Let $X$ be a complex Banach space, $T\in\mathfrak L(X)$ and $\lambda\in\mathbb C$.

Are we able to show $\lambda\in\sigma_p(T)$ (the point spectrum of $T$) if and only if $\overline\lambda\in\sigma_r(T')$ (the residual spectrum of $T':X'\to X'$)?

I'm only able to (partially) prove this claim under the assumption that $X$ is a Hilbert space and $T'$ is replaced by the Hilbert space adjoint $T^\ast:X\to X$:

In that situation, if $S\in\mathfrak L(X)$, then $$\mathcal N(S)={\mathcal R(S^\ast)}^\perp\tag1$$ from which we infer that $S$ is injective if and only if $\mathcal R(S^\ast)$ is dense. Substituting $S=\lambda-T$ and noting that $(\lambda-T)^\ast=\overline\lambda-T^\ast$, it is only left to show that if $\lambda-T$ is not injective (hence $\mathcal R(\overline\lambda-T^\ast)$ is not dense), then $\overline\lambda-T^\ast$ is injective.

How can we show this? And are we able to prove the desired equivalence in the general Banach space case well?


No. If $X_1=X_2$, and $T$ is given by $T(x)=\lambda x$, with $\lambda\in \mathbb C \setminus \mathbb R$, then $$ T'(\phi)=\lambda \phi, \quad\forall \phi\in X_2', $$ so the spectra of both operators coincide with their point spectra and $$ \sigma (T) = \sigma (T')=\{\lambda \} \neq \{\bar \lambda \}. $$

  • $\begingroup$ Thank you for your answer. What about the Hilbert space case? How can we finish the proof? $\endgroup$ – 0xbadf00d Jan 31 at 16:37
  • $\begingroup$ First of all, what do you mean by the spectrum of a linear transformation between different spaces? $\endgroup$ – Ruy Jan 31 at 16:46
  • $\begingroup$ Sorry, I've started with $(1)$, which clearly holds for operators between different Hilbert spaces, but the actual question doesn't make sense, unless these spaces are equal. Fixed that. $\endgroup$ – 0xbadf00d Jan 31 at 16:51
  • $\begingroup$ It is better now, although you have forgotten a few $X_i$ here and there. Regarding your question, have you tried $T =$ identity map? $\endgroup$ – Ruy Jan 31 at 16:54
  • $\begingroup$ Well, it is clear to me that whenever $T$ is normal (which the identity clearly is), then $\lambda\in\sigma_p(A)$ iff $\overline\lambda\in\sigma_p(A^\ast)$. $\endgroup$ – 0xbadf00d Jan 31 at 18:21

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