Jensen, Young, H¨older, Minkowski, and Beyond

7.3 Jensen, Young, H¨older, Minkowski, and Beyond

In questions of science the authority of a thousand is not worth the humble reasoning of a single individual.

Galileo Galilei (1564–1642)

7.3.1. Assume that 1 ≤p<∞ and a , b are positive numbers. Then

(i) inf 1 t 1 /p−1 a + 1 − 1 t 1 /p

b =a

1 /p 1 −1/p

t >0 p

(ii) inf 3t 1 −p a p + (1 − t) 1 −p b p

4 = (a + b) p .

<t<1

Solution. We give two different proofs of this property, cf. Maligranda [74]. The first one relies on basic properties of differentiable functions, while the second proof is based on convexity arguments.

First proof. (i) Let, for t > 0, the function f be defined by

f (t) = t 1 /p−1 a + 1 1 − /p t b .

Then the derivative f ′ satisfies ′ 1 1

f (t) = −1 t 1 /p−2

−1 t 1 /p−2 (a − tb), p p

t 1 /p−1 b =

p p and so f ′ is negative for t <t 0 = a/b, zero for t = t 0 , and positive for t >t 0 . Hence,

f has its minimum at the point t 0 = a/b, and this minimum is equal to

b =a 1 /p b 1 −1/p .

(ii) Let, for 0 < t < 1, the function g be defined by

g (t) = t 1 −p a p + (1 − t) 1 −p b p .

7.3 Jensen, Young, H¨older, Minkowski, and Beyond 295 Then the derivative g ′ satisfies the equation

g ′ (t) = (1 − p)t −p a p − (1 − p)(1 − t) −p b p =0 only when t =t 1 = a/(a + b). Since

g ′′ (t

−p−1

1 ) = (1 − p)(−p)t 1 a p − (1 − p)(−p)(1 − t 1 ) −p−1 b > 0, it follows that g has its local minimum at t 1 = a/(a + b), which is equal to

This local minimum of the function g is equal to its global minimum because g is continuous on (0, 1) and lim t →0 + g (t) = lim t →1 − g (t) = +∞.

Second proof . (i) The function ϕ (u) = exp(u) is convex on R. Thus

1 /p 1 −1/p

≤ exp ln (t 1 /p−1 a ) + 1 exp ln (t 1 − /p b )

= t 1 /p−1

a + 1 1 − /p t b

for every t > 0. For t = a/b we have equality. (ii) The function ψ (u) = u p for p > 1 is convex on [0, ∞). Therefore,

for every 0 < t < 1. For t = a/(a + b) we have equality. If p ∈ (0,1) and we change in the equalities (i) and (ii) the infimum into supre-

mum, then the results remain true. We also point out that the second proof of (i) gives an alternative proof of the Young inequality, as well as a different proof of the AM–GM inequality

a 1 /p b 1 −1/p ≤ a + 1 b .

296 7 Inequalities and Extremum Problems The classical H¨older inequality states that if f and g are continuous functions on

the interval [a, b] and if p and q are positive numbers such that p −1 +q −1 = 1, then " b " b 1 /p " b 1 /q

| f (x)g(x)|dx ≤

| f (x)| p dx

|g(x)| q dx .

Equivalently, if f and g are continuous on [a, b], then

b 1 /q | f (x)| |g(x)| dx ≤

| f (x)|dx

|g(x)|dx .

According to Exercise 7.3.1, the inequality

a 1 /p b 1 /q

t 1 /p−1 a + 1 t 1 ≤ /p − b

holds for all t > 0. It follows that

1 /p 1 /q b 1 1 /p | f (x)| |g(x)| dx ≤

t 1 /p−1 | f (x)| + 1 −

t |g(x)| dx

= t 1 /p−1

1 /p | f (x)|dx + 1 −

t 1 |g(x)|dx.

Taking the infimum for t ∈ (0,∞) and using Exercise 7.3.1 (i), we obtain H¨older’s inequality.

The Minkowski inequality states that if f and g are continuous functions on the interval [a, b] and if p ≥ 1, then

| f (x) + g(x)| dx

≤ p | f (x)| dx + |g(x)| dx . (7.2)

In order to prove (7.2), we apply Exercise 7.3.1 (ii). Thus, by the inequality

we deduce that for all t ∈ (0,1),

| f (x) + g(x)| dx ≤

[| f (x)| + |g(x)|] dx

≤ p | f (x)| + (1 − t) |g(x)| 4 dx

| f (x)| p dx

1 + (1 − t) p −p |g(x)| dx .

Taking the infimum for t ∈ (0,1) and using Exercise 7.3.1 (ii), we obtain (7.2).

7.3 Jensen, Young, H¨older, Minkowski, and Beyond 297 We recall that Young’s inequality asserts that for all p > 1 and x, y > 0,

xy ≤

where p ′ stands for the conjugate exponent of p, that is, 1 /p + 1/p ′ = 1. A simple proof of Young’s inequality is described in what follows. Set x p =e a and y p ′ =e b . Let 1 /p = t and observe that 0 ≤ t ≤ 1. We have to show that

e (1−t)a+tb

a ≤ (1 − t)e b + te .

