# 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).

Not related to the body of the post, but rather the title.

One can provide an explicit example to show the Fourier transform is not surjective: https://mathoverflow.net/a/319225/112504 proves that $$g(t) := 1_{[0\leq t\leq e]}\cdot \frac te + 1_{[t>e]}\cdot \frac{1}{\ln t},$$ extended to be an odd continuous function on $$\mathbb R$$ can not be the Fourier transform of any $$L^1(\mathbb R)$$ function. That answer cites: "this example is taken from the book Classical Fourier Transforms, Springer–Verlag, 1989 by K. Chandrasekharan." For the convenience of the reader, I will copy and paste the proof from the answer's source (M. Thamban Nair's Fourier Analysis notes):

Proposition 3.2.1 If $$f \in L^1(\mathbb{R})$$ is an odd function, then $$\hat{f}$$ is an odd function and there exists $$M>0$$ such that $$\left|\int_r^R \frac{\hat{f}(\xi)}{\xi} d \xi\right| \leq M$$ for all $$r, R$$ with $$0. Proof. Let $$f \in L^1(\mathbb{R})$$ is an odd function. It can be easily seen that $${ }^4$$, $$\hat{f}(\xi)=2 i \int_0^{\infty} f(x) \sin (\xi x) d x$$

Thus, $$\hat{f}$$ is odd. Let $$R \geq r>0$$. Then, we have \begin{aligned} \int_r^R \frac{\hat{f}(\xi)}{\xi} d \xi & =2 i \int_r^R \frac{1}{\xi}\left(\int_0^{\infty} f(x) \sin (\xi x) d x\right) d \xi \\ & =2 i \int_0^{\infty} f(x)\left(\int_r^R \frac{\sin (\xi x)}{\xi} d \xi\right) d x \\ & =2 i \int_0^{\infty} f(x)\left(\int_{r x}^{R x} \frac{\sin (s)}{s} d s\right) d x \end{aligned}

We know that there exists $$M_0>0$$ such that $$\left|\int_a^b \frac{\sin x}{x} d x\right| \leq M_0$$ for all $$(a, b) \subseteq \mathbb{R}$$. Thus, $$\left|\int_r^R \frac{\hat{f}(\xi)}{\xi} d \xi\right| \leq 2 \int_0^{\infty}|f(x)|\left|\int_{r x}^{R x} \frac{\sin (s)}{s} d s\right| d x \leq 2 M_0\|f\|_1 .$$ $$\blacksquare$$

Theorem 3.2.2 The bounded linear operator $$f \mapsto \hat{f}$$ is a bounded linear operator from $$L^1(\mathbb{R})$$ to $$C_0\left(\mathbb{R}\right)$$ is not onto. Proof. By Proposition 3.2.1, it is enough to construct an odd function $$g \in C_0(\mathbb{R})$$ such that $$\left|\int_r^R \frac{g(t)}{t} d t\right| \rightarrow \infty \quad \text { as } \quad R \rightarrow \infty .$$

A candidate for such a function is the odd extension of $$g$$ defined by $$g(t):= \begin{cases}t / e, & 0 \leq t \leq e \\ 1 / \ln (t), & t>e\end{cases}$$

Note that $$\int_e^R \frac{g(t)}{t} d t=\ln (\ln (R)) \rightarrow \infty \quad \text { as } \quad R \rightarrow \infty$$ $$\blacksquare$$

$$\color{red}{\text{That example was based on a C_0(\mathbb R) function decaying slowly to 0.}}$$

Here seems to be one example of a compactly supported continuous function who is not the Fourier transform of any $$L^1(\mathbb R)$$ function: https://mathoverflow.net/a/95678/112504 (it suffices to find an example of a $$g\in C_c(\mathbb R)$$ function whose Fourier transform $$\hat g$$ isn't in $$L^1(\mathbb R)$$, since if we assume for sake of contradiction $$g = \hat f$$ for some $$f\in L^1(\mathbb R)$$, then Fourier inversion would tell us that $$\hat g(s)=f(-s) \in L^1(\mathbb R)$$ (perhaps up to constant factors); contradiction)

