Let's say we have $F(x, y) = x^2 +y^2 - 1$ and we want to apply the implicit function theorem. Then, after some calculations we see that for $y \neq 0$ we can indeed apply the theorem which means that there exist two open sets $U:=]-1, 1[$ of $x$, and $V:=]0, 2[$ of $y$, and a function $g: U \to V$ such that $F(x, g(x)) = 0$ $\forall$ $x \in U$. In this particular case we can explicitly write down $g$ by $g(x)=\sqrt{1-x^2}$ where $g(x)=y$ and $y>0$.
What I don't understand is that even if $y= 0$, $x=1$ and $x=-1$ are allowed $g$, still satisfies $F(x, g(x)) = 0$ $\forall$ $x \in U$. So is it right that if the conditions of the implicit function theorem are not satisfied this does not imply that there doesn't exist such a function at all?
Maybe I didn't understand the theorem correctly... any comments or help are welcome
I'll make two remarks:
It's true that that $F(x,g(x))=0$ for all $x\in [-1,1]$, but the idea of the implicit function theorem is that it gives sufficient conditions for a point in the level set of $F$ to have an open neighborhood over which the level set can be written locally as the graph of a function $g$. In this example, the points $(-1,0)$ and $(1,0)$ do not have any open neighborhoods in which the set $\{F(x,y)=0\}=\{x^2+y^2-1=0\}$ can be written as the graph of a function of $x$. However, $\frac{\partial F}{\partial x}\neq0$ at both of these points, which means there are functions $h$ such that $F(h(y),y)=0$ for neighborhoods of each of these points, respectively.
You are correct that the converse of the implicit function theorem does not necessary hold. For an easy example, the derivative of the function $F(x,y)=x^3-y^3$ vanishes at $(0,0)$, but its zero level set can be written globally as the graph of the function $g(x)=x$ or $g(y)=y$.
In conclusion, you should understand the implicit function as providing a sufficient, but not necessary, condition to be able to express $\{F(x,y)=0\}$ locally as the graph of another smooth function.