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Linear forms and complete metric space 
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#1
Nov1107, 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?? 


#2
Nov1107, 03:25 PM

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If L is not invertible what can happen?



#3
Nov1107, 03:27 PM

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edit: removing my too explicit hint.



#4
Nov1107, 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? 


#6
Nov1107, 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 notationwise. 


#7
Nov1107, 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 squareintegrable functions f on R (with respect to the Lebesgue sigmaalgebra 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]. 


#8
Nov1107, 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.



#9
Nov1107, 09:00 PM

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Apologies!
There is an additional hypothese! L is a nonidentically vanishing functional! 


#10
Nov1107, 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. 


#11
Nov1207, 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! 


#12
Nov1307, 08:35 AM

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Did you manage to do it without Riesz?



#13
Nov1307, 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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