# $\langle S\alpha,\alpha\rangle = \langle T\alpha,\alpha\rangle \Longrightarrow S=T$ for unbounded operators

Like showed for example here, here or here, it is well-known that on a complex Hilbert space $$H$$ (or inner product space), basically by polarisation, for any bounded linear operator $$T : H \to H$$, we have $$\begin{equation} \forall v \in H : \quad \langle Tv,v\rangle = 0 \quad\Longrightarrow\quad T = 0. \end{equation}$$ Applying this equation to the difference, an easy consequence this is that if $$S : H \to H$$ is another bounded linear operator on $$H$$, then $$\begin{equation} \forall v \in H : \quad \langle Sv,v\rangle = \langle Tv,v\rangle \quad\Longrightarrow\quad S = T. \end{equation}$$

Does the result also hold true

• if the dimension of $$H$$ is not finite?
• for possibly unbounded operators or do we must assume that the operator is self-adjoint, normal?

What happens if the operator is not self-adjoint nor even symmetric? For example consider the difference of two covariant derivatives, which is not symmetric in general as $$T$$, and the Hilbert space of equivalence classes of $$L^2$$-Borel $$p$$-forms on a Riemannian manifold.

• Remember that unbounded operators do not usually have domain the entire space. So be careful with the domains, it is possible you have $\langle Tx,x\rangle = \langle Sx,x\rangle$ for all $x\in D(T)\cap D(S)$ but $S\neq T$ (indeed you can have $D(T)\cap D(S) = \{0\}$ without any problem). Before doing any calculation you should demand $D(T) = D(S)$. Oct 25, 2020 at 14:20

By linearity we have $$\forall v\in H:\;\langle (S-T) v, v\rangle = 0.$$ We set $$R:=S-T$$, which is an (possibly unbounded operator $$R:H\to H$$. Note that we have $$\langle Rv,v\rangle =0$$ for all $$v\in H$$.
Since $$\langle R(x+y),x+y\rangle =0$$ that implies : $$\begin{eqnarray} \langle R(x+y),x+y\rangle &=&\langle Rx+Ry,x+y\rangle \\ &=&\langle Rx,x+y\rangle+\langle Ry,x+y\rangle\\ &=& \langle Rx,x\rangle+\langle Rx,y\rangle+\langle Ry,x\rangle+\langle Ry,y\rangle.\\ \end{eqnarray}$$ Then $$\langle R x,y\rangle +\langle Ry,x\rangle=0 \qquad (1)$$ so we replace $$y$$ by $$iy$$ in the last equality we get : $$-i\langle R x,y\rangle +i\langle Ry,x\rangle=0 \qquad (2)$$ multiplying $$(2)$$ by $$i$$ and add to $$(1)$$ we get $$\langle Rx,y\rangle=0 \qquad \forall x,y\in H$$ then we put $$y=Rx$$ we get $$\|Rx\|^2=0$$ for all $$x\in H$$ so $$R=0$$.