MHB Help with functional analysis questions

[simpleton1]

Hi,

Could someone post a solution to the following questions :

1. Let R be the real numbers and A a collection of all groups that are either bound or their complement is bound.
a. Show that A is an Algebra. Is it a sigma algebra?
b. Define measure m by m(B) = {0 , max(on B) x < $\infty$ and 1 , max(not on B) x < $\infty$
Show that m on B is finite sigma additive but not sigma addivite.
2. Show there's a single continuous function from [0,1] to R that makes the equation below for all $0 \le x\le 1$:
f(x) = sinx+$\int_{0}^{1} \frac{f(y)}{exp(x+y+1)} \,dy$
3. Show that in {C}^{n} with the eucledian norm ($ {\left\lVert{X}\right\rVert}^{2} = \sum_{i=1}^{n}{\left| Xi \right|}^{2} $) weak convergion causes strong convergion. Is that true in every Hilbert space?

4. Let B = {${(Xn) \in l2 : {sup}_{n}\left| Xn \right|\le1}$}. Is B a compact sub-group of l2?
5. a. Let A(f) = $\int_{0}^{1}tf(t) \,dt $ . Is A a continuous linear functional in L2[0,1] space?
b. Let A(f) = $\int_{0}^{1}f(t)\frac{2}{t-1} \,dt $ . Is A a continuous linear functional in L2[0,1] space?
No need to prove linearity of A and B.

 

[Euge]

Hi simpleton,

We do not simply post solutions to all problems. Please show what you've tried or what your thoughts are about these exercises.

 

[simpleton1]

1. a. For every group in A , its complement is also in A. Every subgroup of a bound group is also bound.
Using this and the de-morgan rules show that A is closed for intersection, not and finite union.
A is not closed for infinite union because infinite (even countable) union of bound groups of this
type can lead to a group that isn't bound - for example [n,n+1).
b. The measure is finite additive for bound foreign groups because a finite union of bound groups will be finite and
the sum of zeros is zero. Using de-morgan you can show that two groups whose complement is bound cannot
be foreign. this means that there can only be one group whose complement is bound in a finite union of groups
and this will make the whole union unbound and therefore - if there will be such a group the measure of the union
will be 1 and the sum of the groups will be 1.
It is not sigma additive with the same example as part a, the measure of the union will be 1 but the sum of measures
will be 0.
2. I know the answer (using banach fixed point theorem).
3. I have no idea. What is the internal product to use? For some reason they only defined the norm.
4. I claimed that B is closed and bounded and therefore compact. My professor said it is not bound since
ei={0,...,0,1,0,...0} doesn't have a sub-series that converges. I don't understand his answer because why doesn't
ei converge?
5. I think I solved them. a. converges since the operator is bound and b. does not converge since the operator is not bound.

 

[simpleton1]

Some more thoughts :

3. Define Xn a series in ${C}^{n}$ that weakly converges to X , Therefore for every Y in ${C}^{n}$ :
$$\lim_{{k}\to{\infty}}<{X}_{k},Y>=\lim_{{k}\to{\infty}}\sum_{i=1}^{n}<{{X}_{k,i}},{Y}_{i}>=\sum_{i=1}^{n}<{{X}_{i}},{Y}_{i}>$$
We need to prove that $$ \lim_{{k}\to{\infty}}<{X}_{k},{X}_{k}>= \lim_{{k}\to{\infty}}\sum_{i=1}^{n}\left| {{X}_{k,i}}\right|^{2}=\sum_{i=1}^{n}\left| {{X}_{i}}\right|^{2} $$
Then why can't I just use the fact that Xk is also an element of ${C}^{n}$, therefore is part of Y and therefore this
proves the required. But this doesn't make sense to me because it means that weak convergence always causes strong
convergence.

4. Perhaps he meant it doesn't converge if Xn has an infinite dimension, but then ei wouldn't belong to l2.

 

[Evgeny.Makarov]

1. Let R be the real numbers and A a collection of all groups that are either bound or their complement is bound.
a. Show that A is an Algebra. Is it a sigma algebra?
b. Define measure m by m(B) = {0 , max(on B) x < $\infty$ and 1 , max(not on B) x < $\infty$
Show that m on B is finite sigma additive but not sigma addivite.

What do you mean by a group? The word "group" has a special meaning in mathematics.

What does "max(on B) x < $\infty$" mean?

 

[Euge]

Hi Evgeny,

What do you mean by a group? The word "group" has a special meaning in mathematics.

By a "group," simpleton actually means "set" -- I've talked about this with simpleton in another analysis thread.Edit: The definition for $m(B)$ is more unclear than I thought. The Dirac mass I mentioned would make sense if we had

$$m(B) = \begin{cases}0& 0\in B\\1 &0\notin B\end{cases}$$

 

[Evgeny.Makarov]

Really, the definition was supposed to be

$$m(B) = \begin{cases}0 & x\in B\\1 & x\notin B\end{cases}$$

That is, $m$ is the Dirac mass concentrated at $0$.

I am not sure how $m$ is a function on sets since $x$ occurs in the right-hand side but not in the left-hand side. Do you mean $0\in B$ instead of $x\in B$? It does not look much like "max(on B) x < $\infty$"...

 

[Euge]

You replied about the same time I edited. Yes, the $x$ was supposed to be $0$, but anyway, we'll just have to wait until simpleton responds.

 

[Euge]

Ok, considering that the $m$ defined was not supposed to be sigma-additive, I think these changes are to be made: $m(B) = 0$ if $B$ is bounded above, and $m(B) = 1$ if $\Bbb R\setminus B$ is bounded above.

 

[simpleton1]

Hi,

I tried to translate the question as much I could but possibly something got lost on the way.
The definition of the measure is that if a group or set (the professor called it group, possibly he meant
a set) is bound the measure is 0 and if its complement is bound the measure is 1.
But as I've mentioned I think I managed to solve this one on my own.
I really need help with questions 3 and 4.

Thanks,
Simpleton