How to compute the numerical radius of the right shift operator?

Question

Let $T$ be the right shift operator on $\ell^2$ defined by $T(x_1,x_2,\ldots)=(0,x_1,x_2,\ldots)$.

The numerical radius of $T$ is defined by $w(T)=\sup\{|\langle Th,h\rangle|:\, \|h\|=1\}$. It is well known that $r(T)\le w(T)\le \lVert T\rVert$ where $r(T)$ denotes the spectral radius of $T$. Since $T^n$ is isometry for all $n$, $r(T)=\lVert T\rVert=1$. Hence, $w(T)=1$.

But I want to compute $w(T)$ elementarily using the definition. By definition,

$$w(T)=\sup\left\{\left|\sum\limits_{i=1}^\infty \overline{x_{i+1}}x_i\right|:\, \sum\limits_{i=1}^\infty |x_i|^2=1\right\}$$

If I consider the vector $x=(x_1,x_2,\ldots)\in\ell^2$ with $x_i=\frac{1}{2^{i/2}}$. Then $\left|\sum\limits_{i=1}^\infty \overline{x_{i+1}}x_i\right|=\frac{1}{\sqrt{2}}$. This shows that $w(T)\ge \frac{1}{\sqrt{2}}$. But how can I arrive at $1$?

Can anyone help me in this regard? Thanks for your help in advance.

Answer 1

Consider $(0,1,1,...,1,0,0...)/\sqrt N$ where there are $N$ one's. These are unit vectors and $\sum x_{i+1} x_i=\frac {N-1} N \to 1$ as $N \to \infty$.

Answer 2

For any $K > 1$, let $x^K \in \ell^2$ be given by $x^K_i = \sqrt{K-1} \frac{1}{K^{i/2}}$. Then $x^K$ is a unit vector and $\sum_{i=1}^\infty x^K_ix^K_{i+1} = \frac{1}{\sqrt{K}}$ converges to $1$ as $K \to 1$.

Answer 3

For $|z|<1$ let $h_i=\sqrt{1-|z|^2}z^{i-1}.$ Then $\|h\|=1.$ Moreover $$\langle Th,h\rangle =(1-|z|^2)\sum_{i=1}^\infty \overline{z}^iz^{i-1} =\overline{z}$$ Therefore $$ \{z\,:\,|z|<1\}\subset \left \{\langle Th,h\rangle \,:\, \|h\|=1\right \}\subset \{z\,:\,|z|\le 1\}$$ (the second containment follows from $\|T\|=1).$ In particular the numerical radius is equal $1.$

Remark Actually the numerical range of $T$ is equal $\{z\,:\,|z|<1\}$ as $|\langle Th,h\rangle|<1$ for $\|h\|=1.$ This follows from the property that $T$ does not admit eigenvalues, in particular eigenvalues in the unit circle.