# Convergence problem of complex valued functions

Let $$X$$ be a compact subset of the complex plane. Let $$D^1(X)$$ denote the set of all functions $$f: X \to \mathbb C$$ with continuous derivative in $$X$$.

If $$(f_n)$$ is a sequence in $$D^1(X)$$ that converges uniformly to $$f \in D^1(X)$$, is it true that $$f_n '\to f'$$?

My attempt:

Given $$\epsilon > 0$$ and $$h \in \mathbb C, \, h \neq 0$$, exists $$N \in \mathbb N$$ such that $$\|f_n - f\|_X < \epsilon |h| / 2, \, \forall n \geq N$$, where $$\| \cdot \|_X$$ denote the sup norm in $$X$$.

Given $$z_0 \in X$$, let $$(\ast) = \left |\frac{[f_n(z_0 + h) - f_n(z_0)] - [f(z_0 + h) - f(z_0)]}{h}\right|.$$

Then $$(\ast) \leq \frac{1}{|h|}[|f_n(z_0 + h) - f(z_0 + h)| + |f_n(z_0) - f(z_0)|] \leq \frac{2}{|h|} \|f_n - f\|_X \leq \frac{2}{|h|} \left ( \frac{ \epsilon |h|}{2} \right ) = \epsilon,$$

forall $$n \geq N$$. Then, taking $$|h| \to 0$$,

$$\epsilon >\lim_{|h|\to 0} \left |\frac{[f_n(z_0 + h) - f_n(z_0)] - [f(z_0 + h) - f(z_0)]}{h}\right| = |f_n '(z_0) - f'(z_0)|, \, \forall n \geq N.$$

With this I have proven that $$f_n'$$ converges pointly to $$f$$ in every point of $$X$$.

How can I see the uniform convergence?

Help?

I think we can argue as follows, exploiting the Cauchy estimates. But to define the derivative of $$f_j$$ at $$z\in X,\ f_j$$ should be defined in some $$open$$ set $$U\supseteq X.$$ So let's assume this is true and that $$f_j$$ are holomorphic in $$U$$. I will also assume that $$f_j\to f$$ uniformly on $$every$$ compact set in $$U$$.
Now, cover $$X$$ by balls $$B(z,r_z)$$ such that $$\overline B(z,r_z)\subseteq U.$$ This is possible because $$\mathbb C$$ is a normal topological space. Let $$\delta$$ be the Lebesgue number for this cover and pass to a finite subcover $$\mathscr A$$. The union $$K$$, of the closures of the elements of $$\mathscr A$$ is compact and so $$f_j\to f$$ uniformly there.
To finish, note that if $$z\in X,$$ then, $$\overline B(z,\delta/2)$$ is contained in one the original balls, and so, in $$K$$. Now, applying the Cauchy estimate, we have $$|(f'-f_j')(z)|\le\frac{2\|f-f_j\|_{\overline B(z,\delta/2)}}{\delta}\le \frac{\|f-f_j\|_K}{\delta}\to 0$$ uniformly as $$j\to \infty,$$ and we are done.
• Your argument works only for $z$ in the interior of $X$. If we take $z$ in the neighboorhood, we can't find the open set $U$. – user 242964 Oct 19 '18 at 0:11
• Yes, correct. I modified your question because from what I could understand from your proof (for example, "$h\in \mathbb C")$, you are trying to compute a derivative in $\mathbb C$, not $X$. If you are working in $X$ with the subspace topology, you are restricted to $X$. In any case, how would you expect to get the result through a difference quotient, since the result is certainly $not$ true for real-valued functions. At some point, you will need to exploit the fact that $f$ and $f_j$ are complex-valued. – Matematleta Oct 19 '18 at 0:30
• Actually, I'm working in the case that $X$ is the closed unit disk in $\mathbb C$. However, I thought that the result might be true in a general compact in complex plane. – user 242964 Oct 19 '18 at 1:33
• Analytic on the closed unit disk $D$ means analytic on some open set containing $D$, in which case, I believe my proof works. If you mean that $f,f_j$ are defined $only$ on the closed disk, and therefore differentiability on the boundary is not the "usual" definition, then I do not know. Interesting question, though. – Matematleta Oct 19 '18 at 1:57
• Well, you have proven that $f_n - f$ converges uniformly to zero in the open $U$. If $U$ is the open unit disk, we can use an argument about the continuity of $f_n - f$ to see that this function converges to zero in the boundery. I will write this to make sure. – user 242964 Oct 19 '18 at 13:25