5
$\begingroup$

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.$\qquad\qquad\qquad\qquad ////$

Transcribed from this image

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.

$\endgroup$
6
+50
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.