Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $g \in H$?

share|cite|improve this question
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
up vote 1 down vote accepted

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.

share|cite|improve this answer

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 $f_k$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.