Implicit function theorem: from local to global

Suppose that we have $$F(x,y)=0$$ satisfying the usual hypotheses of the IFT at $$(0,0)$$, but such that not only $$F_y(0,0)\neq 0$$, but $$F(x,y)=0$$ and $$F_y(x,y)\neq 0$$ for all $$x\in U\ni0$$, where the open interval $$U\subset\mathbb{R}$$. Then instead of having only a unique local function $$y(x):(-\varepsilon,\varepsilon)\longrightarrow V'$$ such that $$F(x,y(x))=0$$ for some $$V'\subset\mathbb{R}$$, we should be able to assert the existence of a unique global function $$y^*(x):U\longrightarrow V''$$ such that $$F(x,y^*(x))=0$$. The way I was thinking of this was the following: say we have $$y=y_0$$, $$\varepsilon=\varepsilon_0$$ on $$(-\varepsilon_0,\varepsilon_0)$$. Then we can consider $$F(\varepsilon_0,y_0(\varepsilon_0))=0$$, $$F_y(\varepsilon_0,y_0(\varepsilon_0))\neq 0$$ and find another uniques $$y_1:(\varepsilon_0-\varepsilon_1,\varepsilon_0+\varepsilon_1)\longrightarrow V'''$$ and on the overlap by uniqueness $$y_0\equiv y_1$$. So defining $$y^*$$ by cases as $$y_0$$ on $$(\varepsilon_0,\varepsilon_0)$$ and $$y_1$$ on $$[\varepsilon_0,\varepsilon_1)$$ we have, repeating this inductively, a $$y_*$$ globally defined not only on $$(-\varepsilon_0,\varepsilon_1)$$, but on $$U$$. However this can be only if the sequence $$\varepsilon_i$$ does not decrease too fast and the boundary values $$y_i(\varepsilon_i)$$ satisfy $$F_y(\varepsilon_i,y_i(\varepsilon_i))\neq 0$$ and $$F(\varepsilon_i,y_i(\varepsilon_i))=0$$ at each step. The second of these issues is taken care of by the hypothesis that $$F(x,y)=0$$ for every $$x\in U$$, since this ensures that if $$F(\varepsilon_i,y_i(\varepsilon_i))\neq0$$ for some $$i$$, we have covered the whole of $$U$$ (I am thinking only about its right half, the reasoning extends symmetrically). My question is: what are usual hypotheses which on the other hand ensure that the $$\varepsilon_i$$ do not decrease too fast? Are the ones given above already enough? Is it possible to devise one that ensures this? I think perhaps by contradiction one can show that the $$\varepsilon_i$$ can only stop at the boundary of $$U$$, since if, say, they converge at some $$-u, where $$U=(-u,u)$$ for example. Then one can take $$b$$ and apply the IFT there, extending the unique solution a little further, against the hypothesis that it could not be done. Is this the right idea? Thank you all for any help.

• What if $F(x,y)=x - \tan^{-1} y$ (where we take the principal arctangent with range $(-\frac{\pi}{2}, \frac{\pi}{2})$) and $U = (-10, 10)$? To me that looks like it satisfies all your hypotheses, but the solution $y = \tan x$ blows up at $x = \pm \frac{\pi}{2}$ so there is no solution on all of $U$. Commented Apr 8, 2023 at 19:48
• It does not satisfy the first hypothesis $F(x,y)=0$ for all $x\in U$. It only satisfies it at $x=0$.
– xyz
Commented Apr 9, 2023 at 11:19

Yes this argument works quite well provided that you insert one minor edit: you should state explicitly that the open set $$U$$ is an open interval (a connected open subset of the real line.) Your proof boils down to asserting that the largest connected interval on which the solution exists is both open and relatively closed in $$U$$. The only nonempty subset of the connected set $$U$$ that has this topological property is $$U$$ itself.

As a side note, the argument fails if you attempt to extend the theorem to additional dimensions (with more variables), in which $$U$$ is a connected open set in Euclidean space of dimension $$d>1$$. In this multivariable setting, the global form of the theorem can fail if $$U$$ is not also assumed to be simply-connected. This complication occurred for example when Riemann first attempted to invert many-to-one analytic functions of a complex variable: several local inverses exist and are well behaved if we avoid certain obstructions (branch points), but the solutions swirl around one another in too complicated a fashion to fit together to yield a single-valued globally smooth answer. Hence the birth of topology and Riemann surfaces.

