Linear forms and complete metric space

[quasar987]

[SOLVED] Linear forms and complete metric space

Homework Statement

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.

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,

|L(x_k)-L(x_n)|=|L(x_k - x_n)|<\epsilon

Now what??

 

[ZioX]

If L is not invertible what can happen?

 

[morphism]

edit: removing my too explicit hint.

 

[ZioX]

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]

Too late morphism! :D

 

[quasar987]

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]

It's at the core, a problem on measure theory.

Let \mathcal{L}^2(\mathbb{R}) denote the space of square-integrable functions f on R (with respect to the Lebesgue sigma-algebra and Lebesgue measure \lambda) and let \mathcal{L}^2_{\lambda}(\mathbb{R}) 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

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

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

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

||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

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

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

Show that \mathbf{g}_k has a limit in \mathcal{L}^2_{\lambda}(\mathbb{R}).

 

[morphism]

Is L a fixed functional? If so, I don't think this is true. Try constructing a counterexample using the zero functional.

 

[quasar987]

Apologies!

There is an additional hypothese! L is a non-identically vanishing functional!

 

[morphism]

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||_2^2. (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]

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]

Did you manage to do it without Riesz?

 

[quasar987]

Yes, with the help of aforementioned person. :)

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

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

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

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

L(\mathbf{g}_k+\mathbf{g}_{k+p})&gt;2(||L||-\epsilon&#039;)

Combine with equation (*) to obtain an inequality involving ||\mathbf{g}_k+\mathbf{g}_{k+p}||_2 and \epsilon&#039;. Show that to any \epsilon&gt;0, you can find an \epsilon&#039;(\epsilon) such that 4-||\mathbf{g}_k+\mathbf{g}_{k+p}||_2^2&lt;\epsilon^2 for k large enough.