Lecture 3: Quotient Spaces, the Baire Category Theorem and the Uniform Boundedness Theorem

February 23, 2021 Last time, we introduced the space of bounded linear operators between two normed spaces, B(V, W ), and we proved that this space is Banach when W is Banach. Today, we’ll start seeing other ways to get normed spaces from other normed spaces, namely subspaces and quotients. We should recall this definition from linear algebra: Definition 29 Let V be a vector space. A subset W ⊆ V is a subspace of V if for all w1 , w2 ∈ W and λ1 , λ2 ∈ K, we have λ1 w1 + λ2 w2 ∈ W (that is, closure under linear combinations). Proposition 30 A subspace W of a Banach space V is Banach (with norm inherited from V ) if and only if W is a closed subset of V (with respect to the metric induced by the norm). Proof sketch. If W is Banach, the idea is to show that every sequence of elements in W converges (to something in V ) actually converges in W , and we show this by noticing that the sequence must be Cauchy, meaning that (by completeness of W ) there is a convergence point, and then we use uniqueness of limits. For the other direction, if W is closed, then any Cauchy sequence in W is also a Cauchy sequence in V , so it has a limit. Closedness tells us that the limit is in W , so every Cauchy sequence has a limit in W , which proves that it is Banach. Definition 31 Let W ⊂ V be a subspace of V . Define the equivalence relation on V via v ∼ v 0 ⇐⇒ v − v 0 ∈ W, and let [v ] be the equivalence class of v (the set of v 0 ∈ V such that v ∼ v 0 ). Then the quotient space V /W is the set of all equivalence classes {[v ] : v ∈ V }. We can check that the usual conditions for an equivalence relation are satisfied: • Reflexivity: v ∼ v for all v ∈ V (because 0 ∈ W ) • Symmetry: v ∼ v 0 if and only if v 0 ∼ v (because w ∈ W =⇒ −w ∈ W ). • Transitivity: if v ∼ v 0 and v 0 ∼ v 00 , then v ∼ v 00 (because of closure under addition in W ). We will typically denote [v ] as v + W (using the algebra coset notation), since all elements in the equivalence class of v are v plus some element of W . And with this notation, we have (for any v1 , v2 ∈ V ) (v1 + W ) + (v2 + W ) = (v1 + v2 ) + W, and (for any λ ∈ K) λ(v + W ) = λv + W. We do need to check that these operations are well-defined (that is, the resulting equivalence class of the operations is independent of the representative of v + W ), but that’s something that we checked in linear algebra (or can check on our own). We typically pronounce V /W “V mod W ,” and in particular W = 0 + W = w + W for any w ∈ W . We introduced the concept of a seminorm when we defined a normed vector space – basically, seminorms satisfy all of the same assumptions as norms except definiteness (so nonzero vectors can have seminorm 0). Example 32 Consider the norm which assigns the real number sup |f 0 | to a function f : this satisfies homogeneity and the triangle inequality, but it is not a norm because the derivative of any constant function is 0. But the constant functions form a subspace, and this next result is basically talking about how we can mod out by that subspace: Theorem 33 Let || · || be a seminorm on a vector space V . If we define E = {v ∈ V : ||v || = 0}, then E is a subspace of V , and the function on V /E defined by ||v + E||v /E = ||v || for any v + E ∈ V /E defines a norm. Proof. First of all, E is a subspace because (by homogeneity and the triangle inequality) ||λ1 v1 + λ2 v2 || ≤ λ1 ||v1 || + λ2 ||v2 || = 0 for any v1 , v2 ∈ E and λ1 , λ2 ∈ K, and because a seminorm is always nonnegative, we must have ||λ1 v1 + λ2 v2 || = 0 (and thus λ1 v1 + λ2 v2 ∈ E. This means that V /E is indeed a valid quotient space, and now we must show that our function is well-defined (in other words, that it doesn’t depend on the representative from our equivalence class). Formally, that means that if we need to check that if v + E = v 0 + E, then ||v || = ||v 0 ||. And we can do this with the triangle inequality: since v + E = v 0 + E, there exists some e ∈ E such that v = v 0 + e, ||v || = ||v 0 + e|| ≤ ||v 0 || + ||e|| = ||v 0 || by the triangle inequality. But this argument is also true if we swap the roles of v and v 0 , so it’s also true that ||v 0 || ≤ ||v ||, and thus their seminorms must actually be equal. Checking that this function is actually a norm on V /E is now left as an exercise to us: the properties of homogeneity and triangle inequality follow because || · || is already a seminorm, and definiteness comes because everything that evaluates to 0 is in the equivalence class 0 + E. So identifying the subspace of all zero-norm elements gives us a normed space, but we can also start with a normed space V and consider some closed subset W of that normed space. Then V /W is a new normed space – that will be left as an exercise for us. With that, we’ve concluded the “bare-bones” part of functional analysis, and we’re now ready to get into some fundamental results related to Banach spaces. (In other words, the theorems will now have names attached to them, and we should be able to recognize the names.) First, we’ll need a