ for every fixed $x$ .

8.1step 3.4step 7.1F8∎

By step 3.4 the sequence $H ψ_{j}$ converges in $L^{2}$ to a representative of the class $H P_{a}$ ; by [F8] it has a subsequence converging almost everywhere to a representative of $H P_{a}$ , while step 7.1 makes that same subsequence converge to $Q_{a}$ at every point. Hence $Q_{a} = H P_{a}$ almost everywhere, i.e. $Q_{a} = H P_{a}$ in $L^{2} (R; C)$ , which is assertion 3.

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Finite sum of Riesz squares in L2

Statement

Assume Countable Choice and let $n \geq 1$ . Let $R_{1}, \dots, R_{n}$ be the Riesz transforms of Riesz transforms on Euclidean space, the $L^{2} (R^{n}; C)$ operators $R_{j} = F_{2}^{- 1} M_{m_{j}} F_{2}$ with symbols

$m_{j} (ξ) = {\begin{cases} - i ξ_{j} / ∣ ξ ∣, & ξ \neq 0, \\ 0, & ξ = 0. \end{cases}$

Then:

$\sum_{j = 1}^{n} R_{j}^{2} f = - f$ for every $f \in L^{2} (R^{n}; C)$ , the identity holding as $L^{2}$ classes, with the explicit finite symbol computation $\sum_{j = 1}^{n} m_{j} (ξ)^{2} = - 1$ for every $ξ \neq 0$ ;
at $n = 1$ the operator $R_{1}$ is the line Hilbert transform $H$ of The Hilbert transform is an L2 isometry and squares to minus the identity, so the $n = 1$ case of assertion 1 is exactly $H^{2} = - I$ ;
the assigned value $m_{j} (0) = 0$ is immaterial: it is a value on the Lebesgue-null singleton ${0}$ , and the multiplier operator depends only on the almost-everywhere class of its symbol. Unlike the periodic conjugate operator, no zero-mode exception arises here.

This is an $L^{2}$ statement only; no $L^{p}$ bound for $p \neq 2$ is asserted.

Facts & Assumptions

Given: Countable Choice, the dimension $n \geq 1$ , and the Euclidean $L^{2}$ conventions of Complex Lp classes and Euclidean test-function conventions.

[F1]

For $1 \leq j \leq n$ the $j$ -th Riesz transform is $R_{j} = F_{2}^{- 1} M_{m_{j}} F_{2}$ with $m_{j} (ξ) = - i ξ_{j} / ∣ ξ ∣$ for $ξ \neq 0$ and $m_{j} (0) = 0$ ; the symbol is measurable with $∣ m_{j} (ξ) ∣ \leq 1$ everywhere, $R_{j}$ is well-defined and bounded on $L^{2}$ with $∥ R_{j} ∥ \leq 1$ , the assigned value at the origin has no effect on the operator, and the definition asserts only the multiplier description. Riesz transforms on Euclidean space

[F2]

A measurable symbol $m$ with finite essential supremum defines $T_{m} = F_{2}^{- 1} M_{m} F_{2}$ , a bounded operator with $∥ T_{m} ∥ = ∥ m ∥_{\infty}$ , and the operator depends only on the almost-everywhere class of $m$ : values on Lebesgue-null sets, including the single point ${0}$ , do not affect the operator or its norm. Exact L2 Fourier multiplier norm

[F3]

For these Riesz transforms $∥ R_{j} f ∥_{2} \leq ∥ f ∥_{2}$ and $\sum_{j = 1}^{n} R_{j}^{2} f = - f$ for every $f \in L^{2} (R^{n}; C)$ , as $L^{2}$ statements only. Riesz transforms are L2 contractions and square to minus the identity in sum

[F4]

The line Hilbert transform has Schwartz-core symbol $- i sgn (ξ)$ , extends uniquely to a bounded operator $H$ on $L^{2} (R; C)$ with $∥ H f ∥_{2} = ∥ f ∥_{2}$ and $H^{2} f = - f$ ; the point $ξ = 0$ , where the symbol vanishes, is Lebesgue null and creates no zero-mode exception. The Hilbert transform is an L2 isometry and squares to minus the identity

[F5]

Plancherel: $F_{2}$ is a surjective complex-linear isometry of $L^{2} (R^{n}; C)$ , so $F_{2}^{- 1}$ is complex-linear and $F_{2}^{- 1} (- g) = - F_{2}^{- 1} g$ , while $F_{2}^{- 1} (- F_{2} g) = - g$ for every class $g$ . Plancherel theorem

