# 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.

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.

• 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. Mar 6, 2019 at 2:54

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 $$n\to\infty.$$