Analysis

23.1.4 Closed Graph Theorem

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

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

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

Proof: The only thing left to check is that the space is complete. But this follows from the simple observation that

is a Cauchy sequence in X × Y if and only if and are Cauchy sequences in X and Y respectively. Thus if is a Cauchy sequence in X × Y , it follows there exist x and y such that x_n → x and y_n → y. But then from the definition of the norm, →. ■

Proof: If F ⊆ X where X is a Banach space and

is a Cauchy sequence in F, then since X is complete, there exists a unique x ∈ X such that x_n → x. However this means x ∈F = F since F is closed. ■

Note the distinction between closed and continuous. If the operator is closed the assertion that y = Lx only follows if it is known that the sequence

converges. In the case of a continuous operator, the convergence of follows from the assumption that x_n → x. It is not always the case that a mapping which is closed is necessarily continuous. Consider the function f = tan if x is not an odd multiple of and f ≡ 0 at every odd multiple of . Then the graph is closed and the function is defined on ℝ but it clearly fails to be continuous. Of course this function is not linear. You could also consider the map,

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

where the norm is the uniform norm on C

,_∞. If y ∈ D, then

∫ y(x) = xy′(t)dt. 0

Therefore, if

→ f ∈ C and if y_n → y in C it follows that

∫xdyn(t) yn (x) = 0 dx dt ↓ ∫x ↓ y(x) = 0 f (t)dt

and so by the fundamental theorem of calculus f

= y^′ and so the mapping is closed. It is obviously not continuous because it takes y and y + sin to two functions which are far from each other even though these two functions are very close in C. Furthermore, it is not defined on the whole space, C.

The next theorem, the closed graph theorem, gives conditions under which closed implies continuous.

Proof: Let G be the graph of L.G = {(x,Lx) : x ∈ X}. By Lemma 23.1.15 it follows that G is a Banach space. Define P : G → X by P(x,Lx) = x.P maps the Banach space G onto the Banach space X and is continuous and linear. By the open mapping theorem, P maps open sets onto open sets. Since P is also one to one, this says that P⁻¹ is continuous. Thus

≤ K. Hence

∥Lx∥ ≤ max (||x||,||Lx ||) ≤ K ∥x∥

By Theorem 23.1.4 on Page 1590, this shows L is continuous. ■

The following corollary is quite useful. It shows how to obtain a new norm on the domain of a closed operator such that the domain with this new norm becomes a Banach space.

Proof: If

is a Cauchy sequence in D with this new norm, it follows both and are Cauchy sequences and therefore, they converge. Since L is closed, x_n → x and Lx_n → Lx for some x ∈ D. Thus _D → 0. ■