Let $\mathcal{R}$ be the right-shift operator on $\bigoplus_{n=0}^{\infty}\ell^{2}$, that is, $\mathcal{R}(x_{0},x_{1},\ldots):=(0,x_{0},x_{1},\ldots)$ for $x_{j}\in\ell^{2}$ and $\sum_{j}\|x_{j}\|_{\ell^{2}}^{2}<\infty$. Is it true that $\sigma(\mathcal{R})=\overline{\mathbb{D}}$, where $\mathbb{D}$ is the open unit disc in $\mathbb{C}$?
So for the standard right-shift operator $R\colon\ell^{2}\to\ell^{2}$ this is well-known and easy to prove. See for example this post. However, this argument does not seem to work for $\mathcal{R}\colon\bigoplus_{n=0}^{\infty}\ell^{2}\to\bigoplus_{n=0}^{\infty}\ell^{2}$.
Since $\|\mathcal{R}\|=1$, it is easy to see that $\sigma(\mathcal{R})\subset\overline{\mathbb{D}}$. Is the reverse inclusion true?
Best Answer
In place of $\ell^2$ we may consider arbitrary Hilbert space $\mathcal{H}.$ Then we take $\mathcal{H}_\infty =\oplus_{n=0}^\infty \mathcal{H}$ and the operator $$\mathcal{R}(x_0,x_1,\ldots, x_n,\ldots )=(0,x_0,x_1,\ldots, x_{n-1},\ldots )$$ The adjoint operator $\mathcal{R}^*$ is given by $$\mathcal{R}^*(x_0,x_1,\ldots, x_n,\ldots )=(x_1,x_2,\ldots, x_{n+1},\ldots )$$ Fix an element $u\neq 0$ in $\mathcal{H}$ and $|z|<1$ Then the element $v_z:=\{z^nu\}_{n=0}^\infty$ belongs to $\mathcal{H}_\infty$ and $\mathcal{R}^*v_z=zv_z.$ Thus the closed unit disc is contained in $\sigma(\mathcal{R}^*)$ as well as in $\sigma(\mathcal{R})$.