Kenneth Kuttler

68 CHAPTER 2. BASIC TOPOLOGY AND ALGEBRA

In particular, consider the triangle inequality.∥∥L+ L̂∥∥ ≡ sup

∥v∥≤1

∥∥L(v)+ L̂(v)∥∥≤ sup

∥v∥≤1

(∥L(v)∥+

∥∥L̂(v)∥∥)

≤ sup∥v∥≤1

∥Lv∥+ sup∥v∥≤1

∥∥L̂v∥∥≡ ∥L∥+∥∥L̂

∥∥As to the last claim, if v ̸= 0,

∥∥∥L(

v∥v∥

)∥∥∥≤ ∥L∥ and so ∥L(v)∥ ≤ ∥L∥∥v∥. ■

2.9 General Banach SpacesThe above is about linear maps defined on finite dimensional spaces. What if instead, youhave Banach spaces which are just complete normed linear spaces? What then? I will quitwriting functions and vectors in bold face here. It turns out that in this case, you assumethe linear maps are continuous, not just linear.

Theorem 2.9.1 Let X and Y be two normed linear spaces and let L : X→Y be linear(L(ax+by) = aL(x)+bL(y) for a,b scalars and x,y ∈ X). The following are equivalent

a.) L is continuous at 0b.) L is continuousc.) There exists K > 0 such that ∥Lx∥Y ≤ K ∥x∥X for all x ∈ X (L is bounded).

Proof: a.)⇒b.) Let xn → x. It is necessary to show that Lxn → Lx. But (xn− x)→ 0and so from continuity at 0, it follows

L(xn− x) = Lxn−Lx→ 0

so Lxn→ Lx. This shows a.) implies b.).b.)⇒c.) Since L is continuous, L is continuous at 0. Hence ∥Lx∥Y < 1 whenever ∥x∥X ≤

δ for some δ . Therefore, suppressing the subscript on the ∥·∥, it follows that∥∥∥L(

δx∥x∥

)∥∥∥≤1. Hence ∥Lx∥ ≤ 1

δ∥x∥.

c.)⇒a.) follows from the inequality given in c.). ■

Definition 2.9.2 Let L : X→Y be linear and continuous where X and Y are normedlinear spaces. Denote the set of all such continuous linear maps by L (X ,Y ) and define

∥L∥= sup{∥Lx∥ : ∥x∥ ≤ 1}. (2.6)

This is called the operator norm.

Note that from Theorem 2.9.1, ∥L∥ is well defined because of part c.) of that Theorem.The next lemma follows immediately from the definition of the norm and the assump-

tion that L is linear.

Lemma 2.9.3 With ∥L∥ defined in 2.6, L (X ,Y ) is a normed linear space. Also ∥Lx∥ ≤∥L∥∥x∥.

Proof: Let x ̸= 0 then x/∥x∥ has norm equal to 1 and so∥∥∥L(

x∥x∥

)∥∥∥≤ ∥L∥ . Therefore,multiplying both sides by ∥x∥, ∥Lx∥ ≤ ∥L∥∥x∥. This is obviously a linear space. It remains

68 CHAPTER 2. BASIC TOPOLOGY AND ALGEBRAIn particular, consider the triangle inequality.JL+L|| = sup |[L(v) +2 vl < sup, (|Z (v)IIvII<t IIvII<IA\|v||<1AsL (7) Nex < ||L]| and so ||L(v)|| < |[Z]| ||v||.2.9 General Banach SpacesThe above is about linear maps defined on finite dimensional spaces. What if instead, youhave Banach spaces which are just complete normed linear spaces? What then? I will quitwriting functions and vectors in bold face here. It turns out that in this case, you assumethe linear maps are continuous, not just linear.Theorem 2.9.1 Let X andY be two normed linear spaces and let L: X — Y be linear(L(ax + by) = aL(x) + bL(y) for a,b scalars and x,y € X). The following are equivalenta.) Lis continuous at Ob.) Lis continuousc.) There exists K > 0 such that ||Lx||y < K ||x||y for all x € X (L is bounded).Proof: a.)=>b.) Let x, > x. It is necessary to show that Lx, — Lx. But (x, —x) > 0and so from continuity at 0, it followsL(x, —x) = Lx, —Lx > 0so Lx, — Lx. This shows a.) implies b.).b.)=c.) Since L is continuous, L is continuous at 0. Hence ||Lx||, < 1 whenever ||x||y <6 for some 6. Therefore, suppressing the subscript on the ||-||, it follows that i ( Pr) | <1. Hence ||Lx|| < $||x'].c.)=>a.) follows from the inequality given in c.).Definition 2.9.2 Let L: xX 3 Y be linear and continuous where X and Y are normedlinear spaces. Denote the set of all such continuous linear maps by @(X,Y) and define||L|| = sup{||La| : |||] < 1}. (2.6)This is called the operator norm.Note that from Theorem 2.9.1, ||L]| is well defined because of part c.) of that Theorem.The next lemma follows immediately from the definition of the norm and the assump-tion that L is linear.Lemma 2.9.3 With ||L|| defined in 2.6, Y(X,Y) is a normed linear space. Also ||Lx|| <[Ll Ile).Proof: Let x 7 0 then x/ ||x|| has norm equal to 1 and so |e (, ri) | < ||L||. Therefore,