# Convergence with L2 norm functions

## 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 $$f_n \in L^2(a,b)$$ and $$f_n \rightarrow f$$ in norm, then $$<f_n,g> \rightarrow <f,g>$$ for all $$g \in L^2(a,b)$$

## Homework Equations

$$L^2(a,b)$$ is the space of square-integrable functions,
$${f_n}$$ 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:
$$\vert <f_n - f,g> \vert \leq \Vert f_n - f \Vert \Vert g \Vert$$
By the definition of norm convergence, I have
$$\Vert f_n - f \Vert \rightarrow 0$$, which means that
$$\vert <f_n - f,g> \vert \rightarrow 0$$
Since this is absolutely convergent, that means that
$$<f_n,g> - <f,g>$$ is also convergent to 0. So therefore,
$$<f_n,g> -<f,g> \rightarrow 0$$
$$<f_n,g> \rightarrow <f,g>$$

Is that reasonable? Or am I missing something?

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