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5 Norms on Linear Spaces 5 Norms on Linear Spaces 5.1 Introduction In this chapter we study norms on real or complex linear linear spaces. Definition 5.1. Let F = (F, {0, 1, +, −, ·}) be the real or the complex field. A semi-norm on an F -linear space V is a mapping ν : V −→ R that satisfies the following conditions: (i) ν(x + y) ≤ ν(x)+ ν(y) (the subadditivity property of semi-norms), and (ii) ν(ax)= |a|ν(x), for x, y ∈ V and a ∈ F . By taking a = 0 in the second condition of the definition we have ν(0)=0 for every semi-norm on a real or complex space. A semi-norm can be defined on every linear space. Indeed, if B is a basis of V , B = {vi | i ∈ I}, J is a finite subset of I, and x = i∈I xivi, define νJ (x) as P 0 if x = 0, νJ (x)= ( j∈J |aj | for x ∈ V . We leave to the readerP the verification of the fact that νJ is indeed a seminorm. Theorem 5.2. If V is a real or complex linear space and ν : V −→ R is a semi-norm on V , then ν(x − y) ≥ |ν(x) − ν(y)|, for x, y ∈ V . Proof. We have ν(x) ≤ ν(x − y)+ ν(y), so ν(x) − ν(y) ≤ ν(x − y). (5.1) 160 5 Norms on Linear Spaces Since ν(x − y)= |− 1|ν(y − x) ≥ ν(y) − ν(x) we have −(ν(x) − ν(y)) ≤ ν(x) − ν(y). (5.2) The Inequalities 5.1 and 5.2 give the desired inequality. ⊓⊔ Corollary 5.3. If p : V −→ R is a semi-norm on V , then p(x) ≥ 0 for x ∈ V . Proof. Choose y = 0 in the inequality of Theorem 5.2 we have ν(x) ≥ |ν(x)|≥ 0. ⊓⊔ Definition 5.4. Let F = (F, {0, 1, +, −, ·}) be the real or the complex field. A norm on an F -linear space V is a semi-norm ν : V −→ R such that ν(x)=0 implies z = 0 for x ∈ V . The pair (V, ν) is referred to as a normed linear space. 5.2 Vector Norms on Rn R 1 1 Lemma 5.5. Let p, q ∈ −{0, 1} such that p + q =1. Then we have p> 1 if and only if q> 1. Furthermore, one of the numbers p, q belongs to the interval (0, 1) if and only if the other number is negative. Proof. We leave to the reader the simple proof of this statement. ⊓⊔ R 1 1 Lemma 5.6. Let p, q ∈ −{0, 1} be two numbers such that p + q = 1 and p> 1. Then, for every a,b ∈ R≥0, we have ap bq ab ≤ + , p q − 1 where the equality holds if and only if a = b 1−p . xp 1 Proof. By Lemma 5.5, we have q> 1. Consider the function f(x)= p + q −x for x ≥ 0. We have f ′(x)= xp−1 − 1, so the minimum is achieved when x =1 and f(1) = 0. Thus, − 1 f ab p−1 ≥ f(1) = 0, which amounts to p p − p−1 a b 1 − 1 + − ab p−1 ≥ 0. p q p By multiplying both sides of this inequality by b p−1 , we obtain the desired inequality. ⊓⊔ 1 1 Observe that if p + q = 1 and p< 1, then q< 0. In this case, we have the reverse inequality n 5.2 Vector Norms on R 161 ap bq ab ≥ + . (5.3) p q which can be shown by observing that the function f has a maximum in x = 1. The same inequality holds when q< 1 and therefore p< 0. Theorem 5.7 (The H¨older Inequality). Let a1,...,an and b1,...,bn be 1 1 2n nonnegative numbers, and let p and q be two numbers such that p + q =1 and p> 1. We have 1 1 n n p n q p q aibi ≤ ai · bi . i=1 i=1 ! i=1 ! X X X Proof. Define the numbers ai bi xi = 1 and yi = 1 n p p n q q ( i=1 ai ) ( i=1 bi ) for 1 ≤ i ≤ n. Lemma 5.6P applied to xi,yi yieldsP p p aibi 1 ai 1 bi 1 1 ≤ n + n . n p n q p p p q p i ai q i bi ( i=1 ai ) ( i=1 bi ) =1 =1 Adding these inequalities,P P we obtain P P 1 1 n n p n q p q aibi ≤ ai bi i=1 i=1 ! i=1 ! X X X 1 1 because p + q = 1. ⊓⊔ The nonnegativity of the numbers a1,...,an,b1,...,bn can be relaxed by using absolute