I have to show that spectrum for a bounded linear operator and its adjoint on a Banach space are the same. Spectrum is defined as $$ \sigma(T)=\{\lambda\in \mathbb{K}\ :\ T-\lambda I \ \text{is invertible}.\} $$

I have to show $\sigma(T)=\sigma(T^*)$. Let $\lambda \notin \sigma(T)$; then $ (T-\lambda I ) $ is invertible and bounded. This implies $(T-\lambda I)^*$ is also invertible, since $$ (T^*-\lambda I)^{-1}=[(T-\lambda I)^*]^{-1}\implies T^*-\lambda I \ \text{is invertible}. $$ So $\lambda\notin \sigma(T^*).$

I am unable to prove the other part. Can anyone help me please?



$T-\lambda I$ 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 $A^{-1}A=I$.

  • $\begingroup$ I've never heard about this theorem. Anyway, what version are you using? In the wikipedia article I think there are more assumptions than that on the OP. $\endgroup$ – Filburt Nov 28 '17 at 19:13
  • $\begingroup$ This has nothing to do with closed range theorem. Modified answer $\endgroup$ – daw Nov 28 '17 at 20:35
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    $\begingroup$ Could you help me to prove that, if the adjoint $T*$ of a operator is invertible, then $T$ is surjective? $\endgroup$ – Filburt Nov 30 '17 at 15:25

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.

  • $\begingroup$ Very nice proof: maybe it is just the first inequality that is a bit unclear, how can we put the norm of W outside, if it is evaluating at one point (x_n - x_m) (instead of the norm of functions themselves). $\endgroup$ – Philimathmuse Mar 31 at 20:10
  • $\begingroup$ You are definitely right: that equation is wrong. I'll see if it can be saved. $\endgroup$ – Martin Argerami Mar 31 at 20:24
  • $\begingroup$ I see, maybe it is easier to first compose T* with W, instead of W with T*, and then view Wf as g, whose norm is bdd above by that of W. All the estimates of your original argument still hold. $\endgroup$ – Philimathmuse Mar 31 at 20:25
  • $\begingroup$ Yes, you are definitely right. I'll edit that into the answer. Thanks for noticing. $\endgroup$ – Martin Argerami Mar 31 at 20:31

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