Fourier Analasys - Inner Product Spaces

[Ylle]

Homework Statement

I have two assignments I have some problems with.

The first one:

For n > 0, let

f_n(t) = {1, 0 \leq t \leq 1/n
0, otherwise

Show that f_n \rightarrow 0 in L²[0,1]. Show that f_n does NOT converge to zero uniformly on [0,1]


The second one:
Find the L2[-\pi, \pi] orthogonal projection of the function f(x) = x2 onto the space Vn \subset L2[-\pi, \pi] spanned by

{1, sin(jx)/sqrt(\pi), cos(jx)/sqrt(\pi); j =1,...,n}
for n = 1. Repeat this exercise for n= 2 and n = 3.

Homework Equations

The Attempt at a Solution

For the first one, I will take the integral:
|fn|2 = int(0 to 1/n) (1)^2 dt = 1/n - and therefor it converges to 0 for n -> infinite.

It's the second part of this I'm not sure about. I'm not sure how to explain it. Any help ?


For the second one, I first test if the basis' is orthonormal, but first taking the inner product of each of them <1, sin(jx)/sqrt(\pi)>, <1, cos(jx)/sqrt(\pi)> and <sin(jx)/sqrt(\pi), cos(jx)/sqrt(\pi)> to find out they are all = 0.

And then I take the inner product of every basis <1,1>, <sin(jx)/sqrt(\pi), sin(jx)/sqrt(\pi)> and so on, to find out that 1 is not 1, but 2*pi. So I normalize that one by dividing it with the 2*pi, and then I have an orthonormal basis, since the other to are have the length 1.

And then to find the orthogonal projection I say:
<x², 1/(2*pi)>1/2*pi + <x², sin(jx)/sqrt(\pi)>sin(jx)/sqrt(\pi) + <x², cos(jx)/sqrt(\pi)>cos(jx)/sqrt(\pi)

And since I do it for j, and not a 1, 2 or 3, I guess I really don't need to show it for 1, 2 or 3, since it's only some different integrals that make the difference +


Well, I hope you can help me a bit.


Regards

 

[Billy Bob]

For problem 1, showing that the convergence is not uniform is very easy. I don't say that to make you feel bad, but rather as a hint. What is the definition of uniform convergence?

For problem 2, the answer in the n=1 case will be a sum of 3 terms, in the n=2 case a sum of 5 terms, and in the n=3 case a sum of 7 terms. I couldn't tell whether or not you realized that.

Perhaps someone else can comment about whether your use of 1/2π instead of 1 is correct, because I always use the convention 2π=1 (just kidding a little, but not completely). Maybe you should use 1/√(2π).