Choose $$g(x)= \begin{cases} \dfrac{\frac12 -x}{\log(x)},&0

Here's a graph of this function for reference

Here's the proof: assume for sake of contradiction that $$\hat g$$ is in $$L^1$$, so all the integrals in the following line are well-defined: \begin{aligned} \infty > \|\hat g\|_{L^1(\mathbb R)} &= \int_{-\infty}^\infty |\hat g(s)| d s \geq \int_0^\infty |\hat g(s)| d s \geq \int_0^\infty \lvert\operatorname{Im} \hat g(s)\rvert ds \\ &\geq \left|\int_0^\infty \operatorname{Im} \hat g(s)ds\right| \geq \int_0^\infty \operatorname{Im} \hat g(s)ds \end{aligned} In fact, because for $$s\in [0,\infty)$$, $$e^{-\varepsilon s} \operatorname{Im}\hat g(s)$$ is dominated $$|e^{-\varepsilon s} \operatorname{Im}\hat g(s)|\leq |\hat g(s)|$$ we we assumed was in $$L^1(\mathbb R_s)$$, the DCT (dominated convergence theorem) tells us that $$\lim_{\varepsilon \searrow 0} \int_0^\infty e^{-\varepsilon s} \operatorname{Im} \hat g(s)ds = \int_0^\infty \operatorname{Im} \hat g(s)ds.$$ Note that $$\operatorname{Im} \hat g(s) = \int_{-\infty}^\infty g(x) \operatorname{Im}(e^{-ix s}) dx = \int_{0}^{1/2} g(x) \sin(-xs) dx.$$

But \begin{aligned} \int_0^\infty e^{-\varepsilon s} \operatorname{Im} \hat g(s)ds &= \int_0^\infty e^{-\varepsilon s} \int_0^{1/2} (-g(x)) \sin(xs) dx ds \\ &= \int_0^{1/2} (-g(x)) \int_0^\infty e^{-\varepsilon s} \sin (xs) ds dx. \end{aligned} where we used Fubini-Tonelli, which we can use once we check the Tonelli condition: $$\int_0^{1/2} \int_0^\infty |e^{-\varepsilon s} (-g(x)) \sin(xs)| ds dx \leq \int_0^{1/2} \underbrace{|g(x)|}_{\leq 0.2} \underbrace{\int_0^\infty e^{-\varepsilon s} ds}_{=\frac 1{\varepsilon} e^{-\varepsilon}} dx < \infty$$

Finally, it is an easy exercise (using Euler's identity and taking Re/Im parts, or twice integration by parts) to see that $$\int_0^\infty e^{-\varepsilon s} \sin (xs) ds = \lim_{R\nearrow \infty} \left.\left(\frac{e^{-\varepsilon s}}{\varepsilon^2+x^2} (-\varepsilon \sin(xs) - x \cos (xs) ) \right)\right|_{s=0}^{s=R} = \frac{x}{\varepsilon^2+x^2}$$

and so (using that $$-g(x)=|g(x)|\geq -\frac 14\cdot \frac{1}{\log x} \cdot 1_{[0,\frac 14]}$$)

$$\int_0^\infty e^{-\varepsilon s} \operatorname{Im} \hat g(s)ds = \int_0^{1/2} (-g(x)) \cdot \frac{x}{\varepsilon^2+x^2} dx \geq \int_0^{1/4} \frac 14\cdot \frac{1}{-\log x} \cdot \frac{x}{\varepsilon^2+x^2} dx.$$

As $$\varepsilon\searrow 0$$, the RHS goes to $$\int_0^{1/4} \frac 14 \frac{1}{-x\log x} dx$$ by MCT (monotone convergence theorem), which equals $$\infty$$; tracing back all our steps, we get that $$\infty > \|\hat g \|_{L^1(\mathbb R)}\geq \infty$$; contradiction.

$$\color{red}{\text{The key feature that made this chosen function work is its steepness at 0.}}$$ It is only by having such a steep transition from constant $$0$$ on $$(-\infty,0]$$ to something nonzero, that $$(-g(x)) \cdot \frac 1x$$ preserves the non-$$L^1$$-ness of $$\frac 1x$$ at $$0$$.

One corollary of the above example is that one can have a sequence of very nice functions in $$C_c(\mathbb R)$$ (all of which come from Fourier transforms of very nice functions) converging very nicely to a function in $$C_c(\mathbb R)$$ which is not the Fourier transform of any $$L^1(\mathbb R)$$ function.

More precisely, one can take a sequence $$g_n\in C_c^\infty ([0,\frac 12])$$ converging in the sup-norm (and hence in all the $$L^p$$-norms) to $$g \in C_c([0,\frac 12])$$; all the $$g_n$$ (by Fourier inversion on $$C_c^\infty \subseteq \mathcal S$$) come from Schwartz functions $$f_n\in \mathcal S(\mathbb R) \subseteq L^1(\mathbb R)$$, but there is no $$f\in L^1(\mathbb R)$$ s.t. $$\hat f = g$$.

So the $$f_n$$ are infinitely smooth everywhere, faster than any polynomial decay (in $$C_0^\infty(\mathbb R)$$! in every $$L^p(\mathbb R)$$ space!), and the $$g_n$$ are infinitely smooth and compactly supported! And the $$g_n$$ converge in "every possible way" to $$g$$! And yet, we can't even find an integrable $$f$$ whose Fourier transform is $$g$$.

P.S. more examples can be found elsewhere on MSE. I've tried to collect those that I could find here: A Fourier transform of a continuous $L^1$ function (gives an example quite similar to my 2nd example above, the one with compact support)

Some questions I have remaining: