Topics in Modern Hyperbolic PDEs Part 5 of 9

Chapter 5: Littlewood–Paley Theory and Function Spaces

The scattering criterion in Chapter 4 uses the norm \(L_t^3L_x^6\) because

发布于 2026-10-14 1,997 字 · 约 6 分钟

§ 01Why separate scales?

\[ \|u^3\|_{L_t^1L_x^2}=\|u\|_{L_t^3L_x^6}^3. \]

To estimate solutions in such spacetime norms, one often separates their frequency scales and treats each scale using its own propagation and concentration properties. High frequencies describe fine spatial variation; low frequencies describe larger-scale structure. A dyadic decomposition makes that separation systematic.

All constructions in this chapter are on \(\mathbb R^d\), \(d\ge1\). We use the Fourier convention

\[ \widehat f(\xi)=\int_{\mathbb R^d}e^{-ix\cdot\xi}f(x)\,dx, \qquad f(x)=\frac1{(2\pi)^d}\int_{\mathbb R^d}e^{ix\cdot\xi}\widehat f(\xi)\,d\xi. \]

Consequently, \(\widehat{\partial_jf}=i\xi_j\widehat f\) and \(\|f\|_2^2=(2\pi)^{-d}\|\widehat f\|_2^2\). We write \(A\lesssim B\) when \(A\le CB\) for a constant independent of the dyadic index, and \(A\simeq B\) when both inequalities hold.

§ 02A smooth dyadic partition

Choose a radial, radially nonincreasing \(\varphi\in C_c^\infty(\mathbb R^d)\) with

\[ 0\le\varphi\le1,\qquad \varphi(\xi)=1\text{ for }|\xi|\le1,\qquad \varphi(\xi)=0\text{ for }|\xi|\ge2. \]

Define

\[ \psi(\xi)=\varphi(\xi)-\varphi(2\xi),\qquad \psi_k(\xi)=\psi(2^{-k}\xi),\qquad P_kf=\mathcal F^{-1}(\psi_k\widehat f),\quad k\in\mathbb Z. \]

Then \(0\le\psi_k\le1\) and

\[ \operatorname{supp}\psi\subset\{1/2\le|\xi|\le2\},\qquad \operatorname{supp}\widehat{P_kf}\subset\{2^{k-1}\le|\xi|\le2^{k+1}\}. \]

The operators are smooth frequency cutoffs, not idempotent projections: in general \(P_k^2\ne P_k\) because \(\psi_k^2\ne\psi_k\).

The sum telescopes:

\[ \sum_{k=-M}^N\psi_k(\xi) =\varphi(2^{-N}\xi)-\varphi(2^{M+1}\xi). \]

For each \(\xi\ne0\), the right side tends to one as \(M,N\to\infty\). At \(\xi=0\), every annular cutoff vanishes. This exception is harmless for \(L^2\) functions and essential for understanding homogeneous spaces of distributions.

定理 2.1Reconstruction in L²

For every \(f\in L^2(\mathbb R^d)\),

\[ \sum_{k=-M}^NP_kf\longrightarrow f\quad\text{in }L^2 \qquad(M,N\to\infty). \]

Proof

By telescoping,

\[ \widehat{f-\sum_{k=-M}^NP_kf} =\big(1-\varphi(2^{-N}\xi)+\varphi(2^{M+1}\xi)\big)\widehat f. \]

Plancherel and the triangle inequality reduce the error to two tails. They satisfy

\[ \|(1-\varphi(2^{-N}\xi))\widehat f\|_2^2 \le\int_{|\xi|\ge2^N}|\widehat f(\xi)|^2\,d\xi, \]

\[ \|\varphi(2^{M+1}\xi)\widehat f\|_2^2 \le\int_{|\xi|\le2^{-M}}|\widehat f(\xi)|^2\,d\xi. \]

The first tends to zero by integrability at infinity. The second tends to zero by absolute continuity of the integral as the ball shrinks to a measure-zero point. Hence the reconstruction converges in \(L^2\).

This is convergence of symmetric or two-sided truncated sums in a norm. It is not a pointwise assertion for arbitrary distributions.

§ 03Almost orthogonality

The frequency bands overlap only with their neighbors. More precisely,

\[ \psi_k(\xi)\psi_j(\xi)=0\qquad\text{if }|k-j|\ge2. \]

At the possible touching boundaries the smooth cutoffs vanish. Thus the dyadic pieces are not all mutually orthogonal, but separated pieces are.

命题 3.1L² almost orthogonality

For \(f\in L^2\),

\[ \frac13\|f\|_2^2\le\sum_{k\in\mathbb Z}\|P_kf\|_2^2\le\|f\|_2^2. \]

Proof

For finite sums, Plancherel implies \(\langle P_kf,P_jf\rangle=0\) when \(|k-j|\ge2\). Use the complex inner product \(\langle a,b\rangle=\int a\overline b\) when needed. Therefore

\[ \left\|\sum_{k=-M}^NP_kf\right\|_2^2 \le\sum_{\substack{-M\le k,j\le N\\|k-j|<2}}\|P_kf\|_2\|P_jf\|_2 \le3\sum_{k=-M}^N\|P_kf\|_2^2. \]

The last bound uses \(2ab\le a^2+b^2\) and the three possible neighbors. Reconstruction gives the lower inequality in the limit.

For the upper inequality, nonnegativity and the partition identity give \(\sum_k\psi_k(\xi)^2\le\sum_k\psi_k(\xi)=1\) for \(\xi\ne0\). Tonelli and Plancherel yield

\[ \sum_k\|P_kf\|_2^2 =(2\pi)^{-d}\int\sum_k\psi_k(\xi)^2|\widehat f(\xi)|^2\,d\xi \le\|f\|_2^2. \]

Finite overlap gives comparison of squared norms, rather than equality of norms or pointwise magnitudes.

§ 04Spatial kernels and one-scale differentiation

Let \(K=\mathcal F^{-1}\psi\). Fourier inversion gives

\[ P_kf=K_k*f,\qquad K_k(x)=2^{kd}K(2^kx). \]

Since \(\psi\) is smooth and compactly supported, \(K\) is Schwartz. Hence

\[ \|K_k\|_1=\|K\|_1,\qquad \|K_k\|_\infty=2^{kd}\|K\|_\infty, \]

and for every \(A>0\),

\[ |K_k(x)|\le C_A2^{kd}(1+2^k|x|)^{-A}. \]

The kernel has spatial scale \(2^{-k}\) and height of order \(2^{kd}\). It is rapidly localized, not compactly supported in physical space. Young's inequality gives a uniform bound \(\|P_kf\|_p\lesssim\|f\|_p\) for \(1\le p\le\infty\). The multiplier has uniformly bounded height; its kernel does not have uniformly bounded pointwise height.

例 4.1Differentiation of a dyadic piece

Replace the heuristic \(|\nabla P_kf|\sim2^k|P_kf|\) by a precise \(L^2\) statement and prove it.

Solution

On the support of \(\widehat{P_kf}\), \(2^{k-1}\le|\xi|\le2^{k+1}\). Thus

\[ \|\nabla P_kf\|_2^2=(2\pi)^{-d}\int|\xi|^2|\psi_k(\xi)\widehat f(\xi)|^2\,d\xi \]

is bounded below by \(2^{2k-2}\|P_kf\|_2^2\) and above by \(2^{2k+2}\|P_kf\|_2^2\). Consequently,

\[ \tfrac12\,2^k\|P_kf\|_2\le\|\nabla P_kf\|_2\le2\,2^k\|P_kf\|_2. \]

For one coordinate derivative, only the upper bound holds without an angular restriction, since \(\xi_j\) may be arbitrarily small on an annulus. There is no general pointwise two-sided comparison between the function and its gradient.

The same Plancherel argument with \(|\xi|^s\) gives \(\||D|^sP_kf\|_2\simeq2^{ks}\|P_kf\|_2\) for fixed real \(s\).

§ 05Bernstein's inequality

命题 5.1Bernstein at one scale

Let \(1\le p\le r\le\infty\). If \(P_kf\in L^p\), then

\[ \|P_kf\|_r\lesssim2^{kd(1/p-1/r)}\|P_kf\|_p. \]

The constant is independent of \(k\) and \(f\).

Proof using an enlarged cutoff

Choose \(\widetilde\psi\in C_c^\infty\) equal to one on \(\operatorname{supp}\psi\), and let \(\widetilde\psi_k(\xi)=\widetilde\psi(2^{-k}\xi)\). Then

\[ P_kf=\widetilde K_k*P_kf, \qquad\widetilde K_k=\mathcal F^{-1}\widetilde\psi_k. \]

Choose \(a\in[1,\infty]\) with \(1+1/r=1/p+1/a\). Such an \(a\) exists precisely because \(p\le r\). The kernel scaling gives

\[ \|\widetilde K_k\|_a =2^{kd(1-1/a)}\|\widetilde K\|_a =2^{kd(1/p-1/r)}\|\widetilde K\|_a. \]

Young's inequality proves the result, including the endpoints.

