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I'm trying to answer Tu's Introduction to Manifolds exercise 1.3, which says in the first item:

Let $U \subset \mathbb{R}^n$ and $V \subset \mathbb{R}^n$ be open subsets. A $C^\infty$ map $F \colon U \to V$ is called a diffeomorphism is it is bijective and has a $C^\infty$ inverse $F^{-1} \colon V \to U$. Show that the function $f \colon ]\frac{-\pi}2, \frac\pi2[ \to \mathbb{R}$ given by $f(x) = \tan x$ is a diffeomorphism.

By doing some calculations, I arrived at the equality in the title: $\frac{d^2f}{dx^2} = f\frac{df}{dx}$. Note that both $f$ and $\frac{df}{dx}$ are continuous and defined everywhere in the given domain. Is it possible to justify formally that this implies that $f$ is in fact $C^\infty$? It is pretty clear to me that this is in fact true, because differentiating both sides gives $\frac{d^3f}{dx^3}$ in function of $\frac{df}{dx}$ and $f$ again, and this keeps happening forever in a way which preserves continuity for all derivatives. The problem is that I really don't know how to justify this formally.

J.G.
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Lucas Giraldi
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    $\tan$ is known to be analytic (it has a Taylor series representation), so in particular it is also $C^\infty$. – Vercassivelaunos Nov 15 '20 at 10:35
  • Oh, sure, but my point here is not to solve the exercise, but to justify this. I just want to know how one would go to do it so I can have this tool in my pocket. – Lucas Giraldi Nov 15 '20 at 10:37

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From the product rule, we see that if $g$ and $h$ are differentiable functions, then so it the product $gh$.

Thus the equation $$ f''=ff' $$ also shows that $f''$ is differentiable, since $f$ and $f'$ are. Repeat this logic to show that each derivative $f^{(n)}$ of $f$ exists and is differentiable.

Milk
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  • My problem is in fact with the "repeat this logic" part, because $f''' = f'f' + f''f = f'f' + ff'f'$ and $f'''' = f'f'' + f'f'' + f'''f + f''f' = 3f'(f'f) + (f'f' + ff'f')f$. Those functions are way different and I don't know how to generalize the argument. – Lucas Giraldi Nov 15 '20 at 10:46
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    You don't need all your derivatives to be in terms of $f$ and $f'$, you just need that the $n$th derivative $f^{(n)}$ is in terms of the 1st,2nd,...,$(n-1)$st derivatives. Use induction. – Milk Nov 15 '20 at 10:48