# 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 $g \in H$?

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Fix an $\epsilon$. Finite sums of basis vectors can approximate $g$ to any precision, that is, there exists $\{g_{n}\}_{n = 1}^{\infty}$ and an $N$ such that $\|g_{N} - g\| < \epsilon$. Then $\lim_{k \rightarrow \infty}(f_{k}, g_{N}) = (f, g_{N})$. We also have have that $|(f_{k}, g_{N} - g)| \leq \|f_{k}\|\|g_{N} - g\| < B\epsilon$. Therefore $\lim_{k \rightarrow \infty}(f_{k}, g_{N}) = \lim_{k \rightarrow \infty}(f_{k}, g)$. Then $$(f, g_{N}) = \lim_{k \rightarrow \infty}(f_{k}, g_{N}) = \lim_{k \rightarrow \infty}(f_{k}, g).$$ Is this the right track? – QPT Apr 15 '12 at 0:12
@Sam: no, every Hilbert space has an orthonormal basis, whether finite or infinite-dimensional, and whether separable or not. Of course, what is true is that the OP wrote a countable basis, and that is equivalent with the space being separable. – Martin Argerami Apr 15 '12 at 0:23
@QPT: yes, that's what I did in some more detail in my answer. – Martin Argerami Apr 15 '12 at 0:23
@Martin: gah, sorry I omitted the countable detail – Alex R. Apr 15 '12 at 2:02

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 $g-\sum_{j\leq j_0}\alpha_je_j=\sum_{j>j_0}\alpha_je_j$. $$|(f_k-f,g)|=|(f_k-f,\sum_{j\leq j_0}\alpha_je_j)+(f_k-f,\sum_{j>j_0}\alpha_je_j)| \\ \leq|(f_k-f,\sum_{j\leq j_0}\alpha_je_j)|+|(f_k-f,\sum_{j>j_0}\alpha_je_j)| \leq|(f_k-f,\sum_{j\leq j_0}\alpha_je_j)|+\|f_k-f\|\,\|\sum_{j>j_0}\alpha_je_j\| \leq|(f_k-f,\sum_{j\leq j_0}\alpha_je_j)|+(B+\|f\|)\,\varepsilon \leq\sum_{j\leq j_0}|\alpha_j|\,|(f_k-f,e_j)|+(B+\|f\|)\,\varepsilon$$ 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.

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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 $f_k$.

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