12.1. THEOREMS BASED ON BAIRE CATEGORY 299

there exists p∞ such that limn→∞ pn = p∞. Since all but finitely many terms of {pn} are inB(pm,rm), it follows that p∞ ∈ B(pm,rm) for each m. Therefore, p∞ ∈ ∩∞

m=1B(pm,rm) ⊆∩∞

i=1Ui∩B(p,r0). ■The following corollary is also called the Baire category theorem.

Corollary 12.1.6 Let X be a complete metric space and suppose X = ∪∞i=1Fi where

each Fi is a closed set. Then for some i, interior Fi ̸= /0.

Proof: If all Fi has empty interior, then FCi would be a dense open set. Therefore, from

Theorem 12.1.5, it would follow that

/0 = (∪∞i=1Fi)

C = ∩∞i=1FC

i ̸= /0.■

The set D of Theorem 12.1.5 is called a Gδ set because it is the countable intersectionof open sets. Thus D is a dense Gδ set.

12.1.2 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 boundeduniformly. As an example of a situation in which pointwise bounded does not imply uni-formly bounded, consider the functions fα (x)≡X(α,1) (x)x−1 for α ∈ (0,1). Clearly eachfunction is bounded and the collection of functions is bounded at each point of (0,1), butthere is no bound for all these functions taken together. One problem is that (0,1) is not aBanach space. Therefore, the functions cannot be linear. Since the theorem is about linearfunctions, it only applies to linear spaces.

Theorem 12.1.7 Let X be a Banach space and let Y be a normed linear space. Let{Lα}α∈Λ be a collection of elements of L (X ,Y ). Then one of the following happens.

a.) sup{∥Lα∥ : α ∈ Λ}< ∞

b.) There exists a dense Gδ set, D, such that for all x ∈ D,

sup{∥Lα x∥ α ∈ Λ}= ∞.

Proof: For each n ∈ N, define Un = {x ∈ X : sup{∥Lα x∥ : α ∈ Λ}> n}. Then Un is anopen set because if x ∈Un, then there exists α ∈ Λ such that ∥Lα x∥> n. But then, since Lα

is continuous, this situation persists for all y sufficiently close to x, say for all y ∈ B(x,δ ).Then B(x,δ )⊆Un which shows Un is open.

Case b.) is obtained from Theorem 12.1.5 if each Un is dense.The other case is that for some n, Un is not dense. If this occurs, there exists x0 and

r > 0 such that for all x∈B(x0,r), ∥Lα x∥≤ n for all α . Now if y∈B(0,r), x0+y∈B(x0,r).Consequently, for all such y, ∥Lα(x0+y)∥ ≤ n. This implies that for all α ∈Λ and ∥y∥< r,

∥Lα y∥ ≤ n+∥Lα(x0)∥ ≤ 2n.

Therefore, if ∥y∥ ≤ 1,∥∥ r

2 y∥∥< r and so for all α , ∥Lα

( r2 y)∥ ≤ 2n. Now multiplying by r/2

it follows that whenever ∥y∥ ≤ 1, ∥Lα (y)∥ ≤ 4n/r. Hence case a.) holds. ■