5.3: CHAPTER 3

Exercise \(3.1.6\).

Answer

1. Observe that when \(x\) is near \(1 / 2\), \(f(x)\) is near \(1 / 2\) no matter whether \(x\) is rational or irrational. We have \[|f(x)-1 / 2|=\left\{\begin{array}{ll}
   |x-1 / 2|, & \text { if } x \in \mathbb{Q} \text {;}\\
   |1-x-1 / 2|, & \text { if } x \notin \mathbb{Q} \text {.}
   \end{array}\right.\] Thus, \(|f(x)-1 / 2|=|x-1 / 2|\) for all \(x \in \mathbb{R}\).

Given any \(\varepsilon > 0\), choose \(\delta = \varepsilon\). Then \[|f(x)-1 / 2|<\varepsilon \text { whenever }|x-1 / 2|<\delta .\] Therefore, \(\lim _{x \rightarrow 1 / 2} f(x)=1 / 2\).

2. Observe that when \(x\) is near \(0\) and \(x\) is rational, \(f(x)\) is near \(0\). However, when \(f\) is near \(0\) and \(x\) is irrational, \(f(x)\) is near \(1\). Thus, the given limit does not exists. We justify this using the sequential criterion for limits (Theorem 3.1.2). By contradiction, assume that \[\lim _{x \rightarrow 0} f(x)=\ell ,\] where \(\ell\) is a real number. Choose a sequence \(\left\{r_{n}\right\}\) of rational numbers that converges to \(0\). Then \(f\left(r_{n}\right)=r_{n}\) and \(f\left(s_{n}\right)=1-s_{n}\) and, hence, \[\ell=\lim _{n \rightarrow \infty} f\left(r_{n}\right)=0\] and \[\ell=\lim _{n \rightarrow \infty} f\left(s_{n}\right)=\lim _{n \rightarrow \infty}\left(1-s_{n}\right)=1 .\] This is a contradiction.
3. By a similar method to part (b), we can show that \(\lim _{x \rightarrow 1} f(x)\) does not exist.

Solving this problem suggests a more general problem as follows. Given two polynomials \(P\) and \(Q\), define the function \[f(x)=\left\{\begin{array}{ll}
P(x), & \text { if } x \in \mathbb{Q} \text {;}\\
Q(x), & \text { if } x \notin \mathbb{Q} \text {.}
\end{array}\right.\] If \(a\) is a solution of the equation \(P(x)=Q(x)\), i.e., \(P(a)=Q(a)\), then the limit \(\lim _{x \rightarrow a} f(x)\) exists and the limit is this common value. For all other points the limit does not exist.

Similar problems:

1. Determine all \(a \in \mathbb{R}\) at which \(\lim _{x \rightarrow a} f(x)\) exists, where \[f(x)=\left\{\begin{array}{ll}
   x^{2}, & \text { if } x \in \mathbb{Q} \text {;}\\
   x+2, & \text { if } x \in x \notin \mathbb{Q} \text {.}
   \end{array}\right.\]
2. Consider the function \[f(x)=\left\{\begin{array}{ll}
   x^{2}+1, & \text { if } x \in \mathbb{Q} \text {;}\\
   -x, & \text { if } x \notin \mathbb{Q} \text {.}
   \end{array}\right.\]

Prove that \(f\) does not have a limit at any \(a \in \mathbb{R}\).

Exercise \(3.2.5\).

Answer

The given condition implies that if both \(x_{1}\) and \(x_{2}\) are close to \(\bar{x}\), then they are close to each other and, hence, \(f\left(x_{1}\right)\) and \(f\left(x_{2}\right)\) are close to each other. This suggests the use of the Cauchy criterion for limit to solve the problem. Given any \(\varepsilon > 0\), choose \(\delta=\frac{\varepsilon}{2(k+1)}\). If \(x_{1}, x_{2} \in D \backslash\{\bar{x}\}\) with \(\left|x_{1}-\bar{x}\right|<\delta\) and \(\left|x_{2}-\bar{x}\right|<\delta\), then \[\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right| \leq k\left|x_{1}-x_{2}\right| \leq k\left(\left|x_{1}-\bar{x}\right|+\left|x_{2}-\bar{x}\right|\right)<k(\delta+\delta)=2 k \frac{\varepsilon}{2(k+1)}<\varepsilon .\] Therefore, \(\lim _{x \rightarrow \bar{x}} f(x)\) exists.

Exercise \(3.3.8\).

