# Non-unitary isometry and a norm equality

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.

• It suffices to show that $-2\in \sigma(v+v^*).$ Then $\rho+\lambda\in \sigma((\rho^2+\lambda^2)e-\rho\lambda(v+v^*)).$ Commented Mar 17, 2023 at 12:40

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 $$\|\rho v-\lambda\|=\rho+\lambda$$.

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

• A splendid elementary argument! Commented Mar 18, 2023 at 7:01

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:

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 $$-2\in\sigma(v+v^*)$$.$$\quad$$ $$\square$$

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 $$\|\rho v-\lambda e\|\ge \|\rho S-\lambda I\|\ge |\rho|+|\lambda|$$