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B4a Banach spaces Michaelmas 2010 lectured by Bernd Kirchheim based on notes by CJK Batty
Lectures 3 and 4
Recall from the last lecture (X, ‖ · ‖) a normed space, then
- d(x, y) = ‖x − y‖ a metric on X
- converges xn → x ⇐⇒ ‖xn − x‖ → 0
- Y ⊂ X closed ⇔: (xn ∈ Y & xn → x ∈ X ⇒ x ∈ Y )
- T : X → X continuous ⇔: (xn → x ⇒ T xn → T x.
- (λ, x, y) → λx + y continuous from F × X × X → X
4.Corollary Let Y be a linear subspace of a normed vector space X. The closure Y of Y is also a linear subspace of X. Proof Clearly 0X ∈ Y ⊂ Y 6 = ∅. If we take any λ ∈ F and x, y ∈ Y , then by definition of the closure there are sequences (xn), (yn) ∈ Y with xn → x and yn → y. Since Y is a subspace, we have λxn + yn ∈ Y for all n and by continuity of the operations noted above λxx + yn → λx + y, so λx + y ∈ Y again and ∈ Y satisfies subspace criteria.
5.Example Consider sums A + B = {a + b : a ∈ A & b ∈ B} of sets.
- Let Y be a closed subset of a normed vector space X, and A be a compact subset of X. Then
Y + A := {y + a : y ∈ Y, a ∈ A}
is closed in X. Proof Let xn ∈ Y + A and xn → x ∈ X. We need to show x ∈ Y + A. By definition, there are an ∈ A, yn ∈ Y such that xn = an + yn. Since A is compact, we find a subsequence (nk) such that ank → a ∈ A as k → ∞. Therefore, ynk = xnk − an−k → x − a hence y = x − a ∈ Y since Y closed. So x = y + a as required.
- Sums of closed subspaces are not necessarily closed Let X = ℓ^1 , with the subspaces
Y = {y ∈ X : y 2 j = 0 for j ≥ 1 }, Z = {z ∈ X : z 2 j− 1 = jz 2 j all j ≥ 1 }.
Then Y and Z are closed linearsubspaces of X and Y + Z is not closed.
Proof First we show that Y, Z are closed subspaces. For j ≥ 1 let fj (y) = y 2 j ∈ F, then fj is clearly linear, thus Ker(fj ) is a subspace. Is it also closed? Since Ker(fj ) = f −^1 ({ 0 }), we only need that fj is continuous. But
∀x, y ∈ X : |fj (x) − fj (y)| = |x 2 j − y 2 j | = |(x − y) 2 j | ≤ ‖x − y‖ 1 ,
so fj is clearly continuous. Hence Y =
j=1 Ker(fj^ ) is also a closed set and a linear subspace. The argument for Z is the same, using the functions gj (x) = x 2 j− 1 − jx 2 j which satisfy |g(x) − g(y)| ≤ (1 + j)‖x − y‖ 1 and are therefore again continuous. Now consider any point in x = (xj ) ∈ Y + Z, so x = (yj ) + (zj ) with y 2 j = 0 and z 2 j− 1 = jz 2 j. Therefore,
z 2 j = x 2 j , z 2 j− 1 = jx 2 j and y 2 j− 1 = x 2 j− 1 − jx 2 j. (1) {deco}
So, if we find a sequence (αj ) s.t.
j |αj^ |^ <^ ∞^ but^
j j|αj^ |^ =^ ∞^ then
x = (xn) =
0 if n odd αj if n = 2j,
satisfies
j |xj^ |^ <^ ∞^ and^ x^ ∈^ ℓ
(^1) \ (Y + Z), since for the corresponing z ∑ j |z^2 j−^1 |^ =^
j |αj^ |^ =^ ∞^ contradicts^ z^ ∈^ ℓ
(^1). Such (αj ) = 1/j (^2) is easily
found, giving x = (0, 1 , 0 , 14 , 0 , 19 , 0 ,... ). Moreover, we see from (1) that x ∈ Y + Z if xn = 0 for all n large, so all vectors xk^ = (x 1 , x 2 ,... , xk, 0 , 0.. .) ∈ Y + Z.