values. Indeed, we can easily prove the following variant of Theorem 5.7. Theorem 5.8. Let a1,...,an and b1,...,bn be 2n numbers and let p and q 1 1 be two numbers such that p + q =1 and p> 1. We have 1 1 n n p n q p q aibi ≤ |ai| · |bi| . ! ! i=1 i=1 i=1 X X X Proof. By Theorem 5.7, we have 1 1 n n p n q p q |ai||bi|≤ |ai| · |bi| . i=1 i=1 ! i=1 ! X X X The needed equality follows from the fact that 162 5 Norms on Linear Spaces n n aibi ≤ |ai||bi|. i=1 i=1 X X ⊓⊔ We obtain as a special case the Cauchy-Schwarz inequality that was proven in Theorem 7.3. n Corollary 5.9 (The Cauchy-Schwarz Inequality for R ). Let a1,...,an and b1,...,bn be 2n numbers. We have n n n 2 2 aibi ≤ |ai| · |bi| . v v i=1 ui=1 ui=1 X uX uX t t Proof. The inequality follows immediately from Theorem 5.8 by taking p = q = 2. ⊓⊔ Theorem 5.10 (Minkowski’s Inequality). Let a1,...,an and b1,...,bn be 2n nonnegative numbers. If p ≥ 1, we have 1 1 1 n p n p n p p p p (ai + bi) ≤ ai + bi . i=1 ! i=1 ! i=1 ! X X X If p< 1, the inequality sign is reversed. Proof. For p = 1, the inequality is immediate. Therefore, we can assume that p> 1. Note that n n n p p−1 p−1 (ai + bi) = ai(ai + bi) + bi(ai + bi) . i=1 i=1 i=1 X X X 1 1 By H¨older’s inequality for p, q such that p> 1 and p + q = 1, we have 1 1 n n p n q p−1 p (p−1)q ai(ai + bi) ≤ ai (ai + bi) i=1 i=1 ! i=1 ! X X 1 X 1 n p n q p p = ai (ai + bi) . i=1 ! i=1 ! X X Similarly, we can write 1 1 n n p n q p−1 p p bi(ai + bi) ≤ bi (ai + bi) . i=1 i=1 ! i=1 ! X X X Adding the last two inequalities yields n 5.2 Vector Norms on R 163 1 1 1 n n p n p n q p p p p (ai + bi) ≤ a + b (ai + bi) , i i i=1 i=1 ! i=1 ! i=1 ! X X X X which is equivalent to inequality 1 1 1 n p n p n p p p p (ai + bi) ≤ ai + bi . i=1 ! i=1 ! i=1 ! X X X ⊓⊔ n Corollary 5.11. For p ≥ 1, the function νp : R −→ R≥0 defined by 1 n p p νp(x1,...,xn)= |xi| , i=1 ! X n n where x = (x1,...,xn) ∈ R , is a norm on the linear space (R , +, ·). Proof. We must prove that νp satisfies the conditions of Definition 5.1 and that νp(x) = 0 implies x = 0. n Let x = (x1,...,xn), y = (y1,...,yn) ∈ R . Minkowski’s inequality ap- plied to the nonnegative numbers ai = |xi| and bi = |yi| amounts to 1 1 1 n p n p n p p p p (|xi| + |yi|) ≤ |xi| + |yi| . i=1 ! i=1 ! i=1 ! X X X Since |xi + yi| ≤ |xi| + |yi| for every i, we have 1 1 1 n p n p n p p p p (|xi + yi|) ≤ |xi| + |yi| , i=1 ! i=1 ! i=1 ! X X X that is, νp(x + y) ≤ νp(x)+ νp(y). We leave to the reader the verification of n the remaining conditions. Thus, νp is a norm on R . ⊓⊔ n Example 5.12. Consider the mappings ν1, ν∞ : R −→ R given by ν1(x)= |x1| + |x2| + · · · + |xn|, ν∞(x) = max{|x1|, |x2|,..., |xn|}, n n for every x = (x1,...,xn) ∈ R . Both ν1 and ν∞ are norms on R . We will frequently use the alternative notation k x kp for νp(x). n n A special norm on R is the function ν∞ : R −→ R≥0 given by ν∞(x) = max{|xi| | 1 ≤ i ≤ n} (5.4) 164 5 Norms on Linear Spaces n for x = (x1,...,xn) ∈ R . We verify here that ν∞ satisfies the first condition of Definition 5.1. We start from the inequality |xi + yi| ≤ |xi| + |yi|≤ ν∞(x)+ ν∞(y) for every i, 1 ≤ i ≤ n. This in turn implies ν∞(x + y) = max{|xi + yi| | 1 ≤ i ≤ n}≤ ν∞(x)+ ν∞(y), which gives the desired inequality. This norm can be regarded as a limit case of the norms νp. Indeed, let Rn x ∈ and let M = max{|xi| | 1 ≤ i ≤ n} = |xℓ1 | = · · · = |xℓk | for some ℓ1,...,ℓk, where 1 ≤ ℓ1,...,ℓk ≤ n.
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