Advanced Calculus (I)

(1)

Advanced Calculus (I)

WEN-CHINGLIEN

Department of Mathematics National Cheng Kung University

WEN-CHINGLIEN Advanced Calculus (I)

(2)

5.6 Convex Functions

Definition

Let I be an interval andf :I →R

(i) f is said to be convex on I if and only if

f(αx + (1−α)y)≤αf(x) + (1−α)f(y)

for all 0≤α≤1 and allx,y ∈I.

(ii) f is said to beconcaveon I if and only if−f is convex on I.

(3)

5.6 Convex Functions

Definition

f(αx + (1−α)y)≤αf(x) + (1−α)f(y)

for all 0≤α≤1 and allx,y ∈I.

(4)

5.6 Convex Functions

Definition

f(αx + (1−α)y)≤αf(x) + (1−α)f(y) for all 0≤α≤1 and allx,y ∈I.

(5)

5.6 Convex Functions

Definition

f(αx + (1−α)y)≤αf(x) + (1−α)f(y) for all 0≤α≤1 and allx,y ∈I.

(6)

Remark:

Let I be an interval andf :I →R. Then f is convex on I if and only if given any [c,d]⊆I, the chord through the points (c,f(c)),(d,f(d)lies on or above the graph ofy =f(x)for allx ∈[c,d].(See Figure 5.5.)

(7)

Remark:

Let I be an interval andf :I →R. Then f is convex on I if and only if given any [c,d]⊆I, the chord through the points (c,f(c)),(d,f(d)lies on or above the graph ofy =f(x)for allx ∈[c,d].(See Figure 5.5.)

(8)

Proof:

Suppose that f is contex on I andx0∈[c,d]. Choose 0≤α ≤1 such thatx0 =αc+ (1−α)d. The chord from (c,f(c))to (d,f(d)) has slope (f(d)−f(c))

(d −c) . Hence, the point on this chord which has the form(x0,y0)must satisfy y0 =αf(c) + (1−α)f(d). Since f is convex, it follows that f(x0)≤y0; i.e., the point (x0,y0)lies on or above the point (x0,f(x0)). A similar argument establishes the reverse implication.2

(9)

Proof:

(10)

Proof:

(d −c) . Hence,the point on this chord which has the form(x0,y0)must satisfy y0 =αf(c) + (1−α)f(d). Since f is convex, it follows that f(x0)≤y0; i.e., the point (x0,y0)lies on or above the point (x0,f(x0)). A similar argument establishes the reverse implication.2

(11)

Proof:

(12)

Proof:

(d −c) . Hence,the point on this chord which has the form(x0,y0)must satisfy y0 =αf(c) + (1−α)f(d). Since f is convex,it follows that f(x0)≤y0; i.e., the point (x0,y0)lies on or above the point (x0,f(x0)). A similar argument establishes the reverse implication.2

(13)

Proof:

(d −c) . Hence, the point on this chord which has the form(x0,y0)must satisfy y0 =αf(c) + (1−α)f(d). Since f is convex, it follows that f(x0)≤y0;i.e., the point (x0,y0)lies on or above the point (x0,f(x0)). A similar argument establishes the reverse implication.2

(14)

Proof:

(d −c) . Hence, the point on this chord which has the form(x0,y0)must satisfy y0 =αf(c) + (1−α)f(d). Since f is convex,it follows that f(x0)≤y0; i.e., the point (x0,y0)lies on or above the point (x0,f(x0)). A similar argument establishes the reverse implication.2

(15)

Proof:

(d −c) . Hence, the point on this chord which has the form(x0,y0)must satisfy y0 =αf(c) + (1−α)f(d). Since f is convex, it follows that f(x0)≤y0;i.e., the point (x0,y0)lies on or above the point (x0,f(x0)). A similar argument establishes the reverse implication.2

(16)

Proof:

(17)

Proof:

(18)

Remark:

A function f is convex on a nonempty, open interval (a,b) if and only if the slope of the chord always increases on (a,b); i.e.,

a<c <x <d <b implies f(x)−f(c)

x −c ≤ f(d)−f(x) d −x .

