Spectrum of an operator is approximate point spectrum plus spectrum of dual operator

Question

I'm trying to show that given an operator $T \in B(X)$ with $X$ Banach we have $$\sigma(T) = \sigma_{ap}(T) \cup \sigma_p(T') $$

Where $T' \in B(X')$ is the dual operator.

I know that $\sigma(T) = \sigma(T')$ and certainly $\sigma_{ap}(T) \subset \sigma(T)$ so we have $\sigma_{ap}(T) \cup \sigma_p(T') \subset \sigma(T)$ and I'm struggling with the other inclusion.

If we write $R_{\lambda} = {\lambda}I - T$ and now we want to show if $\lambda$ is outside of $ \sigma_{ap}(T) \cup \sigma_p(T')$ then $R_{\lambda}$ is invertible. Here is what I have so far:

• $R_{\lambda}$ has closed image. ($\lambda$ is not in the approx. point spectrum of $T$)
• $R_{\lambda}$ is injective - the point spectrum of $T$ is contained in the approximate point spectrum of $T$.
• I'd like to show that $R_{\lambda}$ has dense image. We know that $\ker(R_{\lambda}')$ is empty - so if $g \in Y'$ is such that $g(R_{\lambda})(x) = 0$ then $g$ must be zero. I'd like to maybe use the Hahn-Banach theorem on an element in $X$ not in the closure of $\operatorname{im} R_{\lambda}$ or perhaps by extending a functional on $\operatorname{im}R_{\lambda}$ but unfortunately I can't seem to get anywhere with this!

Thanks for any help!

Answer 1

CASE 1: If $\mathcal{R}(T-\lambda I)$ is not dense, then $\lambda\in\sigma_{p}(T')$.

CASE 2: $\mathcal{R}(T-\lambda I)=X$.

2a. $T-\lambda I$ is injective, which forces $(T-\lambda I)^{-1}$ to be continuous and, hence, $\lambda\in\rho(T)$; or

2b. $T-\lambda I$ is not injective, which gives $\lambda \in\sigma_{p}(T)$.

CASE 3: $\mathcal{R}(T-\lambda I)$ is dense but $\mathcal{R}(T-\lambda I)\ne X$.

3a. $T-\lambda I$ is not injective and, hence, $\lambda\in\sigma_{p}(T)$.

3b. $T-\lambda I$ is injective. In this case $T-\lambda I$ is not continuously invertible because $\mathcal{R}(T-\lambda I)$ is not dense, which means $\lambda\in\sigma_{ap}(T)$.

That covers all cases: $\sigma(T)\subseteq\sigma_{p}(T')\cup\sigma_{p}(T)\cup\sigma_{ap}(T)$.

Answer 2

By definition $\sigma_{A P}(T) \subset \sigma(T)$, so it is enough to prove

Claim 1: $\sigma_P\left(T'\right) \subset \sigma(T)$
and
Claim 2: $\sigma(T) \setminus \sigma_{A P}(T) \subset \sigma_P\left(T'\right)$

Proof of Claim 1: Let $\lambda \in \sigma_P\left(T'\right)$. Then there exists $f \in X^*$ with $f \neq 0$ so that $T' f=\lambda f$, i.e. so that for every $x \in X$ $$ 0=\left(T' f-\lambda f\right)(x)=f(T x)-\lambda f(x)=f(T x-\lambda x) . $$

Hence the restriction $\left.f\right|_Y$ of $f$ to the image $Y=(T-\lambda \mathrm{I}) X$ of $T-\lambda \mathrm{I}$ is zero, so as $f\ne0$, we must have that $\bar{Y} \neq X$. Thus the image of $T-\lambda \mathrm{I}$ is not dense in $X$ so $\lambda \in \sigma(T)$.

Proof of Claim 2: Let $\lambda \in \sigma(T) \setminus \sigma_{A P}(T)$. Then as $\lambda$ is not an approximate eigenvalue of $T$ we know that $T-\lambda\mathrm{I}$ is bounded below, by the Lemma the image $Y=(T-\lambda \mathrm{I})(X)$ is closed. At the same time $Y$ cannot be all of $X$ as otherwise $T-\lambda \mathrm{I}$ would have a bounded inverse, so $Y$ is a proper closed subspace of $X$. We can thus apply Hahn-Banach, to conclude that there exists some $f \in X^*$ with $f \neq 0$, so that $\left.f\right|_Y=0$. This implies that $T'(f)=\lambda f$ and thus that $\lambda \in \sigma_P\left(T'\right)$ since for every $x \in X$ we have $T x-\lambda x \in Y$ and thus $\left(T'(f)-\lambda f\right)(x)=f(T x-\lambda x)=0$.$\quad\square$

Lemma. Let $X$ be a Banach space. If $S\in B(X)$ is bounded below then $SX ⊂ X$ is closed.
Proof:(When is the image of a linear operator closed?)
Given any sequence $y_n$ in $S X$ which converges $y_n \rightarrow y$ to some $y \in X$, we let $x_n \in X$ be so that $S x_n=y_n$. We then note that as $\left(y_n\right)$ is a Cauchy-sequence, the assumption ($\exists \delta>0$ so that for all $x \in X$ we have $\|S x\| \geq \delta\|x\|$.) implies that $$ \left\|x_n-x_m\right\| \leq \delta^{-1}\left\|S\left(x_n-x_m\right)\right\|=\delta^{-1}\left\|y_n-y_m\right\| \underset{n, m \rightarrow \infty}{\longrightarrow} 0, $$ i.e. that also $\left(x_n\right)$ is Cauchy and thus, as $X$ is complete, that $x_n \rightarrow x$ for some $x \in X$. As $S$ is continuous we thus get that $y=\lim y_n=\lim S x_n=S x \in S X$.