Let f : [1, ∞)→R be the function defined by

5.2.22. Let f : [1, ∞)→R be the function defined by

f (x) = [x] + {x} √ , x

where [x] (resp., {x} ) signifies the integer part (resp., the fractional part) of the real number x .

(a) Find the smallest number z such that f (x) ≤ z , for all x ≥1 . (b) Fix x 0 ≥1 and define the sequence (x n ) by x n = f (x n −1 ) , for all n ≥1 . Prove that lim n → ∞ x n exists.

√ (x) ≤ 1+2 yz /(y+ z) ≤ 2. So 0 ≤ f (x) ≤ 2, for all x ≥ 1. Taking y = 1 we obtain

Solution. (a) Let x = y + z, where y = [x] and z = {x}. Then, by the arithmetic–

geometric means value inequality, f 2 √

that is, sup { f (x); x ≥ 1} = 2. (b) We first observe that 1 ≤x n < 2, for all n ≥ 1. So, without loss of generality,

we can assume that 1 ≤x 0 < 2. If x n = 1, then the sequence is constant and its √ limit equals 1. If not, then f (x) = (1 + x

√ − 1)/ x in (1, 2). At this stage, the

5.2 Introductory Problems 207 key point of the proof is to show that there exists a unique v ∈ (1,2) such that

f (v) = v, f (x) > x if 1 < x < v, and f (x) < x for all v < x < 2. Let us assume for the moment that the result is true. An elementary computation by means of the derivative of f shows that f increases on (1, 2). So, for any 1 < x < v we

have x < f (x) < f (v) = v. Consequently, for all 1 < x 0 < v, the sequence {x n } of the nth iterates of f is bounded and increasing, so it converges. The same conclusion is valid if v < x < 2.

To conclude the proof, it remains to show that f has a unique fixed point v. Let √ x = 1 + u with u > 0. Then f (x) = x if and only if 1 + 2 u + u = 1 + 3u + 3u 2 +u 3 , so u 5 + 6u 4 + 13u 3 + 12u 2 + 4u − 4 = 0. Since the left-hand side is an increasing function with opposite sign values at 0 and 1, we deduce that the equation has a unique root u. Then it is enough to choose v = 1 + u. ⊓ ⊔

The next example is related to the sign of the Taylor polynomial of second order.

5.2.23. Let f :R →(0,∞) be a twice differentiable function. Show that there exists x 0 ∈R such that the Taylor polynomial of second order centered at x 0 is positive on the whole real axis. Note . We recall that the Taylor polynomial of second order centered at x 0 is defined by P f (x 0 ) = f (x 0 )+f ′ (x 0 )(x − x 0 )+[f ′′ (x 0 )] 2 (x − x 0 ) 2 /2 .

Solution. After completing the square of a binomial, we observe that x 0 has the required property if and only if either f ′ (x 0 )=f ′′ (x 0 ) = 0 or f (x 0 )f ′′ (x 0 )>

[f ′ (x 0 )] 2 /2. Let g(x) =

f (x), so g ′ (x) = f ′ (x)/(2g(x)) and

g ′′ (x) =

2 (x) − ( f 3 ′ (x)) /2 (x)). Assuming that there exists x 0 such that g ′′ (x 0 ) > 0, then this is the good x 0 . If not,

then g ′′ (x) ≤ 0 for all x and the graph of g lies entirely under any tangent. Since g is positive, these tangent lines must be horizontal, so g and f are constant, and any

point x 0 has the required property. ⊔ ⊓

A theorem of the German mathematician Hans Rademacher (1892–1969) asserts that Lipschitz functions are “almost” differentiable, in the following sense: If I is an open subset of R and f : I →R is Lipschitz, then f is differentiable almost ev- erywhere in I. If f is only continuous, a sufficient condition that the same property hold is provided in the next problem.

5.2.24. Let I ⊂R be an open interval and consider a continuous function f :I →R satisfying, for all x ∈I , lim y → 0 y −1 [ f (x + y) + f (x − y) − 2 f (x)] = 0 . Prove that the set of points at which f is differentiable is a dense subset of I .

