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Let $f(z)$ be an analytic function in an open set $U\subset\Bbb{C}$. Recall that an analytic continuation of $f$ is a pair $(F,V)$ such that $U\subset V\subset\Bbb{C}$, $F$ is analytic on $V$, and $F(z)=f(z)$ for all $z\in U$.

My question is, how stable is this process? If $\|f-g\|$ is small, are we guaranteed $\|F-G\|$ small in any reasonable sense? If not, are there easy counterexamples? If the answer depends on the choice of norm, I would find that interesting as well.

References gladly accepted in lieu of obvious arguments. Thanks!

icurays1
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2 Answers2

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Suppose that $f_n$ is a sequence of analytic functions converging to $f$ on $U$, say, uniformly. Let $(F,V)$ be an analytic continuation of $(f,U)$ and $(F_n, V_n)$ are analytic continuations of $(f_n,U)$ so that the domains $V_n$ converge to $V$ in the sense Caratheodory. Then the functions $F_n|V$ converge to $F$ uniformly on compacts.

Moishe Kohan
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  • Nice. Our answers would seem to indicate that if you stay away from boundaries where nastiness can happen (by looking at compacts) then things work quite nicely, but if you allow domains going right to some boundary, then extreme behaviour can break things. – Old John Nov 18 '13 at 22:19
  • Ah, I see! Also very useful. – icurays1 Nov 18 '13 at 22:23
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Not a rigorous answer, but I believe the answer is probably no.

Look at the functions

$$f(z) = z + z^2 + z^4 + z^8 + \cdots + z^{32}$$ and $$g(z) = z + z^2 + z^4 + z^8 + z^{16} + z^{32} + \cdots$$

inside the open set $|z| < 1/2$.

Depending on your definition of norm, I suspect that these would be "close", but if you analytically continue them to something like the unit disc, they will not be close by any means, since $f$ is a well-behaved polynomial, but $g$ will be very badly behaved indeed on the unit circle as seen by the results quoted in the second part of my answer here.

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Old John
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  • Quite obvious now with this example, thanks. Do you know if there is a general principle at work here...? – icurays1 Nov 18 '13 at 21:58
  • No general principle that I can think of, I'm afraid. Despite analytic functions being essentially rather well-behaved (in the big scheme of things) I am constantly being surprised at how badly they can behave sometimes! (Especially as you get close to the boundary) – Old John Nov 18 '13 at 22:00
  • If we take the norm $|f-g|=\sup\limits_{|z|<1/2} |f(z)-g(z)|$ (the $L_\infty$ norm), note that for your $f$ and $g$ we then have $|f-g|\simeq$ $5\times 10^{-20}$. You can simply increase the degree of $f$ in the obvious pattern to get the functions even closer. In particular, you'd have a sequence of polynomials converging to $g$ (in $|z|<1/2$), but each of whose continuations will be very distinct from $g$ in the continuation to the unit disc in this norm. – zibadawa timmy Nov 19 '13 at 00:33