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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.

Depends on

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Sources

Loukas Grafakos, Classical Fourier Analysis, third edition (standard reference, not scraped)
Richard S. Laugesen, Harmonic Analysis Lecture Notes (standard reference, not scraped)