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It is known that for a $L^1$ function $f: \mathbb{R} \rightarrow \mathbb{C}$ the Fourier transform vanishes at infinity and is continuous. Does this even mean that $(\hat{f}(n))_{n \in \mathbb{Z}}$ is square-summable?

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Let $$ f_a(x)=\frac{1}{\sqrt{|x|}} \chi_{[-a,a]} $$ Observe that $f_a \in L^1$ but not in $L^2$.

Now just compute the fourier transform and play with $a$. \begin{align} \hat{f_a}(y) &= \int_{-a}^a \frac{1}{\sqrt{|x|}}e^{-iyx} \, dx \\ &=\int_0^a\frac{1}{\sqrt{x}}e^{-iyx}+\frac{1}{\sqrt{x}}e^{iyx} \,dx \\ & =2 \int_0^{\sqrt{a}}e^{-iys^2} + e^{iys^2} \, ds. \end{align}

Ok now compute the first term with this trick: \begin{align} \left( \int_0^{\sqrt{a}}e^{-iys^2} \, ds \right)^2 &=\left( \int_0^{\sqrt{a}}e^{-iys^2} \, ds \right)\left( \int_0^{\sqrt{a}}e^{-iyt^2} \, dt \right) \\ &=\int_0^{\sqrt{a}}\int_0^{\sqrt{a}}e^{-iy(s^2+t^2)} \, dt \,ds \\ &=\int_0^{2\pi} \int_0^\sqrt{a}re^{-iyr^2} \, dr \, d\theta \\ & =2 \pi \frac{1}{-2iy}\int_0^{\sqrt{a}}-2iyre^{-iyr^2} \, dr \\ &=\frac{i\pi}{y}\left(e^{-iy\sqrt{a}} -1\right). \end{align}

In the same way one can compute the second term and we obtain: $$ \hat{f_a}(y)=2\sqrt{\frac{\pi i}{y}}\left( \sqrt{e^{-iya}-1} + \sqrt{1-e^{iya}} \right). $$

If you choose $a=2 \pi$ you obtain that $\hat{f_{2 \pi}}(n)=0$, i.e. $(\hat{f_{2 \pi}}(n))_n$ is square summable.

But if you choose $a=\frac{\pi}{2}$ than you obtain that $\hat{f_{\frac{\pi}{2}}}(n) \sim \frac{1}{\sqrt{n}}$ wich is not square summable.

This means that you can not says anything about $\hat{f}(n)$ if you now only that $f \in L^1$. I think that the problem is namely that the Fourier coefficients describe in an unique way a function if it is in $L^2$. But this is not true in $L^1$.

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At least it does not mean that $\hat{f} \in L^2(\mathbb{R})$, because then by the Plancherel theorem $f$ itself would have to be $L^2$. But this can fail; for instance $f(x)=x^{-2/3} \chi_{(0,1]}(x)$ is $L^1$ but not $L^2$. I would imagine that $\hat{f}(n)$ is not summable on $\mathbb{Z}$ either (at least with an example like mine which is quite far from being $L^2$, rather than $x^{-1/2} \chi_{(0,1]}(x)$ which is "almost" $L^2$).

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