The enlarged cutoff is needed because \(P_k\) is not an idempotent projection. In particular, one cannot justify the proof by writing \(P_kf=K_k*P_kf\) with the original multiplier.

The critical special case is

\[ \|P_kf\|_\infty\lesssim2^{kd/2}\|P_kf\|_2 \simeq\|P_kf\|_{\dot H^{d/2}}. \]

At one frequency scale, the critical estimate holds. Its failure for a whole function will be a failure of summation across scales.

§ 06Sobolev norms as weighted square sums

For \(f\in L^2\) and fixed real \(s\), weighted Plancherel gives, whenever the expressions are finite,

\[ \|f\|_{H^s}^2\simeq \sum_{k\in\mathbb Z}(1+2^k)^{2s}\|P_kf\|_2^2. \]

Indeed, on the support of \(\psi_k\), the weights \((1+|\xi|^2)^s\) and \((1+2^k)^{2s}\) are comparable uniformly in \(k\). At each nonzero frequency, the nonnegative partition has finite overlap and

\[ \tfrac13\le\sum_k\psi_k(\xi)^2\le1. \]

Multiplying this comparison by the Sobolev weight and integrating proves both directions. For the homogeneous expression,

\[ \|f\|_{\dot H^s}^2\simeq \sum_{k\in\mathbb Z}2^{2ks}\|P_kf\|_2^2. \]

For nonnegative integer \(s\), the Fourier Sobolev norm is equivalent to the sum of squared \(L^2\) norms of all weak derivatives of order at most \(s\). The Fourier definition also handles fractional smoothness. Extensions to distributions use the appropriate low-frequency or quotient conventions.

The weights quantify smoothness: \(s\) derivatives cost \(2^{ks}\) at scale \(2^k\). The inhomogeneous norm retains control of undifferentiated low frequencies, while the homogeneous weight can tend to zero there.

§ 07Besov spaces and the low-frequency block

The three Besov indices measure smoothness \(s\), spatial integrability \(p\), and summability across scales \(q\). For \(1\le p,q\le\infty\), the homogeneous expression is

\[ \|f\|_{\dot B^s_{p,q}}= \left(\sum_{k\in\mathbb Z}\big(2^{ks}\|P_kf\|_p\big)^q\right)^{1/q}, \]

with a supremum over \(k\) when \(q=\infty\). On distributions, this is interpreted modulo polynomials or with an explicit dyadic reconstruction convention, because annular cutoffs annihilate the Fourier distributions supported at zero corresponding to polynomials.

For the inhomogeneous space, define the low-frequency block

\[ P_{\le0}f=\mathcal F^{-1}(\varphi\widehat f). \]

Telescoping gives \(f=P_{\le0}f+\sum_{k\ge1}P_kf\), first in \(L^2\). The standard norm is

\[ \|f\|_{B^s_{p,q}}= \|P_{\le0}f\|_p+ \left(\sum_{k\ge1}\big(2^{ks}\|P_kf\|_p\big)^q\right)^{1/q}, \]

again replacing the sum by a supremum when \(q=\infty\).

One can also consider the all-scale weighted expression

\[ \left(\sum_{k\in\mathbb Z}\big((1+2^k)^s\|P_kf\|_p\big)^q\right)^{1/q}. \]

For \(p=q=2\), almost orthogonality makes it equivalent to the Sobolev norm. For general \(p,q\), it must not be substituted for the standard inhomogeneous norm without additional justification: the low frequencies need to be collected into a single block. In particular,

\[ B^s_{2,2}=H^s \]

with equivalent norms.

§ 08Sobolev embedding above the critical index

定理 8.1Sobolev embedding into L∞

If \(s>d/2\), then \(H^s(\mathbb R^d)\hookrightarrow L^\infty(\mathbb R^d)\) and

\[ \|f\|_\infty\lesssim\|f\|_{H^s}. \]

Proof by summing Bernstein estimates

Bernstein gives

\[ \sum_k\|P_kf\|_\infty\lesssim\sum_k2^{kd/2}\|P_kf\|_2. \]

Insert the Sobolev weight and apply Cauchy–Schwarz in \(k\):

\[ \sum_k2^{kd/2}\|P_kf\|_2 \le \left(\sum_k(1+2^k)^{2s}\|P_kf\|_2^2\right)^{1/2} \left(\sum_k\frac{2^{kd}}{(1+2^k)^{2s}}\right)^{1/2}. \]

The first factor is comparable to \(\|f\|_{H^s}\). For \(k\le0\), the second series is bounded by a constant times \(\sum_{k\le0}2^{kd}\). For \(k\ge1\), it is comparable to \(\sum_{k\ge1}2^{-k(2s-d)}\). Both converge when \(s>d/2\).

