- #1

Buri

- 273

- 0

s(t) = integral [t_0, t] ||dg/du|| du.

The text says this is differentiable, so ds/dt = ||dg/du||. But I don't see why. I know that g is smooth, but the norm causes problems and so to apply the fundamental theorem of calculus I would have to know that ||dg/du|| is continuous. If g is also regular then ||dg/du|| is smooth, so it would follow, but I don't see how this follows if g is simply smooth.

Any help?