Let I be an arbitrary interval and consider a convex function f :I →R . Prove that f is locally Lipschitz.

6.3.2. Let I be an arbitrary interval and consider a convex function f :I →R . Prove that f is locally Lipschitz.

Solution. Let K be any compact subset of I. We can choose points a <b<c<d in I such that for any x , y ∈ K with x < y we have a < b < x < y < c < d. Using Exercise 6.1.1 (ii), we obtain

d −c To conclude the proof, it is sufficient to choose

b −a

y −x

f . f M K = max

(b) − f (a) , (d) − f (c) .

b −a

d −c

6.3 Convexity versus Continuity and Differentiability 275

y = f(x) y= f(x 0 )+ p(x −x 0 )

Fig. 6.3 A function is convex if and only if for each point on its graph there is a line through this point that lies below the graph.

A convex function f is characterized by having a support line at each point (see Figure 6.3), that is,

f (x) − f (x 0 ) ≥ p(x − x 0 ),

for all x and x 0 , where p depends on x 0 . In fact, p =f ′ (x 0 ) when f ′ (x 0 ) exists, and p is any number between f ′ (x 0 −) and f ′ (x 0 +) in the countable set where these are different. We prove in what follows this property in the framework of convex differentiable functions. ⊓ ⊔

be a differentiable function. Prove the following properties:

6.3.3. Let I be an arbitrary interval and let f :I →R

(i) f is convex if and only if

f (x) ≥ f (a) + f ′ (a)(x − a) for any x , a ∈ I. (ii) f is strictly convex if and only if

f (x) > f (a) + f ′ (a)(x − a) for any x ,a∈I , x

Solution. (i) We first assume that f is convex. Fix x that λ ∈ (0,1]. Since f is convex, we obtain

f (a + λ (x − a)) = f ((1 − λ )a + λ x ) ≤ (1 − λ ) f (a) + λ f (x) = f (a) + λ ( f (x) − f (a)).

Therefore

f (a + λ (x − a)) − f (a)

· (x − a) ≤ f (x) − f (a).

(x − a)

Passing to the limit as λ ց 0, we obtain f ′ (a)(x − a) ≤ f (x) − f (a).

276 6 Convex Functions

Next, we fix x , y ∈ I and λ ∈ [0,1]. Thus, by hypothesis,

f (x) ≥ f ((1 − λ )x + λ y )+f ′ ((1 − λ )x + λ y )· λ (x − y)

and

f (y) ≥ f ((1 − λ )x + λ y )−f ′ ((1 − λ )x + λ y ) · (1 − λ )(x − y). Multiplying the first inequality by (1 − λ ) and the second inequality by λ and sum-

ming the new inequalities, we obtain (1 − λ ) f (x) + λ f (y) ≥ f ((1 − λ )x + λ y ), which shows that f is convex.

(ii) For necessity, with the same arguments as in (i), we obtain

f (x) − f (a) > (x − a)) .

f (a + λ

But by (i),

f (a + λ (x − a)) − f (a) ≥ f ′ (a)(x − a).

We conclude that f (x) > f (a) + f ′ (a)(x − a). The sufficiency part is argued with the same proof as above.

The following useful test implies that the functions R x , (0, π /2) ∋ x

a (a > 1) are strictly convex, while the mappings

a x (a > 1), (0, π π /2, π are strictly concave. ⊓ ⊔

be a twice differentiable function such that f ′′ (x) ≥ 0 (resp., f ′′ (x) > 0 ) for any x ∈I . Prove that f is a convex (resp., strictly convex) function.

6.3.4. Let I be an arbitrary interval and let f :I →R

Solution. We apply Taylor’s formula. Thus, for any x , a ∈ I, there exists ξ

between x and a such that x −a

(x − a) 2

f (x) = f (a) +

f ′′ ( ξ ) ≥ f (a) + (x − a) f ′ (a) . 1!

f ′ (a) +

At this stage it is enough to apply Exercise 6.2.2. ⊓ ⊔

6.3.5. Let f :I →R

be a convex function. Prove that f has finite one-sided deriva- tives at any interior point of I . Moreover, if x and y , with x ≤y , are interior points of

I , then

(x−) ≤ f ′ (x+) ≤ f (y−) ≤ f (y+) .

Solution. Let a be an interior point of I and choose x 1 ,x 2 ,y ∈ I such that x 1 <x 2 <

a < y. Then, by Exercise 6.1.1 (ii), we have

f (y) − f (a) x

f (x 1 ) − f (a)

f (x 2 ) − f (a)

1 −a

2 −a

y −a

6.3 Convexity versus Continuity and Differentiability 277 Hence, f ′ (a−) = lim x → a − [ f (x) − f (a)]/(x − a) exists, and moreover, this limit is

at most [ f (y) − f (a)]/(y − a); hence f ′ (a−) is finite. A similar argument shows that

f ′ (a+) exists, is finite, and f ′ (a−) ≤ f ′ (a+). The last assertion follows directly from (6.4). ⊓ ⊔ In particular, since one-sided derivatives exist at interior points, we find again that a convex function f : I →R is continuous at interior points of I. Exercise 6.2.4 also shows that if discontinuity points exist, then necessarily they are of the first kind.

A convex function does not have to be smooth, but it is differentiable in massive sets. The original result in this sense is due to Stanislav Mazur (1933) and implies that in a finite-dimensional space, a convex function is differentiable almost every- where (see [36]).

be a convex function. Prove that the set of points where f is not differentiable is at most countable.

6.3.6. Let f :I →R

Solution. Let S denote the set of points x ∈ I such that f is not differentiable at x. We associate to any point x ∈ S that is interior to I the open and nonempty interval I x =(f ′ (x−), f ′ (x+)). According to Exercise 6.2.4. we have I x ∩I y = /0, provided x

x ∈I x ∩ Q. The one-to-one mapping S

x ∈ Q shows that the set S is at most countable. ⊓ ⊔ The following exercise provides an important test for locating the extremum

points of a real function of class C 2 .