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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?

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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.

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  • $\begingroup$ 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 $U$. $\endgroup$ – user 242964 Oct 19 '18 at 0:11
  • $\begingroup$ 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. $\endgroup$ – Matematleta Oct 19 '18 at 0:30
  • $\begingroup$ 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. $\endgroup$ – user 242964 Oct 19 '18 at 1:33
  • $\begingroup$ 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. $\endgroup$ – Matematleta Oct 19 '18 at 1:57
  • $\begingroup$ 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. $\endgroup$ – user 242964 Oct 19 '18 at 13:25

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