(19)

Remark:

A function f is convex on a nonempty, open interval (a,b) if and only if the slope of the chord always increases on (a,b); i.e.,

a<c <x <d <b implies f(x)−f(c)

x −c ≤ f(d)−f(x) d −x .

(20)

Proof:

Fixa<c <x <d <b and letλ(x)be the equation of the chord to f through the points(c,f(c))and(d,f(d)). If f is convex,thenf(x)≤λ(x)(see Figure 5.6). Therefore,

f(x)−f(c)

x −c ≤ λ(x)−λ(c)

x −c = λ(d)−λ(x)

d −x ≤ f(d)−f(x) d −x . Conversely, if f is not convex, thenλ(x)<f(x)for some x ∈(c,d). It follows that

f(x)−f(c)

x −c > λ(x)−λ(c)

x −c = λ(d)−λ(x)

d −x > f(d)−f(x) d −x . Therefore, the slope of the chord does not increase on (a,b). 2

(21)

Proof:

Fixa<c <x <d <b and letλ(x)be the equation of the chord to f through the points(c,f(c))and(d,f(d)). If f is convex, thenf(x)≤λ(x)(see Figure 5.6). Therefore,

f(x)−f(c)

x −c ≤ λ(x)−λ(c)

x −c = λ(d)−λ(x)

f(x)−f(c)

x −c > λ(x)−λ(c)

x −c = λ(d)−λ(x)

(22)

Proof:

Fixa<c <x <d <b and letλ(x)be the equation of the chord to f through the points(c,f(c))and(d,f(d)). If f is convex,thenf(x)≤λ(x)(see Figure 5.6). Therefore,

f(x)−f(c)

x −c ≤ λ(x)−λ(c)

x −c = λ(d)−λ(x)

f(x)−f(c)

x −c > λ(x)−λ(c)

x −c = λ(d)−λ(x)

(23)

Proof:

f(x)−f(c)

x −c ≤ λ(x)−λ(c)

x −c = λ(d)−λ(x)

f(x)−f(c)

x −c > λ(x)−λ(c)

x −c = λ(d)−λ(x)

(24)

Proof:

f(x)−f(c)

x −c ≤ λ(x)−λ(c)

x −c = λ(d)−λ(x)

d −x ≤ f(d)−f(x) d −x . Conversely,if f is not convex, thenλ(x)<f(x)for some x ∈(c,d). It follows that

f(x)−f(c)

x −c > λ(x)−λ(c)

x −c = λ(d)−λ(x)

(25)

Proof:

f(x)−f(c)

x −c ≤ λ(x)−λ(c)

x −c = λ(d)−λ(x)

d −x ≤ f(d)−f(x) d −x . Conversely, if f is not convex,thenλ(x)<f(x)for some x ∈(c,d). It follows that

f(x)−f(c)

x −c > λ(x)−λ(c)

x −c = λ(d)−λ(x)

(26)

Proof:

f(x)−f(c)

x −c ≤ λ(x)−λ(c)

x −c = λ(d)−λ(x)

d −x ≤ f(d)−f(x) d −x . Conversely,if f is not convex, thenλ(x)<f(x)for some x ∈(c,d). It follows that

f(x)−f(c)

x −c > λ(x)−λ(c)

x −c = λ(d)−λ(x)

(27)

Proof:

f(x)−f(c)

x −c ≤ λ(x)−λ(c)

x −c = λ(d)−λ(x)

d −x ≤ f(d)−f(x) d −x . Conversely, if f is not convex,thenλ(x)<f(x)for some x ∈(c,d). It follows that

f(x)−f(c)

x −c > λ(x)−λ(c)

x −c = λ(d)−λ(x)

(28)

Proof:

f(x)−f(c)

x −c ≤ λ(x)−λ(c)

x −c = λ(d)−λ(x)

f(x)−f(c)

x −c > λ(x)−λ(c)

x −c = λ(d)−λ(x)

d −x > f(d)−f(x) d −x . Therefore,the slope of the chord does not increase on (a,b). 2

(29)

Proof:

f(x)−f(c)

x −c ≤ λ(x)−λ(c)

x −c = λ(d)−λ(x)

f(x)−f(c)

x −c > λ(x)−λ(c)

x −c = λ(d)−λ(x)

