# Question on Rudin's Proof of the Fourier Transform Inversion Theorem (Theorem 9.11 in Real and Complex Analysis)

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

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.

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

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 $$C_0$$.