Inner products, operators, equality

[jostpuur]

Is it true, that if in Hilbert space, operators T and S satisfy (Tf|f)=(Sf|f) for all f in H, then T=S?

I think it is clear, that if (Tf|g)=(Sf|g) is true for all f and g in H, then T=S, but I'm not sure if it is sufficient to only allow g=f.

 

[mathwonk]

it implies that for all f, (T-S)f is orthogonal to f, but this does not imply T-S is zero. think of a simple example in R^2.

 

[jostpuur]

Ok. T-S can be made a rotation by angle pi/2.

 

[jostpuur]

I have just learned, that if a bounded operator T in a complex inner product space V satisfies

$$(x,Tx)=0\quad\forall x\in V$$

then T=0. I posted the OP after seeing this being used, but didn't understand what was happening.