Let $C$ be a smooth curve in $\Bbb C$ and suppose that the sequence of continuous functions $f_n$ converges to $f$ uniformly on the curve $C$.

Show that $\int _C f_n(z) dz $ converges to $\int_C f(z) dz$.

Suppose that $f$ is a continuous function on a domain $\Omega $ in $\Bbb C$. Suppose that the continuously differentiable curves $C_n$ converge uniformly to the continuously differentiable curve $C$ in $\Omega$.

Show that $\lim\limits_{n\to \infty} \int _{C_n} f(z) dz\neq \int_C f(z) dz$ for some $C_n$ and $f$ continuous.

Now the first question I solved it like this:

Given $\epsilon >0$ there exists $m\in \Bbb N$ such that

$|f_n(z)-f(z)|<\epsilon $ forall $n\ge m$ and $\forall z\in C$.

Then $\left|\int _C f_n(z) dz-\int_C f(z) dz \right|\le \int _C \left|f_n(z)-f(z)\right|dz $

$< \epsilon \times $ length of the curve $C$ $\forall n\ge m$ and $\forall z\in \Bbb C$.

Is it correct?

I am stuck on the 2nd part and unable to make any progress. How should I find $C_n$?I need some help.

  • $\begingroup$ You meant $\left|\int _C f_n(z) dz-\int_C f(z) dz \right|\le \int _C \left|f_n(z)-f(z)\right| |dz| = \int_a^b |f_n(C(t))-f(C(t))| |C'(t)|dt$. A generalization is to assume that $|C'(t)|$ is only integrable. $\endgroup$
    – reuns
    Mar 6, 2019 at 2:54

1 Answer 1


Your proof is fine. For the second part, with $\Omega=\mathbb C,$ take $\gamma_n(t)=\frac{1}{\sqrt n}e^{2\pi int}$ on $[0,1]$. Then, for each integer $n,\ \gamma_n\in C^1([0,1]).$ Now, $\gamma_n\to \gamma=0$ pointwise and since

$|\gamma_n(t)|= \frac{1}{\sqrt n}$, the convergence is uniform. So, we have $\int_{\gamma}f(z)dz=0$ for $\textit{any}$ continuous $f$.

On the other hand, with $f(z)=\text{Re}z,\ \int_{C_n}f(z)dz=2\pi i\int^1_0\cos 2\pi nt\cdot e^{2\pi int}dt$, which does not converge to zero as $n\to \infty$ because in particular the imaginary part

$2\pi\int^1_0\cos^22\pi nt dt=2\pi \dfrac{\sin\left(4{\pi}n\right)+4{\pi}n}{8{\pi}n}\to \pi$ as $n\to\infty. $


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