Chapter I
Inequalities of convexity
1. The fundamental inequality of convexity
Let X be a set; in the vector space RX of all finite numerical functions1 defined on X, let P be the set of all positive real-valued functions on X. On the other hand, let M be a numerical function,2 finite or not, with values ⩾0, defined on P, such that:
1∘ M(0)=0, and M is positively homogeneous, that is, M(λf)=λM(f) for every real number λ>0.3
2∘ M is increasing in P, in other words the relation f⩽g implies M(f)⩽M(g).
3∘ M is convex in P, in other words (TVS, II, §2, No. 8) satisfies the relation M(f+g)⩽M(f)+M(g).
Example. — Suppose X is a finite set, for example the interval [1,n] of N; denoting by xi (1⩽i⩽n) the coordinates of a vector x∈Rn, the functions
M1(x)=i=1∑nxiandM∞(x)=1⩽i⩽nsupxi satisfy the preceding conditions in the set P of vectors x with coordinates ⩾0.
Remark. — Let S be a pointed convex cone contained in P (that is, a set such that S+S⊂S and λS⊂S for λ⩾0; cf. TVS, II, §2, No. 4); let M be a numerical function (finite or not) with values ⩾0, defined on S and satisfying in S the above conditions 1∘,2∘ and 3∘. Then M can be extended to the entire set P, in such a way that the extended function (which we denote again by M) satisfies the same conditions: it suffices, for every function f∈P, to set M(f)=+∞ if there does not exist any function g∈S such that f⩽g, and M(f)=infg∈S,f⩽gM(g) in the contrary case. This procedure will be applied in Ch. IV, §1 to define the upper integral of a positive function.
(1) Fonction numérique finie—a function with values in R—may also be translated as "real-valued function" (cf. GT, IV, §5, No. 1).
(2) Fonction numérique—"numerical function"—signifies a function with values in R (TG, IV, §5, No. 1). The phrase "finite or not" is sometimes added as a reminder that the function may take on infinite values.
(3) Recall that in R, products such as 0⋅(+∞) are not defined (GT, IV, §4, No. 3).
PROPOSITION 1. — Let φ(t1,t2,…,tn) be a finite numerical function, defined and continuous for ti≥0 (1≤i≤n), such that: 1∘ the relations ti>0 (1≤i≤n) imply φ(t1,t2,…,tn)>0; 2∘ the function φ is positively homogeneous; 3∘ the set K⊂Rn defined by the relations ti≥0 (1≤i≤n), φ(t1,t2,…,tn)≥1 is convex. Under these conditions, if f1,f2,…,fn are n finite functions ≥0 defined on X, such that M(fi)<+∞ for 1≤i≤n, then
M(φ(f1,f2,…,fn))≤φ(M(f1),M(f2),…,M(fn)).
One knows, by the Hahn-Banach theorem (TVS, II, §5), that K is the intersection of the n half-spaces ti≥0 (1≤i≤n) and a family of closed half-spaces (Uι)ι∈I, Uι being defined by a relation of the form
αι1t1+αι2t2+⋯+αιntn−βι≥0,
where the αιk are not all zero. By hypothesis, if t=(ti) is such that ti>0 for 1≤i≤n, then φ(t1,…,tn)>0, therefore there exists a λ0>0 such that the relation λ≥λ0 implies λt∈K; this shows that, for each ι∈I, the relations ti≥0 (1≤i≤n) imply αι1t1+⋯+αιntn≥0 and therefore αιk≥0 for 1≤k≤n; it is then clear that K is also the intersection of the half-spaces ti≥0 (1≤i≤n) and the Uι such that βι≥0; moreover, since the origin does not belong to K, there exists at least one index ι such that βι>0.
Now let C be the convex cone in Rn+1 defined by the relations ti≥0 (1≤i≤n+1), tn+1≤φ(t1,t2,…,tn) (the closure of the convex cone generated in Rn+1 by the convex set K×{1}); it is immediate that C is also defined by the relations ti≥0 (1≤i≤n+1) and
βιtn+1≤αι1t1+⋯+αιntn(ι∈I, βι≥0).
