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

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 $z \in D'$?

• I have a feeling that the answers to the above questions should be 'yes'. – user41481 Aug 3 '15 at 14:05
• @Ramos 1st of all I said $D$ is a domain, so open and connected. Secondly, If $f$ is an analytic function in the interior of the disk and not on the whole disk then $f$ has some singularity on the boundary. How are you going to bound $f$ in that case? – user41481 Aug 3 '15 at 14:57
• Sorry, I've got a bit confused, I mistook harmonic for analytic function, that was silly. But the idea was to try to give a counterexample in this way, as function that 'oscilates' heavily near the boundary. And your statement is not true. There are some analytic functions on the unit disk which are bounded but do not extend to the whole closed disk, as can be seen here: math.stackexchange.com/questions/6208/… – João Ramos Aug 3 '15 at 15:01
• If f is an analytic function in the interior of the disk and not on the whole disk then f has some singularity on the boundary. How are you going to bound f in that case? First, you need to say exactly what you mean by "has a singularity on the boundary". It's not true that such an $f$ must have an isolated singularity on the boundary; exactly what other sort of singularity are you including here?`Second, $f$ can have an isolated singularity on the boundary of the disk and still be bounded in the disk! See my answer... – David C. Ullrich Aug 3 '15 at 15:46

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.

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 $F$.

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.

• In my opinion, if $f$ is analytic on the interior and not on the boundary that means it has a singularity on the boundary. How are you going to bound it then? I still do not get your answer. – user41481 Aug 3 '15 at 15:17
• Well, at least for me it is clear that Jonas Meyer's construction works. If you take a look at his answer, you're going to see that your point of view is wrong, in the sense that the 'singularity' you claim to exist just does not exist! – João Ramos Aug 3 '15 at 15:26

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!!

• In my opinion this is the best answer we've seen so far, because it's so utterly trivial. – David C. Ullrich Aug 4 '15 at 15:27

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.

• Depending on the interpretation of "immediate", it may be immediate. Let $B \in H^\infty$ a Blaschke product such that every point of the unit circle is an accumulation point of zeros. If $B$ were continuous at $z_0 \in S^1$, then a) $B(z_0) = 0$ since $z_0$ is an accumulation point of zeros, so $\lvert B(z)\rvert < \frac{1}{2}$ in an open neighbourhood $V$ (with respect to $\overline{\mathbb{D}}$) of $z_0$, and b) in $V \cap S^1$ there are points where $B$ has a radial limit of modulus $1$, so there are $z\in V$ with $\lvert B(z)\rvert > \frac{1}{2}$, which contradicts a). – Daniel Fischer Aug 4 '15 at 16:43
• @DanielFischer That's certainly immediate. Dave dumb again (used to be it was big news when I said something stupid, but these days the papers just ignore it). I still kind of like the argument I gave, since here we're using (2), which is not quite trivial, while my construction requires no prerequisites at all. – David C. Ullrich Aug 4 '15 at 16:55
• Yes, it's nice that it's a rather elementary construction ["no prerequisites at all" is however a slight exaggeration ;)], while the "everybody knows" (2) is a pretty big thing. – Daniel Fischer Aug 4 '15 at 17:05

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.

• Yes there is such a function. See new answer... – David C. Ullrich Aug 4 '15 at 16:13

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 $\mathbb {D}\cup \{e^{it}\}.$