Theorem 9.13 in Rudin's Real and Complex Analysis

Question

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 $f \in L^{2},\|\hat{f}\|_{2}=\|f\|_{2}$.

(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 $$ \varphi_{A}(t)=\int_{-A}^{A} f(x) e^{-i x t} d m(x) \quad \text { and } \quad \psi_{A}(x)=\int_{-A}^{A} \hat{f}(t) e^{i x t} d m(t), $$ then $\left\|\varphi_{A}-\hat{f}\right\|_{2} \rightarrow 0 \text { and }\left\|\psi_{A}-f\right\|_{2} \rightarrow 0 \text { as } A \rightarrow \infty$.

To prove $(d),$ let $k_{A}$ be the characteristic function of $[-A, A] .$ Then $k_{A} f \in L^{1} \cap L^{2}$ if $f \in L^{2},$ and $$ \varphi_{A}=\left(k_{A} f\right)^{\wedge} $$ since $\left\|f-k_{A} f\right\|_{2} \rightarrow 0$ as $A \rightarrow \infty,$ it follows from $(b)$ that $$ \left\|\hat{f}-\varphi_{A}\right\|_{2}=\|\left(f-k_{A} f\right) \hat{\|}_{2} \rightarrow 0 $$ as $A \rightarrow \infty$.

The other half of $(d)$ is proved the same way.$\qquad\qquad\qquad\qquad ////$

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This is Theorem 9.13 in Rudin's Real and Complex Analysis, and a part of its proof. I have understood the first half of the proof of (d), but I can't see how the other half of (d) follows. Am I missing something? Thanks in advance.

Answer 1

Let $\check{f}:=\mathscr{F}^{-1}(f)$ denote the inverse Fourier transform. By (c) it follows that $||f||_2=||\check{f}||_2$ for all $f\in L^2$. Observe that $k_A \hat{f}\in L^1\cap L^2$ and $\psi_A=(k_A \hat{f})^\vee$ for all $f\in L^2$. It follows that $$||f-\psi_A||_2=||\hat{f}^\vee-(k_A \hat{f})^\vee||_2=||\hat{f}-(k_A \hat{f})||_2\rightarrow 0$$ as $A\rightarrow \infty$.

Before someone jumps at me for forgetting the nuisance factor $2\pi$, please observe that the measure $dm$ in Rudin's book is a scaling of the Lebesgue measure so that the factor is not needed in the argument above.