Basic Analysis I & II: Introduction to Real Analysis, Volumes I & II

Section1.3Absolute value and bounded functions

Note: 0.5–1 lecture
A concept we will encounter over and over is the concept of absolute value. You want to think of the absolute value as the “size” of a real number. Here is the formal definition:
\begin{equation*} \abs{x} \coloneqq \begin{cases} x & \text{if } x \geq 0, \\ -x & \text{if } x < 0 . \end{cases} \end{equation*}
Let us give the main features of the absolute value as a proposition.

Proof.

i: First suppose $$x \geq 0\text{.}$$ Then $$\abs{x} = x \geq 0\text{.}$$ Also, $$\abs{x} = x = 0$$ if and only if $$x=0\text{.}$$ On the other hand, if $$x < 0\text{,}$$ then $$\abs{x} = -x > 0\text{,}$$ and $$\abs{x}$$ is never zero.
ii: If $$x > 0\text{,}$$ then $$-x < 0$$ and so $$\abs{-x} = -(-x) = x = \abs{x}\text{.}$$ Similarly when $$x < 0\text{,}$$ or $$x = 0\text{.}$$
iii: If $$x$$ or $$y$$ is zero, then the result is immediate. When $$x$$ and $$y$$ are both positive, then $$\abs{x}\abs{y} = xy\text{.}$$ As $$xy$$ is also positive, $$xy = \abs{xy}\text{.}$$ If $$x$$ and $$y$$ are both negative, then $$xy=(-x)(-y)$$ is still positive and $$\abs{xy}=xy\text{.}$$ Also, $$\abs{x}\abs{y} = (-x)(-y) = xy\text{.}$$ If $$x > 0$$ and $$y < 0\text{,}$$ then $$\abs{x}\abs{y} = x(-y) = -(xy)\text{.}$$ Now $$xy$$ is negative and $$\abs{xy} = -(xy)\text{.}$$ Similarly when $$x < 0$$ and $$y > 0\text{.}$$
iv: Immediate if $$x \geq 0\text{.}$$ If $$x < 0\text{,}$$ then $$\abs{x}^2 = {(-x)}^2 = x^2\text{.}$$
v: Suppose $$\abs{x} \leq y\text{.}$$ If $$x \geq 0\text{,}$$ then $$x \leq y\text{.}$$ It follows that $$y \geq 0\text{,}$$ leading to $$-y \leq 0 \leq x\text{.}$$ So $$-y \leq x \leq y$$ holds. If $$x < 0\text{,}$$ then $$\abs{x} \leq y$$ means $$-x \leq y\text{.}$$ Negating both sides we get $$x \geq -y\text{.}$$ Again $$y \geq 0$$ and so $$y \geq 0 > x\text{.}$$ Hence, $$-y \leq x \leq y\text{.}$$
On the other hand, suppose $$-y \leq x \leq y$$ is true. If $$x \geq 0\text{,}$$ then $$x \leq y$$ is equivalent to $$\abs{x} \leq y\text{.}$$ If $$x < 0\text{,}$$ then $$-y \leq x$$ implies $$(-x) \leq y\text{,}$$ which is equivalent to $$\abs{x} \leq y\text{.}$$
vi: Apply v with $$y = \abs{x}\text{.}$$
A property used frequently enough to give it a name is the so-called triangle inequality.

Proof.

Proposition 1.3.1 gives $$- \sabs{x} \leq x \leq \sabs{x}$$ and $$- \sabs{y} \leq y \leq \sabs{y}\text{.}$$ Add these two inequalities to obtain
\begin{equation*} - \bigl(\sabs{x}+\sabs{y}\bigr) \leq x+y \leq \sabs{x}+ \sabs{y} . \end{equation*}
Apply Proposition 1.3.1 again to find $$\sabs{x+y} \leq \sabs{x}+\sabs{y}\text{.}$$
There are other often applied versions of the triangle inequality.

Proof.

Let us plug in $$x=a-b$$ and $$y=b$$ into the standard triangle inequality to obtain
\begin{equation*} \abs{a} = \abs{a-b+b} \leq \abs{a-b} + \abs{b} , \end{equation*}
or $$\abs{a}-\abs{b} \leq \abs{a-b}\text{.}$$ Switching the roles of $$a$$ and $$b$$ we find $$\abs{b}-\abs{a} \leq \abs{b-a} = \abs{a-b}\text{.}$$ Applying Proposition 1.3.1, we obtain the reverse triangle inequality.
The second item in the corollary is obtained from the standard triangle inequality by just replacing $$y$$ with $$-y\text{,}$$ and noting $$\abs{-y} = \abs{y}\text{.}$$

Proof.

