Implicit function theorem: from local to global

Question

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 $-u<b<u$, 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.

Answer 1

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 $U$.

Answer 2

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.