Are There a Finite Number of Solutions to $f(x) = y$ in a Bounded Region?

• MHB
• Euge
In summary, a bounded region is a confined and defined area within a larger space, where all points are contained within a specific distance from a center point or boundary. A region is considered bounded if it has a finite area or is enclosed by a boundary. A function cannot have an infinite number of solutions in a bounded region, and the number of solutions can be affected by factors such as the shape and size of the region, the complexity of the function, and constraints or conditions. To prove the finiteness of solutions in a bounded region for a specific function, mathematical methods such as the Intermediate Value Theorem, the Mean Value Theorem, or the Extreme Value Theorem can be used.
Euge
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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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.

1. What is the definition of a bounded region?

A bounded region is a region that has a finite area or volume and is enclosed by a boundary. This boundary can be defined by a set of equations or inequalities.

2. What is the meaning of a finite number of solutions?

A finite number of solutions means that there is a specific and limited number of values for the variable that satisfy the given equation or system of equations. In other words, there is a definite and countable number of solutions.

3. How do you determine if there is a finite number of solutions in a bounded region?

To determine if there is a finite number of solutions in a bounded region, you can graph the equation or system of equations and see how many times the graph intersects with the boundaries of the region. If there are a finite number of intersections, then there is a finite number of solutions. Additionally, you can also use mathematical techniques such as substitution or elimination to solve the equation and determine the number of solutions.

4. What happens if there is an infinite number of solutions in a bounded region?

If there is an infinite number of solutions in a bounded region, it means that the equation or system of equations has an infinite number of values for the variable that satisfy the given conditions. This could happen if the equation is a line that is parallel to the boundary of the region or if the boundary of the region is a curve that the equation intersects an infinite number of times.

5. Can a bounded region have both a finite and infinite number of solutions?

Yes, it is possible for a bounded region to have both a finite and infinite number of solutions. This could happen if the equation or system of equations has multiple solutions that satisfy the given conditions, some of which are finite and some are infinite. In this case, the solutions can be classified into different sets based on their finiteness.

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