Proof

technique · direct

1.1F1algebra

For $ξ \neq 0$ and every $j$ one has $m_{j} (ξ)^{2} = (- i ξ_{j} / ∣ ξ ∣)^{2} = - ξ_{j}^{2} / ∣ ξ ∣^{2}$ , so the finite sum is $s (ξ) : = \sum_{j = 1}^{n} m_{j} (ξ)^{2} = - \sum_{j = 1}^{n} ξ_{j}^{2} / ∣ ξ ∣^{2} = - 1$ , while $s (0) = \sum_{j} m_{j} (0)^{2} = 0$ . Also each $m_{j}$ is measurable, and $∣ m_{j} (ξ) ∣ = ∣ ξ_{j} ∣ / ∣ ξ ∣ \leq 1$ for $ξ \neq 0$ while $m_{j} (0) = 0$ , so $∣ m_{j} ∣ \leq 1$ everywhere.

2.1step 1.1F2

The symbol $s = \sum_{j = 1}^{n} m_{j}^{2}$ of step 1.1 is measurable and satisfies $∣ s (ξ) ∣ = 1$ for $ξ \neq 0$ and $s (0) = 0$ , hence $∣ s ∣ \leq 1$ everywhere; since $s$ agrees with the constant function $- 1$ on the complement of the singleton ${0}$ , which is Lebesgue null, $s$ and $- 1$ have the same almost-everywhere class.

2.2step 1.1F1F2F5

Since $R_{j} = F_{2}^{- 1} M_{m_{j}} F_{2}$ by [F1], the composition of the two bounded operators $F_{2}$ and $F_{2}^{- 1}$ gives $R_{j}^{2} = F_{2}^{- 1} M_{m_{j}} M_{m_{j}} F_{2} = F_{2}^{- 1} M_{m_{j}^{2}} F_{2} = T_{m_{j}^{2}}$ in the notation of [F2], and summing the finitely many bounded operators gives $\sum_{j = 1}^{n} R_{j}^{2} = F_{2}^{- 1} M_{s} F_{2} = T_{s}$ by the complex-linearity of $F_{2}$ and $F_{2}^{- 1}$ in [F1] and [F5].

2.3step 1.1F1F2F4

At $n = 1$ one has $∣ ξ ∣ = \sqrt{ξ^{2}} = ∣ ξ_{1} ∣$ , so for $ξ \neq 0$ the symbol of [F1] is $m_{1} (ξ) = - i ξ / ∣ ξ ∣ = - i sgn (ξ)$ , while $m_{1} (0) = 0 = - i sgn (0)$ as well; hence $m_{1}$ is exactly the signum symbol of [F4] at every point, and by [F2] the operators agree: $R_{1} = T_{m_{1}} = T_{- i sgn} = H$ .

3.1step 2.1step 2.2F2F3F5

By [F2] the operator $T_{s}$ depends only on the almost-everywhere class of $s$ , which by step 2.1 is the class of the constant $- 1$ ; so $T_{s} = T_{- 1}$ , and for $f \in L^{2} (R^{n}; C)$ the isometry and linearity of [F5] give $T_{- 1} f = F_{2}^{- 1} (- F_{2} f) = - F_{2}^{- 1} F_{2} f = - f$ . Combined with step 2.2 this gives $\sum_{j = 1}^{n} R_{j}^{2} f = - f$ for every $f \in L^{2} (R^{n}; C)$ , which is assertion 1 and agrees with the identity recorded in [F3].

4.1step 2.3step 3.1F4

For $n = 1$ , step 2.3 identifies $R_{1}$ with $H$ , so $R_{1}^{2} = H^{2} = - I$ on $L^{2} (R; C)$ by [F4]; this is exactly the case $n = 1$ of the sum identity proved in step 3.1, and it exhibits assertion 2.

5.1step 2.1step 4.1F2F3∎

Finally, the assignment $m_{j} (0) = 0$ is a value on the Lebesgue-null singleton ${0}$ , and the multiplier operator depends only on the almost-everywhere class of its symbol by [F2]; changing that single value therefore changes neither $R_{j}$ nor any identity above. This is the announced contrast with the periodic conjugate operator, whose multiplier is defined on the frequency-zero mode of a finite-measure circle: on Euclidean $L^{2}$ there is no exceptional constant mode attached to the null set ${0}$ , so assertion 3 holds.

Hilbert and Riesz Transforms — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Hilbert transform of an interval indicator

Statement

Facts & Assumptions

Proof

Hilbert transform is not strong type (1,1)

Statement refuted

Facts & Assumptions

Counterexample

Hilbert transform does not map L-infinity to L-infinity

Statement refuted

Facts & Assumptions

Counterexample

Hilbert transform of the line Poisson kernel

Statement

Facts & Assumptions

Proof

Finite sum of Riesz squares in L2

Statement

Facts & Assumptions

Proof

Sources