Spring 2013 - Problem 6

Solution.

1. Write $\hat{f}\p{\xi} = \int_\R e^{i\xi x} f\p{x} \,\diff{x}$. Then if $xi_n \to \xi$ in $\R$, we have

   $$\abs{\hat{f}\p{\xi_n} - \hat{f}\p{\xi}} \leq \int_\R \abs{f\p{x}}\abs{e^{i\xi_nx} - e^{i\xi x}} \,\diff{x}.$$

   The integrand is bounded by $2\abs{f\p{x}} \in L^1\p{\R}$ by assumption, so we may apply dominated convergence and continuity of $e^{i\xi x}$ to see that $\hat{f} \in C\p{\R}$.

   Next, observe

   $$\begin{aligned} \hat{f}\p{\xi} &= -e^{\pi i} \int_\R e^{i\xi x} f\p{x} \,\diff{x} \\ &= -\int_\R e^{i\xi\p{x + \frac{\pi}{\xi}}} f\p{x} \,\diff{x} \\ &= -\int_\R e^{i\xi x} f\p{x - \frac{\pi}{\xi}} \,\diff{x} && \p{x \mapsto x + \frac{\pi}{\xi}} \\ \implies 2\hat{f}\p{\xi} &= \int_\R e^{i\xi x}\p{f\p{x} - f\p{x - \frac{\pi}{\xi}}} \,\diff{x} \\ \implies \abs{\hat{f}\p{\xi}} &\leq \frac{1}{2} \norm{f - f\p{\:\cdot\: - \frac{\pi}{\xi}}}_{L^1}. \end{aligned}$$

   Notice that $f\p{x} - f\p{x - \frac{\pi}{\xi}} \to 0$ as $\abs{\xi} \to \infty$, and $\abs{f\p{x} - f\p{x - \frac{\pi}{\xi}}} \leq \abs{f\p{x}} + \abs{f\p{x - \frac{\pi}{\xi}}} \in L^1\p{\R}$, so by dominated convergence, $\abs{\hat{f}\p{\xi}} \to 0$ as $\abs{\xi} \to \infty$. Thus, $\hat{f} \in C_0\p{\R}$.

2. $C_c^\infty\p{\R}$ is dense in $C_0\p{\R}$ via convolutions against a smooth bump function. Thus, since $C_c^\infty\p{\R}$ are Schwartz functions, the Schwartz space is dense in $C_0\p{\R}$. Since the Fourier transform is a bijection on the Schwartz space, it follows that $X$ contains the Schwartz space, hence dense in $C_0\p{\R}$.

3. Let $\func{\mathcal{F}}{L^1\p{\R}}{C_0\p{\R}}$, $f \mapsto \hat{f}$, where $C_0\p{\R}$ is equipped with the uniform norm. Then $\mathcal{F}$ is a linear function between two Banach spaces. $\mathcal{F}$ is an injective mapping, since it's essentially the Fourier transform, and it is continuous:

   $$\abs{\hat{f}\p{\xi} - \hat{g}\p{\xi}} \leq \int_\R \abs{e^{i\xi x}}\abs{f\p{x} - g\p{x}} \,\diff{x} = \norm{f - g}_{L^1}.$$

   Thus, if $\mathcal{F}$ were surjective, then by the open mapping theorem, $\mathcal{F}$ would be an open map, and it would have continuous inverse. To see why this is impossible, it suffices to find a sequence of functions $\set{g_n}_n \subseteq C_0\p{\R}$ such that $g_n = \mathcal{F}\p{f_n}$ for all $n \geq 1$, $\set{g_n}_n$ is bounded in $C_0\p{\R}$, but $\set{f_n}_n$ is unbounded in $L^1\p{\R}$.

   Let $h_n = \chi_{\br{-n, n}}$, and let $g_n = h_n * h_1$. Observe that

   $$\int_\R e^{-i\xi x}h_n\p{\xi} \,\diff{\xi} = \int_{-n}^n \cos{\xi x} - i\sin{\xi x} \,\diff{x} = \frac{2\sin{nx}}{x},$$

   and it follows via the inverse Fourier transform that

   $$\mathcal{F}\p{\frac{4\sin{nx}\sin{x}}{x^2}} = \mathcal{F}\p{\frac{2\sin{nx}}{x}} * \mathcal{F}\p{\frac{2\sin{x}}{x}} = h_n * h_1 = g_n.$$

   Thus, $g_n \in C_0\p{\R}$ and uniformly bounded via

   $$\abs{g_n\p{\xi}} \leq \norm{h_n}_{L^\infty} \norm{h_1}_{L^1} \leq 2,$$

   but

   $$\begin{aligned} \norm{f_n}_{L^1} &= \int_\R \abs{\frac{4\sin{nx}\sin{x}}{x^2}} \,\diff{x} \\ &\geq 4 \int_\R \abs{\frac{\sin{nx}\sin{x}}{x^2}} \,\diff{x} \\ &= 4n \int_\R \abs{\frac{\sin{x}\sin{\frac{x}{n}}}{x^2}} \,\diff{x} && \p{x \mapsto nx} \\ &\geq \frac{8n}{n\pi} \int_0^{n\pi/2} \abs{\frac{\sin{x}}{x}} \,\diff{x} && (\text{convexity of } \sin{x}) \\ &= \frac{8}{\pi} \int_0^{n\pi/2} \abs{\frac{\sin{x}}{x}} \,\diff{x} \xrightarrow{n\to\infty} \infty. \end{aligned}$$

   Hence, $\mathcal{F}^{-1}$ is unbounded on $C_0\p{\R}$, so $\mathcal{F}$ cannot be surjective, and this completes the proof.