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

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.

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 $\sigma(A)=\overline{\mathbb{D}}$)
• Since the spectrum of an operator is closed, complex numbers $\lambda$ with $|\lambda| = 1$ also belong to the spectrum of $A$. In particular, $1$ is in the spectrum, so $A -I$ is not invertible. Actually, it is easy to check directly $A - I$ is not surjective : for a lot of choices of $y \in \ell^2$, the unique sequence $x$ such that $(A-I)x = y$ does not belong to $\ell^2$. Commented Sep 9, 2014 at 15:30