- #1

Clammierfire20

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- TL;DR Summary
- I've constructed a proof for Placherel's theorem but I would really appreciate it if someone could check it, please. I'm not sure I've implemented the convergence result of the convolution correctly.

I'm trying to prove Plancherel's theorem for functions $$f\in L^1\cap L^2(\mathbb{R})$$. I've included below my attempt and I would really appreciate it if someone could check this for me please, and give me any feedback they might have.

**Note:** I am working with a slightly different definition of the Fourier transform to usual, namely:

$$\widehat{f}(u)=\int_{-\infty}^{\infty}e^{iux}f(x)\,dx$$.

My main worries are that I have not implemented the convergence property of the convolution properly and I'm not sure if I've justified the use of the MCT in the correct way.

## My proof goes as follows:

Firstly, I have previously shown that:

$$(G_\lambda\ast f)(x)\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-iux}e^{-\frac{(u/\lambda)^2}{2}}\widehat{f}(u)\,du. \tag{*}$$

Then, writing $|f(x)|=f(x)\overline{f(x)}$, we have:

$$||{f}||_2^2=\int_{-\infty}^{\infty}f(x)\overline{f(x)}\,dx=\lim_{\lambda\to\infty}\int_{-\infty}^{\infty}f(x)\overline{(G_\lambda \ast f)(x)}\,dx.$$

The last inequality holds since $$||G_\lambda \ast f-f||_2 \to 0$$ (something I've already proved). Implementing (*) and then applying Fubini's theorem (which I have justified separately), we can write:

$$\begin{align*}

||{f}||_2&=\lim_{\lambda \to \infty}\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x)e^{iux}\overline{\widehat{f}(u)}e^{-\frac{(u/\lambda)^2}{2}}\,du\,dx\\

&=\lim_{\lambda \to \infty}\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x)e^{iux}\,dx\overline{\widehat{f}(u)}e^{-\frac{(u/\lambda)^2}{2}}\,du\\

&=\lim_{\lambda \to \infty}\frac{1}{2\pi}\int_{-\infty}^{\infty}\widehat{f}(u)\overline{\widehat{f}(u)}e^{-\frac{(u/\lambda)^2}{2}}\,du\\

&=\lim_{\lambda \to \infty}\frac{1}{2\pi}\int_{-\infty}^{\infty}|\widehat{f}(u)|^2e^{-\frac{(u/\lambda)^2}{2}}\,du.

\end{align*}$$

Define the sequence $$\{x_\lambda\}_{\lambda =1}^\infty$$ by $$x_\lambda=|\widehat{f}(u)|^2e^{\frac{-(u/\lambda)^2}{2}}$$. Clearly, we have $$\lim_{\lambda\to\infty}x_n=|\widehat{f}(u)|$$. We also know $\{x_n\}$ is non-negative and monotone increasing, since $|\widehat{f}(u)|$ is fixed and $e^{\frac{-(u/\lambda)^2}{2}}$ increases in value as $\lambda$ increases. Therefore, by the MCT, we have:

$$||{f}||_2=\frac{1}{2\pi}\int_{-\infty}^{\infty}|\widehat{f}(u)|^2\lim_{\lambda \to\infty}e^{-\frac{(u/\lambda)^2}{2}}\,du=\frac{1}{2\pi}\int_{-\infty}^{\infty}|\widehat{f}(u)|^2\,du=\frac{1}{2\pi}||{\widehat{f}}||_2^2.$$

Any help is really appreciated, thank you in advance!

**Note:** I am working with a slightly different definition of the Fourier transform to usual, namely:

$$\widehat{f}(u)=\int_{-\infty}^{\infty}e^{iux}f(x)\,dx$$.

My main worries are that I have not implemented the convergence property of the convolution properly and I'm not sure if I've justified the use of the MCT in the correct way.

## My proof goes as follows:

Firstly, I have previously shown that:

$$(G_\lambda\ast f)(x)\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-iux}e^{-\frac{(u/\lambda)^2}{2}}\widehat{f}(u)\,du. \tag{*}$$

Then, writing $|f(x)|=f(x)\overline{f(x)}$, we have:

$$||{f}||_2^2=\int_{-\infty}^{\infty}f(x)\overline{f(x)}\,dx=\lim_{\lambda\to\infty}\int_{-\infty}^{\infty}f(x)\overline{(G_\lambda \ast f)(x)}\,dx.$$

The last inequality holds since $$||G_\lambda \ast f-f||_2 \to 0$$ (something I've already proved). Implementing (*) and then applying Fubini's theorem (which I have justified separately), we can write:

$$\begin{align*}

||{f}||_2&=\lim_{\lambda \to \infty}\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x)e^{iux}\overline{\widehat{f}(u)}e^{-\frac{(u/\lambda)^2}{2}}\,du\,dx\\

&=\lim_{\lambda \to \infty}\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x)e^{iux}\,dx\overline{\widehat{f}(u)}e^{-\frac{(u/\lambda)^2}{2}}\,du\\

&=\lim_{\lambda \to \infty}\frac{1}{2\pi}\int_{-\infty}^{\infty}\widehat{f}(u)\overline{\widehat{f}(u)}e^{-\frac{(u/\lambda)^2}{2}}\,du\\

&=\lim_{\lambda \to \infty}\frac{1}{2\pi}\int_{-\infty}^{\infty}|\widehat{f}(u)|^2e^{-\frac{(u/\lambda)^2}{2}}\,du.

\end{align*}$$

Define the sequence $$\{x_\lambda\}_{\lambda =1}^\infty$$ by $$x_\lambda=|\widehat{f}(u)|^2e^{\frac{-(u/\lambda)^2}{2}}$$. Clearly, we have $$\lim_{\lambda\to\infty}x_n=|\widehat{f}(u)|$$. We also know $\{x_n\}$ is non-negative and monotone increasing, since $|\widehat{f}(u)|$ is fixed and $e^{\frac{-(u/\lambda)^2}{2}}$ increases in value as $\lambda$ increases. Therefore, by the MCT, we have:

$$||{f}||_2=\frac{1}{2\pi}\int_{-\infty}^{\infty}|\widehat{f}(u)|^2\lim_{\lambda \to\infty}e^{-\frac{(u/\lambda)^2}{2}}\,du=\frac{1}{2\pi}\int_{-\infty}^{\infty}|\widehat{f}(u)|^2\,du=\frac{1}{2\pi}||{\widehat{f}}||_2^2.$$

Any help is really appreciated, thank you in advance!