Non-unitary isometry and a norm equality

Question

I am looking at a paper which asserts the following equality relating a non-unitary isometry. There is no explanation given for this, and I cannot figure out why this is true:

Here is the proposition: Let $A$ be a unital $C^*$ algebra (some norm closed subalgebra of $B(H)$ where $H$ is a Hilbert space, containing the identity), and let $v$ be a non-unitary isometry. Moreover let $\lambda, \rho$ be positive scalars satisfying $0< \lambda, \rho < 1$. Then we have the following claim,

Claim: $\left|\left|{\rho v - \lambda } \right|\right|= \rho + \lambda$.

I can't seem to find out why this is true (and it certainly does not seem obvious to me).

I cannot even show this for the unilateral shift (i.e. the map that sends $e_i \rightarrow e_{i+1}$ for $i \in \mathbb{N}$ on $\ell^2(\mathbb{N})$).

Here are some things that I tried: First that the inequality $\lvert\lvert {\rho v - \lambda } \rvert\rvert \leq \rho + \lambda$ is obvious. Thus if our $C^*$ algebra is isometrically isomorophic to some subalgebra of $B(H)$ for some Hilbert space $H$, then if we choose normalized $x \in ran(v)^{\perp}$ then $\lvert\lvert {\rho v(x) - \lambda(x) } \lvert \lvert = \sqrt{\rho^2+\lambda^2}$ , but this is not enough to show the equality.

Another hope is to use the $C^{*}$ property and write $\lvert\lvert {\rho v - \lambda } \lvert \lvert ^2 = \lvert\lvert ({\rho v - \lambda })^{*} ({\rho v - \lambda })\lvert \lvert = \lvert\lvert \rho^2+ \lambda^2 - \rho \lambda (v+v^*)\lvert \lvert $ and say something about this quanity and maybe use spectral properties (i.e. use the spectral theorem in a meaningful way) of the self adjoint operator $v+v^*$, but I am unsure how to proceed.

Answer 1

One way to address this is as follows. Since $v$ is a proper isometry, $\ker v^*\ne\{0\}$. Let $e_0\in\ker v^*=(\operatorname{ran} v)^\perp$ be a unit vector. Then the sequence $\{v^ke_0\}_{k=0}^\infty$ is orthonormal, for if $k>h$ then $$ \langle v^ke_0,v^he_0\rangle=\langle v^{k-h}e_0,e_0\rangle=0. $$ Given $\lambda\in\mathbb D$, let $x=\displaystyle\sum_{k=0}^\infty\lambda^kv^ke_0$. The series converges because $\|\lambda^kv^ke_0\|=|\lambda|^k$. We have, since $v^*e_0=0$, $$ v^*x=\sum_{k=1}^\infty\lambda^kv^{k-1}e_0=\lambda x. $$ This shows that $\sigma_p(v^*)=\mathbb D$. You can now choose unit vectors $x_n$ such that $v^*x_n=-(1-\frac1n)x_n$. Then $$ \|\rho v^*x_n-\lambda x_n\|=\Big\|-\Big(1-\frac1n\Big)\rho x_n-\lambda x_n\Big\|=\Big(1-\frac1n\Big)\rho+\lambda. $$ Hence $\|\rho v^*-\lambda\|≥\rho+\lambda$. As $\|\rho v-\lambda\|=\|\rho v^*-\lambda\|$, we have shown that $\|\rho v-\lambda\|=\rho+\lambda$.

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(alternatively, one can use the Wold Decomposition to assume without loss of generality that $v$ is the unilateral shift).

Answer 2

As @Ryszard Swzarc pointed out in their comment if we can prove the following then we are done:

Proposition. Let $v\in B(H)$ be a non-unitary isometry. Then $-2\in\sigma(v+v^*)$.

In order to prove this let us rely on the following basic facts:

• The spectrum of a non-unitary isometry is the closed unit disc
• The boundary of the spectrum is contained in the approximate point spectrum

Proof. Becaus $v$ is a non-unitary isometry $-1\in\partial\sigma(v)$ so there exists a sequence $(x_n)_{n\in\mathbb N}$, $\|x_n\|=1$ such that $\|(v-(-1))x_n\|=\|vx_n+x_n\|\to 0$ as $n\to\infty$. Using again that $v$ is an isometry ($v^*v={\bf 1}$) this implies $$ \|v^*x_n+x_n\|=\|v^*x_n+v^*vx_n\|=\|v^*(x_n+vx_n)\|\leq\|v^*\|\|vx_n+x_n\|\to 0 $$ as $n\to\infty$. Together $vx_n+x_n+v^*x_n+x_n=(v+v^*-(-2))x_n\to 0$ which shows that $-2$ is in the approximate point spectrum of $v+v^*$, hence $-2\in\sigma(v+v^*)$.$\quad$ $\square$

Answer 3

I assume that $\rho$ and $\lambda$ are arbitrary complex numbers.

Let $e_0\perp {\rm Im}\,v.$ Then $v^*e_0=0.$ The property $v^*v=I$ implies that the vectors $e_n=v^ne_0$ are mutually orthogonal and the space $H=\overline{{\rm span}\,e_n}$ is invariant for $v$ and $v^*.$ The operator $v$ restricted to $H$ is unitarily equivalent to the shift operator $S.$ It is well known that $\sigma(S)=\{z\,:\,|z|\le 1\}=\mathbb{D}.$ as the point spectrum of $S^*$ coincides with $\{z\,:\,|z|<1\}.$ Therefore $$\sigma(\rho S-\lambda I)=\rho \mathbb{D}-\lambda$$ Thus the spectral radius of $\rho S-\lambda I$ is equal $|\rho|+|\lambda|.$ In particular $$\|\rho v-\lambda e\|\ge \|\rho S-\lambda I\|\ge |\rho|+|\lambda|$$