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.

  • 1
    $\begingroup$ 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 $D(T) = D(S)$. $\endgroup$
    – s.harp
    Commented Oct 25, 2020 at 14:20

1 Answer 1


I think the claim is still true in complex Hilbert spaces (I am assuming that the operators are still linear, of course, but they do not need to be bounded).

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 $R=0$.

This is mostly copied from an answer from here. As far as I can tell, the boundedness was not used there.


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