Show that if $f_{n}$ $\in$ $L^2(a,b)$ and $f_{n} \to f$ in norm, then $(f_{n}, g) \to (f,g)$ for all $g \in L^2(a,b)$. Hint: Apply the Cauchy-Schwarz inequality to $(f_{n}-f,g)$.
I've tried taking the dot product of $(f_{n}-f,g)$ then distributing out to get $(f_{n},g) - (f,g)$, but then when trying to apply the Cauchy Schwarz inequality I get stuck. Once I apply the Cauchy Schwarz inequality I get
|$(f_{n}-f,g)$| $\leq$ $\parallel f_{n}-f$$\parallel$ $\parallel g\parallel$
Since $f_{n} \to f$ in norm that means that the norm of $f_{n}$ and $f$ will be equal to $0$, hence, we get $0$ in the right side. Thus
|$(f_{n}-f,g)$| = $0$ $\implies$|$(f_{n},g) - (f,g)$|=$0$
Nonetheless, shouldn't I be trying to proof that this will equal 0, since by definition $f_{n} \to f$ in norm implies that the distance between $f_{n} \to f$ is zero?
Suppose $f_n \rightarrow 0$ in norm. You want to show that $\langle f_n,g\rangle \rightarrow \langle f,g\rangle$ for every fixed $g$. To show this, let $\epsilon > 0$ be given, and let $g\in L^2$ be fixed. Then there exists $N$ such that $$ \|f_n-f\| < \frac{\epsilon}{\|g\|+1},\;\;\; \mbox{whenever } n \ge N. $$
Then
\begin{align} |\langle f_n,g\rangle-\langle f,g\rangle|&=|\langle f_n-f,g\rangle| \\ &\le \|f_n-f\|\|g\| \\ &\lt \epsilon\frac{\|g\|}{\|g\|+1} < \epsilon,\;\;\mbox{whenever } n\ge N. \end{align} By the definition of limit, that means $\langle f_n,g\rangle\rightarrow \langle f,g\rangle$ as $n\rightarrow\infty$ for every fixed $g\in L^2$.