A proof of the fact that the Fourier transform is not surjective from $\mathcal{L}^1(\mathbb{R})$ to $C_0( \mathbb{R})$

Question

Let $f_n = \mathbb 1_{[-n,n]}$ for all $n \in \mathbb{N}$

1. Compute explicitly $f_n \star f_1$ for all $n \in \mathbb{N}$.

2. Show that $f_n \star f_1$ is the Fourier transform of $g_n = \frac{ \sin{(2\pi x)} \sin{(2 \pi nx)}}{\pi ^2 x^2}$

3. Show that $\| g_n \|_1 \to \infty$

4. Deduce that the Fourier transform is not surjective from $\mathcal{L}^1(\mathbb{R})$ to $C_0( \mathbb{R})$

Here, $C_0(\mathbb{R})$ is the space of continuous functions $f : \mathbb{R} \to \mathbb{R}$ that tends to $0$ at $-\infty$ and $+\infty$.

I managed to compute 1) and proved 2). But I don't know how to prove 3). Moreover, I don't see how can 1), 2) and 3) prove 4). Can anyone help?

Answer 1

For part (3), $$ ||g_n||_1=\int_{\mathbb{R}}\frac{|\sin(2\pi x)\sin(2\pi nx)|}{\pi^2x^2}\;dx=\frac{2n}{\pi}\int_{\mathbb{R}}\frac{|\sin(x)\sin(\frac{x}{n})|}{x^2}\;dx$$ $$ =\frac{4n}{\pi}\int_0^{\infty}\frac{|\sin(x)\sin(\frac{x}{n})|}{x^2}\;dx\geq \frac{4n}{\pi}\int_0^{n}\frac{|\sin(x)\sin(\frac{x}{n})|}{x^2}\;dx$$

Now $\sin t\geq t-\frac{t^3}{6}\geq \frac{5}{6}t$ on $[0,1]$, so it follows that $\sin(\frac{x}{n})\geq \frac{5}{6}\cdot\frac{x}{n}$ on $[0,n]$. Therefore $$ ||g_n||_1\gg \int_0^n\frac{|\sin x|}{x}\;dx\to\infty $$ as $n\to\infty$.

For part (4), suppose that the Fourier transform $\mathcal{F}$ is a surjective map $L^1(\mathbb{R})\to\mathcal{C}_0(\mathbb{R})$. $\mathcal{F}$ is also injective by the Fourier inversion theorem, so $\mathcal{F}$ is a bounded bijective linear map. It then follows from the open mapping theorem that $\mathcal{F}$ has a bounded inverse, but this contradicts parts (1)-(3).