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Is there any $L_1$ function such that its fourier transform is in $L_1$ and the fourier transform is nowhere differentiable? Or every such Fourier transform must be differentiable a.e.?

Chaos
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  • If the function is continuous and compact supported, then is going to be bounded, so if it is also $L^1$ it will be also $L^2$, then its Fourier Transform is going to be an Analytic function (so a smooth transform), due to Paley–Wiener theorem. So your counter-example, or it will be a discontinuous function, or a continuous function in the whole line as domain. – Joako Feb 13 '22 at 23:47

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Let $f$ be a bounded, nowhere differentiable function on $\mathbb{R}$. Let $g$ be a standard normal pdf. Then $gf$ is nowhere differentiable and in $L^2$. Also $\hat{(gf)} = \hat{g} * \hat{f} $. The fourier transform of a Gaussian is Gaussian, and convolving with a smooth function makes a smooth function, so $\hat{(gf)}$ is smooth and in $L^2$. So this exhibits a smooth function whose Fourier transform is nowhere differentiable. I think this is correct (?). Does it answer your question?

Elle Najt
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  • I guess you should still explain whether/when $\hat g * \hat f$ is $L^1$ – user8268 Feb 11 '17 at 10:36
  • I think $\hat{gf}$ may not be in $L^2$. – Chaos Feb 11 '17 at 16:24
  • @Chaos My reasoning was that $g \in L^2$ (I think), so because $f$ is bounded, $gf \in L^2$. So the Fourier transform operator sends it to another element in $L^2$. (This latter Fourier transform is defined by extending the usual formula from the dense set of smooth compactly supported functions.) Oh but I see that its not clear the Fourier transform is in $L^1$... and perhaps that is not true? – Elle Najt Feb 11 '17 at 17:29
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    @user8268 I think the best I can do is argue that it is in $L^2$. – Elle Najt Feb 11 '17 at 17:30
  • @AreaMan Oh I see it's in $L^2$. But we can't say it's in $L^1$. There are some counterexamples . – Chaos Feb 11 '17 at 19:43