A three-dimensional example: The helicoid surface that is the "graph" of the multivalued function $$z= \theta =Arctan (y/x)$$ can be constructed by setting $$F(x,y,z)=x \sin z - y \cos z =0$$. It satisfies $$F_z\ne 0$$ as long as $$(x,y)\ne (0,0)$$. The removal of this line from $${\mathbb R}^3$$ creates a connected open region $$U$$. However you cannot globally solve for $$z$$ as a single-valued function of $$(x,y)$$ on $$U$$.

• Yes you are correct, I only wrote 'open' and implicitly assumed interval, since later I explicitly write it of the form $(-u,u)$. Interesting, I did not think about the fact that we would need it simply connected in more dimensions, it is true. Thank you so much for this pointer, I actually wanted to see it through in more dimensions next, and your comments answer that as well. Thank you.
– xyz
Commented Apr 8, 2023 at 14:45
• For a finite open set, like usually in the IFT statement, isn't path connectedness enough though? Say I have a hole in my open set. I could still do the argument and cover it (as long as it is bounded) with balls. Why would the holes bother the argument? Unless I am missing some pathological shape allowed by path connectedness...
– xyz
Commented Apr 14, 2023 at 18:54
• The classic example is the function $w=f(z)= e^z$ in the complex plane, which never attains the value $w=0$. If you try to sequentially invert $f$ on a collection of balls that are centered on a parametrized circle $w= e^{ i t}$ that wraps around $w=0$ you discover that the continuous solution along that path is $z(t)=1+ it$ , which at $t=0$ and $t=2\pi$ disagree. Commented Apr 14, 2023 at 23:41
• Yes, I meant the situation when both domain and range space are big. Commented Apr 15, 2023 at 16:51
• Couldn't resist adding a picture of a simple example! Commented Apr 16, 2023 at 0:16

Maybe a different approach will shed a bit more light on the matter: The local function $$\varphi :U \longrightarrow\mathbb{R}$$, which satisfies $$F(x,y)=0 \Longleftrightarrow y=\varphi(x)$$ and $$y_0 = \varphi(x_0)$$, is continuously differentiable and alternatively characterized by the ODE $$\begin{cases}\varphi'(x) &= - F_x(x,\varphi(x))\cdot (F_y(x,\varphi(x))^{-1}\\ \varphi(x_0) &= y_0. \end{cases}$$

Thus, all the usual theory about ODEs can be used to treat $$\varphi$$, including the theory about global existence: With steps similar to what you described, the maximal existence interval of $$\varphi$$ can be extended left and right, as long as there is no "blow-up".

For example, if $$F(x,y) = y-\sqrt{1-x^2}$$, one has $$F_y(x,y) = 1\neq 0$$ for all $$(x,y)\in\mathbb{R}^2$$ and e.g. $$F(0,1)=0$$, but the solution to $$\begin{cases}\varphi'(x) &= -\frac{x}{\varphi(x)}\\ \varphi(0) &=1\end{cases}$$ can not be extended over the interval $$x\in (-1,1)$$, since $$\lim\limits_{x\rightarrow\pm 1} \varphi(x) = \lim\limits_{x\rightarrow\pm 1} \sqrt{1-x^2} = 0$$, i.e. $$\lim\limits_{x\rightarrow\pm1} \varphi'(x)$$ explodes. For more examples, see e.g. this short script.

• I like this connection with ODE's. Then you think the hypotheses I gave are sufficient to have a global solution on the open set $U$ (in your example it would be $(-1,1)$).
– xyz
Commented Apr 8, 2023 at 12:38
• @GGG It's important to be precise about the terminology: A "global" solution means that the open set $U$ is all of $\mathbb{R}$. This is of course not the case if the interval is just $(-1,1)$. So in general, even though your hypotheses are fulfilled, the explicit function does not have to exist globally. However, there are some criterions that prevent a blow up, such as for example if the right hand side of $y'(x) = f(x,y(x)$ is uniformly bounded. Commented Apr 8, 2023 at 13:34
• By global I only meant on $U$, that is the set on which the hypotheses are satisfied, apologies. That it would not work past $U$, I was taking it as trivial.
– xyz
Commented Apr 8, 2023 at 14:40