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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?