Define the function f : (0, 1)→R by f (x) = (1 + x 1 /p ) p + (1 − x 1 /p ) p . (a) For any p ≥2 , prove the following inequalities for any 0 <x<1 :

5.3.6. Define the function f : (0, 1)→R by f (x) = (1 + x 1 /p ) p + (1 − x 1 /p ) p . (a) For any p ≥2 , prove the following inequalities for any 0

(a 1 ) f ′′ (x) ≤ 0 ; (a 2 ) f (x) ≤ f (y) + (x − y) f ′ (y) , for all y ∈ (0,1) ; (a 3 ) f (x) ≤ 2 p −1 (x + 1) ;

(b) If 1 ≤p<2 , show that the above inequalities are true with ≥ instead of ≤ . Solution. (a 1 ) We have

− 1 1 −x /p ≤ 0, for all x ∈ (0,1).

(1−2p)/p

Fig. 5.13 Graph of the function f (x) = (1 + x 2 /5 ) 5 /2 + (1 − x 2 /5 ) 5 /2 .

5.3 The Main Theorems 221 (a 2 ) This property characterizes concave functions. Indeed, it is enough to use the

fact that f ′ is nonincreasing, in conjunction with the Lagrange mean value theorem applied to f on the interval [x, y] ⊂ (0,1).

(a 3 ) A straightforward computation shows that the function g defined by g (x) = (x + 1) −1 f (x) is increasing in (0, 1]. This yields the conclusion. ⊓ ⊔ (b) Apply the same ideas as above. Exercise!

The mean value theorem and beyond! A first step to the true sense of this major result is the following exercise, which can, however, be viewed as a weaker version of the Lagrange mean value theorem. The nice part of our argument is that it avoids using the (somewhat subtle) fact that a continuous function attains its maximum on

a closed interval.

5.3.7. Let f : [a, b]→R

be a function that admits a derivative (not necessarily finite!) at any point of [a, b] . Prove that there exists x 0 ∈ [a,b] such that

I. Halperin Solution. For any integer 0

≤ n ≤ 10, set c (b−a)n n =a+ 10 . We have

| f (b) − f (a)| ≤ ∑ | f (c n +1 − f (c n )|.

n =0

Thus, there exists an integer 0 ≤ k ≤ 9 such that

Let k 1 be the least integer with this property and define a 1 =c k 1 and b 1 =c k 1 +1 . Next, we repeat the same arguments for the interval I 1 : = [a 1 ,b 1 ], and so on. Thus, we obtain a sequence of intervals I n : = [a n ,b n ] such that b n −a n = 10 −n (b − a) and

Since I n are closed intervals, |I n | = (b n −a n )→0, and I 1 ⊃I 2 ⊃··· , it follows by the Cantor principle that there exists a unique element x 0 ∈∩ n ≥1 I n . In particular, this shows that x 0 = lim n → ∞ a n = lim n → ∞ b n . We also observe that in relation (5.9) we can assume that either the a n or b n are replaced by x 0 . This follows from (5.9) combined with the elementary identity

f (b n ) − f (a n )

f (b n ) − f (x 0 ) b n −x 0 f (x 0 ) − f (a n ) x 0 −a n =

. n −a n

b n −x 0 b n −a n

x 0 −a n

b n −a n

222 5 Differentiability In conclusion, we can assume that relation (5.9) holds for infinitely many indices n k

(k ≥ 1) and, say, for a n replaced by x 0 . Taking n k →∞ and using the definition of the derivative of f in x 0 , we obtain the conclusion. ⊓ ⊔ Remark. Under the same assumptions, the above arguments apply to show that there exist x 1 ,x 2 ∈ (a,b) satisfying

Actually, the Lagrange mean value theorem asserts that there exists c ∈ (a,b) for which both of these inequalities hold simultaneously. Furthermore, a general mean value theorem holds for functions with values in arbitrary spaces endowed with

a topology. However, the sense of this result is that it establishes a mean value ine- quality and not an equality, as in the one–dimensional case. The following property was discovered by the French mathematician Arnaud Denjoy in 1915, and it is usu- ally known as the Denjoy–Bourbaki theorem, due especially to the elegance of the proof given in 1949 by the Bourbaki group of French mathematicians. We refer to [75] for a history of the celebrated Bourbaki group. We state in what follows this result.

