# A sequence of analytic functions with converging path integrals converges how?

Suppose you have a sequence $\{f_n\}_{n \in \mathbb Z^+}$ of analytic functions on a compact set $K$ such that for every smooth curve $\gamma$ in $K$ the sequence $\{\int_\gamma f_n\}_{n \in \mathbb Z^+}$ of path integrals over $\gamma$ converges. What can you say about the convergence of $f_n$?

Shouldn't the convergence should be uniform? Is the fact that $K$ is locally convex enough to allow you to define a primitive locally via a path integral and prove local uniform convergence?

## 1 Answer

(First of all, analyticity is defined on open sets; I presume that functions are analytic in some domain $\Omega$ and we are to consider their behavior on a compact subset $K\subset\Omega$.)

The answer is negative in general. Take a sequence of holomorphic functions $F_n$ that converges pointwise to a non-holomorphic function (this can happen). Let $f_n=F_n'$. Then for any curve $\gamma$ we have $\int_\gamma f_n = F_n(b)-F_n(a)$ where $a,b$ are the endpoints of $\gamma$. Thus, $\int_\gamma f_n$ has a limit. However, we do not have locally uniform convergence of $f_n$.

Indeed, suppose that $f_n$ converge locally uniformly. Then so do $F_n$, because $F_n(z_0)$ converge at some fixed point $z_0$, and $F_n(z)-F_n(z_0)$ is found by integrating $f_n$. It follows that the limit $\lim F_n$ is holomorphic, contradicting our choice of $F_n$.