Inner products, operators, equality

1. Nov 26, 2007

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.

2. Nov 27, 2007

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.

3. Nov 27, 2007

jostpuur

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

4. Feb 3, 2008

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.

Last edited: Feb 3, 2008