Limit of Inner Products in Hilbert Space Let $H$ be an infinite dimensional Hilbert space. Then there exists an orthonormal basis $\{e_{i}\}_{i = 1}^{\infty}$. Suppose we know that $\lim_{k \rightarrow \infty}(f_{k}, e_{j}) = (f, e_{j})$ for each $j$ where $\{f_{k}\}$ are such that $\|f_{k}\| = B$ and $f$ some function in $H$. Why is it that $\lim_{k \rightarrow \infty}(f_{k}, g) = (f, g)$ for every $g \in H$?
 A: Write $g=\sum_j\alpha_je_j$. Fix $\varepsilon>0$. Then there exists $j_0$ such that $\|\sum_{j>j_0}\alpha_je_j\|<\varepsilon$. Note that $g-\sum_{j\leq j_0}\alpha_je_j=\sum_{j>j_0}\alpha_je_j$.
\begin{align}
|(f_k-f,g)|&=|(f_k-f,\sum_{j\leq j_0}\alpha_je_j)+(f_k-f,\sum_{j>j_0}\alpha_je_j)| \\ \ \\
&\leq|(f_k-f,\sum_{j\leq j_0}\alpha_je_j)|+|(f_k-f,\sum_{j>j_0}\alpha_je_j)|\\ \ \\
&\leq|(f_k-f,\sum_{j\leq j_0}\alpha_je_j)|+\|f_k-f\|\,\|\sum_{j>j_0}\alpha_je_j\|\\ \ \\
&\leq|(f_k-f,\sum_{j\leq j_0}\alpha_je_j)|+(B+\|f\|)\,\varepsilon\\ \ \\
&\leq\sum_{j\leq j_0}|\alpha_j|\,|(f_k-f,e_j)|+(B+\|f\|)\,\varepsilon
\end{align}
Taking $\limsup_{k\to\infty}$, we get (notice that the sum on the right is finite)
$$
\limsup_{k\to\infty}|(f_k-f,g)|\leq(B+\|f\|)\,\varepsilon.
$$
As $\varepsilon$ was arbitrary, we conclude that the limit exists and is zero, i.e. 
$$
\lim_{k\to\infty}(f_k-f,g)=0.
$$
By the way, we are only using that $\|f_k\|\leq B$ (not equality). Also there's no requirement that the space be infinite-dimensional or separable, i.e. it works for any Hilbert space. 
A: Suppose $g$ is such that $\lim\limits_{k\rightarrow\infty} (f_k,g)\ne(f,g)$. Then, without loss of generality, we may assume that $\Vert g\Vert=1$ and that there is an $\alpha>0$ such that for some  subsequence of $(f_n)$, say $f_{n_k}$ we have $ (  f_{n_k}-f , g   )\ge \alpha$ for all $k$.  
Let $\epsilon>0$. Write $g=\sum\limits_{i=1}^\infty \alpha_i e_i$ and choose $M$ so large that $g_c=\sum\limits_{i=M+1}^\infty \alpha_i e_i$ has norm less than $\epsilon$.  Then for each $k$
$$\eqalign{
\alpha\le  (\,f_{n_k}-f, g\, ) &= \sum_{i=1}^M\alpha_i  ( f_{n_k}-f, e_i ) + ( f_{n_k}-f , g_c ).\cr
 }
$$
Select $K$ so large that  $\sum\limits_{i=1}^M\alpha_i (f_{n_K}-f,e_i) \le \alpha/2$.
Then we have
$$
\alpha/2\le  ( f_{n_K}-f , g_c).
$$ 
This implies the norm of $f_{n_K}-f$ is at least $\alpha/(2\epsilon)$.
But, as  $\epsilon$ was arbitrary, this contradicts the bound on the norms of the $f_k$.
