The mean value theorem relates the average rate of change of a function to its instantaneous rate of change. It is one of the most useful theorems in analysis and will play a central role in the proof of the fundamental theorem of calculus in Chapter 5. The proof of the mean value theorem is based on the existence of local extrema.
4.2.1 Definition. A functionf is said to have alocal maximum (local mini- mum) at ciff is defined on an open intervalI containingc andf(x)≤f(c) (f(x)≥f(c)) for allx∈I. In either case,f is said to have a local extremum
atc. ♦
c1
x f
c2
FIGURE 4.2: Local extrema off.
4.2.2 Local Extremum Theorem. If f has a local extremum atc and iff is differentiable atc, thenf0(c) = 0.
Proof. Suppose thatf has a local maximum atc. LetI be an open interval containingc such thatf(x)≤f(c) for all x∈I. Then
f(x)−f(c) x−c
(≥0 ifx∈I andx < c
≤0 ifx∈I andx > c.
It follows that the left-hand derivative of f at c is ≥0 and the right-hand derivative is≤0, hencef0(c) = 0. The proof for the local minimum case is similar.
4.2.3 Rolle’s Theorem. Letf be continuous on[a, b]and differentiable on (a, b). Iff(a) =f(b), then there exists a pointc∈(a, b)such that f0(c) = 0.
Proof. By the extreme value theorem there exist xm, xM ∈[a, b] such that f(xm)≤f(x)≤f(xM) for allx∈[a, b]. Iff(xm) =f(xM), thenf is a constant function and the assertion of the theorem holds trivially. Iff(xm)6=f(xM), then either xm∈ (a, b) or xM ∈(a, b), and the conclusion follows from the local extremum theorem.
Differentiation onR 81 The following result is the key ingredient in the proof of l’Hospital’s rule in Section 4.5.
4.2.4 Cauchy Mean Value Theorem. Let f andg be continuous on[a, b] and differentiable on(a, b). Then there exists a point c∈(a, b)such that
[f(b)−f(a)]g0(c) = [g(b)−g(a)]f0(c). Proof. The function
h(x) := [f(b)−f(a)]g(x)−[g(b)−g(a)]f(x)
is continuous on [a, b], differentiable on (a, b), and satisfies h(a) = h(b). By Rolle’s theorem, h0(c) = 0 for some c ∈(a, b), which is the assertion of the theorem.
a c b
x y
(f(c), g(c))
(f(b), g(b))
(f(a), g(a))
x y
(a) (b)
FIGURE 4.3: (a) Cauchy mean value theorem. (b) Mean value theorem.
Iff(a)6=f(b) and f0(x)6= 0 on (a, b), then the conclusion of 4.2.4 may be written
g(b)−g(a)
f(b)−f(a) = g0(c) f0(c).
For smooth functions f and g, this equation asserts that at some point f(c), g(c) on the curve given parametrically by the equations x = f(t) andy =g(t), the line through the endpoints (f(a), g(a)) and (f(b), g(b)) is parallel to the line tangent to the curve at f(c), g(c). See Figure 4.3(a).
Takingg(x) =xin the Cauchy mean value theorem yields the standard mean value theorem (Figure 4.3(b)):
4.2.5 Mean Value Theorem. If f is continuous on [a, b]and differentiable on(a, b), then there existsc∈(a, b) such that
f(b)−f(a)
b−a =f0(c).
82 A Course in Real Analysis
4.2.6 Corollary. Letf(x)and g(x) be differentiable on an open interval I such thatf0(x) =g0(x)for allx∈I. Then there exists a constant ksuch that f =g+kon I.
Proof. Leta, b∈I. By the mean value theorem applied toh:=f−g, there existsc∈(a, b) such thath(a)−h(b) =h0(c)(a−b). Sinceh0 = 0,h(a) =h(b).
Sinceaandbwere arbitrary,hmust be constant.