Denjoy–Bourbaki Theorem. Let E be a normed vector space and consider the continuous function f : [a, b]→E . Let ϕ : [a, b]→R

be a continuous nondecreas- ing function. Assume that both f and ϕ admit a right derivative at every point of [a, b) \ A , where the set A is at most countable and, moreover, for all x ∈ [a,b) \ A ,

we have #f ′ (x+)# ≤ ϕ ′ (x+) . Then # f (b) − f (a)# ≤ ϕ (b) − ϕ (a) .

An important consequence of the Denjoy–Bourbaki theorem is the following:

Corollary. Let f : [a, b]→R

be a continuous function that admits a right deriva- tive at every point of [a, b) \ A , where A is at most countable. Assume that for all x ∈ [a,b) \ A we have f ′ (x+) = 0 . Then f is constant on [a, b] .

Are there converses of the mean value theorem?

5.3.8. (i) Let f be a twice differentiable function in [a, b] such that f ′′ (x) > 0 for all x ∈ [a,b] . Prove that for each ξ ∈ (a,b) , there exists x 0 ∈ [a,b] such that either

f ′ ( ξ )=

f (b) − f (x 0 )

or f ′ ( ξ )=

f (a) − f (x 0 )

b −x 0 a −x 0 (ii) Let f be a twice differentiable function in [a, b] such that f ′ (x 0 )=0 , for some

x 0 ∈ (a,b) . Prove that there exists x 1 ∈ [a,b] such that either f (a) = f (x 1 ) or

f (b) = f (x 1 ) . R.S. Luthar, Amer. Math. Monthly, Problem E 2057

Solution. (i) Define the functions

f (x)− f (a)

if x

f ∈ (a,b], (b)− f (x) if x ∈ [a,b),

f ′ (b−) if x =b.

5.3 The Main Theorems 223 Then g, h are continuous on [a, b], and g(b) = h(a). Since f ′ is increasing on [a, b],

we have g (a) < f ′ ( ξ ) < h(b). If g(a) < f ′ ( ξ ) < g(b), then by the continuity of

g , there exists x 0 ∈ (a,b) such that f ′ ( ξ ) = g(x 0 ). Otherwise, we have g(b) =

h (a) ≤ f ′ ( ξ ) < h(b). So, by the continuity of h, there exists x 0 ∈ [a,b] such that

f ′ ( ξ ) = h(x 0 ). Finally, we remark that the case x 0 = a cannot be excluded (see, e.g.,

f (x) = x 2 , [a, b] = [0, 2], ξ = 1). (ii) This follows from (i) if one replaces x 0 in (ii) by ξ , and replaces x 1 in (ii) by x 0 .

As above, the case x 0 = a (or x 0 = b) cannot be excluded (see, e.g., f (x) = x 2 , [a, b] = [−1,1], ξ = 0). ⊓ ⊔

The following problem presents a simple differential inequality that does not follow with elementary arguments.

be a function of class C 3 such that f , f ′ , f ′′ , and f ′′′ are all positive. Moreover, we assume that f ′′′ (x) ≤ f (x) for all x ∈R . Prove that f ′ (x) < 2f (x) for any x ∈R .

5.3.9. Let f :R →R

Putnam Competition, 1999 Solution. We first observe that our hypotheses imply

Indeed, by Lagrange’s mean value theorem, there exists ξ n ∈ (−n 2 , −n) such that

f 2 (−n 2 ) − f (−n) = (n − n )f ′ ( ξ n ) < 0.

