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Can $f\notin L^1$ if its Fourier transform $\hat f \in L^\infty$ ?

This is a question about the endpoint of Pontryagin duality. We know that if a function is in $L^1$, then its Fourier transform lies in $L^\infty$. This is very easy to show. But what about the converse: Can the $L^1$ norm of $f$ be infinite even if the $L^\infty$ norm of its Fourier transform is finite?

I assume that the answer to my question is YES, but I do not see how to handle this case. Any references?

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    Define $f$ as the inverse transform of an $L^\infty$ function $\hat f$. For example, let $\hat f$ be 1 in the interval [-1,1] and zero outside. What happens? I need some help to understand such cases.2011-09-29

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The Fourier transform of the characteristic function of $[-1,1]$ is $\sin(\xi)/\xi$ (times a constant which depends on conventions), which is certainly not in $L^1(\mathbb{R})$.

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    Excellent. I will consider the question answered. I am very grateful to these comments because as a physicist, my math training was not rigorous enough.2011-09-29
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When we do not take distribution theory (i.e., the dirac delta functional) into consideration, we have a constant function as a counter-example, which is in $L^\infty$, but not a Fourier transform of a certain $L^1$ function.