4.2.7 Corollary. Let f be differentiable on an open intervalI.
(a) If f0≥0 (f0>0)on I, thenf is increasing(strictly increasing)onI. (b) If f0≤0 (f0<0)on I, thenf is decreasing(strictly decreasing)on I. Proof. We prove (a) for the strictly increasing case. Let a, b∈I, a < b. By the mean value theorem,f(b)−f(a) =f0(c)(b−a) for somec∈(a, b). Since f0(c)>0,f(b)> f(a).
Exercises
1.S Show that cosx=√
x−1 has exactly one solution xin the interval (0, π/2).
2. Find an interval I such that for each c ∈I, sinx=x2/2 +x+c has exactly one solution xin the interval (0, π/2).
3.SShow thatf(x) =x4−4x3+ 4x2+chas at most one zero in the interval (1,2). For what interval of values ofcdoes f have exactly one zero in (1,2)?
4. Letf havekderivatives andndistinct zeros on an intervalI. Prove that f(k) has at leastn−kdistinct zeros inI.
5. Let f have a continuous second derivative on [−1,3], f(1) = 0, and setg(x) =x2f(x). Prove thatg00 has at least one zero in [−1,2].Hint.
Consider the functiongn(x) :=x(x+ 1/n)f(x).
6. Let P(x) be a polynomial of degreen and let a 6= 0. Prove that the equationeax=P(x) has at mostn+ 1 solutions.
7.S Let P(x) be a polynomial of degree n and let a6= 0. Prove that the equation sin(ax) =P(x) has at mostn+ 1 solutions.
8. Prove Bernoulli’s inequality: (1 +x)r≥1 +rxfor all x≥ −1 and all rational numbersr≥1. (Cf. Exercise 1.5.10.)
9.SLet f andgbe continuous on [a, b] and differentiable on (a, b) such that
|f0| ≤ |g0|. Ifg0 is never zero on (a, b), prove that
|f(x)−f(y)| ≤ |g(x)−g(y)| for allx, y∈[a, b].
Differentiation onR 83 10. Letf andg be differentiable on an open intervalI and leta, b∈I with a < b. Prove that iff(a) =g(a) and f0 > g0 on (a, b), then f > g on (a, b). Use this to show that
(a) lnx < x−1 on the interval (1,+∞).
(b) sinx < xon the interval (0, π/2).
(c) cosx >1−xon the interval (0, π/2).
(d) tanx > xon the interval (0, π/2).
(e) ex > 1 +x+x2/2! +ã ã ã+xn/n! on the interval (0,+∞). (Use induction.)
11.S Show that sinx
x is a decreasing function on (0, π/2).
12. Show that on (0, π/2)
(a) xsinx+ cosx >1. (b) xsinx+pcosx < p, p≥2. (c) x−1(1−cosx) is increasing. (d) x−2(1−cosx) is decreasing. 13. Leta, b, p >0, and forx≥0 define f(x) =ap+xp−(a+x)p. Show
that forx >0,
f0(x)
(>0 if 0< p <1,
<0 ifp >1.
Conclude that
(a+b)p
(< ap+bp if 0< p <1,
> ap+bp ifp >1.
14. Letf andg have derivatives of order non an open interval I and let a∈I. Suppose that
f(j)(a) =g(j)(a) = 0, j= 0, . . . , n−1, and f(j)(x)g(j)(x)6= 0 forx > aandj= 0, . . . , n.
Prove that for anyb∈Iwithb > a there existsc∈(a, b) such that f(b)
g(b) = f(n)(c) g(n)(c).
15. Suppose thatf has a local maximum atc. Prove that lim inf
x→c−
f(x)−f(c)
x−c ≥0≥lim sup
x→c+
f(x)−f(c) x−c .
84 A Course in Real Analysis
16. Letf andgbe continuous on [a, b], differentiable on (a, b) and letf(a) = f(b) = 0. Show that there existsc∈(a, b) such thatf0(c) =g0(c)f(c).
17.S Show that for any polynomialP(x) there exist finitely many intervals with unionRsuch thatP is strictly monotone on each interval.
18. Suppose thatf has the property
|f(x)−f(y)| ≤c|x−y|1+ε for allx, y∈R, wherec, ε >0. Prove that f is constant.
