Convergence with L2 norm functions

tom_rylex
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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 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 &lt;f_n - f,g&gt; \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 &lt;f_n - f,g&gt; \vert \rightarrow 0
Since this is absolutely convergent, that means that
&lt;f_n,g&gt; - &lt;f,g&gt; is also convergent to 0. So therefore,
&lt;f_n,g&gt; -&lt;f,g&gt; \rightarrow 0
&lt;f_n,g&gt; \rightarrow &lt;f,g&gt;

Is that reasonable? Or am I missing something?
 
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I suggest going from <fn - f , g> ---> 0 to <fn , g> - <f , g> ---> 0 more explicitly.
 
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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