Proof of a Lemma guaranteeing the existence of the Borel-measurable functional calculus

Question

In my lecture I had the following Lemma, which guarantees the existence of the Borel-measurable functional calculus:

Le $(H,<,>)$ be a complex Hilbert space and let $q:H\rightarrow \mathbb{C}$ be a function s.t.

1) $\vert q(x)\vert\leq C\cdot\Vert x \Vert^2$, for some $C>0$ and all $x\in H$

2) $q(x+y)+q(x-y)=2q(x)+2q(y)$, for all $x,y\in H$

3) $q(\lambda x)=\vert \lambda \vert^2 q(x)$, for all $x\in H$ and $\lambda\in\mathbb{C}$.

Then there exists a unique bounded operator $T:H\rightarrow H$, s.t. $q(x)=<Tx,x>, \forall x\in H$.

Unfortunately I do not understand the proof given in the lecture. First of all we defined $g(x,y):=\frac{1}{4}\left( q(x+y)-q(x-y)+i\cdot q(x+iy)-i\cdot q(x-iy) \right)$. Then we defined $Tx:=\sum_{i\in I}g(x,e_i)e_i$, whereas $(e_i)_{i\in I}$ is an ONB. Now my professor said, that the following equation is satisfied: $g(x,y)=<Tx,y>, \forall x,y\in H$. But why is this the case? How can I calculate this?

I hope, you can help me.

Answer 1

Suppose $y \in H$, then $$ y = \sum_j \langle y,e_j\rangle e_j \qquad\qquad\text{(1)} $$ and so $$ \langle Tx,y\rangle = \sum_i g(x,e_i)\langle e_i, y\rangle $$ $$ = \sum_i \sum_j g(x,e_i) \overline{\langle y,e_j\rangle} \langle e_i, e_j\rangle $$ $$ = \sum_i g(x,e_i)\overline{\langle y,e_i\rangle} $$ Now if you use the continuity of $g(x,\cdot)$, and (1), then you will see that $$ g(x,y) = \sum_i g(x,e_i)\overline{\langle y,e_i\rangle} $$ (You need to check first that $g(x,\alpha y) = \overline{\alpha}g(x,y)$ for any $\alpha \in \mathbb{C}$)

Remark : Perhaps it is better to phrase this lemma in 2 parts :

1. Given a function $q:H\to \mathbb{C}$ such that the conditions in your question hold, then, $g(x,y)$ given by that equation defines a continuous conjugate linear form on $H$
2. Given a continuous conjugate linear form $g$ on $H$, there is a bounded linear operator $T$ on $H$ such that $$ g(x,y) = \langle Tx,y\rangle $$ The second lemma is then proved using the Riesz Representation theorem, and the entire lemma is perhaps a little more transparent (there is no need to refer to the series expansion as used above).