Question on Rudin's Proof of $\int_{-n}^{n}\hat f(\xi)e^{it\xi}\, d\xi\to f$ in the 2-norm

Question

In Real and Complex Analysis pg. 187, (d) property of Theorem 9.13 says: If $f\in L^2(\mathbb R)$ then: \begin{equation}\int_{-n}^{n}f(t)e^{-it\xi}\, dt\to \hat f\text{ and }\int_{-n}^{n}\hat f(\xi)e^{it\xi}\, d\xi\to f\text{ w.r.t. }\left\|\cdot\right\|_2\end{equation} For the first convergence, as $f\in L^2$, $f_n=f\chi_{[-n,n]}\in L^1\cap L^2$ and $f_n\to f$ in $\left\|\cdot\right\|_2$. Plancherel's Theorem implies \begin{equation}\left\|\hat f-\int_{-n}^{n}f(t)e^{-it\xi}\, dt\right\|_2= \left\|\hat f-\hat f_n\right\|_2=\left\|f-f_n\right\|_2\to 0\end{equation} All goof so far. Then Rudin goes on to write that the other convergence result is proven similarly. My question is how is it proven (similarly)?

Going by the previous result I defined $g_n=\hat f\chi_{[-n,n]}\in L^1\cap L^2$ and $g_n\to \hat f$ in $\left\|\cdot\right\|_2$. But in that case \begin{equation}\int_{-n}^{n}\hat f(\xi)e^{it\xi}\, d\xi=\mathcal F^*(g_n)\end{equation} where $\mathcal F^*$ is the Fourier cotransformation defined for functions in $L^1$ by \begin{equation}\mathcal F^*f=\int_{-\infty}^{\infty}f(\xi)e^{it\xi}\, d\xi\end{equation} and by a density argument for functions in $L^2$. Now I can prove that $\mathcal F^*$ has similar properties to the Fourier Transform, for example if $f\in L^2$ then $\mathcal F^*f\in L^2$ as well. My approach was to use Plancherel's Theorem as before: \begin{equation}\left\|f-\int_{-n}^{n}\hat f(\xi)e^{it\xi}\, d\xi\right\|_2= \left\|f-\mathcal F^*g_n\right\|_2=\left\|\mathcal Ff-\mathcal F\mathcal F^*g_n\right\|_2\end{equation} I would like to say $\mathcal F\mathcal F^*g_n=g_n$ a.e. but this is true if $\mathcal F^*g_n\in L^1$ which I don't know. Any suggestions on how I should proceed (that are preferably similar to Rudin's proof)?

Answer 1

I think what you want to prove is a little trickier than Rudin lets on. At least, it's not as direct as before. Here's the proof I know. It appears in Stein and Weiss's Introduction to Fourier Analysis on Euclidean Spaces.

Using Fubini's theorem, for any $g\in L^2 \cap L^2 $ you have

$$\langle g , \mathcal F^* \mathcal F f \rangle = \langle \mathcal F g , \mathcal F f\rangle.$$

Then, using Plancherel's theorem, this is equal to $\langle g, f\rangle$. Since $F^* \mathcal F f$ and $f$ give the same inner product against any function in $L^2$ (we actually verified it on a dense subset, but same thing), they are equal (as members of $L^2$).

Answer 2

Here's an excerpt from my edition of Rudin. It's an international student edition that predates the third edition, but I'm not sure if it's the second or the first. In any case, it has what you want. He uses a convolution argument to get the function you want to invert into $L^1$, then applies the $L^1$ inversion formula.