Proving Operator Equaility with Polarization Identity

Question

Given linear operators $A,B$ on a Hilbert Space $H$ with $ \langle x,Ax \rangle = \langle x,Bx \rangle $ for all $x$ in $D(A)=D(B)$.

My script says that via the polaritazion identity this implies $A=B$. How would one use the polarization identity to show this result?

Answer 1

I guess you need the domains to be dense. Then for any $y\in D(A)$ $$ \langle y,Ax\rangle=\frac14\,\sum_{k=0}^3i^k\langle y+i^kx,A(y+i^k x)\rangle =\frac14\,\sum_{k=0}^3i^k\langle y+i^kx,B(y+i^k x)\rangle=\langle y,Bx\rangle. $$ As $D(A)=D(B)$ is dense, you get that $\langle y,Ax\rangle=\langle y,Bx\rangle$ for all $y$, which implies $Ax=Bx$.

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So how do you come to the form of the polarization identity you use?

\begin{align} \sum_{k=0}^3i^k\langle y+i ^kx,A( y+i ^kx)\rangle &=\langle y+x,A(y+x)\rangle +i \langle y+i x,A(y+ix)\rangle\\[0.2cm] &\qquad-\langle y-x,A(y-x)\rangle -i \langle y-i x,A(y-ix)\rangle\\[0.2cm] &=\langle y,Ay\rangle+\langle x,Ax\rangle+\langle y,Ax\rangle +\langle x,Ay\rangle\\[0.2cm] &\qquad+i\langle y,Ay\rangle+i\langle x,Ax\rangle+\langle y,Ax\rangle -\langle x,Ay\rangle\\[0.2cm] &\qquad-\langle y,Ay\rangle-\langle x,Ax\rangle+\langle y,Ax\rangle +\langle x,Ay\rangle\\[0.2cm] &\qquad-i\langle y,Ay\rangle-i\langle x,Ax\rangle+\langle y,Ax\rangle -\langle x,Ay\rangle\\[0.2cm] &=4\langle y,Ax \rangle. \end{align}