Answer

1. Observe that \(f(a)=g(a)=h(a)\) and, hence, \[|f(x)-f(a)|=\left\{\begin{array}{ll}
   |g(x)-g(a)|, & \text { if } x \in \mathbb{Q} \cap[0,1] \text {;}\\
   |h(x)-h(a)|, & \text { if } x \in \mathbb{Q}^{c} \cap[0,1] \text {.}
   \end{array}\right.\] It follows that \[|f(x)-f(a)| \leq|g(x)-g(a)|+|h(x)-h(a)| \text { for all } x \in[0,1] .\] Therefore, \(\lim _{x \rightarrow a} f(x)=f(a)\) and, so, \(f\) is continuous at \(a\).
2. Apply part (a).

Exercise \(3.3.9\).

Answer

At any irrational number \(a \in (0, 1]\), we have \(f(a) = 0\). If \(x\) is near \(a\) and \(x\) is irrational, it is obvious that \(f(x) = 0\) is near \(f(a)\). In the case when \(x\) is near \(a\) and \(x\) is rational, \(f(x)=1 / q\) where \(p, q \in \mathbb{N}\). We will see in part (a) that for any \(\varepsilon > 0\), there is only a finite number of \(x \in (0, 1]\) such that \(f(x) \geq \varepsilon\). So \(f(x)\) is close to \(f(a)\) for all \(x \in (0, 1]\) except for a finite number of \(x \in \mathbb{Q}\). Since \(a\) is irrational, we can choose a sufficiently small neighborhood of \(a\) to void such \(x\).

1. For any \(\varepsilon > 0\), \[A_{\varepsilon}=\{x \in(0,1]: f(x) \geq \varepsilon\}=\left\{x=\frac{p}{q} \in \mathbb{Q}: f(x)=\frac{1}{q} \geq \varepsilon\right\}=\left\{x=\frac{p}{q} \in \mathbb{Q}: q \leq \frac{1}{\varepsilon}\right\} .\] Clearly, the number of \(q \in \mathbb{N}\) such that \(q \leq \frac{1}{\varepsilon}\) is finite. Since \(0<\frac{p}{q} \leq 1\), we have \(p \leq q\). Therefore, \(A_{\varepsilon}\) is finite.
2. Fix any irrational number \(a \in (0, 1]\). Then \(f(a) = 0\). Given any \(\varepsilon > 0\), by part (a), the set \(A_{\varepsilon}\) is finite, so we can write \[A_{\varepsilon}=\{x \in(0,1]: f(x) \geq \varepsilon\}=\left\{x_{1}, x_{2}, \ldots, x_{n}\right\} ,\] for some \(n \in \mathbb{N}\), where \(x_{i} \in \mathbb{Q}\) for all \(i=1, \ldots, n\). Since \(a\) is irrational, we can choose \(\delta > 0\) such that \(x_{i} \notin(a-\delta, a+\delta)\) for all \(i=1, \ldots, n\) (more precisely, we can choose \(\delta=\min \left\{\left|a-x_{i}\right|: i=1, \ldots, n\right\}\)). Then \[|f(x)-f(a)|=f(x)<\varepsilon \text { whenever }|x-a|<\delta .\] Therefore, \(f\) is continuous at \(a\).

Now fix any rational number \(b=\frac{p}{q} \in(0,1]\). Then \(f(b)=\frac{1}{q}\). Choose a sequence of irrational numbers \(\left\{s_{n}\right\}\) that converges to \(b\). Since \(f\left(s_{n}\right)=0\) for all \(n \in \mathbb{N}\), the sequence \(\left\{f\left(s_{n}\right)\right\}\) does not converge to \(f(b)\). Therefore, \(f\) is not continuous at \(b\).

In this problem, we consider the domain of \(f\) to be the interval \((0, 1]\), but the conclusion remain valid for other intervals. In particular, we can show that the function defined on \(\mathbb{R}\) by \[f(x)=\left\{\begin{array}{ll}
\frac{1}{q}, & \text { if } x=\frac{p}{q}, p, q \in \mathbb{N}, \text { where } p \text { and } q \text { have no common factors; } \\
1, & \text { if } x=0 \text {;}\\
0, & \text { if } x \text { is irrational } \text {,}
\end{array}\right.\] is continuous at every irrational point, and discontinuous at every rational point.

Exercise \(3.3.10\).

Answer

Consider \[f(x)=\left\{\begin{array}{ll}
\left(x-a_{1}\right)\left(x-a_{2}\right) \cdots\left(x-a_{k}\right), & \text { if } x \in \mathbb{Q} \text {;}\\
0, & \text { if } x \in \mathbb{Q}^{c} \text {.}
\end{array}\right.\]

Exercise \(3.4.6\).

