Let $H$ be a complex, separable Hilbert space, and $T:H \rightarrow H$ a linear, bounded operator. Assume that $$\sigma(T) = \sigma(T^*) = \{ \lambda \in \mathbb{C}: a \leq |\lambda| \leq b \}$$ for $0< a < b$. Now assume that for $a < |\lambda| < b$ we have that $\lambda$ is an eigenvalue for $T^*$ but $T - \lambda I$ is bounded below.

I'm studying a proof that assures that in this case, $(T - \lambda I)^*$ has infinite dimensional kernel. Why? I'm trying to prove that if an operator is bounded below then its adjoint fulfills this property, but I'm stuck trying it.

My first attempt is note that $\ker ( (T - \lambda I)^*) = \overline{ Im(T - \lambda I)}^\perp$ and try to compute the codimension of $Im(T- \lambda I)$, where the fact that $\lambda$ is an eigenvalue may help.

Anyone can help me? Thank you very much.

  • $\begingroup$ Please define what you mean by "bounded below". I have a suspicion that your usage deviates from the usual meaning of that term. $\endgroup$ – MaoWao Mar 21 at 13:17
  • $\begingroup$ I mean that there exists $C > 0$ such that $\| (T- \lambda I)x \| \geq C \| x \|$ for all $x \in H$. $\endgroup$ – Javier González Mar 21 at 13:21
  • $\begingroup$ You should include this in your question (that's not what "bounded below" usually means for operators). $\endgroup$ – MaoWao Mar 21 at 13:22
  • $\begingroup$ @MaoWao: that's exactly what "bounded below" means. $\endgroup$ – Martin Argerami Mar 21 at 19:29
  • $\begingroup$ @MartinArgerami Maybe "usually" was wrong, but in my community one calls an operator bounded below if it's self-adjoint and the spectrum is bounded below (not necessarily by a positive constant). $\endgroup$ – MaoWao Mar 21 at 22:58

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