Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 9: Problem 9

If \(f\) is a differentiable mapping of a connected open set \(E \subset R^{n}\) into \(R^{m}\), and if \(\mathbf{f}^{\prime}(\mathbf{x})=0\) for every \(\mathbf{x} \in E\), prove that \(\mathbf{f}\) is constant in \(E\).

Short Answer

Expert verified
Question: Prove that if a differentiable mapping of a connected open set \(E \subset R^{n}\) into \(R^{m}\) has a derivative equal to 0 for every point in \(E\), then the mapping is constant in \(E\). Answer: To prove that the mapping is constant in \(E\), we used the mean value theorem and properties of connected open sets. Applying the mean value theorem, we found that for any two points \(\mathbf{x}, \mathbf{y} \in E\), there exists a \(\mathbf{z}\) on the line segment connecting these points such that \(\frac{\mathbf{f}(\mathbf{y}) - \mathbf{f}(\mathbf{x})}{\|\mathbf{y} - \mathbf{x}\|} = \mathbf{f}^{\prime}(\mathbf{z})\). Since \(E\) is connected, the line segment between \(\mathbf{x}\) and \(\mathbf{y}\) is contained in \(E\), which implies that \(\mathbf{z} \in E\). As \(\mathbf{f}'(\mathbf{z}) = 0\) for all \(\mathbf{z} \in E\), it follows that \(\mathbf{f}(\mathbf{y}) - \mathbf{f}(\mathbf{x}) = 0\). This relationship holds for any two points \(\mathbf{x}, \mathbf{y} \in E\), implying that the mapping is constant in \(E\).

Step by step solution

01

Use the mean value theorem

Since \(\mathbf{f}\) is a differentiable mapping, we can apply the mean value theorem. For any two points \(\mathbf{x}, \mathbf{y} \in E\), there exists a \(\mathbf{z}\) on the line segment connecting points \(\mathbf{x}\) and \(\mathbf{y}\) such that \[\frac{\mathbf{f}(\mathbf{y}) - \mathbf{f}(\mathbf{x})}{\|\mathbf{y} - \mathbf{x}\|} = \mathbf{f}^{\prime}(\mathbf{z}).\]
02

Use the properties of connected open sets

Since \(E\) is a connected open set and \(\mathbf{x},\mathbf{y} \in E\), the line segment between \(\mathbf{x}\) and \(\mathbf{y}\) is also contained in \(E\). Therefore, \(\mathbf{z} \in E\), and from the exercise statement, we have \(\mathbf{f}'(\mathbf{z}) = 0\).
03

Apply the zero derivative condition

Using the zero derivative condition, we have that \(\frac{\mathbf{f}(\mathbf{y}) - \mathbf{f}(\mathbf{x})}{\| \mathbf{y} - \mathbf{x} \|} = \mathbf{f}'(\mathbf{z}) = 0\). Thus, \[\mathbf{f}(\mathbf{y}) - \mathbf{f}(\mathbf{x}) = 0.\]
04

Prove that \(\mathbf{f}\) is constant in \(E\)

From the previous step, we found that \(\mathbf{f}(\mathbf{y}) - \mathbf{f}(\mathbf{x}) = 0\). This means that \(\mathbf{f}(\mathbf{y}) = \mathbf{f}(\mathbf{x})\) for any two points \(\mathbf{x}, \mathbf{y} \in E\). Thus, \(\mathbf{f}\) is constant in \(E\), which completes the proof.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

The Mean Value Theorem in Real Analysis
The Mean Value Theorem is a foundational concept in real analysis and calculus, setting a vital link between derivatives and integrals. It states that for any differentiable function that continues on a closed interval \[a, b\], there exists at least one point \(c\) in the open interval \(a, b\) at which the derivative of the function is equal to its average rate of change over that interval. In mathematical terms, if \(f\) is continuous on \[a, b\] and differentiable on \(a, b\), then \[ f'(c) = \frac{f(b) - f(a)}{b - a} \] for some \(c \in (a, b)\).

