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Linear forms and complete metric space

by quasar987
Tags: forms, linear, metric, solved, space
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quasar987
#1
Nov11-07, 03:15 PM
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1. The problem statement, all variables and given/known data
Question:

Let L be a linear functional/form on a real Banach space X and let {x_k} be a sequence of vectors such that L(x_k) converges. Can I conclude that {x_k} has a limit in X?

It would help me greatly in solving a certain problem if I knew the answer to that question.

3. The attempt at a solution

The natural approach is to try to show that {x_k} is Cauchy.

Since the sequence of real numbers {L(x_k)} converges, then it is Cauchy, so for n,k large enough,

[tex]|L(x_k)-L(x_n)|=|L(x_k - x_n)|<\epsilon[/tex]

Now what??
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ZioX
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Nov11-07, 03:25 PM
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If L is not invertible what can happen?
morphism
#3
Nov11-07, 03:27 PM
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edit: removing my too explicit hint.

ZioX
#4
Nov11-07, 03:30 PM
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Linear forms and complete metric space

Boo. I'm sure he could have figured it out on his own.

What is the particular quandary?

Edit: Suppose L is invertible! What can you say then?
quasar987
#5
Nov11-07, 03:39 PM
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Too late morphism! :D
quasar987
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Nov11-07, 03:42 PM
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Anyway, guys, in the problem I'm working on, I must show that in a particular situation, the x_k do have a limit in X.

Are you willing to help with the general problem?

If so, I will type it out. Not very long, but a little complicated notation-wise.
quasar987
#7
Nov11-07, 04:00 PM
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It's at the core, a problem on measure theory.

Let [tex]\mathcal{L}^2(\mathbb{R})[/tex] denote the space of square-integrable functions f on R (with respect to the Lebesgue sigma-algebra and Lebesgue measure [itex]\lambda[/itex]) and let [tex]\mathcal{L}^2_{\lambda}(\mathbb{R})[/tex] denote the space of their equivalence classes f where f=g if f=g almost everywhere. [f]+[g]=[f+g] and c[f]=[cf] are well defined.

Now with the norm

[tex]||\mathbf{f}||_2=\int_{-\infty}^{+\infty}|f(x)|^2dx[/tex]

[tex](\mathcal{L}^2_{\lambda}(\mathbb{R}),||||_2)[/tex] is a real Banach space.

A continuous linear form on [tex](\mathcal{L}^2_{\lambda}(\mathbb{R}),||||_2)[/tex] is a linear form [tex]L:\mathcal{L}^2_{\lambda}(\mathbb{R})\rightarrow \mathbb{R}[/tex] such that

[tex]||L||=\sup\left\{\frac{|L(\mathbf{f})|}{||\mathbf{f}||_2}: \mathbf{f}\neq \mathbf{0}\right\}=\sup\left\{|L(\mathbf{f})|:||\mathbf{f}||_2=1\right\ }<+\infty[/tex]

Now consider a sequence [tex]\mathbf{g}_k\in \mathcal{L}^2_{\lambda}(\mathbb{R})[/tex] be such that

[tex]||\mathbf{g}_k||_2=1[/tex] and [tex]\lim_{k\rightarrow\infty}L(\mathbf{g}_k)=||L||[/tex]

Show that [tex]\mathbf{g}_k[/tex] has a limit in [tex]\mathcal{L}^2_{\lambda}(\mathbb{R})[/tex].
morphism
#8
Nov11-07, 08:39 PM
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Is L a fixed functional? If so, I don't think this is true. Try constructing a counterexample using the zero functional.
quasar987
#9
Nov11-07, 09:00 PM
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Apologies!

There is an additional hypothese! L is a non-identically vanishing functional!
morphism
#10
Nov11-07, 09:43 PM
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I'm tempted to use the Riesz representation theorem: we know there is a nonzero f in L^2 such that L(g)=<f,g> for all g in L^2, and ||L||=||f||_2.

Now consider ||f - g_n||[itex]_2^2[/itex]. (Hint: apply the polarization identity, and use the fact that <f,g_n> -> ||f||.) Try to see if you can guess what (g_n) converges to using this.
quasar987
#11
Nov12-07, 12:04 PM
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We are not to use this theorem in this problem, because in a sense, the whole problem sheet comes down to showing explicitely that the Riesz representation theorem hold in the case of L². No prior knowledge of functional analysis should be needed to do this problem.

Someone told me he succeeded in answering this question by effectively proving that the sequence g_k was Cauchy!
morphism
#12
Nov13-07, 08:35 AM
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Did you manage to do it without Riesz?
quasar987
#13
Nov13-07, 04:30 PM
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Yes, with the help of aforementioned person. :)

With the parallelogram identity, we reduce the problem to showing [tex]||\mathbf{g}_k+\mathbf{g}_{k+p}||_2^2 \rightarrow 4[/tex]

Then notice that because ||L|| is the sup,

[tex]\frac{L(\mathbf{g}_k+\mathbf{g}_{k+p})}{||\mathbf{g}_k+\mathbf{g}_{k+p} ||_2}\leq ||L|| \ \ \ \ \ \ (*)[/tex]

On the other hand, write out the facts that L(g_k)-->||L|| and L(g_{k+p})-->||L|| and add the inequalities to obtain

[tex]L(\mathbf{g}_k+\mathbf{g}_{k+p})>2(||L||-\epsilon')[/tex]

Combine with equation (*) to obtain an inequality involving [tex]||\mathbf{g}_k+\mathbf{g}_{k+p}||_2[/tex] and [tex]\epsilon'[/tex]. Show that to any [tex]\epsilon>0[/tex], you can find an [tex]\epsilon'(\epsilon)[/tex] such that [tex]4-||\mathbf{g}_k+\mathbf{g}_{k+p}||_2^2<\epsilon^2[/tex] for k large enough.


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