Inner products, operators, equality

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Discussion Overview

The discussion revolves around the properties of operators in Hilbert spaces, specifically whether the condition (Tf|f) = (Sf|f) for all f in H implies that the operators T and S are equal. Participants explore the implications of this condition and its sufficiency.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant questions whether the condition (Tf|f) = (Sf|f) for all f in H is sufficient to conclude T = S, noting that it only allows for g = f.
  • Another participant points out that the condition implies (T-S)f is orthogonal to f for all f, but argues that this does not necessarily mean T - S is the zero operator, suggesting a counterexample in R^2.
  • A subsequent reply illustrates that T - S could represent a rotation by an angle of π/2, indicating a specific case where the operators are not equal despite the orthogonality condition.
  • Another participant introduces a related concept, stating that if a bounded operator T in a complex inner product space V satisfies (x, Tx) = 0 for all x in V, then T must be the zero operator, expressing a newfound understanding of this result.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether the initial condition is sufficient for T = S, with multiple competing views presented regarding the implications of orthogonality and specific examples.

Contextual Notes

The discussion highlights the nuances of operator equality in Hilbert spaces, with limitations in the assumptions made about the operators and the specific conditions under which the statements hold.

jostpuur
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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.
 
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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.
 
Ok. T-S can be made a rotation by angle pi/2. :redface:
 
I have just learned, that if a bounded operator T in a complex inner product space V satisfies

[tex] (x,Tx)=0\quad\forall x\in V[/tex]

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

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