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I've been studying complex analysis problems, and get stuck on the following:


Let $D \subseteq \mathbb{C}$ be a domain (open connected set) and $z_0 \in D$. Assume that $(f_n)$ is a sequence of analytic functions in $D$ such that $\lim_{n \to \infty} f_n(z_0) = w_0 \in \mathbb{C}$ and that sequence of derivatives $(f'_n)$ converges uniformly on compact subsets of $D$ to a function $g$. Is it true that there exists an analytic function $f$ in $D$ such that $f_n \to f$ uniformly on compact subsets of $D$? Prove or give a counterexample.


I don't even know if this is true. I thought I had a solution (see below) but I realized that I was using the Fundamental Theorem of Calculus on sets which were not necessarily simply connected, which is invalid. However, I have been unsuccessful in finding a counterexample either.

Does anyone know a way to fix the hole in my reasoning, or a counterexample if the statement is false? Thanks.


Suppose $K$ is connected compact subset of $D$ which contains $z_0$ with Lebesgue number $\delta$. Then there exists of an open cover of $K$ by some number $k$ of $\delta$-disks.

For each $m$ there exists $M>0$ such that $|f'_n(z)-g(z)| < 1/m$ and $|f_n(z_0) - w_0| < 1/m$ for all $n > M$ and all $z \in K$. Now let $z \in K$, and let $\gamma$ be a path of minimal length from $z_0$ to $z$ inside the $\delta$-cover of $D$. Then for $n > M$, $$\left | \int_\gamma f'_n(t) - g(t) dt\right| \leq \ell (\gamma) /m \leq 2\delta k /m. $$ Then \begin{align*} \left| f_n(z) - \left(w_0 + \int_\gamma g(t) dt\right ) \right | &= \left| \left(f_n(z)- f_n(z_0) - \int_\gamma g(t) dt \right) + (w_0- f_n(z_0))\right| \\ & \leq \left| f_n(z)- f_n(z_0) - \int_\gamma g(t) dt \right| + |w_0- f_n(z_0)| \\ &\leq (2\delta k+1)/m \\ & \to 0 \text{ as } m \to \infty \end{align*} Thus $f_n$ converges uniformly to $w_0 + \int_{z_0} ^ z g(t) dt $ on compact subsets of $D$ [every compact subset is contained in a connected compact subset which contains $z_0$].

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  • $\begingroup$ By a well-known theorem, $f_n^{(k)}(z_0) \to g^{(k-1)}(z_0)$ for every $k$. Since $f_n(z_0)$ converges, I think that you have all the coefficients of a power-series expansion around $z_0$, and therefore you have to prove that the radius of convergence is positive. $\endgroup$
    – Siminore
    Commented Aug 19, 2012 at 14:31
  • $\begingroup$ Sorry, of course the convergence of the derivatives of order $k$ is uniform on compact sets, not only at $z_0$. $\endgroup$
    – Siminore
    Commented Aug 19, 2012 at 14:53

1 Answer 1

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I haven't checked your argument in detail, but you say

[...] but I realized that I was using the Fundamental Theorem of Calculus on sets which were not necessarily simply connected, which is invalid. [...]

It is true that not every holomorphic function on $D$ need have a primitive. But in this case you already know that each of the $f_n'$ does have a primitive on $D$, namely $f_n$. So you will always have $$\int_\gamma f_n'(z) \, dz = f_n(\gamma(b)) - f_n(\gamma(a))$$ for a curve $\gamma: [a,b]\to D$, by the fundamental theorem of calculus.

You should be able to use this to show $(f_n)$ is uniformly Cauchy when restricted to compact sets. For example you could show

$$|f_n(z) - f_m(z)| \le |f_n(z_0) - f_m(z_0)| + \left|\int_\gamma f_n'(\zeta) - f_m'(\zeta)\, d\zeta \right|$$

where $\gamma$ is any curve in $D$ from $z_0$ to $z$, and go on from there.

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  • $\begingroup$ You missed derivatives in the last integral? $\endgroup$ Commented Aug 19, 2012 at 14:24
  • $\begingroup$ @enzotib: Right, thanks! $\endgroup$
    – Sam
    Commented Aug 19, 2012 at 15:56

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