For every x∈X, we therefore have
βιφ(f1(x),…,fn(x))≤αι1f1(x)+⋯+αιnfn(x)
for all ι∈I. For every index ι such that βι>0, it follows from (4) and the hypotheses on M that M(φ(f1,f2,…,fn)) is finite and
βιM(φ(f1,f2,…,fn))≤αι1M(f1)+αι2M(f2)+⋯+αιnM(fn),
and this relation is also verified in an obvious manner if βι=0. We thus see that the point with coordinates
M(f1),M(f2),…,M(fn),M(φ(f1,f2,…,fn))
belongs to C, which proves the proposition.
2. The inequalities of Hölder and Minkowski
In this No. and in the following one, X and P have the same meaning as in No. 1, and M denotes a function defined on P that satisfies the conditions listed in No. 1.
PROPOSITION 2. — Let α and β be two numbers such that 0<α<1, 0<β<1, α+β=1. If f and g are two finite functions ≥0 defined on X, and if M(f) and M(g) are finite, then
M(fαgβ)≤(M(f))α(M(g))β
(Hölder’s inequality).
By Prop. 1, it all comes down to proving that, in R2, the set defined by the relations t1≥0, t2≥0, t1αt2β≥1 is convex, or again (FRV, I, §4, No. 1, Def. 1) that the function u(t)=t−α/β is convex for 0<t<+∞. Now, setting r=α/β, we have D2u(t)=r(r+1)t−r−2 and, since r>0, D2u(t)>0 on ]0,+∞[, which proves the proposition (FRV, I, §4, No. 4, Cor. of Prop. 8).
COROLLARY. — Let αi (1≤i≤n) be n numbers ≥0 such that
i=1∑nαi=1, and let fi (1≤i≤n) be n functions ≥0 defined on X, such that M(fi) is finite for 1≤i≤n. Under these conditions,
M(f1α1f2α2⋯fnαn)≤(M(f1))α1(M(f2))α2⋯(M(fn))αn.
We can restrict attention to the case that αi>0 for every i. It suffices to argue by induction on n, by applying the inequality (5) to the numbers α=α1 and β=∑i=2nαi, and to the functions f=f1, g=(f2α2f3α3⋯fnαn)1/β.
PROPOSITION 3. — Let p be a real number ≥1. If f and g are two finite functions ≥0 defined on X, then
(M((f+g)p))1/p≤(M(fp))1/p+(M(gp))1/p
(Minkowski’s inequality).
We can restrict attention to the case that M(fp) and M(gp) are finite. By Prop. 1, we are reduced to proving that the set in R2 defined by the relations t1≥0, t2≥0, t11/p+t21/p≥1 is convex, or again that the function u(t)=(1−t1/p)p is convex for 0≤t≤1. Now,
D2u(t)=(1−p1)t1/p−2(1−t1/p)p−2≥0
for 0<t≤1, whence the proposition.
3. The semi-norms Np
Let p be a real number ≥1 and let Fp(X,M) be the set of finite numerical functions f defined on X such that M(∣f∣p) is finite. It is obvious that if g is a function belonging to Fp(X,M) and if ∣f∣≤∣g∣, then f also belongs to Fp(X,M); this remark and Minkowski’s inequality show that the sum of two functions in Fp(X,M) also belongs to this set; taking into account the fact that M is positively homogeneous, we thus see that Fp(X,M) is a linear subspace of the space RX of all finite numerical functions defined on X.
For every number p>0 and every finite numerical function f defined on X, set
Np(f)=(M(∣f∣p))1/p;
then Np(λf)=∣λ∣Np(f) for every scalar λ; moreover, if p≥1 then, by (7),
Np(f+g)≤Np(f)+Np(g),
which proves that Np is a semi-norm on the vector space Fp(X,M) (TVS, II, §1).
Proposition 4. — Let p and q be two real numbers >0 and set 1/r=1/p+1/q. For any finite numerical functions f,g defined on X,
Nr(fg)≤Np(f)Nq(g)
provided that Np(f) and Nq(g) are finite.
For, the relation (9) may be written
M(∣f∣r∣g∣r)≤(M(∣f∣p))r/p(M(∣g∣q))r/q,
which is none other than Hölder’s inequality (5) applied to the numbers α=r/p and β=r/q and to the functions ∣f∣p and ∣g∣q.
COROLLARY. — Suppose that M(1)=1; then, for every finite numerical function f defined on X, the mapping p↦Np(f) is increasing in ]0,+∞[.
Applying the inequality (9) to the case that g=1, we see that Nr(f)≤Np(f) for all q>0; since the number r defined by 1/r=1/p+1/q runs over the set of numbers such that 0<r<p when q runs over the set of numbers >0, the corollary is proved.
PROPOSITION 5. — For every finite numerical function f defined on X, the set I of values of 1/p (p>0) such that Np(f) is finite is either empty or is an interval; if I is not reduced to a point, then the mapping α↦logN1/α(f) is either convex on I or is equal to −∞ on the interior of I.
Let r and s be two distinct numbers >0 such that 1/r and 1/s belong to I; it all comes down to proving that if
p1=rt+s1−t,
with 0<t<1, then
logNp(f)≤t⋅logNr(f)+(1−t)logNs(f),
or, what comes to the same,
Np(f)≤(Nr(f))t(Ns(f))1−t,
a relation that may, by the definition of Np, be written
M(∣f∣p)≤(M(∣f∣r))tp/r(M(∣f∣s))(1−t)p/s.
Setting α=tp/r, we have 1−α=(1−t)p/s by the relation that defines p as a function of t,r,s; whence p=αr+(1−α)s. Hölder’s inequality now yields
M(∣f∣rα∣f∣s(1−α))≤(M(∣f∣r))α(M(∣f∣s))1−α,
which is none other than the inequality (12).
Exercises
1) With the hypotheses of No. 1, show that the set of bounded functions on X such that M(∣f∣) is finite is a subalgebra A of RX, and that the set of bounded functions on X such that M(∣f∣)=0 is an ideal in A. If, moreover, M(1) is finite, show that the mapping f↦M(f) is continuous when A is equipped with the topology of uniform convergence in X.
2) Let X be the interval [0,+∞[ of R, S the convex cone formed by the functions defined on X such that 0≤f(x)≤kx on X (for a finite number k>0 depending on f). Set M(f)=0 for f∈S, and M(f)=+∞ for every positive function f defined on X and not belonging to S. Show that M satisfies the conditions of No. 1, and that M(x)=0, and M(xr)=+∞ for every number r>0 not equal to 1.
3) Give an example where X is a set with two elements, Np(x) is finite for every p>0 and every x∈R2, but where there exist values of p such that the mapping p↦Np(x) is not differentiable at these points.
4) Deduce the inequality (6) from the inequality of the geometric mean
z1α1z2α2⋯znαn≤α1z1+⋯+αnzn(where i=1∑nαi=1)
(FRV, III, §1, No. 1, Prop. 2). (Reduce to the case that M(fi)=1 for 1≤i≤n.)
5) Let α be a real number >1 and let β=1−α<0. Let g be a finite function, defined on X, such that g(x)>0 for all x∈X and such that M(g)>0; show that for every finite function f≥0 defined on X such that M(f) is finite,
M(fαgβ)≥(M(f))α(M(g))β
(apply Hölder’s inequality suitably).
6) Deduce Minkowski’s inequality from Hölder’s inequality (find an upper bound for M((f+g)p−1) with the help of Hölder’s inequality). If one assumes that M(f+g)=M(f)+M(g) for every pair of functions f,g defined and ≥0 on X, deduce similarly from Exer. 5 the inequality
(M((f+g)p))1/p≥(M(fp))1/p+(M(gp))1/p
in the following cases:
a) 0<p<1, f and g finite functions ≥0 defined on X, such that f(x)+g(x)>0 for all x∈X and such that M(fp) and M(gp) are finite;
b) p<0, f and g finite functions defined on X, such that f(x)>0 and g(x)>0 for all x∈X, M(fp) and M(gp) are finite, and M((f+g)p)>0.