# Problem of the Week #300 - May 31, 2022

• MHB
Gold Member
MHB
POTW Director
Here is this week's problem!

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Suppose $f : \Bbb R^n \to \Bbb R^m$ is a differentiable function such that the derivative $Df$ has constant rank $n$. If $\Omega \subset \Bbb R^n$ is bounded, prove that to every $y\in \Bbb R^m$ there corresponds only finitely many solutions to the equation $f(x) = y$ in $\Omega$.
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Gold Member
MHB
POTW Director
GJA has provided a correct proof in the case $f$ is smooth. You can read his solution below.

Suppose there are infinitely many solutions to $f(x)=y$ in $\Omega$. From this collection of infinite solutions, select a sequence $\{x_{n}\}$ of distinct points, noting that $x_{n}\in\Omega$ and $f(x_{n}) = y$ for all $n$.

Since $\Omega$ is a bounded subset of $\mathbb{R}^{n}$, $\text{cl}(\Omega)$ is compact by the Heine-Borel theorem. According to the Bolzano-Weierstrass theorem, $\text{cl}(\Omega)$ is sequentially compact. Hence, $\{x_{n}\}$ admits a convergent subsequence whose limit, say $x$, is in $\text{cl}(\Omega)$. For ease of notation, we denote the convergent subsequence by $\{x_{n}\}$, as the original sequence is no longer needed.

Since $Df$ has constant rank at $x$, the Constant Rank theorem says that there is a diffeomorphism $G$ defined on a neighborhood, say $U$, of $x$ sending $x$ to the origin of $\mathbb{R}^{n}$, and a diffeomorphism $F$ defined on a neighborhood of $f(x) = y$ sending $y$ to the origin of $\mathbb{R}^{m}$ such that $$(F\circ f\circ G^{-1})(x^{1},\ldots, x^{n}) = (x^{1},\ldots, x^{n}, \underbrace{0,\ldots, 0}_{m-n})\qquad (*)$$ Since the sequence of points we chose from the outset are distinct, since $U$ is a neighborhood of $x$, and since $x_{n}\rightarrow x$, there exists $N$ such that $x_{N}\in U$ and $x_{N}\neq x$. Since $x_{N}\neq x$ and since the diffeomorphism $G$ maps $x$ to the origin, we know $G(x_{N})\neq (\underbrace{0,\ldots, 0}_{n})$. Let $(p^{1}, \ldots, p^{n})$ be such that $x_{N} = G^{-1}(p^{1},\ldots, p^{n})$ and note that $(p^{1},\ldots, p^{n})\neq (\underbrace{0,\ldots, 0}_{n})$, as the previous sentence implies.

According to $(*)$, $$(F\circ f\circ G^{-1})(p^{1},\ldots, p^{n}) = (p^{1},\ldots, p^{n}, \underbrace{0,\ldots, 0}_{m-n})\neq (\underbrace{0,\ldots, 0}_{m})\qquad(**),$$ where the $\neq$ follows because $(p^{1},\ldots, p^{n})\neq (\underbrace{0,\ldots, 0}_{n})$.

However, we now have a contradiction because we know $f(x_{N})=y$ and that $F(y) = 0$. Thus, $$(F\circ f\circ G^{-1})(p^{1},\ldots, p^{n}) = (F\circ f)(x_{N}) = F(y) = (\underbrace{0,\ldots, 0}_{m}),$$ contradicting $(**)$.

For general differentiable $f$, you can read my solution below.

Fix $y\in \mathbb{R}^m$, and set $\Sigma = \{x\in \mathbb{R}^n : f(x) = y\}$. Given $a\in \Sigma$, $Df(a)$ is an injective linear map (since $Df(a)$ has rank $n$); thus, there is a linear map $L : \mathbb{R}^m\to \mathbb{R}^n$ such that $L\circ Df(a) = \text{id}$. Since $f$ is differentiable at $a$, there is an open neighborhood $U$ of $a$ such that for all $x\in U$, $|f(x) - y - Df(x)(x - a)| < \frac{1}{2\|L\|}|x - a|$. Hence if $x\in U$, $$|f(x) - y| \ge |Df(x)(x - a)| - |f(x) - y - Df(x)(x - a)| > \frac{1}{\|L\|}|x - a| - \frac{1}{2\|L\|}|x - a| = \frac{1}{2\|L\|}|x - a|$$ In particular, $U \cap \Sigma = \{a\}$. Since $a$ was arbitrary, $\Sigma$ is discrete.

Let $X = \Sigma \cap \overline{\Omega}$. Continuity of $f$ implies $\Sigma$ is closed; thus, $X$ is closed. Boundedness of $\Omega$ forces boundedness of $X$.. By the Heine-Borel theorem, $X$ is compact. As compact discrete sets are finite, $X$ is finite. Hence $\Sigma \cap \Omega$ is finite, as desired.