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The Riesz-transform formula for second derivatives of the Laplacian

Example

Assume Countable Choice. Let n≥1 and u∈S(Rn). Then, as tempered distributions (equivalently, as L2 classes almost everywhere), ∂i∂ju=−RiRj(Δu)=RiRj(−Δu)(i,j∈{1,…,n}), where R1,…,Rn are the Riesz transforms of Riesz transforms on Euclidean space; the symbol of RiRj is −ξiξj/∣ξ∣2 off the origin, and combining the Lp bounds of The Riesz transforms are bounded on Lp gives ∥∂i∂ju∥Lp≤Cn,p2 ∥Δu∥Lp(1<p<∞). For n=1 the identity reads u′′=−R1R1u′′ and is consistent because R12=−id makes both sides equal u′′.

Facts & Assumptions

Given: ACω, n≥1, 1<p<∞, a Schwartz function u∈S(Rn), and the negative-sign 2π-normalized Fourier convention.

[A1]

The only choice assumption is Countable Choice ACω; it enters through the Plancherel and Riesz-transform interfaces. No full Axiom of Choice is used. (The Axiom of Countable Choice (ACω))

[F1]

The Riesz transforms are Rj=F2−1MmjF2 with mj(ξ)=−iξj/∣ξ∣ for ξ≠0 and mj(0)=0; ∥Rj∥L2→L2≤1 and ∑jRj2=−id on L2. For 1<p<∞ each Rj extends boundedly to Lp(Rn;C) with norm at most Cn,p. (Riesz transforms on Euclidean space, Riesz transforms are L2 contractions and square to minus the identity in sum, The Riesz transforms are bounded on Lp, Exact L2 Fourier multiplier norm)

[F2]

F(∂αu)=(2πiξ)αFu on S′(Rn), and F2 is a linear isometry that is injective on L2; two tempered distributions with the same Fourier transform are equal. (Fourier differentiation and multiplication identities on tempered distributions, Plancherel theorem)

Verification

technique · direct
1.1F1F2algebraA1

Fourier multipliers. For u∈S(Rn) and any i,j, [F2] gives F(∂i∂ju)=(2πiξi)(2πiξj)u^=−4π2ξiξju^ and F(Δu)=−4π2∣ξ∣2u^ in S′; since u and its derivatives are Schwartz functions, these are also the L2 Fourier transforms F2 of the corresponding classes. Off the origin mi(ξ)mj(ξ)=(−iξi/∣ξ∣)(−iξj/∣ξ∣)=−ξiξj/∣ξ∣2, so mimj⋅(−4π2∣ξ∣2)=4π2ξiξj and hence F2(RiRj(−Δu))=F2(∂i∂ju) almost everywhere: indeed RiRj(−Δu)=F2−1(mimjF2(−Δu)) and F2(−Δu)=4π2∣ξ∣2u^, while the value of mimj at the single point ξ=0 is immaterial. Injectivity of F2 [F2] gives the L2 identity ∂i∂ju=RiRj(−Δu), hence also the distributional identity and the sign rearrangement RiRj(−Δu)=−RiRj(Δu).

2.1step 1.1F1algebra

Symbol and Lp bound. The multiplier of RiRj is mimj=−ξiξj/∣ξ∣2 off the origin, so its absolute value is at most 1; applying [F1] twice and using the identity of step 1.1, ∥∂i∂ju∥Lp=∥RiRj(−Δu)∥Lp≤Cn,p2∥Δu∥Lp for 1<p<∞, the norms being those of the Lp classes of the Schwartz functions involved.

2.2step 1.1F1algebra

The one-dimensional case. For n=1 one has m1(ξ)=−iξ/∣ξ∣=−isign⁡(ξ) off the origin, so m12=−1 and therefore R12=−id on L2 by [F1]; the identity of step 1.1 then reads u′′=R1R1(−u′′)=−R1R1u′′, whose right-hand side equals u′′, so the two sides agree.

3.1step 1.1step 2.1step 2.2given∎

Conclusion. For every n≥1 and u∈S(Rn) the second derivatives are the composition of the second-order Riesz multiplier with −Δu; the strict range 1<p<∞ is inherited from the Riesz-transform Lp theorem, and the sign convention is the negative-sign 2π-normalized Fourier transform used throughout. No endpoint p=1 or p=∞ bound is asserted.

Remarks

  • The formula identifies the Hessian of u with a bounded combination of Riesz transforms of the Laplacian, which is the multiplier version of the Calderón–Zygmund representation of second derivatives; it is the whole-space model estimate behind the interior W2,p regularity on this page.

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