0

Problem:

Let $H$ be a separable Hilbert space and {$e_n$} a complete orthonormal system of $H$. Prove that, if {$y_k$} is a bounded sequence in $H$, the condition $\lim_{k→∞} (e_n , y_k ) = 0$ for every $n$ implies $\lim_{k→∞} (x, y_k ) = 0$ for every $x \in H$. Provide an example that this is false if {$y_k $} is not bounded.

Any suggestion (above all for the example)?

user250656
  • 13

2 Answers2

0

I'll give you one example: you can take $$ y_k=k\,e_k. $$ Then $(e_n,y_k)=0$ for all $k>n$. And if $x=\sum_n\frac1n \,e_n$, then $$ (x,y_k)=1,\ \ \text{ for all }k. $$

Martin Argerami
  • 205,756
0

For the other part: $x=\sum a_ne_n$ and thus $<x,y_k>=<\sum a_ne_n,y_k>=\sum a_n<e_n,y_k>$

Haha
  • 5,648
  • yes and I think that the limit for $k->\inf$ can pass through the sum because, being $y_k$ bounded, by the completeness of the system $e_n$, $ \sum_n (e_n,y_k)^2< \inf$... – user250656 Jun 25 '15 at 23:37