Show that the map of $L^1(\Bbb R)\cap L^2(\Bbb R)$ into $L^2(\Bbb R)$ defined by $f\mapsto \hat{f}$ is continuous.

99 Views Asked by At

I am doing self-study on Fourier analysis and I have the following questions. Could anyone give an elementary proof? Consider the Fourier transform $\hat{f}$ of $f\in L^1(\Bbb R)$ defined by $$\hat{f}(\xi)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{-i\xi x}dx, \qquad \text{for $\xi\in\Bbb R$}.$$

(1) Show that if $f\in L^1(\Bbb R)\cap L^2(\Bbb R)$ then $\hat{f}\in L^2(\Bbb R)$.

(2) Show that the map of $L^1(\Bbb R)\cap L^2(\Bbb R)$ into $L^2(\Bbb R)$ defined by $f\mapsto \hat{f}$ is continuous with respect to the $L^2$-norm.

1

There are 1 best solutions below

0
On

This is the first step in the proof of the Plancherel Theorem (it's the main step, so much so that people sometimes call this result itself the Plancherel Theorem). As suggested, an excellent reference is Rudin Real and Complex Analysis. Or Folland Real Analysis for a different argument, or any other text that includes beginning Fourier analysis.

The standard proofs are very slick, but they have a fairly high how-would-anyone-think-of-that factor. Here's an outline of another argument, that may or may not make more sense. Note it's just an outline - there are some major details omitted, but regardless it may serve to make the result at least plausible:

For $P>0$ define $$f_P(x)=\sum_{n\in\Bbb Z}f(x+nP).$$Then $f_P$ has period $P$; hence it has a Fourier series. You can figure out the coefficients in terms of $\hat f$; now the $P$-periodic verion of the Parseval Theorem shows that $$\int_{-P/2}^{P/2}|f_P|^2=\frac cP\sum_{k\in \Bbb Z}\left|\hat f\left(\frac kP\right)\right|^2,$$where $c$ is a certain constant, $1/\sqrt\pi$ or $1/\sqrt{2\pi}$ or something. (Different authors define $\hat f$ differently, hence would get different values of $c$ here...)

So far filling in the details is routine. Although the last step is quite plausible, it's where the details become not quite so trivial: If you let $P\to\infty$ you obtain$$||f||_2=c||\hat f||_2.$$