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

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?

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).