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

Question

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

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

Answer 1

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 $C^1$.