302 CHAPTER 12. BANACH SPACES

12.1.4 Closed Graph Theorem

Definition 12.1.10 Let f : D→ E. The graph of f consists of the set of all orderedpairs of the form {(x, f (x)) : x ∈ D}.

Definition 12.1.11 If X and Y are normed linear spaces, make X ×Y into anormed linear space by using the norm ∥(x,y)∥ = max(∥x∥,∥y∥) along with component-wise addition and scalar multiplication. Thus a(x,y)+b(z,w)≡ (ax+bz,ay+bw).

There are other ways to give a norm for X×Y . For example, you could define ∥(x,y)∥=∥x∥+∥y∥

Lemma 12.1.12 The norm defined in Definition 12.1.11 on X ×Y along with the defi-nition of addition and scalar multiplication given there make X ×Y into a normed linearspace.

Proof: The only axiom for a norm which is not obvious is the triangle inequality.Therefore, consider

∥(x1,y1)+(x2,y2)∥ = ∥(x1 + x2,y1 + y2)∥= max(∥x1 + x2∥ ,∥y1 + y2∥)≤ max(∥x1∥+∥x2∥ ,∥y1∥+∥y2∥)

≤ max(∥x1∥ ,∥y1∥)+max(∥x2∥ ,∥y2∥)= ∥(x1,y1)∥+∥(x2,y2)∥ .

It is obvious X×Y is a vector space from the above definition. ■

Lemma 12.1.13 If X and Y are Banach spaces, then X ×Y with the norm and vectorspace operations defined in Definition 12.1.11 is also a Banach space.

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 12.1.14 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 12.1.15 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.