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Define $L_f(a,b)$ denote the arc length of the graph of the function $f$ on $(a,b)$.

For a sequence of functions $f_n(x):D\to \mathbb{R}$ that converges to $f(x)$, even if $f_n$ converge uniformly to $f$, this isn't enough to grantee that that $L_{f_n}(a,b)\to L_f(a,b), \forall (a,b)\subset D$.

Example: Let $$ g(x) = \begin{cases} |x|, & \text{if $|x|\le1$} \\ g(x+2)=g(x), & \end{cases}$$ Now let $f_n(x)=\frac{1}{n}g(nx)$, $f_n$ converge uniformly to $0$, however $L_{f_n}(0,1)=\sqrt{2}, \ \forall n$ and $L_0(0,1)=1$. first four f_n(x)

Another example is The "proof" of $\pi=4$ that use a sequence of curves that converge uniformly to a circle with radius =0.5 with each of $f_n$ having total length of $4$, however uniform convergence doesn't imply the arclength converge to arclength of the function

If $f_n(x)$ are all differentiable at $(a,b)$ then if $f_n\to f$ uniformaly then $L_{f_n}(a,b)=L_f(a,b)$ this is easy to see by using $L_{f_n}(a,b)=\int_a^b\sqrt{1+(f'_n(x))^2}dx$

My questions are:

(1) If $f_n\to f$ pointwise and $L_{f_n}(a,b)\to L_f(a,b)$, must the convergence $f_n\to f$ be uniform?

(2) Is there a natural or standard type of convergence that guarantees $L_{f_n}(a,b)\to L_f(a,b)$?

(3) If $f_n\to f$ uniformly, does $L_{f_n}(a,b)$ have to converge ?

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    $\begingroup$ The last part is definitely false. Consider (for example) $f_n(x) = \frac{1}{n}\sin(\pi n^2 x)$ on $[0, 1]$. $f_n \to 0$ uniformly as $|f_n(x)|=\frac{1}{n}$, but the arc length of $f_n$ is at least $\frac{2}{n}\cdot n^2 = 2n$, because in $[\frac{i}{n}, \frac{i+1}{n}]$ you have $f_n=0$ at the endpoints and $f_n=\pm \frac{1}{n}$ in the middle. Hence $L_{f_n}(0,1)$ diverges. $\endgroup$ Commented yesterday
  • $\begingroup$ @KyanCheung This makes me believes that the convergence of arclength is stronger than uniform convergence. $\endgroup$ Commented yesterday
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    $\begingroup$ Beware that $f_n\to f$ uniformly does not imply that $f'_n\to f$ even pointwise, even if $f$ is differentiable. You might have forgotten primes over $f_n$ and $f$ just before your questions. $\endgroup$ Commented yesterday
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    $\begingroup$ In general just to define arclength you need some conditions on $f$ eg piecewise $C^1$; $\endgroup$ Commented yesterday
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    $\begingroup$ @conrad You can define arclength for rectificable graphs, $\endgroup$ Commented yesterday

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$(1)$ holds for compact intervals $[a,b]$ when $f$ is continuous. I guess you can adapt the proof below for open intervals $(a,b)$ but I think this will obscure the argument, so I will write it down just for compact intervals.

Suppose by contradiction that the convergence is not uniform. Then there is $\epsilon_0 > 0$ such that there is $n_k \rightarrow +\infty$ and $y_k$ such that $|f(y_k)-f_{n_k}(y_k)|> \epsilon_0$.

Given a partition of the interval $[a,b]$

$$\mathcal{P}=\{a=x_0, x_1, \dots ,x_n=b\},$$ with $x_i< x_{i+1}$, define $$L(f,\mathcal{P})= \sum_{i< n} \|(x_{i+1},f(x_{i+1})) - (x_{i},f(x_{i})) \|.$$

Then $$L_f[a,b]= \sup_{\mathcal{P}} L(f,\mathcal{P}),$$ when the sup runs over all partitions $\mathcal{P}$ of $[a,b]$.

Choose a partition $\mathcal{P}$ such that

$$L(f,\mathcal{P}) > L_f[a,b] - \epsilon_0/100.$$

Since $f$ is continuous WLOG we can suppose that $\mathcal{P}$ is fine enough such that

$$\|(w,f(w)) - (z,f(z)) \|< \epsilon_0/1000$$

for every $w,z\in [x_i,x_{i+1}]$ and every $i$.

Since $\lim_n f_n(x_i)= f(x_i)$ and $\lim_n L_{f_n}[a,b]= L_{f}[a,b]$ it follows that for large $k$

$$\|(x_{i+1},f_{n_k}(x_{i+1})) - (x_{i},f_{n_k}(x_{i})) \|< \epsilon_0/500$$ and $$L_f[a,b] - \epsilon_0/100 < L(f_{n_k},\mathcal{P}) < L_f[a,b] + \epsilon_0/100.$$

Consider the new partition $\mathcal{Q}_k= \mathcal{P}\cup \{y_k\}$. Then there is $i_k$ such that

$$x_{i_k} < y_{k} < x_{i_{k}+1}.$$

So

\begin{align} &L(f_{n_k},\mathcal{Q_k})- L(f_n,\mathcal{P}) \\ &= \|(x_{i_k+1},f_{n_k}(x_{i_k+1})) - (y_{k},f_{n_k}(y_k)) \| + \|(x_{i_k},f_{n_k}(x_{i_k})) - (y_{k},f_{n_k}(y_k)) \| - \|(x_{i_k+1},f_{n_k}(x_{i_k+1})) - (x_{i_k},f_{n_k}(x_{i_k})) \|\\ &\geq \| (y_{k},f_{n_k}(y_k)) - (y_{k},f(y_k))\| -\|(x_{i_k+1},f_{n_k}(x_{i_k+1})) - (x_{i_k+1},f(x_{i_k+1}))\| - \|(x_{i_k+1},f(x_{i_k+1})) - (y_{k},f(y_k))\| \\ &+ \| (y_{k},f_{n_k}(y_k)) - (y_{k},f(y_k))\| -\|(x_{i_k},f_{n_k}(x_{i_k})) - (x_{i_k},f(x_{i_k}))\| - \|(x_{i_k},f(x_{i_k})) - (y_{k},f(y_k))\| \\ &- \|(x_{i_k+1},f_{n_k}(x_{i_k+1})) - (x_{i_k},f_{n_k}(x_{i_k})) \| \\ &\geq 2 \| (y_{k},f_{n_k}(y_k)) - (y_{k},f(y_k))\| - \epsilon_0/2 > \epsilon_0. \end{align}

But this implies

$$L_{f_{n_k}}[a,b]\geq L(f_{n_k},\mathcal{Q_k}) > L_f[a,b] - \epsilon_0/2 $$

for every large $k$, that is impossible.

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