19.S Letf have a bounded derivative onR. Prove that for sufficiently large rthe function g(x) :=rx+f(x) is one-to-one and mapsRontoR. 20. Supposef >0 on (1,+∞) and limx→+∞xf0(x)/f(x)∈(1,+∞). Prove
that x/f(x) is decreasing on (b,+∞) for someb >1.
21. Letf be twice differentiable on (0, a), f00 ≥0, and limx→0+f(x) = 0.
Prove that f(x)/x is increasing on (0, a). Show that the conclusion is false if the hypothesisf00≥0 is dropped.
22.SLetg(x) =x2sin(1/x) ifx6= 0 andg(0) = 0. Setf(x) =x+g(x). Show that f0(0)>0 butf is not monotone on any neighborhood of 0.
23. Let limx→+∞f0(x) = 0. Prove that ifg≥c >0 on (a,+∞), then
x→+∞lim
f x+g(x)
−f(x)= 0.
24. Let f be differentiable on R with supx∈R|f0(x)| < 1. Prove that the sequence{xn}defined byxn+1=f(xn) converges, wherex1 is arbitrary.
Conclude that f has a uniquefixed point; that is, there exists a unique x∈Rsuch thatf(x) =x.
25.S Supposef is differentiable on an open intervalI. Show thatf0 has the intermediate value property. Conclude that iff0(x)6= 0 onI, then f is strictly monotone onI.Hint.Apply the extreme value theorem to the functiong(x) =f(x)−y0(x−a),a≤x≤b.
26. Let f be differentiable on I := (1,+∞). Prove that if f0 has finitely many zeros inI, then limx→+∞f(x) exists inR.
27. Letf andg have continuous derivatives on an intervalIwith g0 6= 0 and let aj, bj ∈I withaj < bj, j = 1, . . . , n. Prove that there exists c∈I such that
n
X
j=1
[f(bj)−f(aj)]g0(c) =
n
X
j=1
[g(bj)−g(aj)]f0(c).
Differentiation onR 85 28.S A functionf is said to beuniformly differentiable on an open intervalI
if, givenε >0, there existsδ >0 such that
f(x)−f(y)
x−y −f0(y)
< ε
for all x andy in I with 0 <|x−y| < δ. Prove that f is uniformly differentiable on Iifff0 exists and is uniformly continuous onI. 29. Generalize the preceding exercise as follows: Letf andgbe differentiable
on an open intervalI withg0 6= 0 on I. Prove that f0/g0 is uniformly continuous onIiff, givenε >0, there exists aδ >0 such that
f(x)−f(y)
g(x)−g(y) −f0(y) g0(y)
< ε, for allxandy in Iwith 0<|x−y|< δ.
30. Letf be differentiable on [a,+∞) and suppose that the zeros off0 form a strictly increasing sequencean ↑ +∞. Prove that ifL:= limnf(an) exists inR, then limx→+∞f(x) =L.
31.SProve that a functionf is continuously differentiable on an open interval I iff there exists a continuous functionϕonI2 such that
f(x)−f(y) =ϕ(x, y)(x−y) for allx, y∈I.
32. Letf be continuous on (−r, r) and differentiable on (−r,0)∪(0, r). If limx→0f0(x) exists, prove thatf0(0) exists and f0 is continuous at 0.
*4.3 Convex Functions
4.3.1 Definition. A functionf is said to beconvex on an interval (a, b) if f (1−t)u+tv
≤(1−t)f(u) +tf(v)
for alla < u < v < b and allt∈[0,1].f isconcave if−f is convex. ♦ For example, |x| is convex on R, as is easily established using the triangle inequality.
To see the geometric significance of convexity, letLuv: [u, v]→Rdenote the function whose graph is the line segment from (u, f(u)) to (v, f(v)). Since a typical point on the line segment may be written
(1−t) u, f(u)) +t(v, f(v)= (1−t)u+tv,(1−t)f(u) +tf(v)
, t∈[0,1],
86 A Course in Real Analysis we see that
Luv (1−t)u+tv
= (1−t)f(u) +tf(v).
This shows that f is convex iff the line segment connecting any two points on the graph off lies above the part of the graph between the two points. (See Figure 4.4.)
