The Fourier transform in the sense of tempered distribution is not a function

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Suppose $f\in L^p,p>2$, we can regard it as a tempered distribution. Then we can define the Fourier transform of the tempered distribution as $\hat{f}(\phi)=f(\hat{\phi})$, where $\phi$ is in the Schwartz. I want to prove that there exists $f$ such that $\hat{f}$ can't be represented in the form of locally integrable function.


$p>2$ lies in the fact that if $f\in L^1+L^2$, then $\hat{f}$ is just the standard Fourier transform.