Bounded orthogonal sequence in unitary space.

Question

Let X be an unitary space. a(n) is bounded orthogonal sequence (infinite). Prove that for every point x in X we have convergence: <x, a(n)> -> 0 when n-> infinity.

Answer 1

For each $i$, define $u_i=a(i)/\|a(i)\|$ when $a(i)\ne0$, or $u_i=0$ otherwise. Let $y_k=\sum_{i=1}^k\langle x,u_i\rangle\, u_i$. One can readily verify that $y_k$ is orthogonal to $x-y_k$, i.e., $\langle x-y_k,\,y_k\rangle=0$. Therefore $\|x\|^2=\|x-y_k\|^2+\|y_k\|^2$ and $$\|x\|^2\ge\|y_k\|^2=\sum_{i=1}^k|\langle x,u_i\rangle|^2$$ for every $k$. Hence $\sum_{i=1}^\infty|\langle x,u_i\rangle|^2$ is a convergent series of real numbers and $\lim_{n\to\infty}\langle x,u_i\rangle=0$.

Finally, by assumption, $\|a(i)\|\le C$ for some positive constant $C$. Therefore $$0\le\lim_{n\to\infty}|\langle x,a(i)\rangle|=\lim_{n\to\infty}\|a(i)\||\langle x,u_i\rangle|\le \lim_{n\to\infty}C|\langle x,u_i\rangle|=0$$ and thus $\lim_{n\to\infty}\langle x,a(i)\rangle=0$.

Answer 2

(I'm a bit rusty on the subject thus I hope it's correct)

Recall that a non-negative series might converge if it is a zero-sequence.

Now pick $x\in X$ so that $\|x\|^2=M<\infty$ and complete the sequence $a(n)$ to a bounded orthogonal basis of your space. Then since it's a basis you can decompose

$$X=\sum_n \frac{1}{\|a(n)\|}\langle X,a(n) \rangle\cdot a(n)$$

so that by orthogonality

$$X\cdot X= \sum_n \langle X,a(n)\rangle^2=M <\infty$$ and thus $\langle X,a(n)\rangle$ must be a zero-sequence.