Show there exists $g:(0,1)\to\mathbb{R}^2$ such that $h\circ g$ constant, $g$ continuously differentiable and $g$ injective

Let $h\in C^1(\mathbb{R}^2,\mathbb{R})$. Show: There exists $g:(0,1)\to\mathbb{R}^2$ such that $h\circ g$ constant, $g$ continuously differentiable and $g$ injective.

Dini's theorem:

Let $\Omega \subseteq \mathbb{R}^2$ open, $f \in C^1(\Omega, \mathbb{R})$ and $p \in \Omega$ with $f(p) = 0$.

If $\frac{\partial f}{\partial y}(p) \neq 0$, then there exist open intervals $V_1, V_2$ with $p \in V_1 \times V_2 \in \Omega$ and $g \in C^1(V, \mathbb{R})$ with $$\{(x,y) \in V_1 \times V_2 \mid f(x,y) = 0 \} = \{(x,g(x) \mid x \in V_1\}$$ and $g$ furthermore satisfies $$g' = -\left(\frac{\partial f}{\partial y}(x, g(x))\right)^{-1}\frac{\partial f}{\partial x}(x, g(x)).$$

Proof of existance of $g$:

Let without loss of generality $\frac{\partial f}{\partial y}(p) \gt 0$.

Then there exists $\epsilon \gt 0$ with $\overline{B_\epsilon(p)} \subseteq \Omega$ and $\frac{\partial f}{\partial y} \gt 0$ on $\overline{B_\epsilon(p)}$.

Let $y^- = p - \frac{\epsilon}{2}$ and $y^+ = p + \frac{\epsilon}{2}$. Then there exists $\delta \in (0, \epsilon)$ with $f(x, y^-) \lt 0 \lt f(x, y^+)$ for $x \in (p - \frac{\delta}{2}, p + \frac{\delta}{2}) =: V_1$ and $(y^-, y^+) =: V_2$.

For $x \in V_1$ there exists exactly one $y \in V_2$ with $f(x,y) = 0$ (mean value theorem). Set $y(x) := g$ and the claim follows.

• This might be a stupid question, but why can't $h$ just be constant everywhere? Oh is $h$ given to you? – user142299 May 19 '14 at 15:21
• Are you allowed to use the implicit function theorem? – Etienne May 19 '14 at 15:25
• @NotNotLogical $h$ is just defined to be one time continuously differentiable from $\mathbb{R}^2$ to $\mathbb{R}$, no more information given. I think implicit function theorem has to be used here, buf I don't know how... – sj134 May 19 '14 at 15:26
• So now the answer is in the question? – Etienne May 23 '14 at 5:42
• @Etienne Just the existence follows from the theorem, I am still looking for good explanations on why $h\circ g$ is constant, $g$ is continuously differentiable and injective. – sj134 May 23 '14 at 5:54

Here is a hopefully complete answer. Basically, this is just an elementary proof of the implicit function theorem for functions of 2 variables.

If $h$ is constant, you can take as $g$ any injective $\mathcal C^1$ map from $(0,1)$ into $\mathbb R^2$; for example $g(t)=(t,0)$.

Now, assume that $h$ is not constant. Then the partial derivatives of $h$ cannot be both identically $0$. So we may assume for example that $\frac{\partial h}{\partial y}(x_0,y_0)\neq 0$ for some point $(x_0,y_0)\in\mathbb R^2$; and without loss of generality, we may also assume that $\frac{\partial h}{\partial y}(x_0,y_0)> 0$. Set $$c:= f(x_0,y_0)\, .$$

Since $\frac{\partial h}{\partial y}$ is continuous, one can find an open interval $I_0\ni x_0$ and $\delta >0$ such that $$\frac{\partial h}{\partial y}(x,y)>0\quad {\rm for\; any\;}\; (x,y)\in I_0\times [y_0-\delta,y_0+\delta]\, .$$

The proof will be divided into several steps.

Step 1. One can find an open interval $I$ with $x_0\in I\subset I_0$ such that and the following holds: for every $x\in I$, there is a unique point $y=y(x)\in [y_0-\delta,y_0+\delta]$ such that $h(x,y(x))=c$.

