2; xD0 or xD1;
1 x; 0 < x < 1; C
(a)) is bounded onŒ0; 1, and sup
Therefore,ghas no maximum or minimum onŒ0; 1, since it does not assume either of the values0and1.
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 2.2.11The 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
f .x/D1; inf
0<x<1f .x/D 1:
Theorem 2.2.8Iff is continuous on a finite closed intervalŒa; b;thenf is bounded onŒa; b:
ProofSuppose 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; bis 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; bSniD1 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 2.2.9Suppose thatf is continuous on a finite closed intervalŒa; b:Let
axbf .x/ and ˇDasupxbf .x/:
Then˛andˇare respectively the minimum and maximum off onŒa; bIthat is;there are pointsx1andx2inŒa; bsuch that
f .x1/D˛ and f .x2/Dˇ:
ProofWe 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; bis compact, the Heine–Borel theorem implies that there are finitely many pointst1,
t2, . . . ,tnsuch that the intervalsIt1,It2, . . . ,Itn coverŒa; b. Define ˛1D min
Then, sinceŒa; bSniD1.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
Example 2.2.12We used the compactness ofŒa; bin 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
ProofSuppose 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.