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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 $$ sigma(T)=lambdain mathbbK : T-lambda I textis invertible. $$

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 $$ (T^*-lambda I)^-1=[(T-lambda I)^*]^-1implies T^*-lambda I textis invertible.
$$
So $lambdanotin sigma(T^*).$

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

Thanks.

$T-lambda I$ 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 $A^-1A=I$.

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 $operatornameran T$ is closed. indeed, let $W$ be the inverse of $T^*$, and let $Tx_n$ be a Cauchy sequence. Then
  beginalign
  |x_n-x_m|
  &=supf\ \
  &=sup: fin X^*, \ \
  &=sup(Wf),(Tx_n-Tx_m)\ \
  &leq|Tx_n-Tx_m|,sup\ \
  &=|W|,|Tx_n-Tx_m|.
  endalign
  So $x_n$ is Cauchy; there exists $xin X$ with $x=lim x_n$. As $T$ is bounded, $Tx=lim Tx_n$, and $operatornameran T$ is closed.




• $T$ is injective. Indeed, if $Tx=0$, then for any $fin X^*$ 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 $fin X^*$, and so $x=0$.




• $T$ is surjective. Indeed, if $yin Xsetminus operatornameran T$, using Hahn-Banach (and the fact that $operatornameran T$ is closed) there exists $gin X^*$ 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 $X=$operatornameran T$, and $T$ is surjetive.




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