Spectrum for a bounded linear operator and its adjoint on a Banach space are same.

Solution 1:

We want to prove that if $X$ is a Banach space and $T^*\in B(X^*)$ is invertible, then $T$ is invertible in $B(X)$.

We go through a few steps.

• Note that $\operatorname{ran} T$ is closed. indeed, let $W$ be the inverse of $T^*$, and let $\{Tx_n\}$ be a Cauchy sequence. Then \begin{align} \|x_n-x_m\| &=\sup\{|f(x_n-x_m)|:\ f\in X^*,\ \|f\|=1\}\\ \ \\ &=\sup\{|T^*Wf\,(x_n-x_m)|:\ f\in X^*,\ \|f\|=1\}\\ \ \\ &=\sup\{|(Wf)\,(Tx_n-Tx_m)|:\ f\in X^*,\ \|f\|=1\}\\ \ \\ &\leq\|Tx_n-Tx_m\|\,\sup\{\|Wf\|:\ f\in X^*,\ \|f\|=1\}\\ \ \\ &=\|W\|\,\|Tx_n-Tx_m\|. \end{align} So $\{x_n\}$ is Cauchy; there exists $x\in X$ with $x=\lim x_n$. As $T$ is bounded, $Tx=\lim Tx_n$, and $\operatorname{ran} T$ is closed.

• $T$ is injective. Indeed, if $Tx=0$, then for any $f\in X^*$ we have $f=T^*g$ (since $T^*$ is surjective). Then $$f(x)=T^*g(x)=g(Tx)=g(0)=0.$$ Thus $f(x)=0$ for all $f\in X^*$, and so $x=0$.

• $T$ is surjective. Indeed, if $y\in X\setminus \operatorname{ran} T$, using Hahn-Banach (and the fact that $\operatorname{ran} T$ is closed) there exists $g\in X^*$ with $g(y)=1$, $g(Tx)=0$ for all $x$. But then $0=g(Tx)=T^*g(x)$ for all $x$, and so $T^*g=0$. As $T^*$ is injective, $g=0$; this is a contradiction. So $X=\operatorname{ran} T$, and $T$ is surjetive.

• Finally, since $T$ is bijective and bounded, by the Inverse Mapping Theorem it is invertible.

Solution 2:

$T-\lambda I$ is invertible if and only if $(T-\lambda I)^*=T^*-\lambda I$ is invertible: Since for every linear operator $A$ invertibility of $A$ and of $A^*$ are equivalent, which follows by taking the adjoints of, e.g., $AA^{-1}=I$ and $A^{-1}A=I$.