Spectrum of the left-shift operator on $\ell_2$

Question

I read that the shift operator $A:\ell_2\to\ell_2$, $(x_1,x_2,x_3,\ldots)\mapsto(0,x_1,x_2,\ldots)$ contains $0$ in its spectrum, and that's clear to me. It is also clear to me that it has no eigenvalue. Though, I wonder whether it has got some other complex number $\lambda\ne 0$ in its continuous spectrum. I suspect it does, but I cannot prove it.

Answer 1

This operator is sometimes called unilateral shift. Suppose $Ax=\lambda x$ where $x=(x_1,\cdots,x_n,\cdots).$ If $(0,x_1,x_2,\cdots)=(\lambda x_1,\lambda x_2,\cdots),\,$you can convince yourself that $\lambda$ must be zero. Also notice that obviously $\|A\|=1.$ Now let $0<|\lambda|<1$ and consider $A-\lambda I$. Then if $(1,0,0,\cdots)=(A-\lambda I)(x_1,x_2,\cdots)=(-\lambda x_1,x_1-\lambda x_2,x_2-\lambda x_3,\cdots)$ it follows that $x_1=-\frac{1}{\lambda},\,x_2=-\frac{1}{\lambda^2},\cdots,x_n=-\frac{1}{\lambda^n},\cdots.$ but this is a contradiction since $\sum_{n=1}^{\infty}(\frac{1}{\lambda^n})^2$ diverges. So for all $\lambda$ with $|\lambda|<1$ operator $A-\lambda I$ is not invertible. Its rather easy to show that $A-I$ is also non-invertible,$(1,0,0\cdots)\not\in Rang(A-I)$ and since spectral radious is 1 proof is complete.(we showed $\sigma_p(A)=\emptyset$ and $\sigma(A)=\overline{\mathbb{D}}$)