[SOLVED] matrix-valued complex functions

**choovuck** · September 4th 2008, 03:59 PM

we have a contractive analytic in the unit disk matrix-valued function ( $ff^*\le I$ ), which is unitary on the unit circle ( $ff^*=I$ ), and which in nondegenerate throughout the unit disk. I want to conclude that $f$ is a constant unitary matrix. We can assume that $f$ is analytic in some neighbourhood of the unit disk (rather than analytic inside the unit disk with boundary values which are unitary)

the proof for the 1D case could go as follows -- since f doesn't vanish, we can take $\log|f|$ which is a harmonic function. on the unit circles $\log|f|$ is zero, so by maximum principle for harmonic functions it is zero everywhere, i.e. $|f|$ is constant everywhere, and then so is f. clearly this approach doesn't help me in the matrix case

**Opalg** · September 5th 2008, 02:31 AM

Here's another approach to the scalar-valued case, which looks as though it might generalise more easily to the matrix-valued case. Since the unit disk $\mathbb D$ is simply connected, we can define log(f) in such a way that it is continuous on $\mathbb D$ and analytic in the interior. Use a conformal map taking the unit disk to the upper half-plane $\mathbb H$ to transport log(f) to a bounded analytic function on $\mathbb H$ . Then i*log(f) will be real on the real axis, so by the Schwarz reflection principle it extends to a bounded entire (therefore constant) function.

**ThePerfectHacker** · September 5th 2008, 05:15 AM

Can someone state this problem for the 1-D case?
I just want to see how it looks like - I cannot solve the general one.

**Opalg** · September 5th 2008, 11:00 AM

Originally Posted by ThePerfectHacker

Can someone state this problem for the 1-D case?
I just want to see how it looks like - I cannot solve the general one.

In the scalar-valued case, the result is saying this: Let f be a complex-valued continuous function on the closed unit disc, analytic in the interior of the disc, with no zeros in the disc, and with $|f|=1$ on the boundary of the disc. Then f must be constant.

**choovuck** · September 6th 2008, 08:19 PM

Opalg, I don't like the matrix-valued logarithm, so I slightly modified your idea: instead of taking logarithm and then extending it, we can extend the function f immediately (basically via Schwarz lemma)

for $|z|\ge1$ define $\hat{f}(z)=\left(f(1/\bar{z})^*\right)^{-1}$ , which is analytic and well defined since f doesn't vanish. Then on the unit circle $\hat{f}(e^{i\theta})=\left(f(e^{i\theta})^*\right) ^{-1}=f(e^{i\theta})$ by the unitarity condition, so $f=\hat{f}$ , i.e. f extends to an entire function. It's bounded, so we're happy.

thanks for your help.

on this subject: is it actually true that there exists an analytic logarithm of a nonvanishing analytic function on a simply connected domain for the matrix-valued case? I wasn't able to find it in literature, though it looks like the scalar case proof should go through without problems.

**Opalg** · September 7th 2008, 12:51 AM

Originally Posted by choovuck

Opalg, I don't like the matrix-valued logarithm, so I slightly modified your idea: instead of taking logarithm and then extending it, we can extend the function f immediately (basically via Schwarz lemma)

for $|z|\ge1$ define $\hat{f}(z)=\left(f(1/\bar{z})^*\right)^{-1}$ , which is analytic and well defined since f doesn't vanish. Then on the unit circle $\hat{f}(e^{i\theta})=\left(f(e^{i\theta})^*\right) ^{-1}=f(e^{i\theta})$ by the unitarity condition, so $f=\hat{f}$ , i.e. f extends to an entire function. It's bounded, so we're happy.

Yes, that looks neat.

Originally Posted by choovuck

on this subject: is it actually true that there exists an analytic logarithm of a nonvanishing analytic function on a simply connected domain for the matrix-valued case? I wasn't able to find it in literature, though it looks like the scalar case proof should go through without problems.

I can't give a reference for that. I think it's true (defining the log by the spectral theorem) but I wouldn't want to put money on it.

**ThePerfectHacker** · September 7th 2008, 08:20 PM

This does not help the poster but I had a question.

Let $f$ be analytic on the intetrior of the unit disk and continous on the boundary. With $|f|=1$ on the boundary. Furthermore, $f$ has zeros inside the disk. Then what happens? I know that if $f$ has finitely many zeros $\alpha_1,...,\alpha_n$ (with multiplicity) then $f(z) = \omega \prod_{k=1}^n \frac{z-\alpha_k}{1-\bar \alpha_k z}$ for some $\omega$ on the boundary. But what happens if $f$ has infinitely many zeros?

I just realized the moment I wrote that phrase that it is impossible for it to has infinitely many zeros. And this question was bothering me today for a non-trivial amount of time. How stupid of me. I just leave it here just in case the original poster finds this interesting. Sorry.

**choovuck** · September 7th 2008, 11:46 PM

well, the original question arose in the connection with the so-called inner functions (you can read a bit on this link), which are bounded analytic inside the unit disk (not necessarily continuous on the boundary) functions with boundary values (as limits) of modulus 1. inner function can actually have infinitely many zeros $z_k$ , but in this case $\sum(1-|z_k|)<\infty$ and the infinite Blaschke product (of the form you described) converges and we can factor it out and assume that our inner function is in fact nonvanishing. now, this nonvanishing inner function can be pretty ugly, but if its continuous on the boundary, we just proved that it's not. yay

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