If $|f| \le |g|$, does analytic continuation of $g$ imply analytic continuation of $f$?

Question

Let $f,g$ be two holomorphic functions on a domain $D$ such that $|f(z)| \le |g(z)|$ for all $z \in D$. Further suppose that there is an analytic continuation of $g$ to a bigger domain $D'$. Does that imply that there will be an analytic continuation of $f$ on $D'$? If so, will it be necessarily true that $|f(z)| \le |g(z)|$ for all $z \in D'$?

Answer 1

One can give a very simple explicit counterexample (which I would imagine is a special case of the "complete treatment" mentioned in the other answer).

Note first that if $\Re z<0$ then $\Re(1/z)<0$. Let $\Bbb D$ be the unit disk. If $z\in \Bbb D$ then $\Re(1/(z-1))<0$, hence $$\left|e^{\frac1{z-1}}\right|\le1.$$ And so there you are. Let $f(z)=e^{1/(z-1)}$ and $g(z)=1$. Then $|f|<|g|$ in $\Bbb D$, $g$ extends to a neighborhood of $z=1$ but $f$ does not.

Answer 2

As I've said on a comment above, for such a counterexample, it suffices to consider the domain as the unit disk, and take $f$ to be a generic function in $H^{\infty}$ whose boundary value is not analytic. Then we'll have the original function bounded, but no extension is possible, in such a way that we may take $g$ to be simply a large constant.

A complete treatment of the subject is given on this answer by Jonas Meyer. This kind of idea has to do with the classical theory of Hardy Spaces and Boundary Values of Analytic Functions.

Answer 3

A function can be analytic and bounded on an open disc and continuous on the boundary but not analytically extendable to an open set that contains the whole boundary.For example if $f(z)=(1+z)log(1+z)$ with the principal branch of the complex log, for $abs(z)<1$ The difficulty at $z=-1$ is with the derivative of f. In general I believe there is a bounded analytic function on the open unit disc which cannot be continuously extended to an point on the boundary. I'll return.

Answer 4

Let me explain how to create the most extreme counterexample:

The function $$ f(z)=\sum_{n=0}^\infty z^{n!}, $$ is analytic in the unit disk $\mathbb D$ and it can not be extended analytically to any region $\Omega$, with $\mathbb D\subsetneq \Omega$ (region=open & connected). This makes the unit circle $\partial\mathbb D$ the natural boundary of $f$. Now, integrate $f$ and obtain $$ F(z)=\sum_{n=0}^\infty \frac{z^{n!+1}}{n!+1}. $$ Clearly, $\lvert F(z)\rvert< \mathrm{e}$, for all $z\in\overline{\mathbb D}$, and $F$ cannot be extended analytically to any region, proper superset of the unit disk, since that would imply the same for $f$.

Next, for $g(z)=\mathrm{e}$ observe that $$ \lvert F(z)\rvert<\lvert g(z)\rvert, \quad\text{for all $z\in\mathbb D$}. $$ Now, $g$ is entire while $\partial\mathbb D$ is the natural boundary of $F$.

Answer 5

Such a result would imply that every bounded holomorphic function on the unit disc extends to be entire. Hence things like $1/(z-2)$ would extend from the unit disc to be entire. So ... ahem ... er ... hmm ... NO!!

Answer 6

This is really a comment, much too long to fit in the comment box. user254665 conjectured in a comment that there is a bounded function in the disk which is continuous at no boundary point. Yes there is.

In fact there's a "Blaschke product" with this property. There are whole books on Blaschke products, which must contain this fact. I don't see that it's quite immediate from the things that "everybody" knows about Blaschke products, however; the strongest result that seems immediate is that there is a function continuous at almost no boundary point.

Let $\Bbb D$ be the unit disk. For $a\in \Bbb D$ with $a\ne0$ define $b_a:\Bbb D\to\Bbb D$ by $$b_a(z)=\frac{|a|}a\frac{a-z}{1-\overline az};$$set $b_0(z)=z$. A Blaschke product is a finite or infinite product $$B=\prod_jb_{a_j}.$$Two canonical facts about Blaschke products, neither of which we will be using, are these:

1. The product converges (uniformly on compact subsets of $\Bbb D$) if and only if $\sum(1-|a_j|)<\infty$.

2. A convergent Blaschke product has radial limits of modulus $1$ at almost every boundary point.

We mention (1) and (2) just so we can pose as an exercise: Show how it follows immediately that there exists a $B$ which extends continuously to almost no point of the boundary.

