• No results found

Convex Functions

N/A
N/A
Protected

Academic year: 2022

Share "Convex Functions"

Copied!
22
0
0

Loading.... (view fulltext now)

Full text

(1)

Karlstads universitet 651 88 Karlstad Tfn 054-700 10 00 Fax 054-700 14 60 Information@kau.se www.kau.se Faculty of Technology and Science

Department of Mathematics

Hamid Reza Ghadiri

Convex Functions Konvexa Funktioner

Mathematics

Degree Project 15 ECTC, Bachelor Level

Date/Term: 2011-03-10 Supervisor: Sorina Barza Examiner: Håkan Granath

(2)

(1) Introduction...3 (2) Regularity Properties of Convex Functions...8 (3) Closure under Functional Operations...14

1

(3)

HAMID REZA GHADIRI

Abstract

Abstract:Convexity is a simple and natural notion which can be traced to the ancient times. The theory of convex functions is a part of the general theory on convexity, as convex functions are those whose epi- graph (the set of points above the graph) is a convex set. The thesis presents an elementary introduction in the theory of real convex func- tions. We give some characterizations of convex functions, present some elementary regularity or geometric properties and solve some problems and give some applications. Moreover, we prove that convexity is pre- served under many of the usual functional operations and this offers a way of identifying a more complex convex function.

Sammanfattning:Konvexitet ¨ar ett enkelt och naturligt begrepp som kan sp˚aras till gamla tider. Teorin om konvexa funktioner ¨ar en del av en den allm¨anna konvexitetsteorin eftersom konvexa funktioner ¨ar just de vars epigraf (m¨angden av punkter som ligger ovanp˚a funktionens graf) ¨ar en konvex m¨angd. Detta arbete inneh˚aller en grundl¨aggande introduktion till teorin om reella konvexa funktioner. Vi ger n˚agra ekvivalenta beskrivningar av konvexa funktioner, bevisar n˚agra ele- ment¨ara regularitet- och geometriska egenskaper samt l¨oser n˚agra prob- lem. Ut¨over detta bevisar vi att konvexitetet ¨ar sluten under m˚anga vanliga operationer n˚agot som visar ett s¨att att identifiera en mer kom- plicerad konvex funktion.

2

(4)

1. Introduction

The aim of this thesis is to study some geometric and regularity properties of real-valued convex functions of real variable. Many of the results presented are contained in the books [4], [3] and [5]. However, in this work we give some details of proofs which are omitted in the reference books and also solve some problems posed in them. We start with some definitions which are necessary in our work.

Definition 1. Let f be a function defined on an interval I of the real line. The function f is called convex if and only if the inequality

f (λx1+ (1 − λ)x2) ≤ λf (x1) + (1 − λ)f (x2) is satisfied for any x1 and x2∈I and any 0 ≤ λ ≤ 1.

The function f is called strictly convex if and only if f (λx1+ (1 − λ)x2) < λf (x1) + (1 − λ)f (x2) for any x1 and x2 ∈ I and any 0 < λ < 1 and x1 6= x2.

Remark 1. For example, f (x) = x2, x ∈ R is strictly convex and the function

f (x) =

(x2− 1, |x| ≥ 1 0, |x| < 1 is convex but not strictly convex.

Definition 2. A function f is called concave if

f (λx1+ (1 − λ)x2) ≥ λf (x1) + (1 − λ)f (x2) is satisfied for any x1 and x2∈ I and any 0 ≤ λ ≤ 1.

The function f is called strictly concave if and only if f (λx1+ (1 − λ)x2) > λf (x1) + (1 − λ)f (x2) for any x1 and x2 ∈ I and any 0 < λ < 1 and x1 6= x2.

Remark 2. Observe that a function f is convex if and only if −f is concave. The theory of concave functions may therefore be subsumed under that of convex functions and we shall concentrate our attention on the latter.

Definition 3. A function f is called affine on I if and only if f (x) = mx + b, x ∈ I.

Remark 3. It is clear that any affine function is convex and concave.

(5)

In fact, in the following proposition we show that also the converse is true, i.e. the only functions that can be both convex and concave are the affine ones. The following proposition is stated as an exercise in [4] .

Proposition 1. (1) A function f is affine on R if and only if f (λx + (1 − λ)y) = λf (x) + (1 − λ)f (y),

for all λ ∈ R and x, y ∈ R.

(2) A function f is affine on an interval I if and only if both f and

−f are convex on I.

(3) If f : [a, b] → R is a convex function and there is a single value of λ ∈(0,1) for which

f (λa + (1 − λ)b) = λf (a) + (1 − λ)f (b), then f is affine on [a, b].

