What are the approximate eigenvalues of the right shift operator $R$ on $\ell_\infty$

Question

I have shown that the spectrum of $R=\{z\in C||z|\leq 1\}$.

Also, elements on the boundary of the spectrum are approximate eigenvalues, i.e. $\forall |z|=1$, $z$ is an approx. eigenvalue. However, are these the only ones?

If they are not how do I find the rest?

Thanks.

Answer 1

For all $x \in \ell_\infty$ we have $||Rx||=||x||.$ Thus, for $\lambda$ with $|\lambda | \le 1$ we have

$$(*) ||Rx-\lambda x|| \ge | \quad ||Rx||-|\lambda| ||x|| \quad|=(1-|\lambda|)||x||.$$

Now suppose that $ \lambda $ is an approximate eigenvalue . Then there is a sequence $(x_n)$ in $\ell_\infty$ such that $||x_n||=1$ for all $n$ and $(R-\lambda I )x_n \to 0$.

From $(*)$ we get: $(1-|\lambda|)||x_n|| \to 0.$ Since $||x_n||=1$ for all $n$, we derive $|\lambda|=1.$