Linear forms and complete metric space

In summary, the person said that he was able to solve the problem using the parallelogram identity and the Riesz representation theorem.
  • #1
quasar987
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[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,

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

Now what??
 
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  • #2
If L is not invertible what can happen?
 
  • #3
edit: removing my too explicit hint. :-p
 
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  • #4
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?
 
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  • #5
Too late morphism! :D
 
  • #6
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.
 
  • #7
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].
 
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  • #8
Is L a fixed functional? If so, I don't think this is true. Try constructing a counterexample using the zero functional.
 
  • #9
Apologies!

There is an additional hypothese! L is a non-identically vanishing functional!
 
  • #10
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.
 
  • #11
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!
 
  • #12
Did you manage to do it without Riesz?
 
  • #13
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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FAQ: Linear forms and complete metric space

What is a linear form?

A linear form is a mathematical function that maps a vector space onto its underlying field of scalars. It is a linear mapping from the vector space to the field, meaning that it preserves addition and scalar multiplication.

How is a linear form different from a vector?

A vector is a mathematical object that has both magnitude and direction, while a linear form is a function that operates on vectors to produce a scalar value. Vectors can be thought of as the inputs to a linear form, while the scalar value is the output.

What is a complete metric space?

A complete metric space is a mathematical concept that describes a set of points where every Cauchy sequence converges to a point within that set. In simpler terms, it is a set of points where there are no "missing" points and every possible limit point is included.

How are linear forms used in functional analysis?

In functional analysis, linear forms are used to study the properties of vector spaces and operators. They are essential in understanding the behavior of linear operators, which are important tools in the study of continuous functions and transformations in functional analysis.

Can linear forms be used in other fields of science?

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