(30)

Proof:

f(x)−f(c)

x −c ≤ λ(x)−λ(c)

x −c = λ(d)−λ(x)

f(x)−f(c)

x −c > λ(x)−λ(c)

x −c = λ(d)−λ(x)

d −x > f(d)−f(x) d −x . Therefore,the slope of the chord does not increase on (a,b). 2

(31)

Proof:

f(x)−f(c)

x −c ≤ λ(x)−λ(c)

x −c = λ(d)−λ(x)

f(x)−f(c)

x −c > λ(x)−λ(c)

x −c = λ(d)−λ(x)

(32)

Theorem

Suppose that f is differentiable on a nonempty, open interval I. Then f is convex on I if and only if f’ is increasing on I.

(33)

Theorem

Suppose that f is differentiable on a nonempty, open interval I. Then f is convex on I if and only if f’ is increasing on I.

(34)

Proof:

Suppose that f is convex onI =: (a,b)and that

c,d ∈(a,b)satisfyc <d. Chooseh>0 so small that c+h<d andd+h<b. Then by Remark 5.60,

f(c+h)−f(c)

h ≤ f(d +h)−f(d)

h .

In particular,f⁰(c)≤f⁰(d).

(35)

Proof:

f(c+h)−f(c)

h ≤ f(d +h)−f(d)

h .

(36)

Proof:

f(c+h)−f(c)

h ≤ f(d +h)−f(d)

h .

(37)

Proof:

f(c+h)−f(c)

h ≤ f(d +h)−f(d)

h .

(38)

Proof:

f(c+h)−f(c)

h ≤ f(d +h)−f(d)

h .

(39)

Conversely, let f’ be increasing on (a,b) and let a<c <x <d <b. Use the Mean value Theorem to choosex0 (between c and x) andx1(between x and d) such that

f(x)−f(c)

x −c =f⁰(x0) and f(d)−f(x)

d −x =f⁰(x1).

Sincex0 <x1 it follows thatf⁰(x0)≤f⁰(x1). In particular, we conclude by Remark 5.60 that f is convex on (a,b). 2

(40)

Conversely,let f’ be increasing on (a,b) and let a<c <x <d <b. Use the Mean value Theorem to choosex0 (between c and x) andx1(between x and d) such that

f(x)−f(c)

d −x =f⁰(x1).

(41)

f(x)−f(c)

d −x =f⁰(x1).

(42)

f(x)−f(c)

d −x =f⁰(x1).

(43)

f(x)−f(c)

d −x =f⁰(x1).

(44)

f(x)−f(c)

d −x =f⁰(x1).

(45)

f(x)−f(c)

d −x =f⁰(x1).

(46)

f(x)−f(c)

d −x =f⁰(x1).

(47)

Theorem

If f is convex on some nonempty, open interval I, then f is continuous on I.

(48)

Theorem

If f is convex on some nonempty, open interval I, then f is continuous on I.

(49)

Proof:

Letx0 ∈I =: (a,b). By symmetry, it suffices to show that f(x)→f(x0)asx →x0+. Leta<c <x0 <x <d <b, y =g(x)represent the euqation of the chord through (x0,f(x0)),(d,f(d)). Since f is convex, we have by

Remark 5.59 thatf(x)≤h(x).Sincef(x0)lies on or below the chord from(c,f(c))to(x,f(x)), we also have that g(x)≤f(x). Consequently,

g(x)≤f(x)≤h(x), x ∈(x0,b).

(50)

Proof:

g(x)≤f(x)≤h(x), x ∈(x0,b).

(51)

Proof:

g(x)≤f(x)≤h(x), x ∈(x0,b).

(52)

Proof:

Letx0 ∈I =: (a,b). By symmetry, it suffices to show that f(x)→f(x0)asx →x0+. Leta<c <x0 <x <d <b, y =g(x)represent the euqation of the chord through (x0,f(x0)),(d,f(d)). Since f is convex,we have by

g(x)≤f(x)≤h(x), x ∈(x0,b).

(53)

Proof:

Remark 5.59 thatf(x)≤h(x).Sincef(x0)lies on or below the chord from(c,f(c))to(x,f(x)),we also have that g(x)≤f(x). Consequently,

g(x)≤f(x)≤h(x), x ∈(x0,b).