(4) Let f be a convex function defined on an interval I. Then it is strictly convex there if and only if there is no subinterval of I on which f is affine.

Proof.

(1) Let f (x) = mx + b be an affine function on R and x, y, λ ∈ R.

Then

f (λx + (1 − λ)y) =

= m(λx + (1 − λ)y) + b

= λf (x) + (1 − λ)f (y).

Suppose now that we have the equality

f (λx + (1 − λ)y) = λf (x) + (1 − λ)f (y)

for all λ ∈ R and x, y ∈ R . Let x0 and y0 be two real numbers such that x0 < y0. We denote by M (x) the affine function whose graph contains the points (x0, f (x0)) and (y0, f (y0)).

Hence

M (x) = f (x0) − f (y0)

x0− y0 (x − y0) + f (y0) = mx + b.

Let x be an arbitrary point, x = λx0+ (1 − λ)y0for some λ ∈ R.

Then

M (λx0+ (1 − λ)y0)

= f (x0) − f (y0) x0− y0

(λx0+ (1 − λ)y0− y0) + f (y0)

= λf (x0) − f (y0)

x0− y0 (x0− y0) + f (y0)

(6)

= λf (x0) + (1 − λ)f (y0) = f (λx0 + (1 − λ)y0).

Hence M (x) = f (x) for all x, i.e. f (x) is affine.

(2) Suppose first that f is affine on I i.e. f (x) = mx + b, x ∈ I.

By easy calculations we get

f (λx + (1 − λ)y) = λf (x) + (1 − λ)f (y)

for any x, y ∈ I, which implies that f is convex. Similarly, one can show that −f is convex.

Suppose now that f and −f are convex function. Hence f (λx + (1 − λ)y) ≤ λf (x) + (1 − λ)f (y),

and

(−f )(λx + (1 − λ)y) ≤ λ(−f )(x) + (1 − λ)(−f )(y) or, equivalently

f (λx + (1 − λ)y) ≥ λf (x) + (1 − λ)f (y),

for x, y ∈ I and 0 ≤ λ ≤ 1. By the first and third inequalities we get

f (λx + (1 − λ)y) = λf (x) + (1 − λ)f (y)

for 0 ≤ λ ≤ 1, which means that f is affine. Observe that we proved that the only functions which are both concave and convex are the affine functions.

(3) Suppose that f is not affine. Since f is convex and by unit(1) of this proposition we have

f (λx + (1 − λ)y) < λf (x) + (1 − λ)f (y), λ ∈ [0, 1]

which is in contradiction with the fact that for x = a, y = b we have a single λ for which we have equality. Hence f is affine on [a, b].

(4) We prove first that if f is strictly convex then there is no subin- terval of I on which f is affine. Suppose that there exists an interval I0 where f is affine. Hence

f (λx + (1 − λ)y) = λf (x) + (1 − λ)f (y)

for all x, y, I0 ∈ [a, b] and λ ∈ (0, 1). This means that f is not strictly convex and contradicts the hypothesis. We prove now that if there is no subinterval of I on which f is affine then f is strictly convex. Suppose that f is not strictly convex. Hence there exists x0, y0, x0 6= y0 and λ0 ∈ (0, 1) such that

f (λ0x0+ (1 − λ0)y0) = λ0f (x0) + (1 − λ0)f (y0).

(7)

By (3), f is affine on (x0, y0) so there exists I0 = (x0, y0) ⊂ I on which f is affine, and this contradicts the hypothesis.

The following proposition gives a geometric characterization of convex- ity. It is given as observation in [4, page 2] but present here a detailed proof.

Proposition 2. Let I be an open interval, x1, x2 be two points such that x1 < x2 and M (x) be the linear function whose graph passes through (x1, f (x1)) and (x2, f (x2)). The function f is convex if and only if f (x) ≤ M (x) for all x ∈ [x1, x2] or, equivalently

f (x) − f (x1)

x − x1 ≤ M (x) − M (x1)

x − x1 = M (x2) − M (x)

x2− x ≤ f (x2) − f (x) x2− x for x ∈ (x1, x2).

Proof. Suppose first that f is convex. Let M (x) = f (x2) − f (x1)

x2− x1 (x − x1) + f (x1)

be the linear function whose graph contains the points (x1, f (x1)) and (x2, f (x2)) and let x ∈ [x1, x2]. Then there exists 0 ≤ λ ≤ 1 such that x = λx1+ (1 − λ)x2.

M (λx1+ (1 − λ)x2) = f (x2) − f (x1) x2− x1

(λx1+ (1 − λ)x2− x1) + f (x1)

= λf (x2) − f (x1)

x2− x1 (x2− x1) + f (x1).

