Kenneth Kuttler

114 CHAPTER 5. FUNCTIONS ON NORMED LINEAR SPACES

Now consider whether these dyadics form a linearly independent set. Suppose that∑i,k dikwivk = 0. Are all the scalars dik equal to 0? 0 = ∑i,k dikwivk (vl) = ∑

mi=1 dilwi so,

since {w1, · · · ,wm} is a basis, dil = 0 for each i = 1, · · · ,m. Since l is arbitrary, this showsdil = 0 for all i and l. Thus these linear transformations form a basis and this shows thatthe dimension of L (V,W ) is mn as claimed because there are m choices for the wi and nchoices for the v j. ■

5.2 The Norm of a Linear Map, Operator NormNot surprisingly all of the above holds for a finite dimensional normed linear space. Firsthere is an easy lemma which follows right away from Theorem 3.6.2, the theorem aboutequivalent formulations of continuity.

Lemma 5.2.1 Let (V,∥·∥V ) and (W,∥·∥W ) be two normed linear spaces. Then a linearmap f : V →W is continuous if and only if it takes bounded sets to bounded sets. ( f isbounded) If V is finite dimensional, then f must be continuous.

Proof: =⇒ Consider f (B(0,1)) . If this is not bounded, then there exists ∥vm∥V ≤ 1

but ∥ f (vm)∥W ≥ m. Then it follows that∥∥∥ f(

)∥∥∥W≥ 1 which is impossible for all m

since∥∥∥ vm

∥∥∥≤ 1m and so continuity requires that limm→∞ f

(vm

)= 0 (Theorem 3.6.2). Thus

there exists M such that ∥ f (v)∥ ≤M whenever v ∈ B(0,1). In general, let S be a boundedset. Then S ⊆ B(0,r) for large enough r. Hence, for v ∈ B, it follows that v/2r ∈ B(0,1) .It follows that ∥ f (v/2r)∥W ≤ M and so ∥ f (v)∥W ≤ 2rM. Thus f takes bounded sets tobounded sets.⇐= Suppose f is bounded and not continuous. Then by Theorem 3.6.2 again, there

is a sequence vn → v but f (vn) fails to converge to f (v). Then there exists ε > 0 and asubsequence, still denoted as vn such that ∥ f (vn)− f (v)∥= ∥ f (vn− v)∥ ≥ ε . Then∥∥∥∥ f

(vn− v∥vn− v∥

)∥∥∥∥≥ ε1

∥vn− v∥The right side is unbounded, but the left is bounded, a contradiction.

Consider the last claim about continuity. Let {v1, · · · ,vn} be a basis for V . By Lemma4.4.7, if ym → 0, in V for ym = ∑

nk=1 ym

k vk,then it follows that limm→∞ ymk = 0 and conse-

quently, f (ym)→ f (0) = 0. In general, if ym→ y, then (ym− y)→ 0 and so f (ym− y) =f (ym)− f (y)→ 0. That is, f (ym)→ f (y). ■

Definition 5.2.2 For f : (V,∥·∥V )→ (W,∥·∥W ) continuous, it was just shown thatthere exists M such that ∥ f (v)∥ ≤ M, v ∈ B(0,1) . It follows that, since v

2∥v∥ ∈ B(0,1) ,then ∥ f (v)∥ ≤ 2M ∥v∥. Therefore, letting ∥ f∥ ≡ sup∥v∥≤1 ∥ f (v)∥ it follows that for allv∈V, ∥ f (v)∥ ≤ ∥ f∥∥v∥ .Thus a linear map is bounded if and only if ∥ f∥< ∞ if and only iff is continuous. ‘The number ∥ f∥ is called the operator norm. For X a real normed linearspace, X ′ denotes the space L (X ,R) .

You can show that for L (V,W ) the space of bounded linear maps from V to W,L (V,W ) becomes a normed linear space with this definition. This is true whether V,W arefinite or infinite dimensional. You can also show that if W is complete then so is L (V,W ).This is left as an exercise. Also, when the vector spaces are finite dimensional, Lemma5.2.1 shows that any linear function f is automatically bounded, hence continuous, hence∥ f∥ exists. Here is an interesting observation about the operator norm.

114 CHAPTER 5. FUNCTIONS ON NORMED LINEAR SPACESNow consider whether these dyadics form a linearly independent set. Suppose thatYin dixwive = 0. Are all the scalars dj, equal to 0? O= Yip diwive (v7) = LZ, diwi so,since {w1,:+:,Wm} is a basis, dj; = 0 for each i = 1,--- ,m. Since | is arbitrary, this showsdj; = 0 for all i and /. Thus these linear transformations form a basis and this shows thatthe dimension of &(V,W) is mn as claimed because there are m choices for the w; and nchoices for the v;. ll5.2 The Norm of a Linear Map, Operator NormNot surprisingly all of the above holds for a finite dimensional normed linear space. Firsthere is an easy lemma which follows right away from Theorem 3.6.2, the theorem aboutequivalent formulations of continuity.Lemma 5.2.1 Let (V, ||-||,) and (W, ||-||y) be two normed linear spaces. Then a linearmap f :V — W is continuous if and only if it takes bounded sets to bounded sets. (f isbounded) If V is finite dimensional, then f must be continuous.Proof: = > Consider f (B(0,1)). If this is not bounded, then there exists ||v’"|| <1but ||f(v")||y =m. Then it follows that Ff (*) ly > 1 which is impossible for all m. ynsince || —m| < i and so continuity requires that lim,),.0 f ¥ = 0 (Theorem 3.6.2). Thusthere exists M such that || f (v)|| <M whenever v € B(0,1). In general, let S be a boundedset. Then S C B(0,r) for large enough r. Hence, for v € B, it follows that v/2r € B(0,1).It follows that || f (v/2r)||y <M and so ||f(v)||y <2rM. Thus f takes bounded sets tobounded sets.<= Suppose f is bounded and not continuous. Then by Theorem 3.6.2 again, thereis a sequence v, — v but f(v,) fails to converge to f(v). Then there exists € > 0 andasubsequence, still denoted as v, such that ||f (vn) — f (v)|| = ||f vn —v)|| = €. ThenVa—Vv(ES) > PnThe right side is unbounded, but the left is bounded, a contradiction.Consider the last claim about continuity. Let {v;,--- ,v,} be a basis for V. By Lemma4.4.7, if y” — 0, in V for y” = Yr_ yf vz,then it follows that lim,,,.. yy’ = 0 and conse-quently, f(y”) + f (0) =0. In general, if y” — y, then (y"—y) > 0 and so f (y”—y) =f(y") — f(y) + 0. That is, f(y") > f(y). aDefinition 5.2.2 For f : (Vv, \I-Ily) > (W, ||-Ily) continuous, it was just shown thatthere exists M such that ||f (v)|| <M, v © B(0,1). It follows that, since yp € B(0,1),then ||f (v)|| < 2M ||v||. Therefore, letting ||f\| = supj,\ <1 \|f (v)|| it follows that for allveV, lf ()|| < II ||v|| .Lhus a linear map is bounded if and only if || f || < °° if and only iff is continuous. ‘The number ||f\| is called the operator norm. For X a real normed linearspace, X' denotes the space & (X,R).You can show that for &(V,W) the space of bounded linear maps from V to W,-£ (V,W) becomes a normed linear space with this definition. This is true whether V, W arefinite or infinite dimensional. You can also show that if W is complete then so is Y (V,W).This is left as an exercise. Also, when the vector spaces are finite dimensional, Lemma5.2.1 shows that any linear function f is automatically bounded, hence continuous, hence|| f|| exists. Here is an interesting observation about the operator norm.