Lecture 4: The Open Mapping Theorem and the Closed Graph Theorem

February 25, 2021 Last time, we proved the Uniform Boundedness Theorem from the Baire Category Theorem, and we’ll continue to prove some “theorems with names” in functional analysis today. Theorem 37 (Open Mapping Theorem) Let B1 , B2 be two Banach spaces, and let T ∈ B(B1 , B2 ) be a surjective linear operator. Then T is an open map, meaning that for all open subsets U ⊂ B1 , T (U) is open in B2 . Proof. We’ll begin by proving a specialized result: we’ll show that the image of the open ball B1 (0, 1) = {b ∈ B1 : ||b|| < 1} contains an open ball in B2 centered at 0. (Then we’ll use linearity to shift and scale these balls accordingly.) Because T is surjective, everything in B2 is mapped onto, meaning that B2 = [ T (B(0, n)) n∈N (because any element of B1 is at a finite distance from 0, it must be contained in one of the balls). Now we’ve written B2 as a union of closed sets, so by Baire, there exists some n0 ∈ N such that T (B(0, n0 )) contains an open ball. But T is a linear operator, so this is the same set as n0 T (B(0, 1)) (we can check that closure respects scaling and so on). So we have an open ball inside T (B(0, 1)) – restated, there exists some point v0 ∈ B2 and some radius r > 0 such that B(v0 , 4r ) is contained in T (B(0, 1)) (the choice of 4 will make arithmetic easier later). And we want a point that’s actually in the image of B(0, 1) (not just the closure), so we pick a point v1 = T u1 ∈ T (B(0, 1)) such that ||v0 − v1 || < 2r . (The idea here is that points in the closure of T (B(0, 1)) are arbitrarily close to points actually in T (B(0, 1)).) Now B(v1 , 2r ) is entirely contained in B(v0 , 4r ), which is contained in T (B(0, 1), and now we’ll show that this closure contains an open ball centered at 0 (which is pretty close to what we want). For any ||v || < r , we have 1 1 1 (2v + v1 ) ∈ B(v1 , 2r ) ⊂ T (B(0, 1)) = T (B(0, 12 )), 2 2 2
1
u1 and thus v = −T 2 + 2 (2v + v1 ) is an element of −T u21 + T (B(0, 12 )) (this is not an equivalence class – it’s
the set of elements T (B(0, 12 )) all shifted by −T u21 ), and now by linearity this means that our element v must be

in the set T − u21 + B 0, 12 . But we chose u1 to have norm less than 1, so − u21 and any element of B(0, 12 ) must both have norm at most 1 2 (and their sum has norm at most 1). Thus, this set must be contained in T (B(0, 1)), and therefore the ball of radius r , B(0, r ) (in B2 ) is contained in T (B(0, 1)). But by scaling, we find that B(0, 2−n r ) = 2−n B(0, r ) is contained in 2−n T (B(0, 1)) = T (B(0, 2−n )) (repeatedly using homogeneity), and now we’ll use that fact to prove that B(0, 2r ) is contained in T (B(0, 1)) (finally removing the closure and proving the specialized result). To do that, take some ||v || < 2r ; we know that (plugging in n = 1) v ∈ T (B(0, 21 )). So there exists some b1 ∈ B(0, 12 ) in B1 such that ||v − T b1 || < r 4 (this is the same idea as above that points in the closure are arbitrarily close to points in the actual set). Then taking n = 2, we know that v − T b1 ∈ T (B(0, 14 )), so there is some b2 ∈ B(0, 14 ) such that ||v − T b1 − T b2 || < 8r . Continue iterating this for larger and larger n, so that we have a sequence {bk } of elements in B1 such that ||bk || < 2−k and v− n X T bk < 2−n−1 r. k=1 And now the series P∞ n=1 bk is absolutely summable, and because B1 is a Banach space, that means that the series is summable, and we have b ∈ B1 such that b = P∞ k=1 ||b|| = lim n→∞ bk . And n X bk ≤ lim n→∞ k=1 n X ||bk || k=1 by the triangle inequality, and then we can bound this as = ∞ X ||bk || < k=1 ∞ X 2−k = 1. k=1 Furthermore, because T is a (bounded, thus) continuous operator, ! n n X X T bk = v , T b = lim T bk = lim n→∞ k=1 n→∞ k=1 because we chose our bk so that ||v − T b1 − T b2 − · · · − T bk || converges to 0. Therefore, since b ∈ B(0, 1), v ∈ T (B(0, 1)), and that means the ball B(0, 2r ) in B2 is indeed contained in T (B(0, 1)). We’ve basically shown now that 0 remains an interior point if it started as one, and now we’ll finish with some translation arguments: if a set U ⊂ B1 is open, and b2 = T b1 is some arbitrary point in T (U), then (by openness of U) there exists some ε > 0 such that b1 + B(0, ε) = B(b1 , ε) is contained in U. Furthermore, by our work above, there exists some δ so that B(0, δ) ⊂ T (B(0, 1)). So this means that B(b2 , εδ) = b2 + εB(0, δ) ⊂ b2 + εT (B(0, 1)) = T (b1 ) + εT (B(0, 1)) = T (b1 + B(0, ε)). But b1 + B(0, ε) is contained in U, so indeed we’ve found a ball around our arbitrary b2 contained in T (U), and this proves the desired result. Corollary 38 If B1 , B2 are two Banach spaces, and T ∈ B(B1 , B2 ) is a bijective map, then T −1 is in B(B2 , B1 ). Proof. We know that T −1 is continuous if and only if for all open U ⊂ B1 , the inverse image of U by T −1 (which is T (U)) is open. And this is true by the Open Mapping Theorem. From the Open Mapping Theorem, we get this an almost topological result, which gives sufficient conditions for continuity of a linear operator. But first we need to state another result: Proposition 39 If B1 , B2 are Banach spaces, then B1 × B2 (with operations done entry by entry) with norm ||(b1 , b2 )|| = ||b1 || + ||b2 || is a Banach space. (This proof is left as an exercise: we just need to check all of the definitions, and a Cauchy sequence in B1 × B2 will consist of a Cauchy sequence in each of the individual spaces B1 and B2 . So it’s kind of similar to proving completeness of R2 .) Theorem 40 (Closed Graph Theorem) Let B1 , B2 be two Banach spaces, and let T : B1 → B2 be a (not necessarily bounded) linear operator. Then T ∈ B(B1 , B2 ) if and only if the graph of T , defined as Γ(T ) = {(u, T u) : u ∈ B1 }, is closed in B1 × B2 . This can sometimes be easier or more convenient to check than the boundedness criterion for continuity. And normally, proving continuity means that we need to show that for a sequence {un } converging to u, T un converges and is also equal to T u. But the Closed Graph Theorem eliminates one of the steps – proving that the graph is closed means that given a sequence un → u and a sequence T un → v , we must show that v = T u (in other words, we just need to show that the convergence point is correct, without explicitly constructing one)! Proof. For the forward direction, suppose that T is a bounded linear operator (and thus continuous). Then if (un , T un ) is a sequence in Γ(T ) with un → u and T un → v , we need to show that (u, v ) is in the graph. But   v = lim T un = T lim un = T u, n→∞ n→∞ and thus (u, v ) is in the graph and we’ve proven closedness. For the other direction, consider the following commutative diagram: Γ(T ) π1 π2 T B1 B2 Here, π1 and π2 are the projection maps from the graph down to B1 and B2 (meaning that π1 (u, T u) = u and π2 (u, T u) = T u). We want to construct a map S : B1 → Γ(T ) (so that T = π2 ◦ S), and we do so as follows. Since Γ(T ) is (by assumption) a closed subspace of B1 × B2 , which is a Banach space, Γ(T ) must be a Banach space as well. And now π1 , π2 are continuous maps from the Banach space Γ(T ) to B1 , B2 respectively, so π1 is a bounded linear operator in B(Γ(T ), B1 ), and similarly π2 ∈ B(Γ(T ), B2 ) (we can see this through the calculation ||π2 (u, v )|| = ||v || ≤ ||u|| + ||v || = ||(u, v )||, for example). Furthermore, π1 : Γ(T ) → B1 is actually bijective (because there is exactly one point in the graph for each u), so by Corollary 38, it has an inverse S : B1 → Γ(T ) which is a bounded linear operator. And now T = π2 ◦ S is the composition of two bounded linear operators, so it is also a bounded linear operator. Remark 41. The Open Mapping Theorem implies the Closed Graph Theorem, but we can also show the converse (so the two are logically equivalent). Each of the results so far has been trying to answer a question, and our next result, the Hahn-Banach Theorem, is asking whether the dual space of a general nontrivial normed space is trivial. (In other words, we want to know whether there are any normed spaces whose space of functionals B(V, K) only contains the zero function.) For example, we mentioned that for any finite p ≥ 1, `p and `q are dual is 1 p + q1 = 1, and it’s also true that (c0 )0 = `1 . So Hahn-Banach will tell us that the dual space has “a lot of elements,” but first we’ll need an intermediate result from set theory: Definition 42 A partial order on a set E is a relation  on E with the following properties: • For all e ∈ E, e  e. • For all e, f ∈ E, if e  f and f  e, then e = f . • For all e, f , g ∈ E, if e  f and f  g, then e  g. An upper bound of a set D ⊂ E is an element e ∈ E such that d  e for all d ∈ D, and a maximal element of E is an element e such that for any f ∈ E, e  f =⇒ e = f (minimal element is defined similarly). Notably, we do not need to have either e  f or f  e in a partial ordering, and a maximal element does not need to sit “on top” of everything else in E, because we can have other elements “to the side:” Example 43 If S is a set, we can define a partial order on the powerset of S, in which E  F if E is a subset of F . Then not all sets can be compared (specifically, it doesn’t need to be true that either E  F or F  E). Definition 44 Let (E, ) be a partially ordered set. Then a set C ⊂ E is a chain if for all e, f ∈ C, we have either e  f or f  e. (In other words, we can always compare all elements in a chain.) Proposition 45 (Zorn’s lemma) If every chain in a nonempty partially ordered set E has an upper bound, then E contains a maximal element. We’ll take this as an axiom of set theory, and we’ll give an application of this next lecture. But we can use it to prove other things as well, like the Axiom of Choice. Definition 46 Let V be a vector space. A Hamel basis H ⊂ V is a linearly independent set such that every element of V is a finite linear combination of elements of H. We know from linear algebra that we find a basis and calculate its cardinality to find the dimension for finitedimensional vector spaces. (So a Hamel basis for Rn can be the standard n basis elements, and a Hamel basis for `1 can be (1, 0, 0, · · · ), (0, 1, 0, · · · ), and so on.) And next time, we’ll use Zorn’s lemma to talk more about these Hamel bases!