I have difficulties understanding the reason why when I have a self adjoint linear operator $T : \mathcal{H} \rightarrow \mathcal{H}$, and know that $A\|f\|^2 \leq \langle Tf,f \rangle \leq B\|f\|^2$ with $A,B >0$, I will get a bilinear form $\langle f,g \rangle_T = \langle f,Tg \rangle$ that defines an inner product on $\mathcal{H}$. Furthermore, the inner product will generate a norm $\|f\|_T$ that is equivalent to the original norm on $\mathcal{H}$.

I suspect the reason could be found in functional analysis which I unfortunately have not read. I would appreciate it if someone could explain the reason for this outcome so or at least point me to direction that I should look into so that I can attempt to understand this.

  • $\begingroup$ The $0$ operator is selfadjoint, but it does not generate an inner product. You're missing some assumptions. $\endgroup$ – DisintegratingByParts Jan 9 '16 at 14:54
  • $\begingroup$ I have edited my question. $\endgroup$ – Garcia Jan 9 '16 at 14:59

The fact that $\langle \cdot, \cdot \rangle_T$ is a bilinear form is quite trivial. Then you need to show that it is an inner product, i.e., it meets the following additional requirements:

  1. $\overline{\langle f, g \rangle_T} = \langle g, f \rangle_T$. This holds because $T$ is self-adjoint: $\overline{\langle f, g \rangle_T} = \overline{\langle f, Tg \rangle} = \langle Tg, f \rangle = \langle g, Tf \rangle = \langle g, f \rangle_T$.
  2. $\langle f, f \rangle_T = 0$ implies $f = 0$. This holds because of the hypothesis that $\langle Tf, f \rangle \ge A\|f\|^2$ with $A > 0$.

Finally, the equivalence of norm is just the (square root of) two given inequalities. (We define $\|f\|_T = \sqrt{\langle f, f \rangle_T}$)

  • $\begingroup$ Alright thanks so much, I can understand this. Except how you got from $ \overline \langle f,g \rangle _T = \overline \langle f,Tg \rangle$ $\endgroup$ – Garcia Jan 9 '16 at 15:24
  • $\begingroup$ @Garcia: This was your definition of $\langle\cdot,\cdot\rangle_T$ :) $\endgroup$ – gerw Jan 9 '16 at 18:56
  • $\begingroup$ OK thanks. Silly me. $\endgroup$ – Garcia Jan 10 '16 at 6:56

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