# Can f not be in L1 if its Fourier transform is in L-infinity?

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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How is the Fourier transform $\hat f$ of $f$ even defined, when $f$ is not $\in L^1$? – Sam Sep 29 '11 at 20:42
assuming that the Fourier transform of $f$ is well-defined to begin with? – Mark Sep 29 '11 at 20:42
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. – Gandhi Viswanathan Sep 29 '11 at 20:54

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 $L^1(\mathbb{R})$.
But there is a delicate point here. $\sin(\xi)/\xi$ is not Lebesgue integrable, so its Fourier transform is only defined as a limit of $L^1$ functions. Of course, the OP must have considered this since the question specifically asks about the Fourier Transform a function not in $L^1$. – robjohn Sep 29 '11 at 22:18
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.