0
$\begingroup$

Any hint or demo to prove that the length of a continous parametrized curve defines a continous function in a normed space?

$\endgroup$
1
$\begingroup$

Define $f$ by $x\mapsto x^2\sin(1/x)$ if $x\neq0$ and $x\mapsto0$ if $x=0$. Note that $f$ is differentiable. However, its length $$\int_\Omega\sqrt{1+\left(f'\left(x\right)\right)^2}\text{d}x$$ is undefined on a closed interval $\Omega$ containing the origin and therefore not continuous.

$\endgroup$
  • $\begingroup$ @JackTalion Fine. Think of it as $\gamma(t)=(t,f(t))$. $\endgroup$ – wjm Mar 17 at 22:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.