Since ‖x − xk‖ =
n>k |xn|^ =^
2 j>k j
− (^2) → 0 as k → ∞ we have x ∈
Y + Z \ Y + Z. Remark We will show, that this situation can not occur in finite di- mensions - however you can verify that the sum of the two closed set {x : x 1 x 2 = 1}, {x : x 1 = 0} in R^2 is not closed. [Hint: Show that the sum equals {x ∈ R^2 : x 1 6 = 0}]
It follows that x is bounded, and, therefore, lies in X. Furthermore, from (2), for k > n,
‖xk − x‖∞ = sup{|xk(t) − x(t)| : t ∈ Ω} ≤ ε
and the proof is complete.
- Recall: if Ω is a topological space, then Cb(Ω) is defined as the space of bounded continuous functions on Ω. Based on 2. and our knowledge about uniform convergence it follows now easily that (Cb(Ω), ‖ · ‖∞) is complete. However, (C[0, 1], ‖ · ‖ 1 ) is not complete.
Proof We consider
fn(x) = min{ 1 , max{ 0 ,
)}}, x ∈ [0, 1] & n ≥ 1.
Then each fn ∈ C([0, 1]) since it is max and min of continous functions. Moreover 0 ≤ fn ≤ 1, fn(x) = 0 if x ≤ 12 − (^21) n and fn(x) = 1 if x ≥ 12 + (^21) n. Therefore,
‖fn − fm‖ 1 ≤
2 +^ 21 k 1 2 −^ 21 k
1 ≤ (^1) k for n, m ≥ k,
since then fn(x) = fm(x) if |x − 12 | > (^21) k. So (fk) is Cauchy w.r.t ‖ · ‖ 1. But suppose fk → f ∈ C([0, 1]). Then there is s ∈ { 0 , 1 } s.t. |s − f (^12 )| > 13 , therefore exists ε > 0 such that |f (x) − s| > 13 if |x − 12 | < ε. Thus, if s = 0 then |fk(x)−f (x)| > 13 if 12 −ε < x < 12 − (^21) k which gives ‖f −fk‖ 1 ≥ 13 (ε− (^21) k ) and so ‖f − fk‖ 1 does not converge to 0. Similarly arguing on (^12 + (^21) k , 12 + ε) if s = 1 we see that fk → f is impossible. Thus we established a non convergent Cauchy sequence in C([0, 1]), later we’ll see an easier way to prove non-existence of a limit.
3.Definition Consider a series
1 xn, where^ xn^ ∈^ X. This series^ converges^ to^ x^ ∈^ X if: (^) ∥ ∥ ∥ ∥ ∥
∑^ m
n=
xn − x
→ 0 as m → ∞.
∑ xn is absolutely convergent if
‖xn‖ < ∞.
4.Theorem A normed vector space X is complete if and only if every absolutely convergent series in X is convergent to a limit in X.
