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Inner products, operators, equality

  1. Nov 26, 2007 #1
    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. jcsd
  3. Nov 27, 2007 #2


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    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.
  4. Nov 27, 2007 #3
    Ok. T-S can be made a rotation by angle pi/2. :redface:
  5. Feb 3, 2008 #4
    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. :wink: I posted the OP after seeing this being used, but didn't understand what was happening.
    Last edited: Feb 3, 2008
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