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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.

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    $\begingroup$ $g$ 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 $\endgroup$ – Matematleta Jan 12 '20 at 16:45
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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 $C^3$.

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  • $\begingroup$ 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? $\endgroup$ – Ludwig M Jan 12 '20 at 16:42
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    $\begingroup$ @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. $\endgroup$ – Thomas Jan 12 '20 at 16:45
  • $\begingroup$ 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. $\endgroup$ – Matematleta Jan 12 '20 at 16:47
  • $\begingroup$ @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. $\endgroup$ – Thomas Jan 12 '20 at 16:51
  • $\begingroup$ 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! $\endgroup$ – Matematleta Jan 12 '20 at 16:56

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