(54)

Proof:

Letx0 ∈I =: (a,b). By symmetry, it suffices to show that f(x)→f(x0)asx →x0+. Leta<c <x0 <x <d <b, y =g(x)represent the euqation of the chord through (x0,f(x0)),(d,f(d)). Since f is convex,we have by

g(x)≤f(x)≤h(x), x ∈(x0,b).

(55)

Proof:

Remark 5.59 thatf(x)≤h(x).Sincef(x0)lies on or below the chord from(c,f(c))to(x,f(x)),we also have that g(x)≤f(x). Consequently,

g(x)≤f(x)≤h(x), x ∈(x0,b).

(56)

Proof:

g(x)≤f(x)≤h(x), x ∈(x0,b).

(57)

Proof:

g(x)≤f(x)≤h(x), x ∈(x0,b).

(58)

Proof:

g(x)≤f(x)≤h(x), x ∈(x0,b).

(59)

Both chordsy =g(x)andy =h(x)pass through the point(x0,f(x0)), sog(x)→f(x0)andh(x)→f(x0)as x →x0+. Hence, it follows from the Squeeze Theorem thatf(x)→f(x0)asx →x0+2

(60)

Both chordsy =g(x)andy =h(x)pass through the point(x0,f(x0)),sog(x)→f(x0)andh(x)→f(x0)as x →x0+. Hence,it follows from the Squeeze Theorem thatf(x)→f(x0)asx →x0+2

(61)

(62)

Both chordsy =g(x)andy =h(x)pass through the point(x0,f(x0)), sog(x)→f(x0)andh(x)→f(x0)as x →x0+. Hence,it follows from the Squeeze Theorem thatf(x)→f(x0)asx →x0+2

(63)

(64)

Theorem (Jensen’s Inequality)

Letφ be convex on a closed interval [a,b] and

f : [0,1]→[a,b]. If f andφ◦f are integrable on [0,1], then

φ Z 1

0

f(x)dx

≤ Z 1

0

(φ◦f)(x)dx.

(65)

Theorem (Jensen’s Inequality)

Letφ be convex on a closed interval [a,b] and

f : [0,1]→[a,b]. If f andφ◦f are integrable on [0,1], then φ

Z 1

0

f(x)dx

≤ Z 1

0

(φ◦f)(x)dx.

(66)

Proof:

Set

c = Z 1

0

f(x)dx

and observe that

(21) φ

Z 1

0

f(x)dx

=φ(c) +s Z 1

0

f(x)dx −c

for alls∈R.(Note: Sincea≤f(x)≤b for eachx ∈[0,1], c must belong to the interval [a,b] by the Comparison Theorem for Integrals. Thusφ(c)is defined.)

(67)

Proof:

Set

c = Z 1

0

f(x)dx

and observe that

(21) φ

Z 1

0

f(x)dx

=φ(c) +s Z 1

0

f(x)dx −c

(68)

Proof:

Set

c = Z 1

0

f(x)dx

and observe that

(21) φ

Z 1

0

f(x)dx

=φ(c) +s Z 1

0

f(x)dx −c

(69)

Proof:

Set

c = Z 1

0

f(x)dx

and observe that

(21) φ

Z 1

0

f(x)dx

=φ(c) +s Z 1

0

f(x)dx −c

(70)

Proof:

Set

c = Z 1

0

f(x)dx

and observe that

(21) φ

Z 1

0

f(x)dx

=φ(c) +s Z 1

0

f(x)dx −c

(71)

Proof:

Set

c = Z 1

0

f(x)dx

and observe that

(21) φ

Z 1

0

f(x)dx

=φ(c) +s Z 1

0

f(x)dx −c

(72)

Let

s = sup

x∈[a,c)

φ(c)−φ(x) c−x .

By Remark 5.60,s ≤(φ(u)−φ(c))/(u−c)for all u∈(c,b]; i.e.,

(22) φ(c) +s(u−c)≤φ(u)

for allu ∈[c,b]. On the other hand, ifu∈[a,c), we have by the definition of s that

s≤ φ(c)−φ(u) c−u .

