# Show that $\lambda\in\sigma_p(T)$ iff $\overline\lambda\in\sigma_r(T')$

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 $$\sigma (T) = \sigma (T')=\{\lambda \} \neq \{\bar \lambda \}.$$
• 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. – 0xbadf00d Jan 31 at 16:51
• It is better now, although you have forgotten a few $X_i$ here and there. Regarding your question, have you tried $T =$ identity map? – Ruy Jan 31 at 16:54
• Well, it is clear to me that whenever $T$ is normal (which the identity clearly is), then $\lambda\in\sigma_p(A)$ iff $\overline\lambda\in\sigma_p(A^\ast)$. – 0xbadf00d Jan 31 at 18:21