Fall 2009 - Problem 7

Solution.

1. A linear operator $\func{S}{H}{H}$ is unitary if it is bounded and satisfies $S^*S = SS^* = I$, where $S^*$ is the adjoint of $U$.

2. Since $S$ is invertible, we have $S - \lambda I = S\p{I - \lambda S^{-1}}$. Since $S$ is unitary, it has operator norm $\norm{S} = \norm{S^*} = 1$ and so $\norm{\lambda S^{-1}} \leq \abs{\lambda} < 1$. We claim that $I - \lambda S^{-1}$ is invertible with

   $$\p{I - \lambda S^{-1}}^{-1} = \sum_{n=0}^\infty \p{\lambda S^{-1}}^n = \sum_{n=0}^\infty \lambda^n S^{-n},$$

   where $S^{-n} = S^{-1} \circ S^{-1} \circ \cdots \circ S^{-1}$, $n$ times. Indeed, we have

   $$\p{I - \lambda S^{-1}}\p{I + \lambda S^{-1} + \cdots + \lambda^n S^{-n}} = I - \lambda^{n+1}S^{-\p{n+1}}.$$

   Notice that in the operator norm,

   $$\norm{\sum_{n=N}^M \lambda^n S^{-n}} \leq \sum_{n=N}^M \abs{\lambda}^n\norm{S^{-n}} = \sum_{n=N}^M \abs{\lambda}^n = \frac{\abs{\lambda}^N}{1 - \abs{\lambda}^M} \xrightarrow{N,M\to\infty} 0,$$

   so these operators are Cauchy in the operator norm. Since $H$ is complete, it follows that the bounded linear operators $H \to H$ are also complete with respect to the operator norm, so the series converges to a bounded linear operator $\sum_{n=0}^\infty \lambda^n S^{-n}$.

   Finally, $\norm{\lambda^n S^{-n}} \leq \abs{\lambda}^n \xrightarrow{n\to\infty} 0$, so taking $n \to \infty$ in the equality above,

   $$\p{I - \lambda S^{-1}} \sum_{n=0}^\infty \lambda^n S^{-n} = I.$$

   By repeating the same argument but commuting the terms, we see that

   $$\p{I - \lambda S^{-1}}^{-1} = \sum_{n=0}^\infty \lambda^n S^{-n}.$$

   Thus, $S - \lambda I$ is invertible with inverse

   $$\p{S - \lambda I}^{-1} = \p{I - \lambda S^{-1}}^{-1}S^{-1} = \sum_{n=0}^\infty \lambda^n S^{-n-1}.$$

3. Since inner products are continuous, we may write for $\abs{\lambda} < 1$

   $$\begin{aligned} h\p{\lambda} &= \inner{\p{S + \lambda I}\sum_{n=0}^\infty \lambda^n S^{-n-1}v, v} \\ &= \sum_{n=0}^\infty \lambda^n \inner{S^{-n}v + \lambda S^{-n-1}v, v} \\ &= \sum_{n=0}^\infty \lambda^n \inner{S^{-n}v, v} + \sum_{n=0}^\infty \lambda^{n+1}\inner{S^{-n-1}v, v} \\ &= \inner{v, v} + \sum_{n=1}^\infty 2\inner{S^{-n}v, v}\lambda^n \\ &= \norm{v}^2 + \sum_{n=1}^\infty 2\inner{S^{-n}v, v}\lambda^n. \end{aligned}$$

   so $h\p{\lambda}$ is analytic on the unit disc, which means that $\Re\p{h}$ is harmonic. It remains to show that $\Re\p{h\p{\lambda}}$ is positive. Observe that because $S^* = S^{-1}$,

   $$\begin{aligned} \overline{h\p{\lambda}} &= \overline{\inner{\p{S + \lambda I}\p{S - \lambda I}^{-1}v, v}} \\ &= \inner{v, \p{S + \lambda I}\p{S - \lambda I}^{-1}v} \\ &= \inner{\p{S^{-1} + \conj{\lambda} I}\p{S^{-1} - \conj{\lambda} I}^{-1}v, v}. \end{aligned}$$

   Thus, if we set $A\p{\lambda} = \p{S + \lambda I}\p{S - \lambda I}^{-1} + \p{S^{-1} + \conj{\lambda} I}\p{S^{-1} - \conj{\lambda} I}^{-1}$ and note that everything commutes, we can factor to get

   $$\begin{aligned} A\p{\lambda} &= \p{S - \lambda I}^{-1} \br{\p{S + \lambda I}\p{S^{-1} - \conj{\lambda} I} + \p{S^{-1} + \conj{\lambda}I}\p{S - \lambda I}} \p{S^{-1} - \conj{\lambda}I}^{-1} \\ &= \p{S - \lambda I}^{-1} \p{I + \lambda S^{-1} - \conj{\lambda}S - \abs{\lambda}^2 + I - \lambda S^{-1} + \conj{\lambda}S - \abs{\lambda}^2} \p{S^{-1} - \conj{\lambda}I}^{-1} \\ &= \p{S - \lambda I}^{-1} \p{2I - 2\abs{\lambda}^2} \p{S^{-1} - \conj{\lambda}I}^{-1}. \end{aligned}$$

   Thus, we get

   $$\begin{aligned} \Re\p{h\p{\lambda}} &= \frac{\inner{A\p{\lambda}v, v}}{2} \\ &= \frac{\inner{\p{S - \lambda I}^{-1}\p{2I - 2\abs{\lambda}^2}\p{S^{-1} - \conj{\lambda}I}^{-1}v, v}}{2} \\ &= \inner{\p{S - \lambda I}^{-1}\p{I - \abs{\lambda}^2}\p{S^{-1} - \conj{\lambda}I}^{-1}v, v} \\ &= \p{1 - \abs{\lambda}^2}\inner{\p{S^{-1} - \conj{\lambda}I}^{-1}v, \p{S^{-1} - \conj{\lambda}I}^{-1}v} \\ &= \p{1 - \abs{\lambda}^2}\norm{\p{S^{-1} - \conj{\lambda}I}^{-1}v}^2, \end{aligned}$$

   which is strictly larger than $0$ as long as $v \neq 0$. Indeed, $\abs{\lambda} < 1$, and $\p{S^{-1} - \conj{\lambda}}^{-1}$ is invertible; in particular, $\p{S^{-1} - \conj{\lambda}I}^{-1}v \neq 0$.