** Real Numbers**

**1.3 Absolute value and bounded functions Note: 0.5–1 lecture**

A concept we will encounter over and over is the concept ofabsolute value. You want to think of the absolute value as the “size” of a real number. Let us give a formal definition.

|x|:=

(

x ifx_{≥}0,

−x ifx<0.

Let us give the main features of the absolute value as a proposition. Proposition 1.3.1.

(i) |x| ≥0, and|x|=0if and only if x=0. (ii) |−x|=|x|for all x∈R.

(iii) |xy|=|x||y|for all x,y_{∈}R.

(iv) _{|}x_{|}2=x2for all x_{∈}R.

(v) _{|}x_{| ≤}y if and only if_{−}y_{≤}x_{≤}y.
(vi) _{−|}x_{| ≤}x_{≤ |}x_{|}for all x_{∈}_{R}.

Proof. : Ifx≥0, then|x|=x≥0. Also|x|=x=0 if and only ifx=0. Ifx<0, then|x|=−x>0,

which is never zero.

: Supposex>0, then|−x|=−(−x) =x=|x|. Similarly whenx<0, orx=0.

: Ifxoryis zero, then the result is immediate. Whenxandyare both positive, then|x||y|=xy.
xyis also positive and hence xy=_{|}xy_{|}. If xand y are both negative thenxyis still positive and
xy=_{|}xy_{|}, and_{|}x_{||}y_{|}= (_{−}x)(_{−}y) =xy. Next assumex>0 andy<0. Then_{|}x_{||}y_{|}=x(_{−}y) =_{−}(xy).
Nowxyis negative and hence|xy|=−(xy). Similarly ifx<0 andy>0.

: Immediate ifx_{≥}0. Ifx<0, then_{|}x_{|}2= (_{−}x)2=x2.

: Suppose|x| ≤y. Ifx≥0, thenx≤y. It follows that y≥0, leading to−y≤0≤x. So
−y≤x≤yholds. Ifx<0, then_{|}x_{| ≤}ymeans_{−}x_{≤}y. Negating both sides we getx_{≥ −}y. Again

y_{≥}0 and soy_{≥}0>x. Hence,_{−}y_{≤}x_{≤}y.

On the other hand, suppose−y≤x≤yis true. Ifx≥0, thenx≤yis equivalent to|x| ≤y. If
x<0, then_{−}y_{≤}ximplies(_{−}x)_{≤}y, which is equivalent to_{|}x_{| ≤}y.

: Apply withy=_{|}x_{|}.

A property used frequently enough to give it a name is the so-calledtriangle inequality.
Proposition 1.3.2(Triangle Inequality). _{|}x+y_{| ≤ |}x_{|}+_{|}y_{|}for all x,y_{∈}_{R}.

Proof. gives−|x| ≤x≤ |x|and−|y| ≤y≤ |y|. Add these two inequalities to obtain

−(|x|+|y|)_{≤}x+y_{≤ |}x|+|y|.

There are other often applied versions of the triangle inequality. Corollary 1.3.3. Let x,y∈R

(i) (reverse triangle inequality)

(|x| − |y|)

≤ |x−y|.

(ii) |x−y| ≤ |x|+|y|.

Proof. Let us plug inx=a−bandy=binto the standard triangle inequality to obtain |a|=|a−b+b| ≤ |a−b|+|b|,

or|a| − |b| ≤ |a−b|. Switching the roles ofaandbwe find|b| − |a| ≤ |b−a|=|a−b|. Applying again we obtain the reverse triangle inequality.

The second version of the triangle inequality is obtained from the standard one by just replacing ywith−y, and noting|−y|=|y|.

Corollary 1.3.4. Let x_{1},x2, . . . ,xn∈R. Then

|x1+x2+···+xn| ≤ |x1|+|x2|+···+|xn|.

Proof. We proceed by . The conclusion holds trivially forn=1, and forn=2 it is the standard triangle inequality. Suppose the corollary holds forn. Taken+1 numbersx1,x2, . . . ,xn+1 and first use the standard triangle inequality, then the induction hypothesis

|x1+x2+···+xn+xn+1| ≤ |x1+x2+···+xn|+|xn+1| ≤ |x1|+|x2|+···+|xn|+|xn+1|. Let us see an example of the use of the triangle inequality.

Example 1.3.5: Find a numberMsuch that_{|}x2_{−}9x+1| ≤Mfor all−1≤x≤5.
Using the triangle inequality, write

|x2−9x+1_{| ≤ |}x2_{|}+_{|}9x_{|}+_{|}1_{|}=_{|}x_{|}2+9_{|}x_{|}+1.

The expression|x|2+9|x|+1 is largest when|x|is largest (why?). In the interval provided,|x|is
largest whenx=5 and so_{|}x_{|}=5. One possibility forMis

M=52+9(5) +1=71.

