Let $\{e_n\}$ be a complete orthonormal sequence in an Hilbert space $H$ and let $\{f_n\}$ be an arbitrary sequence of elements in $H$ s.t $$\sum_{n=1}^\infty\|f_n-e_n\|^2<1$$Show that $\{f_n\}$is a complete sequence

First I thought simplifying the last inequality:$$1>\sum_{n=1}^\infty\|f_n-e_n\|^2=\sum_{n,k=1}^\infty|(e_n-f_n,e_k)|^2=\sum_{n,k=1}^\infty|\delta_{n,k}-(f_n,e_k)|^2$$but I was stuck there.

Incorrect Solution: Another option I thought about was using parsevall's generalized identity on $f_n$ means: $$0=(f_n,f)=\sum(f_n,e_n)\overline{(f,e_n)}$$ and since the inner product is non-negative, if $(f,e_n)=0$ then $f\equiv 0$ and we are done and otherwise $(f_n,e_n)=0$ which means that $\forall n\in\mathbb N,f_n=0$ and then by definition of inner product $\{0\}$ is indeed a complete sequence.

What's incorrect in the 'incorrect solution' and how can I use the fact about the inequality?

  • $\begingroup$ Can you give a definition of complete sequence? Do you mean $f_{n}$ is a Cauchy sequence and its limit is in the sequence? Or this en.wikipedia.org/wiki/Complete_sequence? $\endgroup$ – Bombyx mori Oct 11 '14 at 3:58
  • $\begingroup$ The incorrect solution is incorrect because the scalar product is not nonnegative in general (it takes complex values). (You didn't use the assumption, so this cannot possibly be right.) $\endgroup$ – user138530 Oct 11 '14 at 4:10
  • $\begingroup$ That's right. bad habits from $\mathbb{R}$ I guess. thanks again. $\endgroup$ – user65985 Oct 11 '14 at 4:30

I take it by "complete" you mean that the $f_n$ have dense linear span or, equivalently, they have zero orthogonal complement. This we can see as follows. Suppose that $\langle g, f_n\rangle =0$ for all $n$. Then $$ |\langle g, e_n\rangle | =|\langle g, e_n-f_n\rangle | \le \|g\|\, \|e_n-f_n\| . $$ Take squares and sum over $n$. This yields $\|g\|^2 \le S \|g\|^2$ where $S<1$ is the sum from your assumption. Thus $g=0$, as desired.

  • $\begingroup$ why $|<g,e_n>|=|<g,e_n-f_n>|$? $\endgroup$ – user65985 Oct 11 '14 at 4:05
  • $\begingroup$ I assumed that $\langle g, f_n\rangle=0$. $\endgroup$ – user138530 Oct 11 '14 at 4:05
  • $\begingroup$ thanks. So what was incorrect with my answer which didn't use the $S$? $\endgroup$ – user65985 Oct 11 '14 at 4:07
  • $\begingroup$ @CoarguAliquis: I think the inner product at here is the Hermitian inner product. $\endgroup$ – Bombyx mori Oct 11 '14 at 4:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy