Uniform convergence of sequence of functions

Question

Given that $$\gamma_n\rightarrow\gamma$$ uniformly, can we conclude that $$\int^b_a\|\gamma_n'\|\rightarrow\int^b_a\|\gamma'\|$$ uniformly?

I know that we even do not have $$\gamma_n'\rightarrow\gamma'.$$ So it seems to me it is unlikely??? Are there counterexamples? Further, is the $\gamma_n'\rightarrow\gamma'$ sufficient? If so, are there stricter conditions?

Let's keep things simple, like the map goes from closed interval $[a,b]$ to $\mathbb{R}^n$ and the integral has $dx$ missing and so forth.

EDIT I know that if we can find $c$ such that $\gamma_n(c)\rightarrow\gamma(c)$, then we have, $\gamma_n'\rightarrow\gamma'$? How do I argue that this implies $\|\gamma_n'\|\rightarrow\|\gamma'\|$

Final Question when is it true?

Answer 1

If we know $\gamma_n\to\gamma$ uniformly, we don't know if $\gamma_n$ or $\gamma$ is differentiable to begin with, so of course your first question has no as an answer. Your another big misconception is that we know nothing about weather $\gamma_n$ or $\gamma$ being rectifiable.

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If $\gamma'_n$ exists, does it imply $\gamma'$ exists? No. See this.

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If $\gamma'_n,\gamma'$ both exist, does it imply $\gamma_n'\to\gamma'$? No.
Counter example:
let $\gamma_n(x) = 2^{-n}\sin(2^n x)$. $\gamma_n(x)\to 0$ uniformly, but $\gamma_n'(x) = \cos(2^n x)$ does not have a limit.

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From here you can see your EDIT is wrong, we have here $\gamma_n(0)\to \gamma(0) = 0$, but $\gamma_n'$ doesn't tend to a limit.

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If $\gamma_n',\gamma'$ both exists and $\gamma'(x_0)\to\gamma'(x_0)$ for some $x_0$ (note the difference with your EDIT), then $\gamma_n'\to\gamma'$ everywhere is still false.
Counter-example, change the last example to $\gamma_n(x) = 2^{-n}cos(2^nx)$, then we see $\gamma_n'(x) = -\sin(2^nx)$ and $\gamma_n'(0) = 0 \to \gamma'(0) = 0$, but again $\gamma_n'$ does not tend to $\gamma'$

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Final question, when is it true? If $\gamma'_n\to\gamma'$ (exist) and convergent uniformly, then $|\gamma'_n|\to|\gamma|$ uniformly. This is easy to prove. If further we have $\gamma_n'$ being Riemann integrable, then we have $$\int_a^b ||\gamma_n'||dx =\int_a^b ||\gamma'||dx $$ Then further, if we know $\gamma_n$ are continuously differentiable ($\gamma_n'$ continuous, therefore also Riemann integrable for above) then we have the curve length equal to the integral therefore equality as you want.

But if you think twice, you will realize we don't need $\gamma_n\to\gamma$ uniform as a condition at all.
Suppose $\gamma_n$ continuously differentiable, $\gamma$ differentiable and $\gamma'_n\to\gamma'$ uniformly, it follows that $\gamma'$ is continuous, and $$\Lambda(\gamma_n) = \int_a^b ||\gamma_n'||dx \to \int_a^b ||\gamma_n||dx = \Lambda(\gamma)$$
The curves have same length even though they might not converge at all!
Of course if we further have $\gamma_n(x_0)\to\gamma(x_0)$ (one point convergence) then $\gamma_n\to\gamma$ uniformly.

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As you might notice, the condition I gave is quite "weak". I allow $\gamma_n'\to\gamma'$ uniformly and ask $\gamma_n$ to be continuously differentiable. It will be much harder a question if I forbid $\gamma_n$ to be continuously differentiable, then since we don't have a good formula for curve length, it will be very complicated.

Answer 2

No, you can see two geometric counterexamples in this video: a curve $\gamma_n$ can uniformly converge to a second curve $\gamma$ without that $\displaystyle \text{lg}(\gamma_n)=\int_0^1||\gamma_n'(t)||dt$ converges to $\displaystyle \text{lg}(\gamma)=\int_0^1 ||\gamma'(t)|| dt$.

Answer 3

Let $\gamma(t) = t$ for $t \in [0,1]$. Let $\gamma_n(t) = \gamma(\frac{1}{n}\lfloor n t \rfloor)$. Then $\lim_n \gamma_n(t) = \gamma(t)$ for all $t$ and the convergence is uniform, but $\gamma_n'(t)$ doesn't exists for all $t \in [0,1]$. Furthermore, when it does exist, $\gamma_n'(t) = 0$, so $\int \|\gamma_n'\| = 0$, whereas $\int \|\gamma\| = \frac{1}{2}$.