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Let $F:A\subseteq\mathbb{R}^2\to\mathbb{R}$ be a function, where $A$ is open in $\mathbb{R}^2$ and $F\in\mathcal{C}^k(A)$ (we take $\mathbb{R}^2$ and $\mathbb{R}$ for the sake of simplicity), and let $(x_0,y_0)\in A$ be a point such that $F(x_0,y_0)=0$ and at least one partial derivative of $F$ is nonzero in $(x_0,y_0)$, let's assume $F_y(x_0,y_0)\neq0$.

The implicit function theorem says that there exist positive $\varepsilon$ and $\delta$ such that the locus of points $(x,y)$ such that $F(x,y)=0$ can be represented as the graph of a function $\varphi:(x_0-\varepsilon,x_o+\varepsilon)\to(y_0-\delta,y_0+\delta)$, moreover $\varphi\in\mathcal{C}^k$ of this neighbourhood of $x_0$ and we get an expression for the derivatives of $\varphi$ in terms of the derivatives of $F$

Now, if $F$ is $\mathcal{C}^\infty$ of its the domain the same is true for $\varphi$ and I can get a taylor expansion of $\varphi$ around $x_0$, but we didn't discuss the analyticity of $\varphi$ in class, which conditions do I need on $F$ to ensure the analyticity of $\varphi$?

For sure we need $F$ to be infinitely differentiable and I'd guess that $F$ being analytic is required, if $F$ being $\mathcal{C}^\infty$ isn't enough is there a relatively simple counterexample?

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Its the same old example: $$F(x,y):=\cases{y-e^{-1/x^2}\quad&$(x\ne0)$\cr y&$(x=0)$\cr}\ .$$

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  • $\begingroup$ Ah, of course, I should have thought of just considering $y-f(x)$. Is it true though that $F$ analytic implies that the implicit function is analytic? $\endgroup$ – Alessandro Codenotti May 9 '16 at 19:57

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