But (1) and (2) don't give out more subtle result immediately anyway, so we'll ignore them officially.

[EDIT: Daniel Fischer points out that yes it is immediate. Say the zeroes of $B$ accumulate at every boundary point. If $B$ extends continuously to a given boundary point then $|B|<1/2$ in some neighborhood of that point, hence $B$ cannot have radial limits of modulus $1$ at nearby boundary points. That's not so much "immediate" as "blatantly obvious". Sigh. Of course this uses (1) and (2), and (2) is really not quite trivial; the construction below uses nothing.]

We need two easy facts:

Lemma 0. $|b_a|<1$ in $\Bbb D$ and $|b_a|=1$ on $\partial\Bbb D$. (Exercise.)

Lemma 1. $b_a\to1$ uniformly on compact subsets of $\Bbb D$ as $|a|\to1$. (Exercise.)

Now for a number $r_n\in(0,1)$ to be determined later let $$E_n=\{r_ne^{2\pi ik/n}\,:\,k=0,\dots,n-1\},$$which is to say $E_n$ consists of $n$ points evenly spaced on the circle $|z|=r_n$. Let $$B_n=\prod_{a\in E_n}b_a.$$We will show that there exists a sequence $(r_n)$ such that $B=\prod_n B_n$ has the desired property.

Let $r_1=1/2$. Choose $s_1\in(r_1,1)$ so that $$|B_1(z)|>1-\frac1{1+1}\quad(|z|=s_1).$$Now suppose we have chosen $0<r_1<s_1<\dots<r_n<s_n<1$ in such a way that if $f_n=\prod_{j=1}^nB_j$ then $$(*)\quad\quad|f_n(z)|>1-\frac{1}{1+j} \quad(|z|=s_j,1\le j\le n).$$Choose $r_{n+1}\in(s_n,1)$ so close to $1$ that ($*$) holds (for $1\le j\le n$) with $f_{n+1}$ in place of $f_n$. Choose $s_{n+1}\in(r_{n+1},1)$ so close to $1$ that ($*$) holds with $n+1$ in place of $n$.

Now let $B=\prod B_n=\lim f_n$. On the one hand we have $$|B(z)|\ge1-\frac1{1+j}\quad(|z|=s_j).$$ On the other hand the zero set of $B$ accumulates at every boundary point. So $B$ extends continuously to no point on the boundary.

Answer 7

To follow up on David Ullich's "comment" on an $f\in H^\infty(\mathbb {D})$ that cannot be extended continuously to any boundary point, we can also produce such an $f$ this way: Let $t_1,t_2, \dots$ be dense in $[0,2\pi).$ Let $\mu$ be the positive measure on $\partial \mathbb {D}$ that puts mass $1/2^n$ at each $e^{it_n}.$ Let $\lambda$ denote Lebesgue measure on $\partial \mathbb {D}.$

Define $u = P[\mu],$ i.e., $u$ is the Poisson integral of $\mu.$ Then $u$ is positive and harmonic in $\mathbb {D}.$ Here are two well known properties of $u$:

$$(i) \,\,\text {for each} \,n, \lim_{r\to 1}u(re^{it_n}) = \infty.$$

$$(ii) \lim_{r\to 1}u(re^{it})= 0 \,\text {for} \,\lambda\text {-a.e.}\, t\in [0,2\pi).$$

Let $v$ be a harmonic conjugate of $u$ in $\mathbb {D}.$ In $\mathbb {D},$ set

$$f = \exp {-(u+iv)}.$$

Then $f\in H^\infty(\mathbb {D}).$ Following from (i),(ii) above, we see that $f$ has radial limit $0$ at each $e^{it_n},$ and because $f$ has $\lambda$-a.e radial limits, $f$ has radial limits of modulus $1$ at $\lambda$-a.e. $t\in [0,2\pi).$ It follows that for every $e^{it} \in \partial \mathbb {D}, \rho > 0,$ the oscillation of $f$ in $D(e^{it},\rho)\cap \mathbb {D}$ is at least $1.$ Thus $f$ cannot be extended continuously to any $\mathbb {D}\cup \{e^{it}\}.$