My lecture notes seem to use this, although I can't shake the feeling that it's wrong. Let $\phi: [a,b] \to \mathbb{R}^n$ for $n \ge 2$ be smooth (infinitely differentiable) satisfying $|\phi '(t)| \neq 0$ for all $t \in [a,b]$. Then, $|\phi'(t)|$ is also a smooth function.
Does anyone have a nice proof for this? (My first 'counterexample' was $\log$, but it doesn't satisfy the condition $|\phi '(t)| \neq 0$.) Also, is $|\phi(t)|$ a smooth function too?
Although this is from a course in elementary differential geometry, my question is essentially a calculus question…
Best Answer
Let $g(v) = \langle v, v\rangle = \lVert v\rVert^2$ and $f(t) = \sqrt{t}$. Then $\lVert \phi'(t)\rVert = \left(f\circ g\circ\phi'\right)(t)$.
You know that $\phi'$ is smooth and nonvanishing.
It's easy to see that $\nabla_g(v) = 2v$; in other words, $\nabla_g$ is twice the identity, so $g$ is smooth. It follows that the composition $g\circ \phi'$ is smooth.
Moreover, we have that $g(v) = 0 \iff v = 0$, from which we see that the composition $g\circ\phi'$ is also nonvanishing.
Finally, it's a simple univariate calculus exercise to show that $f:[0,+\infty)\longrightarrow \Bbb R$ is smooth away from the origin. Hence, the composition $f\circ (g\circ \phi')$ is smooth.