# Convergence with L2 norm functions

1. Oct 21, 2007

### tom_rylex

1. The problem statement, all variables and given/known data
(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 $$g \in L^2(a,b)$$

2. Relevant 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

3. 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?

2. Oct 21, 2007

### EnumaElish

I suggest going from <fn - f , g> ---> 0 to <fn , g> - <f , g> ---> 0 more explicitly.

3. Oct 22, 2007

### tom_rylex

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)