I got badly stuck on this question.
Let $X$ be a Banach space, $x \in B(X)$, $\|T\| \leq 1$, $x \in X$ not equal to zero. Prove that there exists $z^* \in \ker(I-T^*)$ such that
$$\langle x,z^* \rangle \neq 0$$
I tried using Yosida and rewrite $T(x)$ as $\lim \frac{1}{n} \sum T^j \left(x\right)$, and then I don't know how to proceed.
Can anyone help me on this?
Update. Hi as an update I think I might have made some progress. As the limit of $\frac{1}{n} \sum T^j \left(x\right)$ is the same as its weak limit by Yosida, we can find $y^*$ such that $\langle x,y^*\rangle \neq 0$. Then I let $z^* = \lim T^{*n}(y^*)$. Then I can show
$$ \langle x,z^* \rangle = \left< x,\lim T^{*n} (y^*) \right> = \lim \left< T^n (x),y^* \right> = \langle x,y^* \rangle \neq 0.$$
Then by Yosida again $z^* \in \ker(I-T^*)$.
I am still happy if someone could point out anywhere I can might have made a mistake or confirm if that is correct. Thanks.
By Yosida theorem, If we let $X$ be banach, $T \in B\left(X\right)$, $\|T\| \leq 1$, $x,y \in X$, then the following are equivalent:
- $y\in Ker\left(I-T\right)$ intersect the closure of $conv\left(T^n x, n\in N\right)$
2.$y=\lim \frac{1}{n} \sum_{j=1}^{n} T^j x$