By easy calculations we get

λf (x1) + (1 − λ)f (x2).

Hence

M (x) = λf (x1) + (1 − λ)f (x2), and since f is convex we have

f (x) = f (λx1 + (1 − λ)x2) ≤ M (x).

The second statement of the proposition as well as the reversed part follow easily by the first.

The next proposition is given as an exercise in [5].

Proposition 3. The function f is convex if and only if the determinant

1 x1 f (x1) 1 x2 f (x2) 1 x3 f (x3)

(8)

is nonnegative for any x1 < x2 < x3 in the interval I.

Proof. Suppose that the determinant

1 x1 f (x1) 1 x2 f (x2) 1 x3 f (x3)

is nonnega- tive. Then

x2f (x3) − x3f (x2) − x1(f (x3) − f (x2)) + f (x1)(x3− x2) ≥ 0, which after straightforward calculations leads to

f (x3)(x2− x1) + f (x2)(x1− x3) + f (x1)(x3− x2) ≥ 0 i.e

(x3− x1)f (x2) ≤ (x2− x1)f (x3) + (x3− x2)f (x1), or equivalently,

f (x2) ≤ x2− x1

x3− x1f (x3) + x3− x2 x3− x1f (x1).

Since x1 < x2 < x3, x2 = (1 − λ)x1+ λx3 for some 0 < λ < 1.

Substituting x2 by (1 − λ)x1 + λx3 in the above inequality we get f (λx3+ (1 − λ)x1) ≤ λf (x3) + (1 − λ)f (x1)

which means that f is convex on I.

Suppose now that f is convex. Let x1 < x2 < x3 with x2 = λx1+ (1 − λ)x3

for some 0 < λ < 1. Observe that λ = xx2−x1

3−x1 and 1 − λ = xx3−x2

3−x1. We get

f (x2) ≤ x2− x1

x3− x1f (x3) + x3− x2

x3− x1f (x1), i.e.

(x3− x1)(f (x2) − f (x1)) ≤ (x2− x1)((f (x3) − f (x1)), or,

x2f (x3) − x3f (x2) − x1(f (x3) − f (x2)) + f (x1)(x3− x2) ≥ 0.

Thus

1 x1 f (x1) 1 x2 f (x2) 1 x3 f (x3

≥ 0, and this completes the proof.

(9)

2. Regularity Properties of Convex Functions.

The following lemma will be used in the proof of the Theorem 1.

Lemma 1. Let f : I → R and suppose that f is twice differentiable at x0 ∈ I. Then

f00(x0) = lim

h→0

f (x0+ h) − 2f (x0) + f (x0− h)

h2 .

Proof. Let x0 ∈ I and g(h) = f (x0+ h) − 2f (x0) + f (x0− h), h > 0;

g(0) = 0 and g is differentiable in 0. By L’hospital’s rule we have

h→0lim

f (x0+ h) − 2f (x0) + f (x0 − h)

h2 = lim

h→0

g0(h) 2h . By using the chain rule we get

h→0lim g0(h)

2h = lim

h→0

f0(x0+ h) − f0(x0− h)

2h ,

which equals f00(x0).

Remark 4. The existence of the limit in lim

h→0

f (x0+ h) − 2f (x0) + f (x0− h) h2

does not imply that f is twice differentiable. The function f (x) =

(x2sinx1, x 6= 0

0, x = 0

has no second derivative in 0 but the above limit for x0 = 0 exists and it is equal to 0.

Theorem 1. [1] Let I be an open interval and let f : I → R be a func- tion which has a second derivative on I. Then f is a convex function on I0 if and only if f00(x0) ≥ 0 for all x0 ∈ I.

Proof. Let f be a convex function on I. By Lemma 1, the second derivative is given by the limit

f00(x0) = lim

h→0

f (x0+ h) − 2f (x0) + f (x0− h) h2

for each x0 ∈ I. Given x0 ∈ I, let h be such that x0 + h and x0− h belong to I. Then we have

f (x0) ≤ 1

2f (x0+ h) + 1

2f (x0− h)

(10)

by convexity of f . Therefore, we have f (x0+h)−2f (x0)+f (x0−h) ≥ 0 for any x0 ∈ I. Hence

f00(x0) = lim

h→0

f (x0+ h) − 2f (x0) + f (x0− h)

h2 ≥ 0.

