# About a theorem giving equivalent properties for orthonormal sets in Hilbert space

Good day, I know that: $\left \{ u_{\alpha } \right \}$, ${\alpha }$ in a set A, is orthonormal if $(u_{\alpha },u_{\beta })=\left\{\begin{matrix} 1 \ \alpha =\beta \\ 0 \ \alpha \neq \beta \end{matrix}\right.$

Therefore, to each $x \in H$, it is possible to associate a complex function $\check{x}$, $\check{x}(\alpha)=(x,u_{\alpha })$. My question is: How to obtain the implication III to IV? I understand that one must conclude $4(x,y)=4\sum \widehat{x}(\alpha )\overline{\widehat{y(\alpha )}}=4\sum (x,u_{\alpha })\overline{(y,u_{\alpha })}$. By polarization, I obtain $4(x,y)=4(\sum \left | (x+y,u_{\alpha }) \right |^{2}-\left | (x-y,u_{\alpha }) \right |^{2})+4i\sum (\left | (x+iy,u_{\alpha }) \right |^{2}-\left | (x-iy,u_{\alpha }) \right |^{2})$. Please how I can conclude the demonstration?

A quick verification: Given $x \in H$, we know that $\|x\|^2=\sum_{n=1}^{\infty} |(x,e_n)|^2$. Hence, let $S_n$ be the sequence of partial sums for $\sum_{n=1}^{\infty} (x,e_n) e_n$. Then, applying the pythagorean theorem, we get that $$\|x-S_n\|^2=\|x\|^2-2(x,S_n)+\|S_n\|^2=\|x\|-2(x,S_n)+\sum_{n=1}^{N} |(x,e_n)|^2=\|x\|-(x,S_n),$$ which tends to $0$ by assumption, implying that $x=\sum_{n=1}^{\infty} (x,e_n) e_n$.
Proof of (iv): For ease of notation, let $x_n :=\langle x,e_n \rangle e_n$ and $y_n:=\langle y,e_n\rangle e_n$ respectively. using the continuity of the innter product, we see that: \begin{align*} (x,y) &=\langle\sum_{n=1}^{\infty} x_n e_n, \sum_{n=1}^{\infty} y_n e_n\rangle\\ &=\sum_{n=1}^{\infty}\langle\sum_{n=1}^{\infty} x_n e_n, y_n e_n\rangle\\ &= \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \langle x_n e_n, y_k, e_k\rangle. \end{align*} Noting that all of the summands vanish for $e_n \neq e_k$ by the assumption of orthoganality, we know that since $(e_i,e_i)=1$, this in turn implies that \begin{align*} \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \langle x_n e_n, y_k, e_k\rangle=\sum_{i=1}^{\infty} \langle x_i e_i, y_i e_i\rangle=x_i \overline{y_i} \langle e_i,e_i \rangle=\sum_{i=1}^{\infty} \langle x,e_i \rangle \overline{\langle y,e_i\rangle}, \end{align*} proving the result.
• Minor point: The basis $u_\alpha$ might be uncountable. – copper.hat Nov 19 '16 at 23:16
Define $\phi:H \to l_2(A)$ by $\phi(x)(\alpha) = \langle x , u_\alpha \rangle$, note that $\phi$ is linear and (iii) shows that $\phi$ is an isometry.
Note that $H, l_2(A)$ have inner products that are compatible with the respective norms, hence the inner product is given, in $H,l_2(A)$ respectively by the polarisation identity (again, with the respective norms).
In particular, \begin{eqnarray} \langle {x}, {y} \rangle_H &=& {1 \over 4} \sum_{k=0}^3 (-1)^k \|x + i^ky \|^2_H \\ &=& {1 \over 4} \sum_{k=0}^3 (-1)^k \|\phi(x + i^ky) \|_{l_2(A)} \\ &=& {1 \over 4} \sum_{k=0}^3 (-1)^k \|\phi(x) + i^k\phi(y) \|_{l_2(A)} \\ &=&\langle \phi(x), \phi(y) \rangle_{l_2(A)} \end{eqnarray} Since $\langle \tilde{x}, \tilde{y} \rangle_{l_2(A)} = \sum_\alpha \tilde{x}(\alpha) \overline{ \tilde{y}(\alpha) }$, we have the desired result.