# Local versus global implicit function

Suppose the equation $f\left(x,y\right)=0$, with $x\in I_{1}$ and $y\in I_{2}$, $I_{1}$ and $I_{2}$ being open intervals. Additionally, consider that the conditions required to apply the Implicit Function Theorem (IFT) are verified for all $\left(x_{0},y_{0}\right)\in I_{1}\times I_{2}$. Hence, we can conclude that in a neighborhood containing the point $\left(x_{0},y_{0}\right)$, the equation $f\left(x,y\right)=0$ defines implicitly $y$ as a function of $x$.

And my question is: Since the conditions of IFT hold for all $\left(x_{0},y_{0}\right)\in I_{1}\times I_{2}$, is it true that the equation $f\left(x,y\right)=0$ defines implicitly $y$ as a function of $x$ with the domain of this implicit function being $I_{1}$?

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One of the conditions for the IFT to apply at $(x_0,y_0)$ is that $f(x_0,y_0)=0$. Thus, if we take what you write literally, the answer is that the conditions cannot be fulfilled on all of $I_1\times I_2$, since then $f(x,y)\equiv0$ on all of $I_1\times I_2$ and this function doesn't fulfill the requirement on the partial derivative anywhere.
If you mean that the condition on the partial derivative is fulfilled wherever $f(x_0,y_0)=0$, then the answer is no. For instance, for $f:[0,1]\times [-2,2]$ with $f(x,y)=y^2-1$, there are two different implicit functions, $g(x)=1$ and $g(x)=-1$, and thus no global function.
If you mean that the condition on the partial derivative is fulfilled everywhere but $f(x_0,y_0)=0$ is only fulfilled somwhere, then the answer is still no, since there is no guarantee that for a given value of $x\in I_1$ there is any value of $y\in I_2$ such that $f(x,y)=0$.
However, under the assumption that for each $x\in I_1$ there is at least one $y\in I_2$ such that $f(x,y)=0$, and that the condition on the partial derivative is fulfilled everywhere in $I_1\times I_2$, then in this rather restricted sense the answer is yes, since for given $x$ a continuously differentiable function can't take the value $0$ for two different values of $y$ without the partial derivative with respect to $y$ vanishing somewhere in between, so there is guaranteed to be exactly one value of $y\in I_2$ with $f(x,y)=0$ for each $x\in I_1$.
 Thanks for your answer, Joriki. My setting corresponds to the last one that you identify, and I still have a question. Since the IFT states that the implicit function is unique, is it really necessary to argue with the impossibility of the derivative having two zeros? – Paul Smith Jul 4 '11 at 11:26 @Paul: Yes. The IFT only applies to the points where $f(x,y)=0$. Without using further information about the function at other points, there's no way to exclude counterexamples such as the one I gave, $f:[0,1]\times [-2,2]$ with $f(x,y)=y^2-1$. The IFT holds because the continuous partial derivative is non-zero in some neighbourhood of the point, but to extend it globally you'd need to use the fact that it's non-zero everywhere, not just in neighbourhoods of points with $f(x,y)=0$. – joriki Jul 4 '11 at 12:21 Thanks,Joriki, for the clarification. – Paul Smith Jul 4 '11 at 13:24