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I have the following theorem from the book "Applied Analaysis" by Hunters.

Suppose that $x_n$ is a sequence in a Hilbert space $H$ and $D$ is a dense subset of $H$. Then, $<x_n,y>$ converges to $<x,y>$ for every $y\in H$ iff $\|x_n\|\leq M$ for some constant $M$ and $<x_n,z>$ converges to $<x,z>$ for $z\in D$.

Then, as an illustration of this theorem, the author says that:

If $\{u_{\alpha}\}$ is an orthonormal basis of a Hilbert space, then a sequence $x_n$ converges weakly to $x$ if and only if it is bounded and its coordinate converges, that is, $<x_n,u_{\alpha}> \to <x,u_{\alpha}>$

But, I do not understand this example because an orthonormal basis is not dense in a Hilbert space. What am I missing?

yumiko
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    The span of the orthonormal basis is dense in the Hilbert space. This implies that even if $u_\alpha$ were an element in the span of the o.n.b, it would hold true, by additivity of inner product. – Sarvesh Ravichandran Iyer Jan 05 '17 at 04:57
  • @астонвіллаолофмэллбэрг Do you mind explaining what you mean in more detail? – yumiko Jan 06 '17 at 05:21

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All right. So, let $g$ be an element in the span of $u_{\alpha}$. Then, we can write $g = \sum c_\alpha u_\alpha$, where $c_\alpha$ are elements from the field over which the Hilbert space is constructed (mostly the real or complex numbers).

Note then, that $$ \langle y,g \rangle = \sum c_\alpha \langle y,u_\alpha \rangle $$ holds true for all elements $y$ of the Hilbert space.

Suppose that for all $\alpha$, $\langle x_n,u_\alpha \rangle \to \langle x,u_\alpha \rangle$ is true. Then: $$ \langle x_n,g \rangle=\sum c_\alpha\langle x_n,u_\alpha \rangle \to \sum c_\alpha\langle x,u_\alpha \rangle = \langle x,g \rangle $$

For all $g$ "in the span of $u_\alpha$", which is dense in the Hilbert space. By the given proposition, since $x_n$ are given bounded, we are done. I hope this explanation is sufficient.

Sarvesh Ravichandran Iyer
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  • Why is it that $$ \langle x_n,g \rangle=\sum c_\alpha\langle x_n,u_\alpha \rangle \to \sum c_\alpha\langle x,u_\alpha \rangle = \langle x,g \rangle $$? This is obvious for countable index, but for arbitrary indexed set $\alpha$, how do you show that is still true? – yumiko Jan 07 '17 at 06:14
  • If $\alpha$ is an arbitrary indexed set, then the convergence notion is of NETS, not of sequences, because there is a possibility of uncountability. So you have to read up a little about nets, what are Cauchy nets, convergent nets etc. (I don't want to state definitions, you can look them up, but they are quite complicated to be honest). Once you get through that, this is an easy exercise. – Sarvesh Ravichandran Iyer Jan 07 '17 at 06:20