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Consider this post Fourier transform is uniformly continuous. It establishes that for $L^1$ functions we have that the Fourier transform is uniformly continuous. Now lets suppose that the Fourier transform is also in $L^1$. Then the Fourier Inverse formula applies and we have that, by the same argument verbatim, the original function is uniformly continuous.

However, this argument can not be true. Consider the indicator function $1_{[0,1]}$. Its Fourier transform is $$ \frac{\sin u/2}{u/2}e^{iu/2}. $$ This is uniformly continuous (as expected), and moreover, is $L^1$. So the inversion formula must apply. However, $1_{[0,1]}$ is not uniformly continuous. A contradiction.

Question: Could someone clarify?

Maths Wizzard
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  • $\frac{\sin(x)}{x}$ is not in $L^1,$ because $\int \vert\frac{\sin(x)}{x}\vert dx=\infty.$ – Steen82 May 24 '24 at 11:44
  • Oh. I see. I remembered that the integral itself is $\pi$ and seemed to deduce from that that that it is $L^1$. Thank you for this. Just a point of clarification, how come the integral value seems to exist but it is not $L^1$ integrable meaning the lebesgue integral shouldn't be defined? – Maths Wizzard May 24 '24 at 12:22
  • The integral exists in the extended Riemann sense since there is cancellation of positive and negative areas. The Lebesgue integral requires the areas all to be counted positive and finite for the function to be Lebesgue integrale. – Steen82 May 25 '24 at 12:02

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