21.1. THEOREMS BASED ON BAIRE CATEGORY 539

Proof: The only thing left to check is that the space is complete. But this follows fromthe simple observation that {(xn,yn)} is a Cauchy sequence in X ×Y if and only if {xn}and {yn} are Cauchy sequences in X and Y respectively. Thus if {(xn,yn)} is a Cauchysequence in X ×Y , it follows there exist x and y such that xn → x and yn → y. But thenfrom the definition of the norm, (xn,yn)→ (x,y). ■

Lemma 21.1.16 Every closed subspace of a Banach space is a Banach space.

Proof: If F ⊆ X where X is a Banach space and {xn} is a Cauchy sequence in F , thensince X is complete, there exists a unique x ∈ X such that xn → x. However this meansx ∈ F = F since F is closed. ■

Definition 21.1.17 Let X and Y be Banach spaces and let D ⊆ X be a subspace.A linear map L : D→ Y is said to be closed if its graph is a closed subspace of X ×Y .Equivalently, L is closed if xn→ x and Lxn→ y implies x ∈ D and y = Lx.

Note the distinction between closed and continuous. If the operator is closed the as-sertion that y = Lx only follows if it is known that the sequence {Lxn} converges. In thecase of a continuous operator, the convergence of {Lxn} follows from the assumption thatxn→ x. It is not always the case that a mapping which is closed is necessarily continuous.Consider the function f (x) = tan(x) if x is not an odd multiple of π

2 and f (x)≡ 0 at everyodd multiple of π

2 . Then the graph is closed and the function is defined on R but it clearlyfails to be continuous. Of course this function is not linear. You could also consider themap,

ddx

:{

y ∈C1 ([0,1]) : y(0) = 0}≡ D→C ([0,1]) .

where the norm is the uniform norm on C ([0,1]) , ||y||∞

. If y ∈ D, then y(x) =∫ x

0 y′ (t)dt.Therefore, if dyn

dx → f ∈C ([0,1]) and if yn→ y in C ([0,1]) it follows that

yn (x) =∫ x

0dyn(t)

dx dt↓ ↓

y(x) =∫ x

0 f (t)dt

and so by the fundamental theorem of calculus f (x) = y′ (x) and so the mapping is closed.It is obviously not continuous because it takes y(x) and y(x)+ 1

n sin(nx) to two functionswhich are far from each other even though these two functions are very close in C ([0,1]).Furthermore, it is not defined on the whole space, C ([0,1]).

The next theorem, the closed graph theorem, gives conditions under which closed im-plies continuous.

Theorem 21.1.18 Let X and Y be Banach spaces and suppose L : X →Y is closedand linear. Then L is continuous.

Proof: Let G be the graph of L. G = {(x,Lx) : x ∈ X}. By Lemma 21.1.16 it followsthat G is a Banach space. Define P : G→ X by P(x,Lx) = x. P maps the Banach space Gonto the Banach space X and is continuous and linear. By the open mapping theorem, Pmaps open sets onto open sets. Since P is also one to one, this says that P−1 is continuous.Thus

∥∥P−1x∥∥≤ K ∥x∥. Hence

∥Lx∥ ≤max(∥x∥ ,∥Lx∥)≤ K ∥x∥