To prove this, observe first that the map $y\mapsto f(x_0,y)$ is increasing on $[y_0-\delta,y_0+\delta]$ because $\frac{\partial h}{\partial y}(x_0,y)>0$ on this interval by the choice of $\delta$. Since $f(x_0,y_0)=c$, it follows in particular that $h(x_0, y_0-\delta)<c<h(x_0,y_0+\delta)$. Now the set $$U=\{ x\in\mathbb R;\; h(x,y_0-\delta)<c<h(x,y_0+\delta)\}$$ is an open set in $\mathbb R$ by the continuity of $h$, and $x_0\in U$ by what has just been observed. So one can find an open interval $I_1$ such that $x_0\in I_1\subset U$. If we set $I:=I_0\cap I_1$ then $x_0\in I\subset I_0$ and $$h(x,y_0-\delta)<c<h(x,y_0+\delta)\quad{\rm for\; every\;}x\in I\, .$$ Let us fix any $x\in I$. Then the map $y\mapsto h(x,y)$ is continuous, and it is increasing on $[y_0-\delta, y_0+\delta]$ because $\frac{\partial h}{\partial y}(x,y)>0$ on this interval, and $h(x,y_0-\delta)<c<h(x,y_0+\delta)$. By the intyermediate value theorem, it follows that there exists a unique $y=y(x)\in [y_0-\delta, y_0+\delta]$ such that $h(x,y(x))=c$. This concludes Step 1.

Step 2. The map $x\mapsto y(x)$ is continuous on $I$.

We prove this by contradiction. Assume that this map is not continuous at some point $x\in I$. Then one can find a sequence $(x_n)\subset I$ converging to $x$ and $\varepsilon >0$ such that $\vert y(x_n)-y(x)\vert\geq \varepsilon$ for all $n\in\mathbb N$. Since $y(x_n)\in y_0-\delta,y_0+\delta]$, One can find a subsequence $(y(x_{n_k}))$ and a point $y\in [y_0-\delta,y_0+\delta]$ such that $y(x_{n_k})\to y$ (by Bolzano-Weierstrass). Bu we have $h(x_{n_k},y(x_{n_k}))=c$ for all $k$, so $h(x,y)=c$ because $h$ is continuous and $(x_{n_k},y(x_{n_k}))\to (x,y)$. Since $y\in [x_0-\delta,x_0+\delta]$, it follows that $y=y(x)$; but this is a contradiction since $\vert y(x_{n_k})-y(x)\vert\geq\varepsilon$ for all $k$ and hence $\vert y-y(x)\vert\geq\varepsilon >0$. This concludes Step 2.

Step 3. The map $x\mapsto y(x)$ is $\mathcal C^1$ on $I$.

It is enough to show that this map is differentiable at any point $x\in I$, with $$y'(x)=-\frac{\frac{\partial h}{\partial x}(x,y(x))}{\frac{\partial h}{\partial y}(x,y(x))}\cdot$$ Indeed, since the partial derivatives of $h$ are continuous and the map $x\mapsto y(x)$ is continuous by Step 2, this formula will then show that $y'$ is continuous, i.e. that $y$ is $\mathcal C^1$.

Let us fix $x\in I$. Set $$a:=\frac{\partial F}{\partial x}(x,y(x))\quad{\rm and}\quad b:=\frac{\partial F}{\partial y}(x,y(x))$$ Keep in mind that $b\neq 0$ because $\frac{\partial F}{\partial y}(x,y)>0$ on $I\times [y_0-\delta,y_0+\delta]$.

By step 2, one may write $$y(x+h)=y(x)+\varepsilon(h)\, ,$$ where $\varepsilon(h)\to 0$ as $h\to 0$. Moreover, we also have (by the differentiability of $h$ at $(x,y(x))$)

\begin{eqnarray} h(x+h,y(x+h))&=&h(x+h, y(x)+\varepsilon (h))\\ &=&h(x,y(x))+ a\, h+b\, \varepsilon (h)+R(h)\, , \end{eqnarray} where $R(h)=o(\vert h\vert+ \vert\varepsilon (h)\vert)$ as $h\to 0$, i.e. $$\lim_{h\to 0}\frac{R(h)}{\vert h\vert+ \vert\varepsilon (h)\vert}=0\, .$$ Since $h(x+h,y(x+h))=c=h(x,y(x))$, this can be rewritten as \begin{equation}\varepsilon(h)=-\frac{a}{b}\, h-\frac1b R(h).\end{equation} If $h$ is small enough, we have $\left\vert-\frac1b R(h)\right\vert\leq \frac12\vert h\vert$ because $R(h)=o(\vert h\vert+\vert\varepsilon (h)\vert)$, so that $\vert\varepsilon(h)\vert\leq \left\vert\frac{a}{b}\right\vert \vert h\vert+\frac12(\vert h\vert+ \vert\varepsilon (h)\vert)$ and hence $\vert\varepsilon (h)\vert\leq C\,\vert h\vert$, where $C=2\vert a/b\vert+1$. This gives $\vert\varepsilon (h)\vert+\vert h\vert\leq (C+1)\vert h\vert$, which implies that in fact $$\lim_{h\to 0}\frac{R(h)}{\vert h\vert}=0\, .$$ Since \begin{eqnarray}y(x+h)&=&y(x)+\varepsilon(h)\\ &=&y(x)-\frac{a}{b}\, h-\frac1b R(h)\, ,\end{eqnarray} we conclude that the map $y$ is indeed differentiable at $x$ with $y'(x)=-\frac{a}{b}\cdot$ This concludes Step 3.

