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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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2 Answers

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.

This is probably not what you meant. Then the answer depends on what you did mean.

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 $x\in I_1$.

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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
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@joriki: just to clarify your example $f(x,y)=y^2-1$: the partial derivative condition ($\frac{\partial f}{\partial y}\neq 0$) is not fulfilled at $y=0$, this is why IFT doesn't hold globally in this case. (Maybe I misunderstood you case 2: were you saying that since $y=0$ is not a solution of $f(x,y)=0$, then you would think $\frac{\partial f}{\partial y}\neq 0$? This is indeed not how IFT works.)

In fact, there are further conditions necessary for a global IFT theorem $-$ at least there are for a global inversion theorem [$g(y)=x \Rightarrow y=\phi(x)$], which is closely related: consider $f(x,y)=x-g(y)$. These further conditions involve the topology of the image set of $g$ (simply connected condition) and the requirement that $g^{-1}(\textrm{compact})=\textrm{compact}$.

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