Solution. Fix a , b ∈ I with a < b. It is enough to show that there exists x ∈ (a,b) such that f is differentiable at x. We split the proof into the following two steps. S TEP

1. If f has a local extremum in x 0 ∈ (a,b) then f is differentiable in x 0 and f ′ (x 0 ) = 0. Indeed, since x 0 is a local extremum point of f , then for all t >0

208 5 Differentiability small enough, the quantities t −1 [ f (x 0 + t) − f (x 0 )] and t −1 [ f (x 0 − t) − f (x 0 )] have

the same sign. Next, we use the identity t −1 [ f (x 0 + t) + f (x 0 − t)− 2 f (x 0 )]= t −1 [ f (x 0 + t)− f (x 0 )]+t −1 [ f (x 0 − t)− f (x 0 )] . So, by hypothesis,

Thus, f ′ (x 0 ) = 0. S TEP

2. There exists a local extremum point x 0 ∈ (a,b) of f . Indeed, consider the linear function g (x) = cx + d such that g(a) = f (a) and g(b) = f (b). Then the function h : = f − g satisfies the same assumption as f , that is, for all x ∈ (a,b),

h lim (x + y) + h(x − y) − 2h(x) =0.

Since h is continuous and h (a) = h(b) = 0, there exists a local extremum point x 0 ∈ (a,b) of h. According to the first step, h is differentiable in x 0 , and hence f has the same property. ⊓ ⊔

Functions with bounded first-order derivative are Lipschitz. Is the converse true?

be a differentiable function for which there exists a positive constant M such that for all x , y ∈ (0,1) , | f (x) − f (y)| ≤ M |x − y| . Prove that for any x ∈ (0,1) , |f ′ (x)| ≤ M .

5.2.25. Let f : (0, 1)→R

Solution. Fix x ∈ (0,1) and consider t ∈ R \ {0} such that x + t ∈ (0,1). Thus, by hypothesis,

f (x + t) − f (x) |f (x)| ≤

Passing now at the limit as t →0, we conclude the proof. ⊓ ⊔ We have seen that convex functions are continuous at interior points. What

about the set of differentiability points of a convex function? The following prob- lem shows that there are many such points. The complementary deep theorem due to Denjoy shows that the result is true for larger classes of functions.

be an interval and assume that f :I →R is a convex function. Prove that

5.2.26. Let I ⊂R

(a) f has one-sided derivatives at any interior point x of I , and moreover, f ′ (x−) ≤

f ′ (x+) ; (b) the set of points in I where f is not differentiable is at most countable.

5.2 Introductory Problems 209 Solution. (a) Fix x 0 ∈ IntI. For convexity reasons, the function g : I \ {x 0 }→R

defined by g (x) = ( f (x) − f (x 0 )) /(x − x 0 ) is nonincreasing and continuous. Hence there exists f ′ (x 0 −) and f ′ (x 0 +). Moreover, since for all x, y ∈ I with x < x 0 < y we have g (x) ≤ g(y), it follows that the lateral limits of f in x 0 are finite and f ′ (x 0 −) ≤

f ′ (x 0 +). (b) Set A : = {x ∈ IntI; f ′ (x−) < f ′ (x+)}. For any x ∈ A, define

I x =(f ′ (x−), f ′ (x+)). In order to prove that the set A is at most countable, it is enough to show that I x ∩I y this implies that A is a disjoint union of open intervals, so A is at most countable.

Fix x , y ∈ A with x < y. Then

f (y) − f (x) (x+) ≤ ′

≤f (y−).

−x

By the definition of I x we deduce that I x ∩I y = /0, and the proof is concluded. ⊓ ⊔ Remark. The French mathematician Arnaud Denjoy (1884–1974) proved the following result related to the set of points at which a function is differentiable.

be a function that admits one-sided derivatives at any point of I \A , where A is at most countable. Then f admits a derivative at any point of I , excepting a set that is at most countable. Part (a) of the above exercise asserts that a convex function f : I →R has one- sided derivatives at every interior point of I. So, according to Denjoy’s theorem, we obtain the result established in (b).

Denjoy’s Theorem. Let f :I →R

The sign of the derivative ensures monotony properties of the function. Does the result remain true for the symmetric derivative? We recall that a function f : I →R (where I ⊂ R an interval) has a symmetric derivative at x 0 ∈ IntI if