I understood the whole theorem proof, except for the very last part.
Here is the theorem
9.13 Theorem One can associate to each $f\in L^2$ a function $\hat{f}\in L^2$ so that the following properties hold:
(a) if $f\in L^1\cap L^2$, then $\hat{f}$ is the previously defined Fourier transform of $f$
(b) for every $f\in L^2\quad,\quad||\hat{f}||_2=||f||_2$
(c) The mapping $f\rightarrow \hat{f}$ is a Hilbert space isomorphism of $L^2$ onto $L^2$.
(d) The following symmetric relation exists between $f$ and $\hat{f}$ : If $$\varphi_A(t)=\int_{-A}^{A}f(x)e^{-ixt}dm(x)\quad\text{and}\quad\psi_A(t)=\int_{-A}^{A}\hat{f}(x)e^{ixt}dm(x)$$ then $||\varphi_A-\hat f||_2\rightarrow 0\quad\text{and}\quad||\psi_A- f||_2\rightarrow 0$
The part I miss is the last (d) dealing with $\psi_A$ ; Rudin wrote "The other half of (d) is proved the same way" (with no detail at all: the "first half" deals with $\varphi_A$).
One answer that I read here assumes that the following function $\mathscr F^{-1} : L^1\rightarrow L^\infty$ is the inverse Fourier transform $$\mathscr F^{-1}(f)(t)=\int_{-\infty}^{+\infty}f(x)e^{ixt}dm(x)\qquad\qquad(\star)$$
But in Rudin's "Real and Complex Analysis", the inversion theorem (9.11) gives this formula only when the function and its Fourier transform are in $L^1$ :
9.11 The Inversion Theorem If $f\in L^1\quad\text{and}\quad \hat f\in L^1$ and if $$g(x)=\int_{-\infty}^{+\infty}\hat f(x)e^{ixt}dm(t)\qquad (x\in R^1)$$
then $g\in C_0\quad\text{and}\quad g(x)=f(x)$ a.e.
If $k_A$ is the characteristic function of $[-A,A]$, then $\psi_A$ seems to be the inverse Fourier transform of $k_A\hat f$, assuming that the inverse Fourier transform is given by $(\star)$.
"Seems to be" : this is exactly the point I miss : although obviously $k_A\hat f\in L^1$, nothing proves that $\psi_A$ is in $L^1$ (I daresay the contrary, since generally $k_A\hat f$ is not equals a.e. to a continuous function, whereas the Fourier transform of a $L^1$ function is coninuous), so Rudin's 9.11 theorem can't be applied.
Hence my question : how can this "second part of (d)" be resolved ?
Best Answer
For $ {\lambda} > 0$, let
\begin{equation}{h}_{{\lambda}} \left(x\right) = \sqrt{\frac{2}{{\pi}}} \frac{{\lambda}}{{{\lambda}}^{2}+{x}^{2}} , \qquad {H}_{{\lambda}} \left(t\right) = {e}^{{-{\lambda}} \left|t\right|}\end{equation}
as in Rudin's paragraph 9.7. It is easy to check that
\begin{equation}{\hat{h}}_{{\lambda}} = {H}_{{\lambda}} , \qquad {\hat{H}}_{{\lambda}} = {h}_{{\lambda}}\end{equation}
Now we have the following
Lemma A: if $ g \in {L}^{1} \cap {L}^{2}$ and $ {g}_{{\lambda}} = g \ast {h}_{{\lambda}}$ then $ {g}_{{\lambda}} \in {L}^{1} \cap {L}^{2}$ and $ {\hat{g}}_{{\lambda}} \in {L}^{1} \cap {L}^{2}$.
Indeed, by theorem 9.10 in Rudin we have $ {g}_{{\lambda}} \in {L}^{1} \cap {L}^{2}$ (see also exercise 8.4) and by theorem 9.2, we have $ {\hat{g}}_{{\lambda}} = \hat{g} {H}_{{\lambda}}$, but $ \hat{g} \in {C}_{0} \subset {L}^{\infty }$ (theorem 9.6) and $ {H}_{{\lambda}} \in {L}^{1} \cap {L}^{2}$.
Lemma B: The set $ X = \left\{g \ast {h}_{{\lambda}} \ \mid \ g \in {L}^{1} \cap {L}^{2} , {\lambda} > 0\right\}$ is dense in $ {L}^{2}$.
Indeed, let $ f \in {L}^{2}$ and $ {\varepsilon} > 0$. There exists $ A > 0$ such that $ {\left\|{f}_{A}-f\right\|}_{{L}^{2}} < {\varepsilon}$ where $ {f}_{A} = f\, \mathbf{1}_{\left[{-A} , A\right]}$. We have $ {f}_{A} \in {L}^{1} \cap {L}^{2}$ and by theorem 9.10, there exists some $ {\lambda} > 0$ such that $ {\left\|{f}_{A} \ast {h}_{{\lambda}}-{f}_{A}\right\|}_{{L}^{2}} < {\varepsilon}$. Hence $ {\left\|{f}_{A} \ast {h}_{{\lambda}}-f\right\|}_{{L}^{2}} < 2 {\varepsilon}$ and $ {f}_{A} \ast {h}_{{\lambda}} \in X$.
Let $ {\Phi}$ be the Fourier transform defined on $ {L}^{1}$ and also on $ {L}^{2}$ by the density argument given by Rudin in the proof of parts a) b) c) of theorem 9.13, and let $ \mathcal{R}$ be the mapping $ \mathcal{R} \left(f\right) \left(x\right) = f \left({-x}\right)$.
It is clear that $ \mathcal{R} {\Phi} {\Phi} \left(f\right) = f$ for every function $ f \in X$ because by lemma A, these functions satisfy the conditions of theorem 9.11. As $ \mathcal{R} {\Phi} {\Phi}$ is an isometry of $ {L}^{2}$, the relation $ \mathcal{R} {\Phi} {\Phi} \left(f\right) = f$ extends to all $ f \in {L}^{2}$ by continuity.
Now in the proof of part d) of theorem 9.13 we have that $ {\left\|{{\psi}}_{A}- \mathcal{R} {\Phi} {\Phi} \left(f\right)\right\|}_{{L}^{2}} \rightarrow 0$, which proves the result since $ \mathcal{R} {\Phi} {\Phi} \left(f\right) = f$.