438 CHAPTER 17. BANACH SPACES

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. This proves the lemma.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.

Theorem 17.1.7 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||.

This proves the theorem.

17.1.2 Uniform Boundedness TheoremThe next big result is sometimes called the Uniform Boundedness theorem, or the Banach-Steinhaus theorem. This is a very surprising theorem which implies that for a collectionof bounded linear operators, if they are bounded pointwise, then they are also bounded