Kenneth Kuttler

21.1. THEOREMS BASED ON BAIRE CATEGORY 535

Note that from Theorem 21.1.4 ||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 21.1.6 With ∥L∥ defined in 21.1, 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 remainsto verify the operator norm really is a norm. First of all, if ∥L∥ = 0, then Lx = 0 for all∥x∥ ≤ 1. It follows that for any x ̸= 0,0 = L

(x∥x∥

)and so Lx = 0. Therefore, L = 0. Also,

if c is a scalar,∥cL∥= sup

∥x∥≤1∥cL(x)∥= |c| sup

∥x∥≤1∥Lx∥= |c|∥L∥ .

It remains to verify the triangle inequality. Let L,M ∈L (X ,Y ) .

∥L+M∥ ≡ sup∥x∥≤1

∥(L+M)(x)∥ ≤ sup∥x∥≤1

(∥Lx∥+∥Mx∥)

≤ sup∥x∥≤1

∥Lx∥+ sup∥x∥≤1

∥Mx∥= ∥L∥+∥M∥ .

This shows the operator norm is really a norm as hoped. ■For example, consider the space of linear transformations defined on Rn having values

in Rm. The fact the transformation is linear automatically imparts continuity to it. Youshould give a proof of this fact. Recall that every such linear transformation can be realizedin terms of matrix multiplication.

Thus, in finite dimensions the algebraic condition that an operator is linear is sufficientto imply the topological condition that the operator is continuous. The situation is not sosimple in infinite dimensional spaces such as C (X ;Rn). This explains the imposition of thetopological condition of continuity as a criterion for membership in L (X ,Y ) in additionto the algebraic condition of linearity. Here is an example which shows that this extraassumption cannot be eliminated.

Example 21.1.7 Let V denote all linear combinations of functions of the form e−αx2for

α > 0. Thus a typical element of V is an expression of the form

∑k=1

β ke−αkx2,αk > 0.

Let L : V → C be given by L f ≡∫R f (x)dx. For a norm on V,∥ f∥ ≡max{| f (x)| : x ∈ R} .

Of course V is not complete, but it is a normed linear space and you could consider itscompletion if desired, in terms of equivalence classes of Cauchy sequences, similar to theconstruction of R from Q. Recall that

∫∞

−∞e−x2

dx =∫

∞

−∞

1n e−(x2/n2) =

√π where here

n ∈ N. Consider the sequence of functions fn (x) ≡ 1n e−(x2/n2). Its maximum value is 1/n

and so ∥ fn∥ → 0 but L fn fails to converge to 0. Thus L is not continuous although it islinear.

Theorem 21.1.8 If Y is a Banach space, then L (X ,Y ) is also a Banach space.

21.1. THEOREMS BASED ON BAIRE CATEGORY 535Note that from Theorem 21.1.4 ||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 21.1.6 With ||L|| defined in 21.1, Y(X,Y) is a normed linear space. AlsoI|Lxl] < [LI] lla.Proof: Let x 4 0 then x/ ||x|| has norm equal to 1 and so ie (4) | < ||L||. Therefore,multiplying both sides by ||x||, ||Z|| < ||Z|| ||x||. This is obviously a linear space. It remainsto verify the operator norm really is a norm. First of all, if ||L|| = 0, then Lx = 0 for all\|x|| < 1. It follows that for any x 40,0 = L (47) and so Lx = 0. Therefore, L = 0. Also,IT]if c is a scalar,|cL|| = sup ||cL(x)|] = |e] sup ||Zx|] = |c||[Z\).I|x||<1 ||| <1It remains to verify the triangle inequality. Let L,M €¢ £ (X,Y).|JL+M|| = sup |\(L+M) (a) < sup (|[Lx|| + ||-l])I|xl|<1 I|xllS1S< sup ||Lx||+ sup ||Mx|] = ||L|| + [MIi|xl|<1 I|xllS1This shows the operator norm is really a norm as hoped.For example, consider the space of linear transformations defined on R” having valuesin R”. The fact the transformation is linear automatically imparts continuity to it. Youshould give a proof of this fact. Recall that every such linear transformation can be realizedin terms of matrix multiplication.Thus, in finite dimensions the algebraic condition that an operator is linear is sufficientto imply the topological condition that the operator is continuous. The situation is not sosimple in infinite dimensional spaces such as C (X;1R"). This explains the imposition of thetopological condition of continuity as a criterion for membership in Y (X,Y) in additionto the algebraic condition of linearity. Here is an example which shows that this extraassumption cannot be eliminated.Example 21.1.7 Let V denote all linear combinations of functions of the form eno fora > 0. Thus a typical element of V is an expression of the formn 2y B,e 5A > 0.k=1Let L:V — C be given by Lf = J f (x) dx. For anorm on V, || f|| = max {|f (x)|:x € R}.Of course V is not complete, but it is a normed linear space and you could consider itscompletion if desired, in terms of equivalence classes of Cauchy sequences, similar to theconstruction of R from Q. Recall that {~., e* dx = fe 1e-("/ m) /n where here2 [2n €N. Consider the sequence of functions f, (x) = 1e-(x /™”) Its maximum value is 1 /nand so || fn|| > 0 but Lf, fails to converge to 0. Thus L is not continuous although it islinear.Theorem 21.1.8 IfY is a Banach space, then (X,Y) is also a Banach space.