# Functional Analysis HMWK

1. Nov 12, 2012

### Kindayr

1. The problem statement, all variables and given/known data
Let $e_{n}(t)= \frac{1}{ \sqrt{2\pi}}\cdot e^{int}$ for $n\in\mathbb{Z}$ and $-\pi\le t\le\pi$.

Show that for any $f\in L^{2}[-\pi,\pi]$ we have that $$(f,e_{n})=\int_{-\pi}^{\pi}f(t)\cdot e^{-int}dt\to0$$ as $|n|\to \infty$.

3. The attempt at a solution
I want to use dominant convergence, but unfortunately measure theory isn't a prerequisite for this course. Any help will be awesome!

2. Nov 12, 2012

### Kindayr

Wait, if ${e_n}$ is complete, then $\lim_{n\to\infty}\sum_{m=-n}^{n}(f,e_m)e_m$ converges to $f$ absolutely, so the coefficients necessarily converge to zero. Would this work?

3. Nov 12, 2012

### micromass

Staff Emeritus
Have you seen and proved Bessel's inequality??

4. Nov 12, 2012

### Kindayr

Yep we have. Could I say that:

Since $\{e_n\}$ is complete it follows that $f=\lim_{n\to\infty} \sum_{m=-n}^{n}(f,e_m)e_m$. Thus, the latter sum converges and hence $\lim_{n\to\infty}\sum_{m=-n}^{n}|(f,e_m)|^{2}<\infty$. Thus $|(f,e_m)|^2 \to 0$ so $|(f,e_m)|\to 0$ and whence $(f,e_m)\to 0$, as required.

Does this work?

5. Nov 12, 2012

### micromass

Staff Emeritus
You don't really need $\{e_n\}$ to be complete for that. Bessel's inequality works fine.
Also, the limit you mention is $L^2$-convergence. You might want to be careful with that.

6. Nov 12, 2012

### Kindayr

What should I be careful with? Does $L^2$-convergence not imply absolute convergence? Sorry, just a little confused by your statement.

7. Nov 12, 2012

### Kindayr

What if I re-index so that $n\in \mathbb{N}$ so that $d_0=e_0, d_1=e_1, d_2=e_{-1},\dots$. Then by Bessel's inequality we have $\sum_{n=0}^{\infty}|(f,d_n)|^2\le \parallel f\parallel ^{2}<\infty$. Hence $\sum_{n=0}^{\infty}|(f,d_n)|^2$ converges absolutely and whence $(f,d_n)\to 0$ and thus $(f,e_n)\to\ 0$ as $|n|\to\infty$.

Would this be better?

I'm far from an analyst as you can probably tell hahahah.