Suppose $f_n$ is a sequence of holomorphic functions taking $\mathbb{D} \to \mathbb{C}$ where $\mathbb{D}$ is the unit disk. Further, $f_n$ has a continuous extension to $\overline{\mathbb{D}}$. We can assume $\sum_{n=0}^\infty f_n$ converges normally on compact subsets of $\mathbb{D}$ to a holomorphic function $f$.

Additionally, for each $f_n$ we know $\int_C |f_n| < \infty$ for any contour $C \subset \overline{\mathbb{D}}$. Less strictly, so we don't have enough to use the monotone convergence theorem, we know that $\sum_{n=0}^\infty |\int_C f_n| < \infty$. But additionally, for any closed contour $C^*$ in $\overline{\mathbb{D}}$ we know $\int_{C^*}f_n = 0$.

Must it follow

$$\int_C f = \sum_{n=0}^\infty \int_C f_n$$

I'm asking because all the naive instances where the monotone convergence theorem fail are exempt from these criterion. I think there's something more subtle in this instance.

I've been able to strengthen the condition to a proof that

$$\int_C f = \sum_{n=0}^\infty \int_C f_n \,\,\Leftrightarrow\,\, \sum_{n=0}^\infty \sup_{C\subset \overline{\mathbb{D}}}|\int_Cf_n| < \infty$$

Or rephrase it to

$$\sum_{n=0}^\infty |\int_C f_n| < \infty\,\, \Leftrightarrow \,\,\,f \, \text{can be continuously extended to}\,\overline{\mathbb{D}}$$

Those are the two avenues I've taken. None have really given me an answer.

Any suggestions, comments, questions are greatly appreciated.

  • $\begingroup$ How are you even defining $\int_C f $ if $C$ is a contour in $\overline{D}$? $f$ is not necessarily defined on the boundary of $D$. $\endgroup$ – Robert Israel Oct 31 '17 at 8:09
  • $\begingroup$ @Robert Israel: "Further $f_n$ has a continuous extension to $\overline{\mathbb{D}}$" $\endgroup$ – js21 Oct 31 '17 at 8:50
  • $\begingroup$ @js21 But $f$ does not. $\endgroup$ – Robert Israel Oct 31 '17 at 18:11

Take $f_n(z) = 3z^{3n} - z^n$ and $C = \{ e^{i\theta} \ | \ \theta \in (0,\pi) \}$. It satisfies all of your hypotheses with $$ f(z) = \frac{2+z}{1+z+z^2} $$ and $\int_C f_n =0$ for $n \geq 1$. But $\int_C f$ is not a convergent integral.

Of course the stronger condition $\sum_n \int_C |f_n| < \infty$ is sufficient for your conclusion to hold.

  • $\begingroup$ Even simpler example is $f_n=z^n$. $\endgroup$ – Alexandre Eremenko Oct 31 '17 at 13:15
  • $\begingroup$ @Alexandre Emerenko: The OP required $\sum_{n=0}^\infty |\int_C f_n| < \infty$ $\endgroup$ – js21 Oct 31 '17 at 13:18
  • $\begingroup$ This holds trivially because $\int_Cf_n=0$ for every analytic function. $\endgroup$ – Alexandre Eremenko Oct 31 '17 at 13:22
  • $\begingroup$ @Alexandre Emerenko: OP's terminology confused me as well. If I understand correctly by "contour" he or she means "path" and by "closed contour" he or she means what most of us would simply call "contour". Besides that, you are right, I do not know why I absolutely wanted a non-closed path of integration in my counterexample. $\endgroup$ – js21 Oct 31 '17 at 15:41
  • $\begingroup$ I agree that the question is very confusing. First of all it is not explained what is $\int f$. Is this $\int fdz$ or $\int f |dz|$ :-) $\endgroup$ – Alexandre Eremenko Oct 31 '17 at 19:10

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