Luv
a u b
f
v x
FIGURE 4.4: Convex function.
Now letx∈(u, v). Then for some t∈(0,1),
x= (1−t)u+tv=t(v−u) +u= (1−t)(u−v) +v, hence
t= (x−u)/(v−u) and 1−t= (v−x)/(v−u). It follows thatf is convex on (a, b) iff
f(x)≤Luv(x) =f(u)v−x
v−u+f(v)x−u
v−u for alla < u < x < v < b. (4.3) 4.3.2 Theorem. If f : (a, b) → R has an increasing derivative, then f is convex. In particular,f is convex if f00≥0.
Proof. Leta < u < x < v < b. By the mean value theorem applied tof on each of the intervals [u, x] and [x, v], there exist pointsy∈ u, x
andz∈ x, v such that
f(x)−f(u)
x−u =f0(y)≤f0(z) = f(v)−f(x) v−x . Solving the inequality forf(x) yields (4.3).
Thusx2n is convex onRfor anyn∈N, ln(x) is concave on (0,+∞), and xp is convex on (0,+∞) ifp≥1 and concave if p <1.
There is a partial converse to 4.3.2. For this we need following lemma.
4.3.3 Lemma. Iff is convex anda < u < x≤y < v < b, then (a) f(x)−f(u)
x−u ≤f(y)−f(u)
y−u ≤f(v)−f(y) v−y , and (b) f(v)−f(x)
v−x ≤f(v)−f(y) v−y .
Differentiation onR 87 Proof. Referring to Figure 4.5, for (a) we have
f(x)−f(u)
x−u ≤ Luy(x)−f(u)
x−u by convexity, sinceu < x < y,
= f(y)−f(u)
y−u by equality of slopes onLuy,
≤ Luv(y)−f(u)
y−u by convexity, sinceu < y < v,
= Luv(v)−Luv(y)
v−y by equality of slopes onLuv,
≤ f(v)−f(y)
v−y by convexity sinceu < y < v. Luv
Luy Lxv
v
u x y
f
FIGURE 4.5: Convex function inequalities.
A similar calculation verifies (b):
f(v)−f(y)
v−y ≥ Lxv(v)−Lxv(y)
v−y =Lxv(v)−Lxv(x)
v−x = f(v)−f(x) v−x . 4.3.4 Theorem. If f is convex, then fr0 and f`0 exist, are increasing, and f`0(x)≤fr0(x).
Proof. Let a < u < x ≤ y < v < b. By (a) of the lemma, the difference quotients [f(x)−f(u)]/(x−u) decrease asx→u+, sofr0(u) exists inRand
fr0(u)≤f(v)−f(y)
v−y <+∞.
Lettingv→y+shows thatfr0(u)≤fr0(y). Therefore,fr0 is increasing. Similarly, by (b) the difference quotients [f(v)−f(y)]/(v−y) increase asy → v− so f`0(v) exists inRand
f`0(v)≥f(v)−f(x)
v−x >−∞.
Takingx=yin (a) of the lemma, we have f(x)−f(u)
x−u ≤f(v)−f(x) v−x .
88 A Course in Real Analysis
Letting u↑ x andv ↓ x, we obtain f`0(x) ≤fr0(x). In particular, f`0(x) and fr0(x) are finite.
4.3.5 Corollary. A convex functionf is continuous.
Proof. By the theorem,f has finite left-hand and right-hand derivatives and hence is left and right continuous.
4.3.6 Theorem. If a convex functionf is differentiable atx∈(u, v), then f0(x)(t−x) +f(x)≤f(t) for allt∈(u, v).
That is, the tangent line at (x, f(x))lies below the graph off on(u, v).
Proof. Since the difference quotients
f(t)−f(x)
/(t−x) decrease as t↓x, fr0(x)≤f(t)−f(x)
t−x , t > x.
The same difference quotients increase ast↑x, hence fl0(x)≥ f(t)−f(x)
t−x , t < x.
Therefore, iff0(x) exists, then f0(x)(t−x) +f(x)≤f(t) for all t.