Proof Note that for any series
1 xn^ in any normed space^ X^ the partial sums sk =
∑k n 1 xn^ satisfy the estimate
‖sk − sl‖ = ‖
∑^ k
n=l+
xn‖ ≤
∑^ k
n=l+
‖xn‖ ≤
∑^ ∞
n=l+
‖xn‖,
if k ≥ l. So given an absolutely convergent series and ε > 0 we find m such that
n=m ‖xn‖^ < ε^ and hence ‖sk − sl‖ < ε if k, l ≥, m
in other words the sequence of partial sums (sk) is Cauchy. Now, if X is complete, then (sk) must be convergent, and the series
1 xn^ as well. Conversely, suppose each absolutely convergent series in X is convergent and take any Cauch sequence (yn) in X. For each k ≥ 1 we find nk > nk− 1 such that ‖yl − ym‖ < 2 −k^ if l, m ≥ nk. We consider the subsequence sk = ynk =
∑k l=1 xl^ where^ x^1 =^ s^1 and the^ xk^ =^ sk^ −^ sk−^1 satisfy^ ‖xk‖^ < 2 −k+1^ if k > 1. Therefore, the series
k xk^ is absolute convergent. By our assumption on X, the series must converge in X. In other words there exists s = lim k → ∞sk = limk ynk ∈ X. The fact that now the whole Cauchy sequence converges to the limit of any subsequence should be known from mods and finishes the proof. (Or recall: ∀ε > 0 ∃l : n, m > l ⇒ ‖yn −ym‖ < ε. Clearly, there is also k such that nk > l and ‖ynk − s‖ < ε, therefore ‖yn − s‖ ≤ ‖yn − ynk ‖ − ‖ynk − s‖ < 2 ε for all n > k. )
5.Example The spaces Lp(R), p ∈ [1, ∞] are complete. For the case 1 ≤ p < ∞ see the Integration course a4 (e.g. Section 8 of A. Etheridge’s Lecture notes). Simarly, all the spaces ℓp, p ∈ [1, ∞], are Banach spaces. For the case p ∈ [1, ∞) Theorem 4 together with Minkowski’s inequality are crucial. We dicuss here L∞(R), and by Theo 4 we only need to show if ∑ n fn^ is an absolutely convergent sequence in this space, then it converges. We notice that by defintion of ‖ · ‖∞ find for each n a Lebesgue null set An ⊂ R such that ∀x ∈ R \ An : |fn(x)| ≤ ‖f ‖∞ + 2−n. Clearly, A =
n An is also a null set and we define
gn(x) =
fn(x) x ∈ R \ A, 0 x ∈ A.
such that i(X) = X˜. Moreover, X˜ is unique up to norm-preserving bijection. (ˆi : X → Xˆ any other completion, then i◦ˆi−^1 has a norm preserving bijective
extension I : Xˆ → X˜, i.e. I is an isometric linear isomorphism.)
For example, if X = (C[0, 1], ‖ · ‖ 1 ), then
X^ ˜ = (L^1 [0, 1], ‖ · ‖ 1 ) , i(f ) = [f ].
We clearly see that the
fn(x) = min{ 1 , max{ 0 ,
)}}, x ∈ [0, 1] & n ≥ 1
from Exam 2.3 are in X but converge in X˜ to χ[ 1 2 ,1]^
∈/ X, this gives another
explanation why (fn) can not have a limit in X (though the fact that f = χ[ 1 2 ,1]^
∈/ X would need to be proven carefully, i.e. we have to show that no
g ∈ X˜ with g = f a.e is continuous!)
In general, if X is a Banach space and Y a linear subspace, then the closure Y¯ in X is a completion of Y.
We are not going to prove any of these general statements now in detail, but the ideas behind them will be clear at the end of the course.
7.Definition Two norms ‖ · ‖, ‖ · ‖′^ on the same space X are said to be equivalent if there are constants a > 0, b such that
a‖x‖ ≤ ‖x‖′^ ≤ b‖x‖ for all x ∈ X.
Then B(x, ε) ⊃ B′(x, aε), B′(x, ε) ⊃ B(x, ε/b),
where B(x, ε) = {y ∈ X : ‖y − x‖ < ε}
etc. So
- the topologies coincide.
- The two norms have the same convergent sequences.
- They also have the same Cauchy sequences!!
- Thus if ‖ · ‖ is complete, so is any equivalent norm.
Remark Equivalence of norms is a stronger concept than equivalence of metrics! If two metrics ̺ ,σ are equivalent, i.e. induce the same topology, then in general the quotient ̺ (x, y)/σ(x, y) is not bounded away from zero (or infinity). For example, the metrics
̺ (x, y) = |x − y|, σ(x, y) = | arctan(x) − arctan(y)| on R
both induce the same (standart) topology. (R, σ(x, y)) is, however, not even complete. Indeed xn = n is a Cauchy sequence (since σ(xn, xm) < ε if n, m > tan(π− 2 ε)) which is not convergent - completness is not a topological property( not perserved under homeomorphisms)