result from metric space theory: Theorem 34 (Baire Category Theorem) Let M be a complete metric space, and let {Cn }n be a collection of closed subsets of M such that M = S n∈N Cn . Then at least one of the Cn contain an open ball B(x, r ) = {y ∈ M : d(x, y ) < r }. (In other words, at least one Cn has an interior point.) (This theorem doesn’t have anything to do with category theory, despite the name.) Sometimes in applying this theorem, we take Cn to not necessarily be closed, and then the result is that one of their closures must contain an open ball. In other words, we can’t have all of the Cn be nowhere dense. Remark 35. This theorem is pretty powerful – it can actually be used to prove that there is a continuous function which is nowhere differentiable. Proof. Suppose for the sake of contradiction that there is some collection of closed subsets Cn that are all nowhere S dense such that n Cn = M. We’ll prove that there’s a point not contained in any of the Cn s, using completeness, below. To do this, we’ll construct a sequence inductively. Because M contains at least one open ball, and C1 cannot contain an open ball, this means that M 6= C1 , and thus there is some p1 ∈ M \ C1 . Because C1 is closed, M \ C1 is open, and thus there is some ε1 > 0 such that B(p1 , ε1 ) ∩ C1 = ∅. Now, B(p1 , ε31 ) is not contained in C2 (because the closed set C2 is assumed to not contain any open ball), and thus there exists some point p2 ∈ B(p1 , ε31 ) such that p2 6∈ C2 . Because C2 is closed, we can then find some ε2 < ε1 3 such that B(p2 , ε2 ) ∩ C2 = ∅. More generally (we’ll be explicit this time but cover this in less detail in the future), suppose we have constructed points p2 , · · · , pk and constants ε1 , · · · , εk such that εk < pj ∈ B(pj−1 , εj−1 3 ), εk−1 3 < ··· < ε1 3k−1 , and with the constraint that B(pj , εj ) ∩ Cj = ∅ for all j. Then we construct pk+1 as follows: because B(pk , ε3k ) is not contained in Ck+1 , there exists an element pk+1 ∈ B(pk , ε3k ) such that pk+1 6∈ Ck+1 . Then we can pick some εk+1 < εk 3 so that B(pk+1 , εk+1 ) ∩ Ck+1 = ∅ (because M \ Ck+1 is open). So by induction we get a sequence of points {pk } in M and a sequence of numbers εk ∈ (0, ε1 ), such that the two boxed statements above are satisfied. This sequence is Cauchy, basically because we’ve made our εs decrease fast enough: for all k, ` ∈ N, repeated iterations of the triangle inequality gives us d(pk , pk+` ) ≤ d(pk , pk+1 ) + d(pk+1 , pk+2 ) + · · · + d(pk+`−1 , pk+` ). And now by the first boxed statement, we can bound this as < εk+1 εk+`−1 ε1 ε1 εk + + ··· + < k + · · · + k+` . 3 3 3 3 3 This sum can be bounded by the infinite geometric series < ε1 ∞ X 1 ε1 −k+1 = ·3 , 3m 2 m=k and thus making k large enough bounds this independently of `. So the sequence of points {pk } is Cauchy, and because M is complete, there exists some p ∈ M such that pk → p. And now we can show that p doesn’t lie in any of the Ck s (which is a contradiction) by showing that it lives in all of the balls B(pj , εj ) – this is because for all k ∈ N, we have  d(pk+1 , pk+1+` ) < εk+1  ∞ X 1 1 εk+1 1 + 2 + · · · + ` < εk+1 3−n = . 3 3 3 2 n=1 So taking the limit as ` → ∞, we have d(pk+1 , p) ≤ εk+1 εk < , 2 6 and thus 1 1 εk + εk < εk . 3 6 S So p ∈ B(pk , εk ) for each k, and each of these balls is disjoint from Ck . So p is not in any Ck , meaning p ∈ 6 k Ck = M, d(pk , p) ≤ d(pk , pk+1 ) + d(pk+1 , p) ≤ which is a contradiction. And we can use this to prove some results in functional analysis now: Theorem 36 (Uniform Boundedness Theorem) Let B be a Banach space, and let {Tn } be a sequence in B(B, V ) (of linear operators from B into some normed space V ). Then if for all b ∈ B we have supn ||Tn b|| < ∞ (that is, this sequence is pointwise bounded), then supn ||Tn || < ∞ (the operator norms are bounded). Proof. For each k ∈ N, define the subset Ck = {b ∈ B : ||b|| ≤ 1, sup ||Tn b|| ≤ k}. n This set is closed, because for any sequence {bn } ⊂ Ck with bn → b, we have ||b|| = limn→∞ ||bn || = 1, and for all m ∈ N, we have ||Tm b|| = lim ||Tm bn || n→∞ (using the fact that these operators are bounded and thus continuous). And now ||Tm bn || ≤ k because bn ∈ Ck , so the limit point must also be at most k. But we have {b ∈ B : ||b|| ≤ 1} = [ Ck , k≤n because for any b ∈ B, there is some k such that supm ||Tm b|| ≤ k (by assumption). And now the left-hand side is a complete metric space, because it is a closed subset of M, and thus by Baire’s theorem, one of the Ck s contains an open ball B(b0 , δ0 ). So now for any b ∈ B(0, δ0 ) (meaning that ||b|| < δ0 ), we know that b0 + b ∈ B(b0 , δ0 ) ⊂ Ck , so sup ||Tn (b0 + b)|| ≤ k. n But then sup ||Tn b|| = sup || − Tn b0 + Tn (b0 + b)|| ≤ sup ||Tn b0 || + sup ||Tn (b0 + b)|| ≤ k + k, n n n n because b0 , b0 + b are both in B(b0 , δ0 ). So for any b in the open ball B(0, δ0 ) satisfies supn ||Tn b|| < 2k, and rescaling means that for any n ∈ N and for all b ∈ B with ||b|| = 1, we have   4k δ0 Tn b ≤ 2k =⇒ ||Tn b|| ≤ , 2 δ0 meaning that the operator norm of Tn is at most 4k δ0 for all n, and thus supn ||Tn || ≤ 4k δ0 , and we’re done.