# A question about differentiability of implicit functions

Let $$U \subseteq \mathbb{R}^2$$ be an open neighborhood of $$(0,0) \in \mathbb{R}^2$$. Let $$F: U \to \mathbb{R}$$ be a three times continuous differentiable function with $$D_2F(0,0) \neq 0$$. It exists a function $$g:(- \varepsilon, \varepsilon) \to \mathbb{R}$$ for a $$\varepsilon > 0$$ with $$g(0) = 0$$ and $$F(x,g(x)) = 0$$ for all $$x \in (- \varepsilon, \varepsilon)$$.

Now I have to show that $$g$$ is three times continuous differentiable in a neigborhood of $$0 \in \mathbb{R}$$ and I have to find formulas for $$g', g''$$ and $$g'''$$ that depend only on $$g$$ and the derivatives of $$F$$.

The last part should be easy, since we can write $$g'(x)=- \frac{D_1F(x,g(x))}{D_2F(x,g(x))}$$. $$g''$$ and $$g'''$$ can be calculated by differentiation of the previous derivative.

So my question is how to prove that $$g$$ is three times continuous differentiable in a neighborhood of $$0 \in \mathbb{R}?$$ I would appreciate some hints/ help.

• $g$ is $C^3$ because $F$ is. If you differentiate the formula implicitly, solve for $g'$, you will find the the RHS is $again$ differentiable (because $F$ is). Then, rinse and repeat! Conclusion: $g$ is as many times differentiable as $F$ is. For a formal proof by induction, see Pugh p 286 – Matematleta Jan 12 '20 at 16:45

Differentiability of $$g$$ in a neighbourhood of $$0$$ is one of the implications of the implicit function theorem.

So you can differentiate $$F(x,g(x)) =$$ in a whole neighbourhood of $$x=0$$:

$$F_1(x,g(x)) =- F_2(x,g(x)) g^\prime(x)$$ By assumption and continuity of $$D_2F$$ and $$g(x)$$, $$D_2F(x,g(x)) \neq 0$$ in a neighbourhood of $$x=0$$, so this can be solved for $$g^\prime(x)$$ in a neighbourhood of $$0$$. If you do that you will arrive at an equation $$g^\prime(x) =$$ something of class $$C^1$$ (it's not difficult, you already did it, but I want to leave something to do for you...).

So $$g$$ is $$C^2$$ in a neighbourhood of $$0$$. Differentiate once more to conclude it's even $$C^3$$.

• So the clue is to assume that $g$ is differentiable and to show that we can write it as something of class $C^1$ so it really is? – Ludwig M Jan 12 '20 at 16:42
• @lambda As far as I'm concerned the clue is to fully exploit the power of the implicit function theorem, which tells you something about a whole neighbourhood of the the point in question (including differentiability of $g$). Once you know that $g$ is $C^1$ and that you can solve for $g$ (which is also only true since the conclusion of the implicit function theorem holds in a whole neighbourhood) the rest is just formal reasoning. – Thomas Jan 12 '20 at 16:45
• The point is that when you solve for $g'$ the result is $again$ differentiable, so $g''$ exists. $g$ is as many times differentiable as $F$ is. Pugh has a nice proof by induction of this fact. – Matematleta Jan 12 '20 at 16:47
• @Matematleta well, partly. If you look closely at the question of the OP you will note that he already solved for $g^\prime$. The point here, imho, is that people are often ignorant about the fact that the implicit function theorem is a statement which is valid in a whole neighbourhood of $0$. Without this observation you cannot conclude differentiability of $g^\prime$, because for differentiability you (usually) need an open set. – Thomas Jan 12 '20 at 16:51
• Yes, indeed. But this follows directly from the proof of the theorem. My point was only that $g$ is $C^r$ whenever $F$ is. Rudin only proves the $C^1$ case. My favorite proof is the one in Loomis' book. And of course, as you point out. you always need to observe that $g$ is defined in some open nbhd of $0.$ In fact, otherwise $g$ would not be interesting at all! – Matematleta Jan 12 '20 at 16:56