This theorem is crucial when dealing with questions about the behavior of functions within an interval and is pivotal in the proof provided in the exercise, serving to show that if the derivative of \(f\) is zero at every point in an interval, then the function must be constant on that interval. To make this concept more digestible, think of the derivative as a speedometer reading of a car. If the speedometer reads zero throughout the trip, that means the car hasn't moved - similarly, if the derivative is zero, the function has not changed its value, implying constancy.
Understanding Connected Open Sets
Connected open sets are an essential concept in topology, a branch of mathematics focused on the properties of space that are preserved under continuous transformations. An open set is a collection of points with the specific attribute that each point has a neighborhood entirely contained within the set. In the realm of real analysis, it usually refers to intervals or areas without including their boundaries. A set is connected if there are no separate 'pieces' of it can be distinguished that are not connected to each other.

To illustrate with an example, consider an open interval on the number line, like \(0, 1\). This interval is both open (it doesn’t include the endpoints 0 and 1) and connected (there’s no way to split the interval into two non-overlapping open intervals that make up the original interval). When solving problems, knowing that a set is connected helps us deduce that any property holding at one point, under the right conditions, must hold throughout the entire set, just as we use in the given exercise to conclude that a zero derivative throughout an open connected set ensures the function is constant.
The Derivative Condition and Function Behavior
The derivative of a function at a point offers deep insight into the function's behavior around that point. When we talk about the 'derivative condition,' we refer to the requirement a function must meet at its derivatives for certain properties to hold. If a function \(f\) meets the derivative condition that \(f'(x) = 0\) for all \(x\) in some domain 'E,' then the function does not change - its graph is flat, and it represents a constant function over 'E'.

Returning to our speedometer analogy, if a car's speedometer shows a constant zero, the car is not moving. Similarly, if the derivative (the mathematical 'speedometer') of \(f\) is constantly zero throughout a connected domain, it indicates that \(f\) is not 'moving' or changing in value; it is staying constant. The exercise we are dissecting uses this derivative condition to establish that since the function's derivative is zero at all points in a connected open set, the function itself must be constant throughout that set. This concept is a cornerstone in proving many other theorems in real analysis, as it relates the local behavior of a function to its global behavior.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For \(t \geq 0\), put $$ \phi(x, t)=\left\\{\begin{array}{ll} x & (0 \leq x \leq \sqrt{t}) \\ -x+2 \sqrt{t} & (\sqrt{t} \leq x \leq 2 \sqrt{t}) \\ 0 & (\text { otherwise }), \end{array}\right. $$ and put \(\varphi(x, t)=-\varphi(x,|t|)\) if \(t<0 .\) Show that \(\varphi\) is continuous on \(R^{2}\), and $$ \left(D_{2} \varphi\right)(x, 0)=0 $$ for all \(x .\) Define $$ f(t)=\int_{-1}^{1} \varphi(x, t) d x $$ Show that \(f(t)=t\) if \(|t|

Define \(f(0,0)=0\) and $$ f(x, y)=\frac{x^{3}}{x^{2}+y^{2}} \quad \text { if }(x, y) \neq(0,0) $$ (a) Prove that \(D_{1} f\) and \(D_{2} f\) are bounded functions in \(R^{2}\). (Hence \(f\) is continuous.) (b) Let u be any unit vector in \(R^{2}\). Show that the directional derivative \(\left(D_{w} f\right)(0,0)\) exists, and that its absolute value is at most 1 . (c) Let \(\gamma\) be a differentiable mapping of \(R^{1}\) into \(R^{2}\) (in other words, \(\gamma\) is a differentiable curve in \(\left.R^{2}\right)\), with \(\gamma(0)=(0,0)\) and \(\left|\gamma^{\prime}(0)\right|>0 .\) Put \(g(t)=f(\gamma(t))\) and prove that \(g\) is differentiable for every \(t \in R^{1}\). If \(\gamma \in \mathscr{C}^{\prime}\), prove that \(g \in \mathscr{8}\) '. (d) In spite of this, prove that \(f\) is not differentiable at \((0,0)\). Hint: Formula (40) fails.

