# Example 2.2.10 The function

In document Introduction to Real Analysis (Page 69-72)

g.x/D (1

2; xD0 or xD1;

1 x; 0 < x < 1; C

(Figure2.2.4

### (a)

) is bounded onŒ0; 1, and sup

0x1

g.x/D1; inf

0x1g.x/D0:

Therefore,ghas no maximum or minimum onŒ0; 1, since it does not assume either of the values0and1.

The function

h.x/D1 x; 0x1;

which differs fromgonly at0and1(Figure2.2.4

### (b)

), has the same supremum and infi- mum asg, but it attains these values atxD0andxD1, respectively; therefore,

max 0x1h.x/D1 and 0minx1h.x/D0: 2 1 1 1 y x 1 1 y x (a) (b) y = g(x) y = 1 − x

### Example 2.2.11

The function

f .x/Dex.x 1/sin 1

x.x 1/; 0 < x < 1;

oscillates between˙ex.x 1/infinitely often in every interval of the form.0; /or.1 ; 1/, where0 < < 1, and

sup 0<x<1

f .x/D1; inf

0<x<1f .x/D 1:

### Theorem 2.2.8

Iff is continuous on a finite closed intervalŒa; b;thenf is bounded onŒa; b:

### Proof

Suppose thatt 2Œa; b. Sincef is continuous att, there is an open intervalIt containingtsuch that

jf .x/ f .t /j< 1 if x2It\Œa; b: (2.2.6) (To see this, setD1in (2.2.1), Theorem2.2.2.) The collectionH D ˚Itˇˇat b is

an open covering ofŒa; b. SinceŒa; bis compact, the Heine–Borel theorem implies that there are finitely many pointst1,t2, . . . , tnsuch that the intervalsIt1,It2, . . . , Itn cover Œa; b. According to (2.2.6) witht Dti,

jf .x/ f .ti/j< 1 if x2Iti \Œa; b: Therefore, jf .x/j D j.f .x/ f .ti//Cf .ti/j jf .x/ f .ti/j C jf .ti/j 1C jf .ti/j if x2Iti \Œa; b: (2.2.7) Let M D1C max 1injf .ti/j: SinceŒa; bSniD1 Iti \Œa; b

, (2.2.7) implies thatjf .x/j M ifx2Œa; b.

This proof illustrates the utility of the Heine–Borel theorem, which allows us to choose M as the largest of afiniteset of numbers.

Theorem2.2.8and the completeness of the reals imply that

iff is continuous on a finite closed intervalŒa; b, thenf has an infimum and a supre- mum on Œa; b. The next theorem shows that f actually assumes these values at some points inŒa; b.

### Theorem 2.2.9

Suppose thatf is continuous on a finite closed intervalŒa; b:Let

˛D inf

axbf .x/ and ˇDasupxbf .x/:

Then˛andˇare respectively the minimum and maximum off onŒa; bIthat is;there are pointsx1andx2inŒa; bsuch that

f .x1/D˛ and f .x2/Dˇ:

### Proof

We show thatx1exists and leave it to you to show thatx2exists (Exercise2.2.24). Suppose that there is no x1 inŒa; b such thatf .x1/ D ˛. Thenf .x/ > ˛ for all

x2Œa; b. We will show that this leads to a contradiction. Suppose thatt 2Œa; b. Thenf .t / > ˛, so

f .t / >f .t /C˛ 2 > ˛:

Sincef is continuous att, there is an open intervalIt abouttsuch that

f .x/ > f .t /C˛

2 if x2It\Œa; b (2.2.8)

(Exercise2.2.15). The collectionH D˚Itˇˇat b is an open covering ofŒa; b. Since

Œa; bis compact, the Heine–Borel theorem implies that there are finitely many pointst1,

t2, . . . ,tnsuch that the intervalsIt1,It2, . . . ,Itn coverŒa; b. Define ˛1D min

1in

f .ti/C˛

2 :

Then, sinceŒa; bSniD1.Iti \Œa; b/, (2.2.8) implies that f .t / > ˛1; at b:

But˛1> ˛, so this contradicts the definition of˛. Therefore,f .x1/D˛for somex1in

Œa; b.

### Example 2.2.12

We used the compactness ofŒa; bin the proof of Theorem 2.2.9

when we invoked the Heine–Borel theorem. To see that compactness is essential to the proof, consider the function

g.x/D1 .1 x/sin1 x;

which is continuous and has supremum2on the noncompact interval.0; 1, but does not assume its supremum on.0; 1, since

g.x/1C.1 x/ ˇ ˇ ˇ ˇsin 1 x ˇ ˇ ˇ ˇ 1C.1 x/ < 2 if 0 < x1: As another example, consider the function

f .x/De x;

which is continuous and has infimum0, which it does not attain, on the noncompact interval .0;1/.

The next theorem shows that iff is continuous on a finite closed intervalŒa; b, thenf assumes every value betweenf .a/andf .b/asxvaries fromatob(Figure2.2.5, page64).

### Theorem 2.2.10 (Intermediate Value Theorem)

Suppose that f is con- tinuous onŒa; b; f .a/ ¤ f .b/;andis betweenf .a/andf .b/:Thenf .c/ D for somecin.a; b/:

a x b x y

y = f(x)

y = µ

### Proof

Suppose thatf .a/ < < f .b/. The set

S D˚xˇˇaxb and f .x/

is bounded and nonempty. Letc D supS. We will show thatf .c/ D . Iff .c/ > , thenc > a and, sincef is continuous atc, there is an > 0 such that f .x/ > if c < x c (Exercise 2.2.15). Therefore, c is an upper bound for S, which contradicts the definition ofcas the supremum ofS. Iff .c/ < , thenc < band there is an > 0such thatf .x/ < forcx < cC, socis not an upper bound forS. This is also a contradiction. Therefore,f .c/D.

The proof for the case wheref .b/ < < f .a/can be obtained by applying this result to f.

In document Introduction to Real Analysis (Page 69-72)