There are, of course, otherMthat work. The bound of 71 is much higher than it need be, but we didn’t ask for the best possibleM, just one that works.

The last example leads us to the concept of bounded functions.

Definition 1.3.6. Suppose f: D_{→}_{R}is a function. We say f isboundedif there exists a number
Msuch that|f(x)| ≤Mfor allx∈D.

inff(D) supf(D) M −M f(D) D

Figure 1.3:Example of a bounded function, a boundM, and its supremum and infimum.

In the example we provedx2_{−}_{9}_{x}_{+}_{1 is bounded when considered as a function on}_{D}_{=}_{{}_{x}_{:}
−1≤x≤5}. On the other hand, if we consider the same polynomial as a function on the whole
real lineR, then it is not bounded.

For a function f: D_{→}Rwe write (see for an example)

sup

x∈Df(x):=sup f(D), inf

x∈Df(x):=inf f(D).

We also sometimes replace the “x∈D” with an expression. For example if, as before, f(x) =
x2_{−}_{9}_{x}_{+}_{1, for}_{−}_{1}_{≤}_{x}_{≤}_{5, a little bit of calculus shows}

sup

x∈Df(x) =−1sup≤x≤5(x

2_{−}_{9}_{x}_{+}_{1}_{) =}_{11}_{,} _{inf}

x∈Df(x) =−1inf≤x≤5(x

2_{−}_{9}_{x}_{+}_{1}_{) =}_{−}_{77}_{/}_{4}_{.}
Proposition 1.3.7. If f: D_{→}_{R}and g: D_{→}_{R}(D nonempty) are bounded functions and

f(x)_{≤}g(x) for all x_{∈}D,

then

sup

x∈Df(x)≤supx∈Dg(x) and xinf∈Df(x)≤xinf∈Dg(x). (1.1) You should be careful with the variables. The x on the left side of the inequality in ( ) is different from thexon the right. You should really think of the first inequality as

sup

x∈Df(x)≤supy∈Dg(y).

Let us prove this inequality. Ifbis an upper bound forg(D), then f(x)≤g(x)≤bfor allx∈D,
and hencebis an upper bound for f(D). Taking the least upper bound we get that for allx_{∈}D

f(x)_{≤}sup
y∈Dg(y).

Therefore supy∈Dg(y)is an upper bound for f(D)and thus greater than or equal to the least upper bound of f(D).

sup

x∈Df(x)≤supy∈Dg(y).

The second inequality (the statement about the inf) is left as an exercise. A common mistake is to conclude

sup

x∈Df(x)≤yinf∈Dg(y). (1.2) The inequality ( ) is not true given the hypothesis of the proposition above. For this stronger inequality we need the stronger hypothesis

f(x)_{≤}g(y) for allx_{∈}Dandy_{∈}D.
The proof as well as a counterexample is left as an exercise.

### 1.3.1

### Exercises

Exercise1.3.1: Show that|x−y|<ε if and only if x−ε <y<x+ε.

Exercise1.3.2: Show: a)max_{{}x,y_{}}= x+y+_{2}|x−y| b)min_{{}x,y_{}}=x+y−|_{2}x−y|

Exercise1.3.3: Find a number M such that_{|}x3_{−}x2+8x| ≤M for all−2≤x≤10.

Exercise1.3.4: Finish the proof of . That is, prove that given any set D, and two bounded

functions f:D→Rand g:D→Rsuch that f(x)≤g(x)for all x∈D, then

inf

x∈Df(x)≤xinf∈Dg(x).

Exercise1.3.5: Let f:D_{→}Rand g:D→Rbe functions (D nonempty).

a) Suppose f(x)_{≤}g(y)for all x∈D and y∈D. Show that
sup

x∈Df(x)≤xinf∈Dg(x).

b) Find a specific D, f , and g, such that f(x)≤g(x)for all x∈D, but sup

x∈Df(x)>xinf∈Dg(x).

Exercise1.3.6: Prove 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:D→Rand g:D→Rare bounded functions.

a) Show sup

x∈D f(x) +g(x)

≤sup

x∈Df(x) +supx∈Dg(x) and xinf∈D f(x) +g(x)

≥ inf

x∈Df(x) +xinf∈Dg(x).

Exercise1.3.8: Suppose f:D→Rand g:D→Rare bounded functions andα∈R.

a) Show thatαf:D→Rdefined by(αf)(x):=αf(x)is a bounded function.

b) Show that f+g:D_{→}Rdefined by(f+g)(x):= f(x) +g(x)is a bounded function.

Exercise1.3.9: Let f:D_{→}Rand g:D→Rbe functions,α ∈R, and recall what f+g andαf means
from the previous exercise.

a) Prove that if f+g and g are bounded, then f is bounded.

b) Find an example where f and g are both unbounded, but f+g is bounded.

c) Prove that if f is bounded but g is unbounded, then f+g is unbounded.