(73)

Let

s = sup

x∈[a,c)

By Remark 5.60,s ≤(φ(u)−φ(c))/(u−c)for all u∈(c,b];i.e.,

(22) φ(c) +s(u−c)≤φ(u)

s≤ φ(c)−φ(u) c−u .

(74)

Let

s = sup

x∈[a,c)

(22) φ(c) +s(u−c)≤φ(u)

for allu ∈[c,b]. On the other hand, if u∈[a,c), we have by the definition of s that

s≤ φ(c)−φ(u) c−u .

(75)

Let

s = sup

x∈[a,c)

By Remark 5.60,s ≤(φ(u)−φ(c))/(u−c)for all u∈(c,b];i.e.,

(22) φ(c) +s(u−c)≤φ(u)

for allu ∈[c,b]. On the other hand,ifu∈[a,c), we have by the definition of s that

s≤ φ(c)−φ(u) c−u .

(76)

Let

s = sup

x∈[a,c)

(22) φ(c) +s(u−c)≤φ(u)

for allu ∈[c,b]. On the other hand, if u∈[a,c),we have by the definition of s that

s≤ φ(c)−φ(u) c−u .

(77)

Let

s = sup

x∈[a,c)

(22) φ(c) +s(u−c)≤φ(u)

for allu ∈[c,b]. On the other hand,ifu∈[a,c), we have by the definition of s that

s≤ φ(c)−φ(u) c−u .

(78)

Let

s = sup

x∈[a,c)

(22) φ(c) +s(u−c)≤φ(u)

for allu ∈[c,b]. On the other hand, ifu∈[a,c),we have by the definition of s that

s≤ φ(c)−φ(u) c−u .

(79)

Let

s = sup

x∈[a,c)

(22) φ(c) +s(u−c)≤φ(u)

s≤ φ(c)−φ(u) c−u .

(80)

Thus (22) holds for allu ∈[a,b]. Applying (22) to u=f(x),we obtain

φ(c) +s(f(x)−c)≤(φ◦f)(x).

Integrating this inequality as x runs from 0 to 1, we obtain

φ(c) +s Z 1

0

f(x)dx −c

≤ Z 1

0

(φ◦f)(x)dx.

Combining this inequality with (21), we conclude that (20) holds.2

(81)

Thus (22) holds for allu ∈[a,b]. Applying (22) to u=f(x), we obtain

φ(c) +s(f(x)−c)≤(φ◦f)(x).

Integrating this inequality as x runs from 0 to 1, we obtain

φ(c) +s Z 1

0

f(x)dx −c

≤ Z 1

0

(φ◦f)(x)dx.

(82)

Thus (22) holds for allu ∈[a,b]. Applying (22) to u=f(x),we obtain

φ(c) +s(f(x)−c)≤(φ◦f)(x).

Integrating this inequality as x runs from 0 to 1,we obtain φ(c) +s

Z 1

0

f(x)dx −c

≤ Z 1

0

(φ◦f)(x)dx.

(83)

φ(c) +s(f(x)−c)≤(φ◦f)(x).

Integrating this inequality as x runs from 0 to 1, we obtain φ(c) +s

Z 1

0

f(x)dx −c

≤ Z 1

0

(φ◦f)(x)dx.

(84)

φ(c) +s(f(x)−c)≤(φ◦f)(x).

Integrating this inequality as x runs from 0 to 1,we obtain φ(c) +s

Z 1

0

f(x)dx −c

≤ Z 1

0

(φ◦f)(x)dx.

Combining this inequality with (21),we conclude that (20) holds.2

(85)

φ(c) +s(f(x)−c)≤(φ◦f)(x).

Z 1

0

f(x)dx −c

≤ Z 1

0

(φ◦f)(x)dx.

(86)

φ(c) +s(f(x)−c)≤(φ◦f)(x).

Z 1

0

f(x)dx −c

≤ Z 1

0

(φ◦f)(x)dx.

Combining this inequality with (21),we conclude that (20) holds.2

(87)

φ(c) +s(f(x)−c)≤(φ◦f)(x).

Z 1

0

f(x)dx −c

≤ Z 1

0

(φ◦f)(x)dx.

(88)

Advanced Calculus (I)