2.9. GENERAL BANACH SPACES 69

to 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. ■As a review, consider the space of linear transformations defined on Rn having values

in Rm. The fact the transformation is linear automatically imparts continuity to it. Toshow this, you might recall that every such linear transformation can be realized in termsof 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.

Theorem 2.9.4 If Y is a Banach space, then L (X ,Y ) is also a Banach space.

Proof: Let {Ln} be a Cauchy sequence in L (X ,Y ) and let x ∈ X .

∥Lnx−Lmx∥ ≤ ∥x∥ ∥Ln−Lm∥.

Thus {Lnx} is a Cauchy sequence. Let

Lx = limn→∞

Lnx.

Then, clearly, L is linear because if x1,x2 are in X , and a,b are scalars, then

L(ax1 +bx2) = limn→∞

Ln (ax1 +bx2)

= limn→∞

(aLnx1 +bLnx2)

= aLx1 +bLx2.

Also L is continuous. To see this, note that {∥Ln∥} is a Cauchy sequence of real numbersbecause |∥Ln∥−∥Lm∥| ≤ ∥Ln−Lm∥. Hence there exists K > sup{∥Ln∥ : n ∈ N}. Thus, ifx ∈ X ,

∥Lx∥= limn→∞∥Lnx∥ ≤ K∥x∥.■

Definition 2.9.5 More generally, given Banach spaces X ,Y,L (X ,Y ) is the spaceof continuous linear maps from which map X to Y .