# Thread: Some Fourier Analysis

1. ## Some Fourier Analysis

I have to proof the following:

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

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

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

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

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

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

2. Originally Posted by brouwer
I have to proof the following:

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

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

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

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

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

I'm rather clueless on how to prove this one..
You should try to use the hint. If $\displaystyle g$ is a constant, this is trivial. Assume now that $\displaystyle g(x)=e^{ikx}$ for some $\displaystyle k\in\mathbb{N}^*$. Then $\displaystyle \frac{1}{2\pi}\int_0^{2\pi} f(\theta)g(n\theta)d\theta = \widehat{f}(-kn)$, 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 $\displaystyle \widehat{g}(0)=0$.

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

3. Thanks for your reply, a great help and insight!

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

Thanks I got it completely now!

Cheers.