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I am reading the proof of the Plancherel Theorem in Folland. But I am quite confused about one of his claims.

Suppose $f,g \in L^2(T)$ and $\hat{f}\in L^1$ ($\hat{f}$ is the Fourier transform). then $f$ is in $L^{\infty}$.

How can we show it? Thank you!

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The reason is as follows: by the Fourier inversion theorem, if $f,\widehat{f}\in L^1$, then we know that $f$ agrees almost everywhere with a continuous function $f_0$. Moreover, $f_0 = (f^{\vee})^{\wedge}$ so, since $f^{\vee}$ is in $L^1$ (being that it is related to $\widehat{f}$ via reflection), $f_0$ is bounded by the Riemann-Lebesgue Lemma. Hence $f$ is bounded almost everywhere, i.e. $f\in L^{\infty}$.

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