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In Walter Rudin's Real and Complex Analysis, he gives a proof of the Fourier transform inversion theorem as follows:

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Theorem 9.6 states that if $ f \space \epsilon \space L^1$, then $\hat{f}\space \epsilon \space C_0$ and $\Vert \hat{f} \Vert_\infty = \Vert f \Vert_1$. Here $\hat{f}$ is the Fourier transform of $f$ given by $$ \hat{f}(t) = \frac{1}{\sqrt{2\pi}}\int^{\infty}_{-\infty} f(x)e^{-ixt}dx$$

I don't understand how the fact that $g \space \epsilon \space C_0$ follows from Theorem 9.6. My understanding is that $g$ is the inverse Fourier transform of $f$ in this case. Does Theorem 9.6 imply that the inverse Fourier transform of $\hat{f}$ is in $C_0$ if $\hat{f} \space \epsilon \space L^1$. If so, how? Any help is appreciated.

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  • $\begingroup$ Continuity follows from the dominated convergence theorem and the $C_0$ part from the Riemann Lebesgue lemma. $\endgroup$
    – copper.hat
    Commented Sep 9, 2023 at 20:02

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Just observe that $$ g(x)=\sqrt{2\pi}\hat{\hat {f}}(-x). $$ So it differs from the transform of an $L^1$- function by a scaling factor and a reflection. Both operators preserve belonging to $C_0$.

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  • $\begingroup$ Thanks for the explanation. That makes much more sense now. $\endgroup$ Commented Sep 9, 2023 at 20:11

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