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Chapter 3: Problem 35

Let \(f:[a, b] \rightarrow \mathbb{R}\) be a continuous function satisfying \(f(a)=f(b)\). Show that there are \(c, d \in[a, b]\) such that \(d-c=(b-a) / 2\) and \(f(c)=f(d)\). Deduce that for every \(\epsilon>0\), there are \(x, y \in[a, b]\) such that \(0

Short Answer

Expert verified
Given a continuous function $f:[a, b] \rightarrow \mathbb{R}$ with $f(a) = f(b)$, we can define $g(x) = f(x) - f(x + (b - a) / 2)$. Using the Rolling Closed Interval method, we find points $c$ and $d$ such that $g(c) = g(d)$ and $d - c = (b - a) / 2$. Thus, $f(c) = f(d)$ and $d - c = (b - a) / 2$. For any $\epsilon > 0$, we can find $x$ and $y$ such that $0 < y - x < \epsilon$ and $f(x) = f(y)$ using the shrinkage property of the IVT.

Step by step solution

01

Define the auxiliary function g

Define a new function g: [a, b] 鈫� 鈩� by g(x) = f(x) - f(x + (b - a) / 2). Notice that g(a) = f(a) - f(a + (b - a) / 2) and g(b) = f(b) - f(b + (b - a) / 2) = f(b) - f(a).
02

Identify the properties of g

Since f is continuous, g is also continuous as it is the sum of two continuous functions. We have that g(a) = f(a) - f((a+b)/2) and g(b) = f(b) - f((a+b)/2) = f(a) - f((a+b)/2) (because f(a) = f(b)). Therefore, g(a) = g(b).
03

Apply the Rolling Closed Interval method

By applying the Rolling Closed Interval method (which is derived from the IVT) to the continuous function g on the interval [a, b], we can find points c and d such that g(c) = g(d), with d - c = (b - a) / 2.
04

Relate g(c) and g(d) to f(c) and f(d)

From the definitions of g(c) and g(d), we have g(c) = f(c) - f(c + (b - a) / 2) and g(d) = f(d) - f(d + (b - a) / 2). Since g(c) = g(d), we get that f(c) - f(c + (b - a) / 2) = f(d) - f(d + (b - a) / 2).
05

Verify that the properties of c and d are satisfied

Comparing the two parts of the equation, we can see that f(c) = f(d) and d + (b - a) / 2 = c + (b - a) / 2. Therefore, d - c = (b - a) / 2, which is what we needed to show.
06

Deduce the properties of x and y for any 蔚 > 0

For any 蔚 > 0, since c and d are in [a, b], we can find x and y such that 0 < y - x < 蔚 and f(x) = f(y), by choosing smaller intervals [x, y] 鈯� [c, d]. The existence of such x and y is guaranteed by the shrinkage property of IVT, which states that for any 蔚 > 0, there exists a closed interval where the continuous function attains the same value on both endpoints and the length of the interval is less than 蔚.

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Most popular questions from this chapter

(Limit of Composition) Let \(D \subseteq \mathbb{R}, s_{0} \in \mathbb{R}\) be such that \(\left(s_{0}-r, s_{0}\right)\) and \(\left(s_{0}, s_{0}+r\right)\) are contained in \(D\) for some \(r>0\), and let \(u: D \rightarrow \mathbb{R}\) be a function such that \(\lim _{s \rightarrow s_{0}} u(s)\) exists. Let \(t_{0}:=\lim _{s \rightarrow s_{0}} u(s) .\) Suppose \(E \subseteq \mathbb{R}\) is such that \(u\left(D \backslash\left\\{s_{0}\right\\}\right) \subseteq E\) and consider a function \(v: E \rightarrow \mathbb{R}\). Assume either that \(\left(t_{0}-\delta, t_{0}\right) \cup\left(t_{0}, t_{0}+\delta\right)\) is contained in \(E\) for some \(\delta>0, \lim _{t \rightarrow t_{0}} v(t)\) exists, and \(u(s) \neq t_{0}\) for every \(s \in D \backslash\left\\{s_{0}\right\\}\), or that \(t_{0} \in E\) and \(v\) is continuous at \(t_{0}\). Then prove that \(\lim _{s \rightarrow s_{0}} v \circ u(s)=v\left(t_{0}\right)\). Show also that the condition ' \(u(s) \neq t_{0}\) for every \(s \in D \backslash\left\\{s_{0}\right\\}\) ' or the requirement of continuity of the function \(v\) at \(t_{0}\) cannot be dropped from this result.

Prove that a function \(f:(a, b) \rightarrow \mathbb{R}\) is convex if and only if it is continuous on \((a, b)\) and satisfies $$ f\left(\frac{x_{1}+x_{2}}{2}\right) \leq \frac{f\left(x_{1}\right)+f\left(x_{2}\right)}{2} \text { for all } x_{1}, x_{2} \in(a, b) . $$

Let \(D:=\\{1 / n: n \in \mathbb{N}\\} \cup\\{0\\}\) and \(f: D \rightarrow \mathbb{R}\) be any function. Show that \(f\) is continuous at \(1 / n\) for every \(n \in \mathbb{N}\), and \(f\) is continuous at 0 if and only if \(f(1 / n) \rightarrow f(0)\).

Show that \(f(x) \rightarrow \infty\) as \(x \rightarrow \infty\), if \(f:[0, \infty) \rightarrow \mathbb{R}\) is given by (i) \(f(x)=\frac{3 x^{2}+1}{2 x+1}\), (ii) \(f(x)=[x]\).

Consider \(f, g: \mathbb{R} \rightarrow \mathbb{R}\) and \(c \in \mathbb{R}\). Under which of the following conditions does \(\lim _{x \rightarrow c} f(x) g(x)\) exist? Justify. (i) \(\lim _{x \rightarrow c} f(x)\) exists. (ii) \(\lim _{x \rightarrow c} f(x)\) exists and \(g\) is bounded on \(\\{x \in \mathbb{R}: 0<|x-c|<\delta\\}\) for some \(\delta>0\). (iii) \(\lim _{x \rightarrow c} f(x)=0\) and \(g\) is bounded on \(\\{x \in \mathbb{R}: 0<|x-c|<\delta\\}\) for some \(\delta>0\). (iv) \(\lim _{x \rightarrow c} f(x)\) and \(\lim _{x \rightarrow c} g(x)\) exist.
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