Functional analysis, ortho basis, weakly convergent

Fellowroot
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Homework Statement



problem%205.26_zpsxlq56zgm.png


This is a problem from Haim Brezis's functional analysis book.

Homework Equations

The Attempt at a Solution



I'm assuming (e)n is the vectors like (e)1 = (1,0,0), (e)2=(0,1,0) and so on.

We know every hilbert space has an orthonormal basis.

I also need to know the difference between l^2 and L^2 spaces. I know that this might not be much of a solution, but I'm really stuck.
 
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You could start by stating clearly what weak convergence implies (here in the context of a Hilbert space).

Once you have a clear understanding of weak convergence, 1. and 2. shouldn't be too difficult.
As (##e_n)## is an orthornormal basis of the Hilbert space H, you can use ## \forall x \in H: x= \sum_{i=1}^\infty <x,e_i>e_i## (series converging in the norm of the Hilbert space).
And also ##\forall x \in H:|x|²=\sum_{i=1}^\infty |<x,e_i>|²##.
 
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Thanks.

I think I now have the answer to part 1 because you can use an idea from Parseval which automically shows that en weakly converges to zero. But I'm still not sure about what to do with an and un. I also think the boundedness of an plays a role.

For part 2 I know that this is showing strongly convergence, but I don't really get how to show abs(un). Do I just pretend that the summation is really a sequence?
 
Fellowroot said:
I think I now have the answer to part 1 because you can use an idea from Parseval which automically shows that en weakly converges to zero.
Yes, that is correct.
Fellowroot said:
For part 2 I know that this is showing strongly convergence, but I don't really get how to show abs(un). Do I just pretend that the summation is really a sequence?
You can compute an upper bound for ##|u_n|##, making use of the fact that ##(a_n)## is bounded, and ##(e_n)## orthonormal.
You can use that, in general, you have:
##|\sum_{i=1}^n b_ie_i|²=\sum_{i=1}^n |b_i|²## (where ##(b_i)## are any numbers), because ##(e_n)## is orthonormal.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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