Now suppose that f is twice differentiable on I and f00 ≥ 0. We will use Taylor’s Theorem to prove that f is convex. Let x1,x2 be two arbitrary points of I. Let 0<t <1 and x0 := (1 − t)x1+ tx2 be an arbitrary point between x1 and x2. By Taylor’s Theorem there exists c1 between x0 and x1 such that

f (x1) = f (x0) + f0(x0)(x1− x0) + 1

2f00(c1)(x1− x0)2 and c2 between x0 and x2 such that

f (x2) = f (x0) + f0(x0)(x2− x0) + 1

2f00(c2)(x2− x0)2. Since f00 is nonnegative on I the term

R := 1

2(1 − t)f00(c1)(x1− x0)2+ 1

2tf00(c2)(x2− x0)2 is also nonnegative. Thus

(1 − t)f (x1) + tf (x2) =

= f (x0) + f0(x0)((1 − t)x1 + tx2− x0)+

1

2(1 − t)f00(c1)(x1− x0)2+ 1

2tf00(c2)(x2− x0)2

= f (x0) + R ≥ f (x0) = f ((1 − t)x1+ tx2).

Hence, f is a convex function on I and the proof is complete.

Remark 5. An easier proof is given in the second part of Corollary 2.

Theorem 2. [5]Let I be an open interval. If f : I → R is convex function (strictly convex), then the left derivative f0 (x) and the right derivative f+0 (x) exist and are increasing (strictly increasing) on I.

Proof. We prove first that the right derivative exists. Consider (see Proposition 2) the function g(h) = f (x+h)−f (x)

h , h > 0. If h1 < h2, we have by the convexity of f that

f (x + h1) − f (x) h1

≤ f (x + h2) − f (x) h2

,

(11)

which means that g is increasing. Thus

h→0lim+

f (x + h) − f (x)

h = inf

h>0

f (x + h) − f (x) h

is −∞ or finite. We prove that inf

h>0

f (x + h) − f (x)

h can not be −∞.

Let x and x0 two fixed points and x0 < x. By Proposition 2 we have f (x) − f (x0)

x − x0 ≤ f (x + h) − f (x) h

for any h > 0 such that x + h ∈ I. Hence f (x) − f (x0)

x − x0 ≤ inf

h>0

f (x + h) − f (x) h

which means that

lim

h→0+

f (x + h) − f (x) h

is finite, i.e. the right derivative f+0 (x) exists.

Now we show that the right derivative is an increasing function. Let x, y ∈ I and x < y. Choose h > 0 such that x + h and y + h ∈ I. By proposition 2 we have

f (x + h) − f (x)

h ≤ f (y + h) − f (y) h

Passing to the limit as h → 0+we get f+0 (x) ≤ f+0 (y) which means that f+0 is an increasing function.

We prove now that the left derivative exists. The proof is similar but for the sake of completeness we give the details also in this case. Consider the function

g(h) = f (x) − f (x − h)

h ,

for h > 0. If 0 < h1 < h2 we have by convexity of f that f (x) − f (x − h2)

h2 ≤ f (x) − f (x − h1) h1

which means that g is decreasing. Thus lim

h→0+

f (x) − f (x − h)

h = sup

h>0

f (x) − f (x − h) h

is +∞ or finite. We will prove that sup

h>0

f (x) − f (x − h)

h can not be

+∞.

(12)

Let x and x0 two fixed points and x0 < x . By the convexity of f we have f (x) − f (x0)

x − x0 ≥ f (x) − f (x − h) h

for any h > 0 such that x − h ∈ I. Hence f (x) − f (x0)

x − x0 ≥ sup

h>0

f (x) − f (x − h) h

which means that

h→0+lim

f (x) − f (x − h) h

is finite, i.e. the left derivative f0(x) exists.

Now we show that the left derivative is an increasing function. Let x, y ∈ I and x < y. Choose h > 0 such that x − h and y − h ∈ I. By convexity we have

f (x) − f (x − h)

h ≤ f (y) − f (y − h) h

Passing to the limit as h → 0+ we get f0 (x) ≤ f0 (y), which means that f0 is an increasing function.

Corollary 1. Any convex function defined on an open interval I is continuous.

Proof. Let x0 ∈ I. We want to show that f (x0) = limx→x0+f (x) = limx→x0f (x). We have

x→xlim0+f (x) = lim

x→x0+

f (x) − f (x0)

x − x0 .(x − x0) + f (x0)

= f+0(x0).0 + f (x0) = f (x0),

since by theorem 2 the right derivative exists. Similarly

x→xlim0f (x) = f (x0).

Hence, f is continuous in x0. Since x0 was arbitrary, f is continuous on I.

Proposition 4. Let f be a convex function on an open interval I.

