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.


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.

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  • $\begingroup$ How do you know $\psi_A = (k_A \hat{f})^\vee$? In other words, it is equivalent to show $\hat{\psi}_A = k_A \hat{f}$. Since we don't know $\psi_A \in L^1$, we might not even be able to use $\hat{\psi}_A = \int_{-\infty}^{\infty} \psi_A(x)e^{-ixt}\,dm(x)$. $\endgroup$ – withgrace1040 Jun 23 at 12:28
  • $\begingroup$ @withgrace1040: I use that $k_A \hat{f}\in L^1$. This implies that the inverse Fourier transform hereof, that is $x\mapsto\big(k_A \hat{f}\big)^\vee(x)$, is given by the integral $\int_{-\infty}^\infty k_A(t) \hat{f}(t) e^{itx} dt$. $\endgroup$ – StarBug Jun 23 at 14:16
  • $\begingroup$ I think if you use the inverse Fourier transform, then you cannot use $f = \hat{f}^\vee$ in the first equality, which is the one this theorem aims to show eventually. $\endgroup$ – withgrace1040 Jun 23 at 14:50
  • $\begingroup$ @withgrace1040: Observe that I am not actually using the identity $f(x)=\int_{-\infty}^\infty \hat{f}(t) e^{itx} dm(t)$, which I agree with you is not correct (at least not in general for $f\in L^2$). Instead, I am using (c) from Rudin's Theorem to justify $f=\hat{f}^\vee$ $\endgroup$ – StarBug Jun 23 at 22:07
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    $\begingroup$ @withgrace1040: At the outset, the inverse Fourier transform is defined as the operator $\mathcal{F}^{-1}: L^1\rightarrow L^\infty$ with $\mathcal{F}^{-1}(f)(t):=f^\vee(t):=\int f(x) e^{itx}dx$. Now (c) tells me that this operator extends to an isomorphism $\mathcal{F}^{-1}:L^2\rightarrow L^2$. Since I only know $f\in L^2$ (and not necessarily in $L^1$), I can only apply the extension of $\mathcal{F}^{-1}$ to $f$. In contrast, I know that $k_A\hat{f}\in L^1$, which means I can use the original integral representation of $\mathcal{F}^{-1}$ to $k_A\hat{f}$. $\endgroup$ – StarBug Jun 24 at 12:01

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