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Spring 2016 - Problem 5

Fourier analysis

For $f \in C_0^\infty\p{\R^2}$ define $u\p{x, t}$ by

u\p{x, t} = \int_{\R^2} e^{ix \cdot \xi} \frac{\sin\p{t\abs{\xi}}}{\abs{\xi}} f\p{\xi} \,\diff{\xi}, \quad x \in \R^2,\ t > 0.

Show that $\lim_{t\to\infty} \norm{u\p{\:\cdot\:, t}}_{L^2} = \infty$ for a set of $f$ that is dense in $L^2\p{\R^2}$ .

Solution.

Observe that $u$ is defined as the inverse Fourier transform of

\xi \mapsto \frac{\sin\p{t\abs{\xi}}}{\abs{\xi}} f\p{\xi}.

Hence, by Plancherel,

\begin{aligned} \norm{u\p{\:\cdot\:, t}}_{L^2}^2 &= \norm{\xi \mapsto \frac{\sin\p{t\abs{\xi}}}{\abs{\xi}} f\p{\xi}}_{L^2}^2 \\ &= \int_{\R^2} \p{\frac{\sin\p{t\abs{\xi}}}{\abs{\xi}}}^2 \abs{f\p{x}}^2 \,\diff\xi \\ &\geq \int_{\set{0 \leq t\abs{\xi} \leq \frac{\pi}{2}}} \p{\frac{2t\abs{\xi}}{\pi\abs{\xi}}}^2 \abs{f\p{\xi}}^2 \,\diff\xi \\ &= \frac{4t^2}{\pi^2} \int_{\set{\abs{\xi} \leq \frac{\pi}{2t}}} \abs{f\p{\xi}}^2 \,\diff\xi. \end{aligned}

Thus, if $\lim_{\xi\to0} \abs{f\p{\xi}} = \infty$ , then this tends to $\infty$ . It remains to show that $\set{f \in L^2\p{\R^2} \mid \lim_{\xi\to0} \abs{\p{\xi}} = \infty}$ is dense in $L^2\p{\R^2}$ .

Let $\phi\p{\xi} = \abs{\xi}^{-1/2}\chi_{B\p{0,1}}$ so that

\norm{\phi}_{L^2}^2 = \int_{B\p{0,1}} \frac{1}{\abs{\xi}} \,\diff\xi = \int_0^1 \int_0^{2\pi} \diff\theta \,\diff{r} = 2\pi.

Thus, given $f \in C_c\p{\R^2}$ and $\epsilon > 0$ ,

\lim_{\xi\to0}\,\p{f\p{\xi} + \epsilon\phi\p{\xi}} = \infty \quad\text{and}\quad \norm{f - \p{f - \epsilon\phi}}_{L^2} = \epsilon\norm{\phi}_{L^2},

which completes the proof.