Thus the dyadic series converges absolutely in \(L^\infty\). Its limit agrees with the \(L^2\) reconstruction: on each bounded set, \(L^\infty\) convergence also gives \(L^2\) convergence, so the two limits coincide almost everywhere. The displayed norm bound follows.

At \(s=d/2\), the high-frequency terms in the last series stay comparable to one. This explains why this proof fails at the endpoint. Failure of a proof alone is not a counterexample; the next construction establishes failure of the embedding itself.

§ 09A critical Sobolev counterexample

Choose a nonzero real, even, nonnegative \(\chi\in C_c^\infty\) supported in \(\{1<|\xi|<3/2\}\), and set \(h=\mathcal F^{-1}\chi\). Then \(h\) is real and Schwartz, and

\[ h(0)=(2\pi)^{-d}\int\chi(\xi)\,d\xi>0. \]

Let

\[ f_N(x)=\sum_{j=1}^N\frac1j h(2^jx). \]

The Fourier supports of the summands are disjoint annuli. Scaling gives

\[ \|h(2^j\cdot)\|_{\dot H^{d/2}}^2=\|h\|_{\dot H^{d/2}}^2, \qquad \|h(2^j\cdot)\|_2^2=2^{-jd}\|h\|_2^2. \]

All these supports have \(|\xi|>2\), where the inhomogeneous and homogeneous critical weights are comparable. Orthogonality therefore yields

\[ \|f_N\|_{H^{d/2}}^2\lesssim\sum_{j=1}^N\frac1{j^2}\lesssim1. \]

But

\[ f_N(0)=h(0)\sum_{j=1}^N\frac1j\to\infty. \]

Each \(f_N\) is continuous, so its essential supremum is at least \(|f_N(0)|\). There can be no uniform estimate \(\|f\|_\infty\le C\|f\|_{H^{d/2}}\). Hence the usual continuous Sobolev embedding fails at the critical index.

§ 10The Besov endpoint remedy

命题 10.1Critical Besov control

If

\[ \sum_{k\in\mathbb Z}2^{kd/2}\|P_kf\|_2<\infty, \]

the dyadic reconstruction defines a bounded representative and satisfies

\[ \left\|\sum_kP_kf\right\|_\infty \lesssim\|f\|_{\dot B^{d/2}_{2,1}}. \]

For \(f\in L^2\), this representative equals \(f\) almost everywhere. With the stated reconstruction convention, this is the embedding \(\dot B^{d/2}_{2,1}\hookrightarrow L^\infty\).

Proof

Sum the one-scale Bernstein estimate:

\[ \sum_k\|P_kf\|_\infty \lesssim\sum_k2^{kd/2}\|P_kf\|_2. \]

The series converges absolutely in \(L^\infty\). If the original function is in \(L^2\), the same uniqueness-of-limits argument as above identifies the reconstruction with it. For a homogeneous distribution class, the assertion chooses this reconstruction as its representative; it is not an estimate for every polynomial added to that representative.

Similarly, \(B^{d/2}_{2,1}\hookrightarrow L^\infty\): the low block satisfies \(\|P_{\le0}f\|_\infty\lesssim\|P_{\le0}f\|_2\) by an enlarged ball cutoff, and the high blocks are summed as above.

Space at critical smoothness	Summation of band sizes	Consequence
\(H^{d/2}=B^{d/2}_{2,2}\)	Square summation	Does not control the sum of amplitudes
\(B^{d/2}_{2,1}\)	Absolute summation	Controls the \(L^\infty\) norm
\(\dot B^{d/2}_{2,1}\)	Absolute summation over all scales	Controls the bounded dyadic representative

§ 11Exercise: concentration and scaling

例 11.1A narrowing profile

For \(g\in C_c^\infty(\mathbb R^d)\) with \(\int g=1\), set \(g_\varepsilon(x)=\varepsilon^{-d}g(x/\varepsilon)\). Show distributional convergence to a delta and compute the \(L^p\) scaling.

Solution

For every test function \(a\),

\[ \int g_\varepsilon(x)a(x)\,dx =\int g(y)a(\varepsilon y)\,dy\to a(0). \]

A change of variables gives

\[ \|g_\varepsilon\|_p=\varepsilon^{-d(1-1/p)}\|g\|_p, \]

including \(p=\infty\) with the usual convention. The total integral is preserved, but the \(L^p\) norms for \(p>1\) grow. Also \(\widehat{g_\varepsilon}(\xi)=\widehat g(\varepsilon\xi)\), so concentration broadens the relevant frequency range.

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