How to construct an operator with an infinite spectrum?

Question

Let $X$ be an infinite dimensional Banach space. How to construct a bounded operator $T\in B(X)$ such that $\sigma(T)$ is infinite, that is $ card (\sigma(T))= \infty$ ? Thank you.

Answer 1

Fix $x_1\in X$ with $\|x_1\|=1$. By the Hahn-Banach theorem, there exists $f_1\in X^*$ with $\|f_1\|=1$ and $f_1(x_1)=1$. Now fix $x_2\in\ker(f_1)$ with $\|x_2\|=1$. By the Hahn-Banach theorem, there is some $f_2\in X^*$ with $\|f_2\|=1$, $f_2(x_2)=1$, and $f_2(x_1)=0$.

Proceeding by induction, for each $n$ obtain $x_n\in\bigcap_{k=1}^{n-1}\ker(f_n)$ with $\|x_n\|=1$ and some $f_n\in X^*$ with $\|f_n\|=1$, $f_n(x_n)=1$ and $f_n(x_k)=0$ for $1\leq k<n$.

For each $n$, define $T_n\in B(X)$ by $$T_n(x)=\sum_{k=1}^n2^{-k}f_k(x)x_k.$$ For each $x$, the sequence $\{T_n(x)\}$ is clearly convergent, so by the Banach-Steinhaus the sequence $\{T_n\}$ is pointwise convergent to some $T\in B(X)$. For each $k$, the sequence $\{T_n(x_k)\}$ is eventually $2^{-k}x_k$, so $T(x_k)=2^{-k}x_k$. Thus $$\{2^{-k}:k\in\mathbb N\}\subset\sigma(T),$$ and therefore $\sigma(T)$ is infinite.