0
$\begingroup$

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 $\check{x}$, $\check{x}(\alpha)=(x,u_{\alpha })$.

enter image description here

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Minor point: The basis $u_\alpha$ might be uncountable. $\endgroup$ – copper.hat Nov 19 '16 at 23:16
0
$\begingroup$

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.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.