Then f0 (x) ≤ f+0 (x), x ∈ I.

Proof. Let h > 0, such that x+h and x−h are in I. By Proposition 2 we have

f (x) − f (x − h)

h ≤ f (x + h) − f (x) h

for any h. By letting h → 0 we get f0 (x) ≤ f+0 (x).

(13)

Corollary 2. Let f be defined on an open interval I.

(1) If f is differentiable on I then f is convex if and only if f0 is increasing.

(2) If f is twice differentiable on I then f is convex if and only if f00 ≥ 0 on I.

Proof.

(1) Suppose first that f is convex and differentiable on I. Then f0 = f+0 = f0 on I. By Theorem 2, f0 is increasing.

Conversely suppose now that f0 is increasing. Choose x1 <

x2 < x3 such that x1, x2, x3 ∈ I. By Lagrange’s mean value theorem we have that

f (x2) − f (x1)

x2− x1 = f0(c1), for some x1 < c1 < x2 and

f (x3) − f (x2)

x3− x2 = f0(c2),

for some x1 < c2 < x3. By the monotonicity of f0 we have that f (x2) − f (x1)

x2− x1 ≤ f (x3) − f (x2) x3− x2 i.e. f is convex.

(2) Suppose that f is convex. By (1) it follows that f00 ≥ 0. Con- versely, suppose now that f00 ≥ 0. Then f0 is increasing and by (1) f is convex.

Proposition 5. Let f be a convex function, differentiable on an inter- val I. Then the graph of f lies above any tangent line on that interval.

Proof. Let x0 ∈ I and y = T (x), T (x) = f0(x0)(x − x0) + f (x0) be the tangent line in the point x0 to the graph of the function y = f (x).

Denote by

h(x) = f (x) − T (x)

We have that h(x0) = 0, h0(x) = f0(x) − T0(x) = f0(x) − f0(x0) and hence h0(x0) = 0. By Corrollary 2, f0 is increasing. Hence h0(x) ≤ 0, for x ≤ x0 and h0(x) ≥ 0 for x ≥ x0, x0 is a local min point, which means that h(x) ≥ 0 and therefore, f (x) ≥ T (x).

Theorem 3. Let f be a convex function defined on an open interval I. Then f is differentiable except on a countable set.

(14)

Proof. Let x0 be a point where f+0 is continuous, i.e. f+0(x0) = limx→x0f+0 (x).

We show first that f is differentiable in x0, i.e. f+0(x0) = f0 (x0). By Proposition 4 we have f0(x0) ≤ f+0(x0) so we just have to prove that f0(x0) ≥ f+0(x0).

By the convexity of f , we have f0 (x0− h) ≤ f+0 (x0) and by the conti- nuity of f+0 we get the desired inequality.

Since f+0 is an increasing function it has mostly a countable number of discontinuities and hence f has countable many points where it is not differentiable, see e.g. [1, Theorem 5.6.4].

Proposition 6. [4] A convex function defined on a closed interval [a, b]

is bounded.

Proof. Let M = max(f (a), f (b)) and z = λa + (1 − λ)b for some λ∈[0,1]. Then

f (z) ≤ λf (a) + (1 − λ)f (b) ≤ (λ + (1 − λ))M = M, i.e. f is bounded from above.

Now we show that the function is also bounded from below. We con- sider an arbitrary point written in the form a+b2 +t, where t ∈ [a−b2 ,b−a2 ].

Then by convexity of f we have f (a + b

2 ) ≤ 1

2f (a + b

2 + t) + 1

2f (a + b 2 − t).

Since

−f (a + b

2 − t) ≥ −M we get

f (a + b

2 + t) ≥ 2f (a + b

2 ) − M = m for any t∈[a−b2 , b−a2 ].

Hence, f (x) ≥ m, for all x ∈ [a, b] which means that f is bounded from below.

Remark 6. One can see that a convex function may not be continuous at the boundary point of its domain as it might have upward jump there (see Proposition 7). Considering a convex function defined on an open interval one can see that it must not be bounded from above. For instance, the function f (x) = 1−x12, x ∈ (−1, 1) is a convex function not bounded from above.

Proposition 7. Let f be a convex function defined on the interval [a,b]

and suppose that f is not continuous in a. Then f (a) > limx→a+f (x) = f (a+).

(15)

Proof. Clearly, limx→a+f (x) exists since f is monotone on a neigh- borhood of a, see e.g. [3, Proposition 1.3.4]. Suppose that f (a) ≤ limx→a+f (x). Since f is convex we have for λ = 12 that

f (a + x

2 ) ≤ f (a) + f (x)

2 ,

for x ∈ (a, b]. Hence,

x→alim+

f (a + x

2 ) −f (x)

2 ≤ f (a) 2 i.e

x→alim+

(f (x) − f (x)

2 ) ≤ f (a) 2 , since

x→alim+

f (a + x

2 ) = lim

x→a+

f (x).