We proceed by induction. The conclusion holds trivially for $$n=1\text{,}$$ and for $$n=2$$ it is the standard triangle inequality. Suppose the corollary holds for $$n\text{.}$$ Take $$n+1$$ numbers $$x_1,x_2,\ldots,x_{n+1}$$ and first use the standard triangle inequality, then the induction hypothesis
\begin{equation*} \begin{split} \sabs{x_1 + x_2 + \cdots + x_n + x_{n+1}} & \leq \sabs{x_1 + x_2 + \cdots + x_n} + \sabs{x_{n+1}} \\ & \leq \sabs{x_1} + \sabs{x_2} + \cdots + \sabs{x_n} + \sabs{x_{n+1}} . \qedhere \end{split} \end{equation*}
Let us see an example of the use of the triangle inequality.

Example1.3.5.

Find a number $$M$$ such that $$\sabs{x^2-9x+1} \leq M$$ for all $$-1 \leq x \leq 5\text{.}$$
Using the triangle inequality, write
\begin{equation*} \sabs{x^2-9x+1} \leq \sabs{x^2}+\sabs{9x}+\sabs{1} = \sabs{x}^2+9\sabs{x}+1 . \end{equation*}
The expression $$\sabs{x}^2+9\sabs{x}+1$$ is largest when $$\abs{x}$$ is largest (why?). In the interval provided, $$\abs{x}$$ is largest when $$x=5$$ and so $$\abs{x}=5\text{.}$$ One possibility for $$M$$ is
\begin{equation*} M = 5^2+9(5)+1 = 71 . \end{equation*}
There are, of course, other $$M$$ that work. The bound of 71 is much higher than it need be, but we didn’t ask for the best possible $$M\text{,}$$ just one that works.
The last example leads us to the concept of bounded functions.

Definition1.3.6.

Suppose $$f \colon D \to \R$$ is a function. We say $$f$$ is bounded if there exists a number $$M$$ such that $$\abs{f(x)} \leq M$$ for all $$x \in D\text{.}$$
In the example, we proved $$x^2-9x+1$$ is bounded when considered as a function on $$D = \{ x : -1 \leq x \leq 5 \}\text{.}$$ On the other hand, if we consider the same polynomial as a function on the whole real line $$\R\text{,}$$ then it is not bounded.

For a function $$f \colon D \to \R\text{,}$$ we write (see Figure 1.4 for an example)
\begin{equation*} \sup_{x \in D} f(x) \coloneqq \sup\, f(D) \qquad \text{and} \qquad \inf_{x \in D} f(x) \coloneqq \inf\, f(D) . \end{equation*}
We also sometimes replace the “$$x \in D$$” with an expression. For example if, as before, $$f(x) = x^2-9x+1\text{,}$$ for $$-1 \leq x \leq 5\text{,}$$ a little bit of calculus shows
\begin{equation*} \sup_{x \in D} f(x) = \sup_{-1 \leq x \leq 5} ( x^2 -9x+1 ) = 11, \qquad \inf_{x \in D} f(x) = \inf_{-1 \leq x \leq 5} ( x^2 -9x+1 ) = \nicefrac{-77}{4} . \end{equation*}
Be careful with the variables. The $$x$$ on the left side of the inequality in (1.1) is different from the $$x$$ on the right. You should really think of, say, the first inequality as
\begin{equation*} \sup_{x \in D} f(x) \leq \sup_{y \in D} g(y) . \end{equation*}
Let us prove this inequality. If $$b$$ is an upper bound for $$g(D)\text{,}$$ then $$f(x) \leq g(x) \leq b$$ for all $$x \in D\text{,}$$ and hence $$b$$ is also an upper bound for $$f(D)\text{,}$$ or $$f(x) \leq b$$ for all $$x \in D\text{.}$$ Take the least upper bound of $$g(D)$$ to get that for all $$x \in D$$
\begin{equation*} f(x) \leq \sup_{y \in D} g(y) . \end{equation*}
Therefore, $$\sup_{y \in D} g(y)$$ is an upper bound for $$f(D)$$ and thus greater than or equal to the least upper bound of $$f(D)\text{.}$$
\begin{equation*} \sup_{x \in D} f(x) \leq \sup_{y \in D} g(y) . \end{equation*}
The second inequality (the statement about the inf) is left as an exercise (Exercise 1.3.4).
A common mistake is to conclude
$$\sup_{x \in D} f(x) \leq \inf_{y \in D} g(y) .\tag{1.2}$$
The inequality (1.2) is not true given the hypothesis of the proposition above. For this stronger inequality we need the stronger hypothesis
\begin{equation*} f(x) \leq g(y) \qquad \text{for all } x \in D \text{ and } y \in D. \end{equation*}
The proof as well as a counterexample is left as an exercise (Exercise 1.3.5).

