$\rho(T)=\rho(T^*) $ if $T$ is defined between Banach Spaces.

Question

Let $T:E\rightarrow E$ be a continuous linear map between Banach spaces. We define $T^*:E^*\rightarrow E^*$ by $T^*(e^*)(e)=e^*(T(e))$. Under these conditions prove that the resolvent set $\rho(T)=\{\lambda \:| T-\lambda I \: \text{ is invertible}\}$ satisfies $\rho(T)=\rho(T^*)$.

That $\rho(T)\subseteq \rho(T^*)$ is straightforward from the definition.

Indeed, if $T-\lambda I $ is invertible, then in particular it is surjective and this implies that:

$$f_1((T-\lambda)(x))=f_2((T-\lambda)(x))\: \forall x\in E\Rightarrow f_1=f_2$$

But this implies $T^*-\lambda I^*$ is injective. For surjectivity it is enough to consider $g=f\circ (T-\lambda I)^{-1}\in E^*$, because $(T^*-\lambda I^*)(g)=f$.

For the other inclusion I am having trouble. Suppose we have $\lambda \in \rho(T^*)$, then $T^*-\lambda I^*$ is invertible.

Take $x\in \ker(T-\lambda I)$, then $(T-\lambda I)(x)=0$ and so $(T^*-\lambda I^*)(f)(x)=0$ for all $f\in E^*$. But because of surjectivity, this means $f(x)=0$ for all $f\in E^*$ and so by Hahn Banach, $\lVert x \rVert=0$ which implies $x=0$. Hence, $T-\lambda I$ is injective.

So far I haven't been able to prove $T-\lambda I$ is surjective. I wanted to suppose by way of contradiction $E\setminus (T-\lambda I)(E)\not=\emptyset$ and build a nonzero function which is zero in $(T-\lambda I)(E)$, but I cannot use geometric Hahn-Banach, because $(T-\lambda I)(E)$ is not necessarily closed.

Answer 1

Boils down to : $T$ is invertible if and only if $T^{*}$ is. L to R is immediate, so let's sketch R to L.

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1. Assume $T^{*}$ is injective. Let's show that $T$ has dense image.

For let $\psi\in F^{*}$ zero on $\operatorname{Im} T$, that is $\psi(T e)=0$ for all $e \in E$. Then $T^{*}(\psi) e = 0$ for all $e$ and this means $T^(\psi) = 0$. We conclude $\psi =0$. Again, by Hahn-Banach, this means $\operatorname{Im} T$ is dense in $F$.

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2. Assume $T^{*}$ surjective. Let us show that $\|T e\| \ge c \cdot \|e\|$ for some $c>0$.

Indeed we have

$$\|T e\| = \sup_{\psi\ne 0} \frac{|\psi( T e)|}{\|\psi\|}= \frac{|T^*(\psi)e|}{\|\psi\|}\ge c \cdot \frac{|T^*(\psi)e|}{\|T^* \psi\|} $$

Now use $T^{*}$ surjective, so instead of $T^{*} \psi$ can be any non-zero $\phi\in E^*$.

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Combining 1. and 2. we get that $T$ is invertible ( take $Te_n \to f$, then use 2. and get $e_n$ Cauchy, so convergent, since $E$ Banach, $\&$c)

Answer 2

Take $\lambda \in \rho(T^*)$. Then $T^*-\lambda I^*$ is invertible and $T-\lambda I$ is injective by the original post. It suffices to prove that $T-\lambda I$ is surjective to conclude $\lambda \in \rho(T)$. Let us prove surjectivity:

$(T-\lambda I)(E)$ is dense in $E$.

Indeed, if $e\in E\setminus \overline{(T-\lambda I)(E)}$, by geometric Hahn Banach, we have $f\in E^*$ such that $f(e)>0$ and $f|_{(T-\lambda I)(E)}=0$. Then, $(T^*-\lambda I^*)(f)=0$ and this contradicts the injectivity of $T^*-\lambda I^*$.

$(T-\lambda I)(E)$ is closed and hence $T-\lambda I$ is surjective.

$T^*-\lambda I^*$ is open, so $cB_{E^*}\subseteq (T^*-\lambda I^*)(B_{E^*}) $, but this implies:

$$ \lVert (T-\lambda I)(x)\rVert=\sup_{f\in B_E^*}|f\circ (T-\lambda I)(x)|=\sup_{f\in B_{E^*}}|(T-\lambda I)^*(f)(x)|\geq \sup_{g\in c B_{E}^* } |g(x)|=c\lVert x \rVert$$

If $(T-\lambda I)(x_n)\rightarrow y$, then the above inequality shows $x_n$ is Cauchy and so $x_n\rightarrow x$. This in turn implies by continuity that $(T-\lambda I)(x)=y$ and so $(T-\lambda I)(E)$ is closed!