Therefore,

x→alim+

f (x)

2 ≤ f (a) 2 thus,

x→alim+

f (x) ≤ f (a) which is contrary to the fact that f (a) ≤ f (a+).

3. Closure under Functional Operations and Applications Proposition 8. [4]If f and g are convex functions defined on an inter- val I, then any linear combination of αf + βg is also convex provided α and β are nonnegative real numbers.

Proof. If we consider the function

h(x) = αf (x) + βg(x), by using the convexity of f and g we get

h(λx + (1 − λ)y) = αf (λx + (1 − λ)y) + βg(λx + (1 − λ)y)

≤ α (λf (x) + (1 − λ)f (y)) + β (λg(x) + (1 − λ)g(y))

= λ(αf (x) + βg(x)) + (1 − λ)(αf (y) + βg(y))

= λh(x) + (1 − λ)h(y) i.e. h is a convex function.

Proposition 9. If f and g are convex functions and g is increasing, then the composition g ◦ f is also convex on I.

(16)

Proof. Suppose that I is an open interval and a, b ∈ I and λ∈[0,1].

Let h(x) = g(f (x)) and x be an arbitrary point in the interval I such that x = λa + (1 − λ)b. Since f is convex,

f (x) ≤ λf (a) + (1 − λ)f (b) Since g is convex and increasing, we have that

h(x) = g(f (x)) ≤

≤ g(λf (a) + (1 − λ)f (b))

≤ λg(f (a)) + (1 − λ)g(f (b))

= λh(a) + (1 − λ)h(b) i.e. h = g ◦ f is a convex function.

Remark 7. In the above-mentioned proposition g has to be increasing.

For example, consider f (x) = x1 and g(x) = 1x, x > 0. The function g(f (x)) =√

x is concave although both f and g are convex functions.

Remark 8. We can summarize convexity and concavity of the com- posite function g ◦ f in this way: if f : I → R, g : J → R and range f ⊆ J , then we have

(1) If f and g are both convex and g is increasing, then g ◦ f is convex.

(2) If f is concave and g is concave and decreasing, then g ◦ f is convex.

(3) If f and g are both concave and g is increasing, then g ◦ f is concave.

(4) If f is convex and g is concave and decreasing, then g ◦ f is concave.

Corollary 3. Let f be a convex function on an open interval I. Then ef (x) is also convex on I.

Proof. Apply Proposition 9 with g(x) = ex.

Proposition 10. If f and g are both non-negative, either decreasing or increasing and convex functions, then h(x) = f (x)g(x) is also convex.

Proof. Let x and y be two arbitrary points such that x < y and λ ∈ [0, 1]. Observe that for x < y we have

[f (x) − f (y)][g(y) − g(x)] ≤ 0 i.e.

f (x)g(y) + f (y)g(x) ≤ f (x)g(x) + f (y)g(y)

(17)

Since f and g are convex functions, we have

f (λx+(1−λ)y)g(λx+(1−λ)y) ≤ (λf (x) + (1 − λ)f (y)) (λg(x) + (1 − λ)g(y)) i.e.

f (λx + (1 − λ)y)g(λx + (1 − λ)y) ≤

≤ λ2f (x)g(x) + λ(1 − λ)[f (x)g(y) + f (y)g(x)] + (1 − λ)2f (y)g(y).

By easy calculations we get

f (λx + (1 − λ)y)g(λx + (1 − λ)y) ≤ λf (x)g(x) + (1 − λ)f (y)g(y) =

= λh(x) + (1 − λ)h(y).

The following theorem will be used in the proof of Propositions 11 and 12.

Theorem 4. (see e.g. [3]) Let f : I → R be a continuous function and x, y ∈ I. Then f is convex if and only if f is midpoint convex i.e.

f (x + y

2 ) ≤ f (x) + f (y)

2 .

Proof. The fact that f convex implies f midpoint convex is trivial.

We need to prove the sufficiency part. Suppose that f is not convex.

Then there exists a subinterval [a, b] such that the graph of f is not under the chord (a, f (a)) and (b, f (b)), which is the function

ϕ(x) = f (x) −f (b) − f (a)

b − a (x − a) − f (a) ≥ 0, x ∈ [a, b].

It is easy to see that

γ = sup{ϕ(x)|x ∈ [a, b]} > 0.

