An orthonormal sequence in an inner product space converges weakly to $0$.

Question

It is a well-known theorem that "An orthonormal sequence in a Hilbert space converges weakly to $0$". The proof uses the Bessel's inequality and the Rise representation theorem for which completeness of the space is a necessary condition.

However, I somehow proved the same theorem for any inner product space. If my proof were correct, then textbooks would use that stronger result instead of the current one.

Theorem: Let $E$ be an inner product space, $\{ x_n\}$ an orthonormal sequence in an inner product space, and . Then $\{ f(x_n)\}$ converges to $0$ for any $f\in E'$.

Proof: Let $f\in E'$. Assume to the contrary that $\{f(x_n)\}$ does not go to $0$. Then there is some $\epsilon >0$ and a subsequence $\{ x_{k_n}\}$ such that $|f(x_{k_n})|\geq \epsilon $. But then infinitely many of $f(x_{k_n})$'s are positive or they are negative. So without loss of generality, assume that $f(x_{k_n})$'s are all positive. Then $n\cdot \epsilon \leq f(x_{k_1})+...+f(x_{k_n})=| f(x_{k_1})+...+f(x_{k_n})|=|f(x_{k_1}+...+x_{k_n})|\leq ||f||\cdot ||x_{k_1}+...+x_{k_n}||=||f||\cdot \sqrt{n}$ where the last equality follows from the Pythagorean theorem. But we then obtain $||f||\geq \epsilon \cdot \sqrt{n}$, which contradicts with boundedness of $f$. So $f(x_n)\rightarrow 0$. $\square$

I couldn't see the issue here.

Answer 1

Your proof is corrrect. You can even generalize it to the complex case by adding a factor $\epsilon_{k_i}$ of modulus $1$ to $x_{k_i}$ so that $f(\epsilon_{k_i}x_{k_i}) > 0$. There’s just not much point, since you could just complete the inner product space to a Hilbert space, so the result for Hilbert spaces implies the result for general inner product spaces anyway.

Answer 2

It is possible to show, basing just on the Pythagorean theorem, that $f(e_n)$ is square summable, hence it tends to $0.$ Indeed, we have $$\left |\sum_{k=1}^nx_kf(e_k)\right |^2= \left |f\left (\sum_{k=1}^nx_ke_k\right )\right |^2\\ \le \|f\|^2\sum_{k=1}^n|x_k|^2$$ Let $x_k=\overline{f(e_k)}.$ Then $$\left (\sum_{k=1}^n|f(e_k)|^2\right )^2\le \|f\|^2\sum_{k=1}^n|f(e_k)|^2$$ Hence $$\sum_{k=1}^n|f(e_k)|^2\le \|f\|^2$$ As $n$ is arbitrary we get $$\sum_{k=1}^\infty |f(e_k)|^2\le \|f\|^2$$