Hincin, 1927 Solution. (a) Arguing by contradiction, let x 1 ,x 2 ∈ IntI be such that x 1 <x 2 and
A. Hincin, 1927 Solution. (a) Arguing by contradiction, let x 1 ,x 2 ∈ IntI be such that x 1
f (x 1 ) > f (x 2 ). Set A x 1 ,x 2 : = [ f (x 1 ) + f (x 2 )] /2 and observe that f (x 1 )>A x 1 ,x 2 . This enables us to define
c : = inf{x ∈ [x 1 ,x 2 ]; f (x) < A x ,x 2 }.
210 5 Differentiability Then f (c) = A c ,x 2 , and by the definition of c, there exists a sequence of positive
numbers (t n ) n ≥1 such that for all positive integers n,
Thus, for all n ≥ 1,
f (c + t n ) − f (c − t n ) < 0, 2t n
which implies that the symmetric derivative of f in c is nonpositive, a contradiction. (b) We will argue by contradiction. Replacing f (x) with ± f (±x) if necessary, we
may assume there are x 1 ,x 2 ∈ IntI such that x 1 <x 2 and f (x 1 ) < f (x 2 ). Fix
0 < ε < [ f (x 2 ) − f (x 1 )] /(x 2 −x 1 ) and define g(x) := f (x) − ε (x − x 0 ). Then, by hypothesis, the symmetric derivative of g is negative in Int I. So, by (a), g is nondecreasing in Int I. On the other hand, by the choice of ε ,g (x 1 ) > g(x 2 ).
This contradiction shows that f (x 1 ) = f (x 2 ). ⊓ ⊔
If I ⊂ R is an (unbounded!) interval, then a function f : I→R is said to be subad- ditive if f (x+ y) ≤ f (x)+ f (y) for all x,y ∈ I. Are there any really useful subadditive functions? The answer is definitively yes, because the distance function given by the absolute value is a major representative of this class. What happens if differentia- bility properties are fulfilled?
5.2.28. Let f : (0, ∞)→R be a differentiable function. (a) Prove that f (x)/x is decreasing on (0, ∞) if and only if f ′ (x) < f (x)/x on (0, ∞) .
(b) Deduce that if f ′ (x) < f (x)/x on (0, ∞) , then f is subadditive. (c) Prove that if f is subadditive on (0, ∞) and if f (x) ≤ − f (−x) for all x >0 , then
f ′ is nonincreasing on (0, ∞) . Solution. (a) The proof is immediate on observing that
xf ′
(x) − f (x)
(b) Follows from (a) and the observation that if f (x)/x is nonincreasing, then f is subadditive.
5.2 Introductory Problems 211 (c) Fix x , y ∈ (0,∞). Thus, for any t > 0,
f (x + y + t) − f (x + y)
f (x + y + x + t − x) − f (x + y) t
f (x + y) + f (x + t) + f (−x) − f (x + y)
f (x + t) − f (x)
Taking now t ց 0 we obtain f ′ (x + y) ≤ f ′ (x). ⊓ ⊔ How much larger is (|a| + |b|) p than |a| p + |b| p ? The ratio of course, should
depend on p.
5.2.29. Define the function f : [0, 1]→R by f (x) = (1 + x) p /1 + x p , p ≥1 . Find the maximum and minimum values of f . Apply this result to prove the following inequalities:
f or all a , b ∈ R. Solution. A straightforward computation shows that for all x ∈ [0,1], f ′ (x) ≥ 0.
|a| p + |b| p ≤ (|a| + |b|) p ≤2 p −1 (|a| p + |b| p ),
So, f is increasing in [0, 1] and the extreme values of f are 1 and 2 p −1 . Next, without loss of generality, we can assume that
1 ≤ f (x) ≤ 2 p −1 and replacing x by |a|/|b|, we deduce the conclusion. ⊓ ⊔ We prove below a very useful inequality.
5.2.30. (Young’s Inequality) q . Let p >1 , q = p/p − 1 , and consider the function defined by f (x) = |a| p /p + x q − |a|x for all x ≥0 , a ∈R . Find the minimum of f on [0, ∞) and deduce that
|a| p
|ab| ≤ q + |b| /q, for all a , b ∈ R.
Solution. The minimum of f is achieved in x = |a| p −1 and equals 0. In particular, this implies f (|b|) ≥ 0, which implies the conclusion. ⊓ ⊔