I am reading Invariant Subspaces by H. Radjavi and P. Rosenthal. I need help understanding the proof of Theorem $1.9$. I see why $\Phi(x,x)$ is bounded for every $x\in \mathcal H$, but how does this imply $\Phi(x,y)$ is bounded for all $x,y\in \mathcal H$?
Theorem. If $f$ is a bounded, Borel measurable function on $\Bbb C$ and $E$ is a spectral measure in $\mathcal H$, then there exists a unique $A\in \mathcal B(\mathcal H)$ such that $\langle Ax,y\rangle = \int f(\lambda) d\langle E_\lambda x,y\rangle$ for all $x,y\in \mathcal H$.
Proof. Define $\Phi$ by $\Phi(x,y) = \int f(\lambda) d\langle E_\lambda x,y\rangle$ for each fixed $x,y$. Clearly $\Phi$ is linear in $x$, and conjugate linear in $y$. Moreover, $|\Phi(x,x)| \le \|f\|_\infty\|x\|^2$. Thus, $\Phi(x,y)$ is a bounded bilinear functional, and therefore there is a unique $A$ such that $\langle Ax,y\rangle = \Phi(x,y)$ for all $x,y$.
I thought of using $$\langle E_\lambda (x+iy), x+iy\rangle = \langle E_\lambda x,x\rangle + \langle E_\lambda y,y\rangle - i \langle E_\lambda x,y\rangle + i \overline{\langle E_\lambda x,y\rangle}$$ for all $x,y\in \mathcal H$, but I'm not sure how to proceed. Thanks!
Image for Reference:
Definition of Spectral Measure:
Note that $$\nu(\triangle) := \langle E(\triangle)x,y\rangle$$ defines a complex valued measure $\nu$, and the notation $d\langle E_\lambda x,y\rangle$ above is the same as $d\nu$.
If $f$ is not positive, then $\Phi$ will in general not be postive semi-definite, so that the Cauchy-Schwarz argument from the comments does not apply directly.
But one does not need positivity here. If $\Phi$ is a sesquilinear form on a complex vector space, then $$ \Phi(x,y)=\sum_{k=0}^3 i^k \Phi(x+i^ky,x+i^ky). $$ This is known as polarization identity and can be checked by direct, albeit tedious calculation.
If the underlying vector space is an inner product space and there exists a constant $C>0$ such that $|\Phi(x,x)|\leq C\|x\|^2$, then $$ |\Phi(x,y)|\leq\sum_{k=0}^3|\Phi(x+i^ky,x+i^ky)|\leq C\sum_{k=0}^3\|x+i^ky\|^2\leq 8C(\|x\|^2+\|y\|^2). $$ This implies that $\Phi$ is continuous at $(0,0)$, and you can use sesquilinearity to show that it is continuous at arbitrary $(x,y)$.