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I came across this question and I am not sure how to prove it.

Show that the arc length integral is continuous in $C^1$.

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Could you be a little more specific? The $C^1$ norm on what exactly? Presumably the space $C^1([a,b], \Omega)$ where $\Omega \subset \mathbb R^n$? – JSchlather Mar 7 '13 at 2:53
I assume you tried to write down $\operatorname{length}(\gamma_1)-\operatorname{length}(\gamma_2)$ as the difference of two integrals? Which can be combined into integral of the difference? Where the integrand somewhat resembles the expression we see in the Mean Value theorem? – user53153 Mar 7 '13 at 2:55
Unfortunately I didn't have room in my schedule to take a calculus of variations so I have been teaching it to myself and found this question in a textbook. I mean the norm commonly used in C^1 which measures the size of the function and its derivative – user65439 Mar 7 '13 at 3:01

Let $f,g$ be two $C^1$ functions on $[a,b]$.

We have $$ \sqrt{1+f'^2}-\sqrt{1+g'^2}=\frac{f'^2-g'^2}{\sqrt{1+f'^2}+\sqrt{1+g'^2}}. $$ So $$ |\sqrt{1+f'^2(x)}-\sqrt{1+g'^2(x)}|\leq\frac{|f'^2(x)-g'^2(x)|}{2}\leq\frac{|f'(x)-g'(x)|(|f'(x)|+|g'(x)|)}{2} $$ for all $x\in(a,b)$.

Now fix $f$ and take $g$ such that $\|g'-f'\|_\infty\leq 1$. Then $\|g'\|_\infty\leq 1+\|f'\|_\infty$, so $$ |\sqrt{1+f'^2(x)}-\sqrt{1+g'^2(x)}|\leq\frac{(1+2\|f'\|_\infty)}{2}\|f'-g'\|_\infty $$ for all $x\in(a,b)$.

Denote $$ I(f):=\int_a^b\sqrt{1+f'^2(x)}dx $$ the arclength function.

We have, with $f$ and $g$ as above, $$ |I(f)-I(g)|\leq \frac{(b-a)(1+2\|f'\|_\infty)}{2}\|f'-g'\|_\infty. $$

So $I(f)$ is locally Lipschitz with respect to $\|f'\|_\infty$. A fortiori, with respect to the usual $C^1$ norm $\|f\|_\infty+\|f'\|_\infty$.

Hence it is continuous at $f$ for all $f$ in $C^1$.

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