Answer

Let \(\alpha=\min \{f(x): x \in[a, b]\}\) and \(\beta=\max \{f(x): x \in[a, b]\}\). Then \[\frac{f\left(x_{1}\right)+f\left(x_{2}\right)+\cdots+f\left(x_{n}\right)}{n} \leq \frac{n \beta}{n}=\beta .\] Similarly, \[\alpha \leq \frac{f\left(x_{1}\right)+f\left(x_{2}\right)+\cdots+f\left(x_{n}\right)}{n} .\] Then the conclusion follows from the Intermediate Value Theorem.

Exercise \(3.4.7\).

Answer

1. Observe that \[|f(1 / n)| \leq 1 / n \text { for all } n \in \mathbb{N} .\]
2. Apply the Extreme Value Theorem for the function \(g(x)=\left|\frac{f(x)}{x}\right|\) on the interval \([a, b]\).

Exercise \(3.4.8\).

Answer

First consider the case where \(f\) is monotone decreasing on \([0, 1]\). By Exercise 3.4.5, \(f\) has a fixed point in \([0, 1]\), which means that there exists \(x_{0} \in[0,1]\) such that \[f\left(x_{0}\right)=x_{0}\] Since \(f\) is monotone decreasing, \(f\) has a unique fixed point. Indeed, suppose that there exists \(x_{1} \in[0,1]\) such that \(f\left(x_{1}\right)=x_{1}\). If \(x_{1}<x_{0}\), then \(x_{1}=f\left(x_{1}\right) \geq f\left(x_{0}\right)=x_{0}\), which yields a contradiction. It is similar for the case where \(x_{1}>x_{0}\). Therefore, \(x_{0}\) is the unique point in \([0, 1]\) such that \(f\left(x_{0}\right)=x_{0}\).

Since \(f(g(x))=g(f(x))\) for all \(x \in [0, 1]\), we have \[f\left(g\left(x_{0}\right)\right)=g\left(f\left(x_{0}\right)\right)=g\left(x_{0}\right) .\] Thus, \(g\left(x_{0}\right)\) is also a fixed point of \(f\) and, hence, \(g\left(x_{0}\right)=x_{0}=f\left(x_{0}\right)\). The proof is complete in this case.

Consider the case where \(f\) is monotone increasing. In this case, \(f\) could have several fixed points on \([0, 1]\), so the previous argument does not work. However, by Exercise 3.4.5, there exists \(c \in [0, 1]\) such that \(g(c) = c\). Define the sequence \(\left\{x_{n}\right\}\) as follows: \[\begin{aligned}
x_{1} &=c \text {,}\\
x_{n+1} &=f\left(x_{n}\right) \text { for all } n \geq 1 \text {.}
\end{aligned}\] Since \(f\) is monotone increasing, \(\left\{x_{n}\right\}\) is a monotone sequence. In fact, if \(x_{1} \leq x_{2}\), then \(\left\{x_{n}\right\}\) is monotone increasing; if \(x_{1} \geq x_{2}\), then \(\left\{x_{n}\right\}\) is monotone decreasing. Since \(f\) is bounded, by the monotone convergence theorem (Theorem 2.3.1), there exists \(x_{0} \in[0,1]\) such that \[\lim _{n \rightarrow \infty} x_{n}=x_{0} .\] Since \(f\) is continuous and \(x_{n+1}=f\left(x_{n}\right)\) for all \(n \in \mathbb{N}\), taking limits we have \(f\left(x_{0}\right)=x_{0}\).

We can prove by induction that \(g\left(x_{n}\right)=x_{n}\) for all \(n \in \mathbb{N}\). Then \[g\left(x_{0}\right)=\lim _{n \rightarrow \infty} g\left(x_{n}\right)=\lim x_{n}=x_{0} .\] Therefore, \(f\left(x_{0}\right)=g\left(x_{0}\right)=x_{0}\).

Exercise \(3.5.2\).

Answer

1. Let \(f: D \rightarrow \mathbb{R}\). From Theorem 3.5.3 we see that if there exist two sequences \(\left\{x_{n}\right\}\) and \(\left\{y_{n}\right\}\) in \(D\) such that \(\left|x_{n}-y_{n}\right| \rightarrow 0\) as \(n \rightarrow \infty\), but \(\left\{\left|f\left(x_{n}\right)-f\left(y_{n}\right)\right|\right\}\) does not converge to \(0\), then \(f\) is not uniformly continuous on \(D\). Roughly speaking, in order for \(f\) to be uniformly continuous on \(D\), if \(x\) and \(y\) are close to each other, then \(f(x)\) and \(f(y)\) must be close to each other. The behavior of the graph of the squaring function suggests the argument below to show that \(f(x)=x^{2}\) is not uniformly continuous on \(\mathbb{R}\).