Suppose that \(f\) is a real-valued function defined in an open set \(E \subset R^{n}\), and that the partial derivatives \(D_{1} f, \ldots, D_{n} f\) are bounded in \(E\). Prove that \(f\) is continuous in \(\bar{E}\) Hint: Proceed as in the proof of Theorem 9.21.

Let \(f \in \mathscr{f}^{(m)}(E)\), where \(E\) is an open subset of \(R^{n}\). Fix a \(\in E\), and suppose \(x \in R^{n}\) is so close to 0 that the points $$ \mathbf{p}(t)=\mathbf{a}+\boldsymbol{r} \mathbf{x} $$ lie in \(E\) whenever \(0 \leq t \leq 1\). Define $$ h(t)=f(\mathrm{p}(t)) $$ for all \(t \in R^{1}\) for which \(\mathrm{p}(t) \in E\). (a) For \(1 \leq k \leq m\), show (by repeated application of the chain rule) that $$ h^{(k)}(t)=\sum\left(D_{i_{1} \ldots i_{k}} f\right)(\mathrm{p}(t)) x_{i_{1} \ldots} x_{i_{k}} . $$ The sum extends over all ordered \(k\) -tuples \(\left(i_{1}, \ldots, i_{k}\right)\) in which each \(i\), is one of the integers \(1, \ldots, n\). (b) By Taylor's theorem (5.15), $$ h(1)=\sum_{k=0}^{m-1} \frac{h^{(k)}(0)}{k !}+\frac{h^{(m)}(t)}{m !} $$ for some \(t \in(0,1) .\) Use this to prove Taylor's theorem in \(n\) variables by showing that the formula $$ f(\mathbf{a}+\mathbf{x})=\sum_{k=0}^{m-1} \frac{1}{k !} \sum\left(D_{i 1} \ldots, f\right)(\mathbf{a}) x_{i_{1}} \ldots x_{1,}+r(\mathbf{x}) $$ represents \(f(\mathrm{a}+\mathbf{x})\) as the sum of its so-called "Taylor polynomial of degree \(m-1, "\) plus a remainder that satisfies $$ \lim _{x \rightarrow 0} \frac{r(\mathbf{x})}{|x|=-1}=0 $$ Each of the inner sums extends over all ordered \(k\) -tuples \(\left(i_{1}, \ldots, i_{k}\right)\), as in part \((a)\); as usual, the zero-order derivative of \(f\) is simply \(f\), so that the constant term of the Taylor polynomial of \(f\) at a is \(f(\mathbf{a})\). (c) Exercise 29 shows that repetition occurs in the Taylor polynomial as written in part ( \(b\) ). For instance, \(D_{113}\) occurs three times, as \(D_{113}, D_{131}, D_{311}\). The sum of the corresponding three terms can be written in the form $$ 3\left(D_{1}^{2} D_{3} f\right)(a) x_{1}^{2} x_{3} $$ Prove (by calculating how often each derivative occurs) that the Taylor polynomial in \((b)\) can be written in the form $$ \sum \frac{\left(D_{1}^{21} \cdots D_{n}^{s n} f\right)(a)}{s_{1} ! \cdots s_{n} !} x_{1}^{21} \cdots x_{n}^{s_{n}} $$ Here the summation extends over all ordered \(n\) -tuples \(\left(s_{1}, \ldots, s_{n}\right)\) such that each \(s_{1}\) is a nonnegative integer, and \(s_{1}+\cdots+s_{n} \leq m-1\).

If \(f\) and \(g\) are differentiable real functions in \(R^{n}\), prove that $$ \nabla(f g)=f \nabla g+g \nabla f $$ and that \(\nabla(1 / f)=-f^{-2} \nabla f\) wherever \(f \neq 0\).
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free