Show that if $||T||$ is an eigenvalue for $T$, it is also an eigenvalue for $T^*$

Question

Suppose $T \in B(X)$ is a bounded linear operator on a Banach space $X$. Define $T^*: X^* \rightarrow X^*$ by $T^* (\phi) = \phi \circ T$.

First, let $\lambda$ be an eigenvalue of $T$. Then for some $x \in X$, $Tx = \lambda x$.

If we then consider the operator $\sum_{n=1}^N T^* \phi$, we see that for $s$, $$\sum_{n=1}^N \left(T^{*n} \phi\right) x = \phi \circ Tx + \phi \circ T^2 x + \cdots + \phi \circ T^N x = ||T|| \phi (x) + ||T||^2 \phi(x) + \cdots + ||T||^n \phi(x) = \sum_{n=1}^N ||T||^n \phi(x).$$

Thus for some $\phi \in X^*$, $\sum_{n=1}^\infty T^{*n} \phi = \sum_{n=1}^\infty ||T||^n \phi$ for all $n$. Therefore, $||T||$ is in the point spectrum of $T^*$.

Is this a sufficient proof?

Answer 1

This is not a sufficient proof. All that you've shown is that $(T^*\phi)(x)=\|T\|\phi(x)$ for all $\phi\in X^*$. But how does this show that $(T^*\phi)(y)=\|T\|\phi(y)$ for all $y\in X$?

To show that $\|T\|$ is an eigenvalue of $T^*$, you have to show that there exists some non-zero $\phi\in X^*$ such that $(T^*\phi)(y)=\|T\|\phi(y)$ for all $y\in X$.