Define two sequences \(\left\{x_{n}\right\}\) and \(\left\{y_{n}\right\}\) as follows: \(x_{n}=n\) and \(y_{n}=n+\frac{1}{n}\) for \(n \in \mathbb{N}\). Then \(\left|x_{n}-y_{n}\right|=\frac{1}{n} \rightarrow 0\) as \(n \rightarrow \infty\). However, \[\left|f\left(x_{n}\right)-f\left(y_{n}\right)\right|=\left(n+\frac{1}{n}\right)^{2}-n^{2}=2+\frac{1}{n^{2}} \geq 2 \text { for all } n \in \mathbb{N} .\] Therefore, \(\left\{\left|f\left(x_{n}\right)-f\left(y_{n}\right)\right|\right\}\) does not converge to \(0\) and, hence, \(f\) is not uniformly continuous on \(\mathbb{R}\). In this solution, we can use \(x_{n}=\sqrt{n+\frac{1}{n}}\) and \(y_{n}=\sqrt{n}\) for \(n \in \mathbb{N}\) instead.

2. Use \(x_{n}=\frac{1}{\pi / 2+2 n \pi}\) and \(y_{n}=\frac{1}{2 n \pi}\), \(n \in \mathbb{N}\).
3. Use \(x_{n}=1 / n\) and \(y_{n}=1 /(2 n)\).

It is natural to ask whether the function \(f(x)=x^{3}\) is uniformly continuous on \(\mathbb{R}\). Following the solution for part (a), we can use \(x_{n}=\sqrt[3]{n+\frac{1}{n}}\) and \(y_{n}=\sqrt[3]{n}\) for \(n \in \mathbb{N}\) to prove that \(f\) is not uniformly continuous on \(\mathbb{R}\). By similar method, we can show that the function \(f(x)=x^{n}\), \(n \in \mathbb{N}\), \(n \geq 2\), is not uniformly continuous on \(\mathbb{R}\). A more challenging question is to determine whether a polynomial of degree greater than or equal to two is uniformly continuous on \(\mathbb{R}\).

Exercise \(3.5.7\).

Answer

Hint: For part (a) use Theorem 3.5.5. For part (b) prove that the function can be extended to a continuous function on \([a, b]\) and then use Theorem 3.5.5.

Exercise \(3.5.8\).

Answer

1. Applying the definition of limit, we find \(b > a\) such that \[c-1<f(x)<c+1 \text { whenever } x>b .\] Since \(f\) is continuous on \([a, b]\), it is bounded on this interval. Therefore, \(f\) is bounded on \([a, \infty]\).
2. Fix any \(\varepsilon > 0\), by the definition of limit, we find \(b > a\) such that \[|f(x)-c|<\frac{\varepsilon}{2} \text { whenever } x>b .\] Since \(f\) is continuous on \([a, b + 1]\), it is uniformly continuous on this interval. Thus, there exists \(0 < \delta < 1\) such that \[|f(u)-f(v)|<\frac{\varepsilon}{2} \text { whenever }|u-v|<\delta, u, v \in[a, c+1] .\] Then we can show that \(|f(u)-f(v)|<\frac{\varepsilon}{2} \text { whenever }|u-v|<\delta, u, v \in[a, c+1]\) whenever \(|u-v|<\delta, u, v \in[a, \infty)\).
3. Since \(\lim _{x \rightarrow \infty} f(x)=c>f(a)\), there exists \(b > a\) such that \[f(x)>f(a) \text { whenever } x>b .\] Thus, \[\inf \{f(x): x \in[a, \infty)\}=\inf \{f(x): x \in[a, b]\} .\] The conclusion follows from the Extreme Value Theorem for the function \(f\) on \([a, b]\).

Exercise \(3.7.4\).

Answer

Since \(\inf \{f(x): x \in[a, \infty)\}=\inf \{f(x): x \in[a, b]\}\), there exists \(a > 0\) such that \[f(x) \geq f(0) \text { whenever }|x|>a .\] Since \(f\) is lower semicontinuous, by Theorem 3.7.3, it has an absolute minimum on \([ -a, a]\) at some point \(\bar{x} \in [-a, a]\). Obviously, \[f(x) \geq f(\bar{x}) \text { for all } x \in[-a, a] .\] In particular, \(f(0) \geq f(\bar{x})\). If \(|x| > a\), then \[f(x) \geq f(0) \geq f(\bar{x}) .\] Therefore, \(f\) has an absolute minimum at \(\bar{x}\).

Observe that in this solution, we can use any number \(\gamma\) in the range of \(f\) instead of \(f(0)\). Since any continuous function is also lower semicontinuous, the result from this problem is applicable for continuous functions. For example, we can use this theorem to prove that any polynomial with even degree has an absolute minimum on \(\mathbb{R}\). Since \(\mathbb[R}\) is a not a compact set, we cannot use the extreme value theorem directly.TST9