Observe that ϕ is continuous and ϕ(a) = ϕ(b) = 0. A direct calculation shows that ϕ is also midpoint convex. Put =¸ inf{x|ϕ(x) = γ}; then ϕ(c) = γ and c ∈ (a, b). By definition of c, for every h > 0 for which c + h, c − h ∈ (a, b) we have ϕ(c − h) < ϕ(c) and ϕ(c + h) ≤ ϕ(c), so that

ϕ(c) > ϕ(c − h) + ϕ(c + h)

2 ,

which is in contradiction to the fact that ϕ is midpoint convex.

Remark 9. The above theorem is not true without the continuity as- sumption. There exist midpoint convex functions which are not contin- uous but such an example is far away the aim of this thesis.

As an application of Theorem 4 we formulate the following geometric inequality called Hermitte-Hadamard inequality. The proof is our own.

(18)

Proposition 11. [[3]] Let f : (a, b) → R be a continuous function.

Then f is convex if and only if 1

t − s Z t

s

f (x)dx ≤ 1

2[f (s) + f (t)]

for all a < s < t < b.

Proof. Suppose first that f is a convex function and x = λs+(1−λ)t, 0 < λ < 1. Hence

λ = x − t s − t. Thus, by a change of variables we get

1 t − s

Z t s

f (x)dx = Z 1

0

f (λs + (1 − λ)t)dλ.

Since f is a convex function, we have 1

t − s Z t

s

f (x)dx = Z 1

0

f (λs + (1 − λ)t)dλ ≤ Z 1

0

[λf (s) + (1 − λ)f (t)]dλ

= 1

2[f (s) + f (t)]

and this completes the proof of the first part of the proposition.

Now we suppose that 1 t − s

Z t s

f (x)dx ≤ 1

2[f (s) + f (t)]

for any s < t in (a, b).

We assume that the function is not convex, i.e. by Theorem 4, there exist s and t, s < t in the interval (a, b) such that

f s + t 2



> f (s) + f (t)

2 .

Consider the set

C = {x ∈ (s, t) : f (x) > f (s) + f (t) − f (s)

t − s (x − s)}.

The set C is not empty. One can easily verify that s+t2 ∈ C. Since f is continuous the set C is an open set. Let (y, z) be a maximal connected component of the set C containing the point s+t2 . Since (y, f (y)) and (z, f (z)) are on the graph of the function g(x) = f (s) +f (t)−f (s)

t−s (x − s) we have:

f (y) = f (s) + f (t) − f (s)

t − s (y − s)

(19)

and

f (z) = f (s) + f (t) − f (s)

t − s (z − s).

It implies that

f (y) + f (z)

2 = f (s) + f (t) − f (s) t − s

 y + z 2 − s

 . Therefore

1 z − y

Z z y

f (x)dx > 1 z − y

Z z y

(f (s) + f (t) − f (s)

t − s (x − s))dx

= f (s) +f (t) − f (s) t − s (y + z

2 − s)

= f (y) + f (z) 2 which contradicts our assumption.

Proposition 12. [2] Let I be a closed interval of the form either [0, a]

or [0, ∞). Suppose that f is a continuous function which satisfies f (0) = 0. Then f is convex if and only if

n

X

i=1

(−1)i−1f (xi) ≥ f (

n

X

i=1

(−1)i−1xi),

for any n ≥ 2 and any n points x1 ≥ x2 ≥ .... ≥ xn−1 ≥ xn in the interval I.

Proof. We prove the statement by induction. Suppose that f is convex on the closed interval I. Let x1 > x2 > x3 be arbitrary points in the interval I and let λ > 0 such that x2 = λx1 + (1 − λ)x3. Since f is convex we have

f (x2) = f (λx1+ (1 − λ)x3) ≤ λf (x1) + (1 − λ)f (x3), and

f (x1− x2+ x3) = f ((1 − λ)x1+ λx3) ≤ (1 − λ)f (x1) + λf (x3).

Therefore we get

f (x2) + f (x1− x2+ x3) ≤ f (x1) + f (x3),

which is also valid for any x1 ≥ x2 ≥ x3 in I by continuity of f . If we take x3 = 0, then by f (0) = 0 we have

f (x1− x2) ≤ f (x1) − f (x2).

Hence we have the inequality in the proposition for the n = 2 and n = 3.

(20)

Suppose now that the inequality is valid for n = m ≥ 2. Then for any n = m + 2 points x1 ≥ x2 ≥ .... ≥ xm+2 in the interval I we have,

m+2

X

i=1

(−1)i−1f (xi) = f (x1) − f (x2) + f (

m+2

X

i=3

(−1)i−1xi)

≥ f (x1) − f (x2) + f (

m+2

X

i=3

(−1)i−1xi),

≥ f (

m+2

X

i=1

(−1)i−1xi)

which means that the inequality is valid for every n ≥ 2.

