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Fall 2015 - Problem 6

Fourier analysis

Let $u \in L^2\p{\R^d}$ and let us say that $u \in H^{1/2}\p{\R^d}$ (a Sobolev space) if

\p{1 + \abs{\xi}^{1/2}} \hat{u}\p{\xi} \in L^2\p{\R^d}.

Here $\hat{u}$ is the Fourier transform of $u$ . Show that $u \in H^{1/2}\p{\R^d}$ if and only if

\iint \frac{\abs{u\p{x + y} - u\p{x}}^2}{\abs{y}^{d+1}} \,\diff{x} \,\diff{y} < \infty.

Solution.

First, recall that

\hat{u}\p{\xi} = \frac{1}{\p{2\pi}^{d/2}} \int_{\R^d} e^{-i\xi \cdot x} u\p{x} \,\diff{x},

and by the change of variables $x \mapsto x + y$ , we have $\hat{u}\p{\xi} = e^{-i\xi \cdot y} \hat{u}_y\p{\xi}$ , where $u_y\p{x} = u\p{x + y}$ .

Suppose $u \in H^{1/2}\p{\R^d}$ . By Plancherel,

\begin{aligned} \iint \frac{\abs{u\p{x + y} - u\p{x}}^2}{\abs{y}^{d+1}} \,\diff{x} \,\diff{y} &= \int_{\R^d} \frac{1}{\abs{y}^{d+1}} \int_{\R^d} \abs{e^{i\xi \cdot y} \hat{u}\p{\xi} - \hat{u}\p{\xi}}^2 \,\diff\xi \,\diff{y} \\ &= \int_{\R^d} \abs{\hat{u}\p{\xi}}^2 \int_{\R^d} \frac{\abs{e^{i\xi \cdot y} - 1}^2}{\abs{y}^{d+1}} \,\diff{y} \,\diff\xi, \end{aligned}

by Fubini-Tonelli (everything is non-negative). It remains to establish bounds for the inner integral.

" $\implies$ "

Note that because the Plancherel transform is an isometry, it follows that $\hat{u}\p{\xi} \in L^2\p{\R^d}$ , so

\abs{\xi}^{1/2} \hat{u}\p{\xi} = \br{\p{1 + \abs{\xi}^{1/2}} \hat{u}\p{\xi}} - \hat{u}\p{\xi} \in L^2\p{\R^d}.

By the fundamental theorem of calculus,

\abs{e^x - 1} = \abs{\int_0^x -e^t \,\diff{t}} \leq \abs{x}e^x.

Thus, on $\abs{x} \leq 1$ , we have $\abs{e^x - 1} \leq C\abs{x}$ . If $\abs{\xi}\abs{y} \leq 1$ , then by Cauchy-Schwarz, $\abs{\xi \cdot y} \leq 1$ , so

\begin{aligned} \int_{\R^d} \frac{\abs{e^{i\xi \cdot y} - 1}^2}{\abs{y}^{d+1}} \,\diff{y} &= C^2 \int_{\set{\abs{\xi}\abs{y} \leq 1}} \frac{\abs{\xi \cdot y}^2}{\abs{y}^{d+1}} \,\diff{y} + 4 \int_{\set{\abs{\xi}\abs{y} \geq 1}} \frac{1}{\abs{y}^{d+1}} \,\diff{y} \\ &= C^2\abs{\xi}^2 \int_0^{1/\abs{\xi}} \int_{S^{d-1}} \,\diff\sigma \,\diff{r} + 4 \int_{1/\abs{\xi}}^\infty \int_{S^{d-1}} \frac{1}{r^2} \,\diff\sigma \,\diff{r} \\ &\leq C^2 \sigma\p{S^{d-1}} \abs{\xi} + 4 \sigma\p{S^{d-1}} \abs{\xi} \\ &= A\abs{\xi}, \end{aligned}

and so

\iint \frac{\abs{u\p{x + y} - u\p{x}}^2}{\abs{y}^{d+1}} \,\diff{x} \,\diff{y} \leq A \int_{\R^d} \abs{\xi}\abs{\hat{u}\p{\xi}}^2 \,\diff\xi < \infty.

" $\impliedby$ "

By another application of the fundamental theorem of calculus, we quickly see that

\abs{e^x - 1} \geq \abs{x}

for all $x \in \R$ . Let $A \in \SO\p{d}$ be the rotation which maps $\frac{\xi}{\abs{\xi}} \mapsto \p{0, \ldots, 0, 1} = e_n$ . Then for a fixed $\xi$ ,

\xi \cdot Ay = \abs{\xi}\p{A^\trans\frac{\xi}{\abs{\xi}} \cdot y} = \abs{\xi}\p{e_n \cdot y}.

By continuity of the inner product, there exists an open neighborhood $U \subseteq S^{d-1}$ such that $\abs{e_n \cdot y} \geq \frac{1}{2}$ . Since $A$ has determinant $1$ , a change of variables yields

\begin{aligned} \int_{\R^d} \frac{\abs{e^{i\xi \cdot y} - 1}^2}{\abs{y}^{d+1}} \,\diff{y} &\geq \int_{\set{\abs{\xi}\abs{y} \leq 1}} \frac{\abs{\xi \cdot y}^2}{\abs{y}^{d+1}} \,\diff{y} \\ &= \abs{\det{A}} \int_{\set{\abs{\xi}\abs{y} \leq 1}} \frac{\abs{\xi \cdot Ay}^2}{\abs{y}^{d+1}} \,\diff{y} \\ &= \abs{\xi}^2 \int_0^{1/\abs{\xi}} \int_{S^{d-1}} \abs{e_n \cdot s}^2 \,\diff\sigma\p{s} \,\diff{r} \\ &\geq \abs{\xi}^2 \int_0^{1/\abs{\xi}} \int_U \frac{1}{2} \,\diff\sigma\p{s} \,\diff{r} \\ &\geq C \abs{\xi}. \end{aligned}

Indeed, $U$ does not depend on $\xi$ , so $C$ is independent of $\xi$ . Finally, we see

C \int_{\R^d} \abs{\xi}\abs{\hat{u}\p{\xi}}^2 \,\diff\xi \leq \iint \frac{\abs{u\p{x + y} - u\p{x}}^2}{\abs{y}^{d+1}} \,\diff{x} \,\diff{y} < \infty,

so $\abs{\xi}^{1/2}\hat{u}\p{\xi} \in L^2$ . Since $u \in L^2$ , we automatically have $\hat{u} \in L^2$ , and so $\p{1 + \abs{\xi}^{1/2}} \hat{u}\p{\xi} \in L^2$ , which completes the proof.