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Convergence with L2 norm functions

  • Thread starter tom_rylex
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  • #1
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Homework Statement


(I'm posting this because my proofs seem to be lousy. I want to see if I'm missing anything.)
Show that if [tex] f_n \in L^2(a,b) [/tex] and [tex] f_n \rightarrow f [/tex] in norm, then [tex] <f_n,g> \rightarrow <f,g> [/tex] for all [tex] g \in L^2(a,b) [/tex]

Homework Equations


[tex] L^2(a,b) [/tex] is the space of square-integrable functions,
[tex] {f_n} [/tex] is a finite sequence of piecewise continuous functions, and
< , > is the inner product


The Attempt at a Solution



I started with a linear combination of inner products and applied the Cauchy Schwarz inequality:
[tex] \vert <f_n - f,g> \vert \leq \Vert f_n - f \Vert \Vert g \Vert [/tex]
By the definition of norm convergence, I have
[tex] \Vert f_n - f \Vert \rightarrow 0 [/tex], which means that
[tex] \vert <f_n - f,g> \vert \rightarrow 0 [/tex]
Since this is absolutely convergent, that means that
[tex] <f_n,g> - <f,g> [/tex] is also convergent to 0. So therefore,
[tex] <f_n,g> -<f,g> \rightarrow 0[/tex]
[tex] <f_n,g> \rightarrow <f,g> [/tex]

Is that reasonable? Or am I missing something?
 

Answers and Replies

  • #2
EnumaElish
Science Advisor
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I suggest going from <fn - f , g> ---> 0 to <fn , g> - <f , g> ---> 0 more explicitly.
 
  • #3
13
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As in:
<fn-f,g> --> 0 (absolutely convergent series are convergent)
<fn,g> - <f,g> --> 0 (linearity property wrt the first variable for inner products)
 

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