Spring 2013 - Problem 8

Solution.

1. Since $A$ and $B$ are positive definite, we can express $A = S_AD_AS_A^{-1}$ and $B = S_BD_BS_B^{-1}$, where $D_A$ and $D_B$ are diagonal matrices with positive entries on the diagonal. Thus, we may define $A^z = D_A^z$ and $B^z = D_B^z$, where we raise each entry in the diagonal matrix to the $z$-th power. Thus, when expanding out $\inner{y, B^zA^{-z}x}$, we get a finite linear combination of functions of the form $a^z$ with $a > 0$, hence entire.

2. We wish to apply Hadamard's three-lines theorem. If $\Re{z} = 0$, then observe that $B^0$ and $A^0$ have eigenvalues $1$, so they have operator norm $1$. Thus, by Cauchy-Schwarz,

   $$\abs{\inner{y, B^zA^{-z}x}} \leq \norm{x}\norm{y}.$$

   If $\Re{z} = 1$, write $z = 1 + bi$. The eigenvalues of $A^{-ib}, B^{ib}$ take the form $\lambda^{ib} = e^{ib\log{\lambda}}$, which have absolute value $1$. Hence,

   $$\norm{B^zA^{-z}} = \norm{B^{ib}BA^{-1}A^{-bi}} \leq \norm{B^{ib}}\norm{BA^{-1}}\norm{A^{-bi}} \leq 1$$

   by assumption, so by Cauchy-Schwarz again,

   $$\abs{\inner{y, B^zA^{-z}x}} \leq \norm{B^zA^{-z}x}\norm{y} \leq \norm{x}\norm{y}.$$

   Since $f_{x,y}\p{z} = \inner{y, B^zA^{-z}x}$ is entire, it is in particular holomorphic near the strip $S = \set{z \mid 0 \leq \Re{z} \leq 1}$, so by the Hadamard three-lines theorem, if we set $M\p{t} = \sup_{b}\,\abs{f_{x,y}\p{t + ib}}$ and $\theta = 1 - t$ for $\theta \in \br{0, 1}$, we obtain

   $$M\p{\theta} \leq M\p{0}^t M\p{1}^{1-t} \leq \norm{x}\norm{y}.$$

   Hence, in particular, $\abs{\inner{y, B^\theta A^{-\theta}x}} \leq \norm{x}\norm{y}$. Taking the supremum over $\norm{y} = 1$, we obtain

   $$\norm{B^\theta A^{-\theta}x} \leq \norm{x},$$

   which completes the proof.