Let f

a (t) = (1 −t)e 2 + te b −e (1−t)a+tb . Then f ′′ (t) = −(a − b) e (1−t)a+tb ≤ 0. Thus,

f is concave and f (0) = f (1) = 0. This forces f (t) ≤ 0 for any t ∈ [0,1]. ⊓ ⊔ The next exercise is a refinement of Young’s inequality.

7.3.2. Let 1 <p≤2 and let p ′

be its conjugate exponent. Then for all x ,y≥0 ,

1 ′ ′ (x

1 ′ x p y p p ′ /2 p /2

−y Solution. If p =2=p ′ the result is trivial, so assume 1 < p < 2. We prove the first

inequality because the second can be obtained via an essentially identical argument by interchanging the roles of p and p ′ and of x and y. If either x = 0 or y = 0, the first inequality is obviously true. Fix 1 < p < 2, x > 0, and suppose y > 0. Expanding the square and simplifying, we see that it is enough to check the following inequality:

p ′ x p /2 y p ′ /2 − xy ≥ 0. Now y =x p −1 is the unique solution of f ′ (y) = 0. Since f ′′ > 0, f (x p −1 ) = 0 is the

f (y) := −p x p

global minimum of f . Jensen’s inequality asserts that if f : (a, b)→R is a convex function, then

f ( λ 1 x 1 + λ 2 x 2 + ··· + λ n x n )≤ λ 1 f (x 1 )+ λ 2 f (x 2 ) + ··· + λ n f (x n ), for any integer n ≥ 2, where x 1 ,x 2 ,...,x n ∈ (a,b) are arbitrary points, and λ i ≥0

(i = 1, . . . , n) are constrained by λ 1 + ··· + λ n = 1.

We give in what follows a direct proof of Jensen’s inequality, whose main idea relies on the property that any convex function has a support line, in the sense that through each point of its graph, a line can be drawn such that the graph of the func- tion does not lie below the line. We first observe that if ϕ (x) = ax + b, where a and

b are real numbers with b

ϕ ∑ λ i x i = ∑ λ i ϕ (x i ) if and only if ∑ λ i = 1.

i =1

i =1

i =1

298 7 Inequalities and Extremum Problems Indeed, relation (7.3) reduces to

a ∑ λ i x i +b=a ∑ λ i x i + ∑ λ i b ,

which is true if and only if ∑ n i =1 λ i = 1. Returning to the proof of Jensen’s inequality, fix x 1 ,x 2 ,...,x n ∈ (a,b), λ i ∈R (i = 1, . . . , n), and set ξ =∑ n i =1 λ i x i . Let ϕ (x) = ax + b be a support line of f at ξ with f ( ξ )= ϕ ( ξ ). Thus, by (7.3),

f ∑ λ i x i = ∑ λ i ϕ (x i ) if and only if ∑ λ i = 1.

But ϕ (x) ≤ f (x) for all x ∈ (a,b). Hence

∑ λ i ϕ (x i )≤ ∑ λ i f (x i ) if and only if λ i ≥ 0.

Using relations (7.4) and (7.5), we obtain Jensen’s inequality. Equality holds either if x 1 = ··· = x n or if f is linear. ⊓ ⊔

Examples. (i) Taking f (x) = −lnx in Jensen’s inequality, we obtain the follow- ing generalized arithmetic–geometric means inequality: for any x 1 ,x 2 ,...,x n >0 and all λ

i ≥ 0 (1 ≤ i ≤ n) with ∑ i =1 λ i = 1,

λ 1 x 1 + ··· + λ n x

n ≥x 1 ··· x n .

(ii) Taking f (x) = x q /p in Jensen’s inequality, we obtain the power means inequal- ity: if p < q, then M p ≤M q , where

with a i > 0 and ∑ n i =1 λ i = 1, λ i ≥ 0. In Figure 7.1 we illustrate this property for n = 2 and M r defined by M

−∞ = min(a, b) =: m, M −1 = 1/(p/a + q/b), M 0 =a b ,M 1 = pa + qb, M 2 = pa 2 + qb 2 ,M ∞ = max(a, b) =: M, where a, b ∈ (0,∞), p ∈ (0,1),

and q = 1 − p. We analyze in what follows a different framework, corresponding to λ 1 = ··· =

λ n = 1, n ≥ 2. In this case, for positive numbers a 1 ,...,a n and p

1 /p

i =1

7.3 Jensen, Young, H¨older, Minkowski, and Beyond 299

Fig. 7.1 The power mean M r is a monotone increasing function.

It was shown (see [93]) by the German mathematician Alfred Pringsheim (1850–1941), who attributed his proof to L¨uroth, and by Jensen [51, p. 192], that

M p >M q for any 0 < p < q.

This inequality can be obtained directly as follows:

1 /p = ∑

Let m = min{a 1 ,...,a n } and M = max{a 1 ,...,a n }. It follows that for any p > 0,

On the other hand, since lim p

0 + ∑ j =1 a j = n, we obtain lim M

p → 0 + p = +∞ .

Hence M p decreases from +∞ to M as p increases from 0+ to +∞. Moreover, since

M −p =M p ,

300 7 Inequalities and Extremum Problems it follows that M p decreases from m to 0 as p increases from −∞ to −0. By contrast,

the mean value functions M r defined in (7.6) increase continuously from m to M, or

remain constant if a 1 = ··· = a n , as r increases from −∞ to +∞.