Conversely suppose the inequality holds. Then we get in particular that

f (x2) + f (x1− x2+ x3) ≤ f (x1) + f (x3), for any x1 ≥ x2 ≥ x3 in I. By taking x2 = x1+x2 3 we get

f (x1+ x3

2 ) ≤ f (x1) + f (x3)

2 ;

hence f is convex on I, by Theorem 4. The following proposition is formulated as an exercise in [4].

Proposition 13. Let f be a bijection between two intervals I and J . Then f is convex and increasing if and only if its inverse f−1 is in- creasing and concave.

Proof. Suppose that f is convex. Then we have f (λx + (1 − λ)t) ≤ λf (x) + (1 − λ)f (t).

Considering f (x) = y and f (t) = u and by changing variables we get f (λf−1(y) + (1 − λ)f−1(u)) ≤ λy + (1 − λ)u

By taking inverse from both sides and using the monotonicity of f−1 we have

λf−1(y) + (1 − λ)f−1(u) ≤ f−1(λy + (1 − λ)u), which means that f−1 is concave.

Remark 10. Observe that if f is convex and decreasing then f−1 is convex and decreasing as well.

(21)

Theorem 5. [3] (the Discrete Jensen’s Inequality). Let f be convex function on the open interval I and let xi ∈ I. If λi > 0 andPk

i=1λi = 1, then

f

k

X

i=1

λixi

!

k

X

i=1

λif (xi)

Proof. In order to prove this theorem we use induction. If λ1+ λ2 = 1 then for x1 and x2 we have

λ1f (x1) + λ2f (x2) ≥ f (λ1x1+ λ2x2).

This is true by definition of convexity. Now, suppose that the theorem is true with k − 1 values. Let λ0i = (1−λλi

k) for i = 1, 2, ..., k − 1, then we have

k

X

i=1

λif (xi) = λkf (xk) + (1 − λk)

k−1

X

i=1

λ0if (xi)

≥ λkf (xk) + (1 − λk)f

k−1

X

i=1

λ0ixi

!

≥ f λkxk+ (1 − λk)

k−1

X

i=1

λ0ixi

!

= f

k

X

i=1

λif (xi)

! .

Hence, by the principle of induction the inequality is true for any k ∈ N.

Remark 11. Observe that in fact Jensen’s inequality is equivalent with the notion of convexity. Take λ1 = λ, λ2 = 1 − λ, k = 2 in the above inequality. Jensen’s inequality has a lot of applications in mathematical analysis and elsewhere.

(22)

References

[1] R. G. Bartle and D. R Sherbert, Introduction To Real Analysis, John Wiley

& Sons, Inc, 2000.

[2] Masayoshi Hata, Problems and Solutions in Real Analysis, World Scientific Publishing Co. Pte. Ltd. 2007.

[3] C. P. Niculescu and L. E Persson, Convex Functions, Universitaria Press, 2003.

[4] A. W. Roberts and D. E Varberg, Convex Functions, New York and London, Academic Press, 1973.

[5] B. S. Thomson and J. B. Bruckner and A. M. Bruckner, Elementary Real Analysis, 2008.

References

Related documents

Another approach is taken by Betke and Gurvits [4], who also estimate the robot's pose from erroneous measurements given the correct matching. Instead of intersecting a number

It can be interesting to study what parameters (k, α, γ) are feasible for a strongly regular graph, as there are clearly combinations that are not possible.. The following theorem is

• Lecture 1: Basics of Entropy and Relative Entropy, with an application to Reputations In this lecture, I will introduce entropy and relative entropy, describe the relevant

Stability results for limit cycles with a first- order sliding mode as well as fast switchings close to a second-order sliding mode were de- rived in Johansson et al.. (

Identication and control of dynamical sys- tems using neural networks. Nonparametric estima- tion of smooth regression functions. Rate of convergence of nonparametric estimates

In Paper II, we generalise Burger’s method to study the equidistribution of translates of pieces of horospherical orbits in homogeneous spaces Γ\G for arbitrary semisimple Lie groups

Chemical oxidations and reductions of the azo group in aromatic azo compounds Both aliphatic and aromatic azo compounds are commonly used as reaction intermediates in organic

In Paper IV we consider Problem 2.7 for line graphs of balanced complete bipar- tite graphs; that is, we consider the problem of list edge coloring complete bipartite graphs from