Rudin's RCA Theorem $4.18$.

Question

There is the definition which we need for the proof:

There is the theorem which we need for the $4.18$:

There is $4.18$:

Let {$u_\alpha : \alpha$ $\in$ $A$} be an orthonormal set in $H$. Each of the following four conditions on {$u_\alpha$} implies the other three:

$(1)$ ${u_\alpha}$ is a maximal orthonormal set in $H$.

$(2)$ The set $P$ of all finite linear combinations of members of {$u_\alpha$} is dense in $H$.

$(3)$ The equality

$\sum_{\alpha \in A}$ $|\hat x(\alpha)|^2$ $=$ $||x||^2$

holds for every $x$ $\in$ $H$.

$(4)$ The equality

$\sum_{\alpha \in A}$ $\hat x(\alpha)$ $\overline {\hat y(\alpha)}$ $=$ $(x,y)$

holds for all $x$. $\in$ $H$ and $y$ $\in$ $H$.

We shall prove that $(1)$ $\to$ $(2)$ $\to$ $(3)$ $\to$ $(4)$ $\to$ $(1)$ .

I understand that $(1)$ $\to$ $(2)$.

In the book it's written that if $(2)$ holds, so does $(3)$ by Theorem $4.17$.

I don't understand how does $(2)$ imply $(3)$ by Theorem $4.17$ ?

Also there's written that the implication $(3)$ $\to$ $(4)$ follows from the easily proved Hilbert space identity

$4(x,y)$ $=$ $||x+y||^2$ $-$ $||x-y||^2 $ $+$ $i||x+iy||^2$ $-$ $i||x-iy||^2$.

How does it follow from this identity?

Any help would be appreciated.

Answer 1

The relevant part of 4.17 is the following:

Let $\{ u_\alpha : \alpha \in A \}$ be an orthonormal set and let P be the space of all finite linear combinations of the vectors $u_\alpha$. Then the map $x \to \hat x$ is an isometry of $\bar P$ onto $l^2(A)$.

Let's digest this.

$\hat x$ is a function $A \to \mathbb C$, defined by $\hat x (\alpha) := (x, u_\alpha)$. When we say that the mapping $x \to \hat x$ is an isometry on $\hat P$, what we mean is that for any $x \in \hat P$, $$ \|x\| = \| \hat x \|_{l^2(A)} $$

The thing on the left hand side is the norm of $x$, with $x$ viewed as an element of the Hilbert space $H$. The thing on the right hand side is the $l^2$-norm of $\hat x$, with $\hat x$ viewed as a function $A \to \mathbb C$.

So what is the $l^2$-norm of $\hat x$? It is the following: $$ \| \hat x \|_{l^2(A)} = \left(\int_A |\hat x |^2 d\mu \right)^{\tfrac 1 2} = \left(\sum_{\alpha \in A} |\hat x (\alpha)|^2 \right)^{\tfrac 1 2}.$$ $\mu$ is the counting measure on $A$. The second equality follows from the discussion in your previous question.

Thus $$ \|x\| = \left(\sum_{\alpha \in A} |\hat x (\alpha)|^2 \right)^{\tfrac 1 2}$$ holds for all $x \in \bar P$.

Now, returning to 4.18, we want to prove that (ii) implies (iii). Statement (ii) says that $P$ is dense in $H$, which is the same as saying that the $\bar P$ is the whole of $H$. But then, $$ \|x\| = \left(\sum_{\alpha \in A} |\hat x (\alpha)|^2 \right)^{\tfrac 1 2}$$ holds for all $x \in H$. This is precisely statement (iii).

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Regarding the additional question in your edited post, we're requirement to show that statement (iii) implies statement (iv). The logic goes like this:

\begin{align} 4 (x, y) & = \| x + y\|^2 - \|x - y\|^2 + i \|x + iy\|^2 - i \|x - iy\|^2 \\ & = \sum_{\alpha \in A} |\widehat {(x + y)} (\alpha) |^2 - \sum_{\alpha \in A} |\widehat {(x - y)} (\alpha) |^2 +i \sum_{\alpha \in A} |\widehat {(x + iy)}(\alpha)|^2 -i \sum_{\alpha \in A} |\widehat {(x -i y)} (\alpha)|^2 \\ & = 4 \sum_{\alpha \in A} \hat x (\alpha) \overline {\hat y (\alpha)} \end{align}

• The first equality follows from the identity $4 (x, y) = \| x + y\|^2 - \|x - y\|^2 + i \|x + iy\|^2 - i \|x - iy\|^2 $.
• The second equality follows by applying statement (iii) to $x + y$, $x - y$, $x + iy$ and $x - iy$.
• The third equality follows from the identity $4 (\hat x, \hat y)_{l^2(A)} = \| \hat x + \hat y\|^2_{l^2(A)} - \|\hat x - \hat y\|^2_{l^2(A)} + i \|\hat x + i\hat y\|^2_{l^2(A)} - i \|\hat x - i\hat y\|^2_{l^2(A)} $.