If $f_n$ uniformly converges to $f$, the arc length of graph of $f_n$ converges to the arc lenth of graph of $f$? If $f$ and $f_n$ are continuously differentiable functions on $[0,1]$, and $f_n$ uniformly converges to $f$, the the arc length of graph of $f_n$ converges to the arc lenth of graph of $f$ ?
I feel this proposition is wrong, but I cannot came up with some counter-example. In this question, answer said if every $f_n$ is convex, the lengths of graphs converge but this is sufficient condition.
I would be happy to give you a counter-example.
 A: Counterexample. Take $f_n(x)=\frac{\sin(nx)}{n}$ then $f_n$ converges uniformly to $f=0$ in $[0,1]$. On the other hand, the limit of the arclength of $f_n$ over $[0,1]$ is
$$
\begin{align}L_n&=\int_0^1\sqrt{1+f'(x)^2}\,dx=\int_0^1\sqrt{1+\cos^2(nx)}\,dx\\
&=\frac{1}{n}\int_0^n\sqrt{1+\cos^2(t)}\,dt\to \frac{1}{\pi}\int_0^{\pi}\sqrt{1+\cos^2(t)}\,dt>1
\end{align}$$
which strictly greater than $1$, the arclenght of $f$ over $[0,1]$.
P.S. For the limit of $L_n$ we applied  Proving a question about the average of a periodic function formula
A: Just for fun, another nerd counterexample.
Consider $I=[-1,1]$.
Take $f_1:I\to \Bbb R$ to be the function whose graph is the upper circle of diameter $I$. Its clear that its length is $\pi$ and it also holds that $|f_1(x)|\le 1$.
Take $f_2:I\to\Bbb R$ to be the function whose graph is

*

*the upper semicircle of diameter $[-1,0]$

*the upper semicircle of diameter $[0,1]$.

It is also clear that the length of its graph is, another time $\pi$.
Also holds that $|f_2(x)|\le\frac{1}{2}$.
Take $f_3:I\to\Bbb R$ to be the function whose graph is:

*

*The upper semicircle of diameter $[-1,-1/2]$

*The upper semicircle of diameter $[-1/2,0]$

*The upper semicircle of diameter $[0,1/2]$

*The upper semicircle of diameter $[1/2,1]$
The length of the graph is, another time $\pi$ and $|f_3(x)|\le \frac{1}{2^2}$.
Following inductively this construction, we get a sequence $(f_n(x))_n$ such that the length of its graphs is always $\pi$ but $|f_n(x)|\le\frac{1}{2^{n-1}}$, so $f_n\to0$ uniformly on $I$.
BUT the length of the graph of the limit function isn't $\pi$, but $2$.
PS: One can use this example to make a FALSE proof of the fact that $\pi=2$.