Step 4. There is a one-to-one $\mathcal C^1$ map $g_0:I\to\mathbb R^2$ such that $h(g_0(x))=c$ for all $x\in I$.

Just define $g_0(x)=(x,y(x))$. This map is $\mathcal C^1$ by Step 3, it is clearly one-to-one because of its first coordinate, and $h(g_0(x))=c$ for all $x\in I$ by the very definition of $y(x)$.

Step 5. Conclusion.

Write $I=(\alpha,\beta)$ and define $g:(0,1)\to\mathbb R^2$ by $g(t)=g_0(\alpha+ t(\beta-\alpha))$. Then $g$ has the required properties.

• Thank you very much for this answer! The only question I have is, what $\alpha$ and $\beta$ in step 5 stand for? – sj134 May 23 '14 at 21:16
• @sj134 I is an interval; $\alpha$ and $\beta$ are its endpoints. – Etienne May 25 '14 at 10:13

First of all, if $h$ is constant, you can take $g$ to be any smooth injective curve in $R^2$, e.g., $g(x)=(x,0)$. Therefore, in what follows I assume that $h$ is nonconstant.

The way to solve the problem is to use the implicit function theorem; however, one has to be careful with the regularity assumptions: For a $C^1$-smooth function $h: R^2\to R$ and a point $p\in R^2$ such that $\nabla h(p)\ne 0$, $q=h(p)$, the implicit function theorem states that (after swapping the $x$ and $y$ coordinates in $R^2$ if necessary) there exists a $C^1$-function $f: R\to R$ and a neighborhood $U$ of $p$, such that the set $V=U\cap h^{-1}(q)$ is the graph of the function $f$ (restricted to some open interval $I\subset R$). The graph of $f|I$ admits the $C^1$-parameterization $$g(x)=(x, f(x)), x\in I.$$ Clearly, $g: I\to V$ is injective and $C^1$-smooth. It is also clear that $h\circ g$ is constant, equal $q\in R$. It remains, therefore, to find a point $p\in R^2$ such that $\nabla h(p)\ne 0$. By the mean value theorem, if $\nabla h=0$ everywhere on $R^2$, then $h$ is constant (the mean value theorem requires only $h$ to be differentiable). Thus, the point $p$ with $\nabla h(p)\ne 0$ exists unless $h$ is constant, which is already ruled out.

A side remark. Whitney (see here) constructed an example of a $C^1$-function $h: R^2\to R$ such that there exists a (non-rectifiable) arc $\alpha$ in $R^2$ such that:

1. $h|\alpha$ is nonconstant.

2. $\nabla h$ vanishes everywhere on $\alpha$.

Therefore, in the above argument, one cannot take $q$ to be a "generic" real number. However, if we assume in addition that $h\in C^2(R^2,R)$ then Sard's theorem applies and one can take $q$ to be generic (in the measure-sense) and $p\in h^{-1}(q)$ to be arbitrary.

• Thanks for you detailed answer, however I am not quite familiar with the notations you are using. I added Dini's theorem (I think it is a $\mathbb{R}^2$ restriction of the implicit function theorem) with the proof of existence in my question. Could you edit your post using the given notations and in general use the theorem I gave? – sj134 May 21 '14 at 19:43
• What notations are you unfamiliar with? Gradient? Dini's theorem is just the implicit function theorem in 2d case. – Moishe Kohan May 21 '14 at 19:49
• The gradient, your definition of the graph and your definition of the intervals (neighborhood of $p$) I do not really understand... – sj134 May 21 '14 at 19:52

Assume $h$ is not constant. Suppose $dh$ does not vanish at $x\in \mathbb{R}^2$. Try to use the implicit function theorem.

• $h(x_1,x_2)=y$ with implicit function theorem we find $h(x_1,g(x_1))=y$. – user151873 May 19 '14 at 15:33
• More generaly, a submersion defines a differential variety. – user151873 May 19 '14 at 15:34

This is not a full answer but it may provide some visual intuition for the problem.

If you visualize the graph of $h: \mathbb{R}^2 \rightarrow \mathbb{R}$ as a surface in the usual way, and imagine a plane of the form $z=c$ intersecting that surface, then what you want to do is choose a path $g:(0,1) \rightarrow \mathbb{R}^2$ that traces along part of the level curves where the plane cuts the surface.

• This does not even begin the answer the question, as it is a priori possible that all level curves of $h$ are fractals, and contain no smooth arcs. – Moishe Kohan May 21 '14 at 16:59