Some Fourier Analysis

**brouwer** · April 15th 2010, 11:53 PM

I have to proof the following:

If $f:\mathbb{T}\rightarrow\mathbb{C}$ Riemann integrable and $g:\mathbb{T}\rightarrow\mathbb{C}$ continuous, then:

$\lim_{n\to\infty}\frac{1}{2\pi}\int_{0}^{2\pi}f(\t heta)g(n\theta)d\theta = \hat{f}(0)\hat{g}(0)$

Where a Fourier coefficient $\hat{f}(n)$ is given by $\frac{1}{b-a}\int_{a}^{b}f(x)e^{-2\pi inx/L}dx, \qquad n \in \mathbb{Z}$

A hint is given to first proof it for $g$ a trigonometric function and then in general for continuous functions.

What might help in solving this is a property of convolutions. Given two $2\pi$ -periodic Riemann integrable functions $f$ and $g$ on $\mathbb{R}$ , then their convolution $f\ast g$ on $[-\pi,\pi]$ is defined by
$(f\ast g)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(y)g(x-y)dy$
Our property of (probable) interest is: suppose $f$ and $g$ are $2\pi$ -periodic Riemann integrable functions. Then:
$\widehat{f\ast g}=\hat{f}(n)\hat{g}(n)$ .

I'm rather clueless on how to prove this one..

**Laurent** · April 16th 2010, 04:58 AM

Originally Posted by brouwer

I have to proof the following:

If $f:\mathbb{T}\rightarrow\mathbb{C}$ Riemann integrable and $g:\mathbb{T}\rightarrow\mathbb{C}$ continuous, then:

$\lim_{n\to\infty}\frac{1}{2\pi}\int_{0}^{2\pi}f(\t heta)g(n\theta)d\theta = \hat{f}(0)\hat{g}(0)$

Where a Fourier coefficient $\hat{f}(n)$ is given by $\frac{1}{b-a}\int_{a}^{b}f(x)e^{-2\pi inx/L}dx, \qquad n \in \mathbb{Z}$

A hint is given to first proof it for $g$ a trigonometric function and then in general for continuous functions.

What might help in solving this is a property of convolutions. Given two $2\pi$ -periodic Riemann integrable functions $f$ and $g$ on $\mathbb{R}$ , then their convolution $f\ast g$ on $[-\pi,\pi]$ is defined by
$(f\ast g)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(y)g(x-y)dy$
Our property of (probable) interest is: suppose $f$ and $g$ are $2\pi$ -periodic Riemann integrable functions. Then:
$\widehat{f\ast g}=\hat{f}(n)\hat{g}(n)$ .

I'm rather clueless on how to prove this one..

You should try to use the hint. If $g$ is a constant, this is trivial. Assume now that $g(x)=e^{ikx}$ for some $k\in\mathbb{N}^*$ . Then $\frac{1}{2\pi}\int_0^{2\pi} f(\theta)g(n\theta)d\theta = \widehat{f}(-kn)$ , and maybe you know that the Fourier coefficients in this case converge to 0 at infinity (this is also called Lebesgue's lemma) hence you find that the limit is zero, which is ok since $\widehat{g}(0)=0$ .

Then use uniform approximation of a general function $g$ by trigonometric ones, i.e. linear combinations of functions like the previous one (it comes from Stone-Weierstrass theorem).

**brouwer** · April 16th 2010, 05:11 AM

Thanks for your reply, a great help and insight!

One quick question though, what do you mean by $\mathbb{N}^*$ ? Never mind, I figured it's just $\mathbb{N}_{>0}$

Thanks I got it completely now!

Cheers.

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