I am trying to solve a problem, its statement is:
Let $\Omega \subset \mathbb C$ an open set and $f_n,f:\Omega \to \mathbb C$. Show that if $f_n \rightrightarrows f$ on a curve $\int_{\gamma} \subset \Omega$, then $_{\gamma} f_n(z)dz \to \int_{\gamma} f(z)dz$.
I don't have any idea how can I prove this. Moreover, I have doubts related to the conditions of the statement: wouldn't the functions $f_n$ have to be holomorphic or at least continuous? and $\gamma$ piecewise-smooth differentiable. What conditions do I need for this statement and what would be the idea or sketch of a solution (hints).
Best Answer
Exactly what conditions you need on $\gamma$ will depend on how you've defined the integral, but integrability of the $f_n$ (and thus $f$) should be enough. Let's assume $\gamma$ is smooth. The integral $\int_\gamma f_n$ is by (the usual) definition $\int_0^1 f_n(\gamma(t))\gamma'(t)\,dt $, taking $\gamma:[0, 1]\to\Omega$. Break the integral into real and imaginary parts, and use the corresponding result for the real case: For a compact interval $I\subset {\mathbb{R}}$ and integrable $g_n:I\to {\mathbb{R}}$ with $g_n\to g$ uniformly, the function $g$ is also integrable with $$\int_I g_n\to \int_I g.$$