Subsection1.3.1Exercises

Exercise1.3.1.

Show that $$\abs{x-y} < \epsilon$$ if and only if $$x-\epsilon < y < x+\epsilon\text{.}$$

Exercise1.3.2.

Show:        a) $$\max \{x,y\} = \frac{x+y+\abs{x-y}}{2}$$        b) $$\min \{x,y\} = \frac{x+y-\abs{x-y}}{2}$$

Exercise1.3.3.

Find a number $$M$$ such that $$\sabs{x^3-x^2+8x} \leq M$$ for all $$-2 \leq x \leq 10\text{.}$$

Exercise1.3.4.

Finish the proof of Proposition 1.3.7. That is, prove that given a set $$D\text{,}$$ and two bounded functions $$f \colon D \to \R$$ and $$g \colon D \to \R$$ such that $$f(x) \leq g(x)$$ for all $$x \in D\text{,}$$ then
\begin{equation*} \inf_{x\in D} f(x) \leq \inf_{x\in D} g(x) . \end{equation*}

Exercise1.3.5.

Let $$f \colon D \to \R$$ and $$g \colon D \to \R$$ be functions ($$D$$ nonempty).
1. Suppose $$f(x) \leq g(y)$$ for all $$x \in D$$ and $$y \in D\text{.}$$ Show that
\begin{equation*} \sup_{x\in D} f(x) \leq \inf_{x\in D} g(x) . \end{equation*}
2. Find a specific $$D\text{,}$$ $$f\text{,}$$ and $$g\text{,}$$ such that $$f(x) \leq g(x)$$ for all $$x \in D\text{,}$$ but
\begin{equation*} \sup_{x\in D} f(x) > \inf_{x\in D} g(x) . \end{equation*}

Exercise1.3.6.

Prove Proposition 1.3.7 without the assumption that the functions are bounded. Hint: You need to use the extended real numbers.

Exercise1.3.7.

Let $$D$$ be a nonempty set. Suppose $$f \colon D \to \R$$ and $$g \colon D \to \R$$ are bounded functions.
1. Show
\begin{equation*} \sup_{x\in D} \bigl(f(x) + g(x) \bigr) \leq \sup_{x\in D} f(x) + \sup_{x\in D} g(x) \qquad \text{and} \qquad \inf_{x\in D} \bigl(f(x) + g(x) \bigr) \geq \inf_{x\in D} f(x) + \inf_{x\in D} g(x) . \end{equation*}
2. Find an example (or examples) where we obtain strict inequalities.

Exercise1.3.8.

Suppose $$D$$ is nonempty, $$f \colon D \to \R$$ and $$g \colon D \to \R$$ are bounded functions, and $$\alpha \in \R\text{.}$$
1. Show that $$\alpha f \colon D \to \R$$ defined by $$(\alpha f) (x) \coloneqq \alpha f(x)$$ is a bounded function.
2. Show that $$f+g \colon D \to \R$$ defined by $$(f+g) (x) \coloneqq f(x) + g(x)$$ is a bounded function.

Exercise1.3.9.

Let $$f \colon D \to \R$$ and $$g \colon D \to \R$$ be functions with $$D$$ nonempty, $$\alpha \in \R\text{,}$$ and recall what $$f+g$$ and $$\alpha f$$ means from the previous exercise.
1. Prove that if $$f+g$$ and $$g$$ are bounded, then $$f$$ is bounded.
2. Find an example where $$f$$ and $$g$$ are both unbounded, but $$f+g$$ is bounded.
3. Prove that if $$f$$ is bounded but $$g$$ is unbounded, then $$f+g$$ is unbounded.
4. Find an example where $$f$$ is unbounded but $$\alpha f$$ is bounded.
The boundedness hypothesis is for simplicity, it can be dropped if we allow for the extended real numbers.
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