By our assumptions, the limits of f and f ′ at −∞ exist, and moreover, they are nonnegative. Arguing by contradiction, let us suppose that lim x → −∞ f ′ (x) > 0. Pass- ing to the limit at −∞ in the above relation we obtain 0 = −∞, contradiction. Hence lim x −∞ f → ′ (x) = 0. A similar argument shows that lim x → −∞ f ′′ (x) = 0.

We shall apply the following elementary result several times in the proof. ⊓ ⊔

Lemma. Let f :R →R be a differentiable function such that lim x → −∞ f (x) ≥ 0

and f ′ (x) > 0 , for all x ∈R . Then f (x) > 0 , for all x ∈R .

Proof of the lemma. If there is some x 0 ∈ R such that f (x 0 ) < 0, then by our assumption that f ′ > 0, we deduce f (x) < f (x 0 ), for all x < x 0 . Therefore lim x → −∞ f (x) ≤ f (x 0 ) < 0, contradiction.

We continue the proof of our problem. By our assumption f ′′′ (x) ≤ f (x) we obtain

f ′′ (x) f ′′′ (x) ≤ f ′′ (x) f (x) < f ′′ (x) f (x) + f ′2 (x), for all x ∈R. Using this inequality and applying the above lemma to the function f (x) f ′ (x)

− 1/2( f ′′ (x)) 2 , we deduce that

(f ′′ (x)) 2 < f (x) f ′ (x), for all x

2 ∈ R.

224 5 Differentiability On the other hand, by f > 0 and f ′′′ > 0 it follows that 2f ′ (x) f ′′ (x) < 2 f ′ (x) f ′′ (x) + 2 f (x) f ′′′ (x), for all x ∈ R. Applying the lemma again, we obtain

f ′2 (x) ≤ 2 f (x) f ′′ (x), for all x ∈ R. (5.11) Combining relations (5.10) and (5.11), we obtain

1 2 (x)

< (f ′′ (x)) 2 < f (x) f ′ (x),

2 2f (x)

that is, f ′3 (x) < 8 f 3 (x), for all x ∈ R. It follows that f ′ (x) < 2 f (x), for all x ∈ R. We remark that it is possible to prove a stronger inequality than that in our state- ment, more precisely, with a constant less than 2. Indeed, adding 1 /2 f ′ (x) f ′′ (x) to both sides of (5.10) and applying then the hypothesis f ′′′ (x) ≤ f (x), we have

1 [f ′ (x) f ′′′ (x) + ( f ′′

2 (x)) 1 ] < f (x) f ′

2 2 (x) ≤ f (x) f 2 (x). By the above lemma we deduce

(x) + f ′ (x) f ′′′

2 4 ∈ R. Multiplying here by f ′ (x) and applying again the lemma, we deduce that

f ′ (x) f ′′

(x) < f 2 (x), for all x

f ′3

(x) < f 3 (x), for all x ∈ R.

6 4 Hence f ′ (x) < (3/2) 1 /3 f (x) < 2 f (x).

We do not know at this stage the best constant satisfying the inequality in our statement. Using the fact that the function f (x) = e x satisfies our hypotheses, we can only assert that the best constant cannot be less than 1, so 1 is not a sharp bound. Let us define f (x) = 1 for x ≤ 0 and f (x) = 1/3e x + 2/3e −x/2

√ cos ( 3x /2) for x ≥ 0.

Then f ′′′ has a singularity at x = 0, but “smoothing” (this is a standard procedure in higher analysis, say, by means of “mollifiers”) a little removes it without changing the basic result below. On x ≥ 0, this f satisfies f ′′′ = f , so it is easy to check that f and its first three derivatives are nonnegative and that f ≥f ′′′ . However, one can compute that f ′ (x)/ f (x) = 1.01894 . . . at x = 3.01674 . . .. Since other related examples can be built, it is difficult to think that the constant 1 .01894 . . . is sharp.

A qualitative property involving the symmetric derivative and the mean value theorem is stated below.