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As the implicit function theorem shows, if

(i)Function F is continuous in the region D$\subseteq R^2$;

(ii)F($x_0,y_0)=0,P_0(x_0,y_0)\in$D;

(iii)There is a continuous partial derivative $F_y$(x,y)=0 in the region D;

(iv)$F_y(x_0,y_0)\neq$0;

Then there uniquely exists the function y=f(x) defined in the interval ($x_0-\alpha$,$x_0+\alpha$),that

1 f(x$_0$)=$y_0$,(x,f(x))$\in$U(P$_0$)when x$\in$(x$_0$-$\alpha$,x$_0$+$\alpha$)and F(x,f(x))$\equiv$0;

2 f(x) is continuous in ($X_0-\alpha$,$X_0+\alpha$).

And what if the conditions are weaker than those above? That is (i)Function F is continuous in the region D$\subseteq R^2$;

(ii)F($x_0,y_0)=0,P_0(x_0,y_0)\in$D;

(iii)There is a continuous partial derivative $F_y$(x,y)=0 in the region D;

In other words,what is the conclusion for the existence of implicit function in the branch or the subset of $R^2$?

There have been a series conclusions in complex space，who can list some of them and what's the relationship with the situation of complex space and real space?

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  • $\begingroup$ What are U and P_0? $\endgroup$
    – Thierry Zell
    Commented Aug 14, 2010 at 14:32
  • $\begingroup$ U stands for neighborhood，and $P_0=(x_0,y_0)$ $\endgroup$
    – DarkLight
    Commented Aug 14, 2010 at 14:53
  • $\begingroup$ The condition (iii) $F_y(x,y)=0$ in the region would simply imply that F is a function of x only. Maybe what you wanted instead to replace (iii) by $F_y(x_0,y_0)=0$? Anyway, it's easy to see that there's no way to generalize the IFT in that case (try $x^2+y^2=1$). $\endgroup$
    – Thierry Zell
    Commented Aug 14, 2010 at 15:31
  • $\begingroup$ In the first statement of the implicit function theorem, don't (iii) and (iv) contradict each other? The implicit function theorem I know has condition (iv) only and no zero condition such as (iii). $\endgroup$
    – Deane Yang
    Commented Aug 14, 2010 at 15:45
  • $\begingroup$ And, what about $F(x,y)=x^2+y^2$? Or $F\equiv0$ identically? Or $F=dist((x,y),K)^2$ where dist is the euclidean distance and $K is any closed set? $\endgroup$
    – Piero D'Ancona
    Commented Aug 14, 2010 at 18:05

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