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Analysis Solutions: Problem Set 3 Math 201A, Fall 2006
Problem 1. Let X be the space of all real sequences of the form
x = (x 1 , x 2 , x 3 ,... , xn, 0 , 0 ,.. .) , xi ∈ R
whose terms are zero from some point on. Define
‖x‖∞ = max i∈N
|xi|.
(a) Show that (X, ‖ · ‖∞) is a normed linear space.
(b) Show that X is not complete.
(c) Give a description of the completion of X as a space of sequences.
Solution.
- (a) X is a linear space under component-wise addition and scalar mul- tiplication: if λ ∈ R, x = (xi), y = (yi) then
λx = (λxi) , x + y = (xi + yi).
The 0-vector is the sequence (0, 0 , 0 ,.. .).
- We have: ‖x‖ ≥ 0 and ‖x‖ = 0 if and only if xi = 0 for every i ∈ N, or x = 0; for any λ ∈ R,
‖λx‖∞ = max i∈N |λxi| = |λ| max i∈N |xi| = |λ| ‖x‖∞;
and
‖x + y‖∞ = max i∈N
|xi + yi|
≤ max i∈N {|xi| + |yi|}
≤ max i∈N |xi| + max i∈N |yi|
≤ ‖x‖∞ + ‖y‖∞.
- (b) Define a sequence (xn) in X by
xn =
n
For m > n, we have
‖xm − xn‖∞ =
n + 1
which implies that the sequence is Cauchy.
- If y ∈ X, then y = (y 1 , y 2 ,... , yN , 0 , 0.. .) for some N ∈ N. If n > N , then ‖xn − y‖∞ ≥
N + 1
so (xn) does not converge to y as n → ∞. Therefore, the sequence (xn) has no limit in X, and X is not complete.
- (c) The completion of X is the space c 0 of all real sequences (xi) such that xi → 0 as i → ∞. The norm on c 0 is ‖ · ‖∞.
- The inclusion map φ : X → c 0 , where φ(x) = x, is an isometric imbed- ding of X into c 0 , so by the uniqueness part of the completion theorem we just need to show that X is dense in c 0 and c 0 is complete.
- Given y = (y 1 , y 2 ,... , yi,.. .) ∈ c 0 let
xn = (y 1 , y 2 ,... , yn, 0 , 0.. .) ∈ X.
Since yi → 0 as i → ∞, we have
‖xn − y‖∞ = max i>n |yi| → 0 as n → ∞.
Thus, every point in c 0 is a limit of a sequence in X, so X is dense in c 0.
- Suppose that (xn)∞ n=1 is a Cauchy sequence in c 0. Let xn = (xn,i)∞ i=1. For each i ∈ N we have
|xn,i − xm,i| ≤ ‖xn − xm‖∞,
so (xn,i)∞ n=1 is a Cauchy sequence in R. Since R is complete, there exists yi ∈ R such that xn,i → yi as n → ∞. Let y = (yi)∞ i=1. We will show that xn → y with respect to ‖ · ‖∞ and y ∈ c 0.
Problem 2. Suppose that (X, dX ) and (Y, dY ) are metric spaces and (Y, dY ) is complete. If D is a dense subset of X and f : D → Y is uniformly continuous on D, prove that there exists a unique continuous function F : X → Y such that F |D = f.
Solution.
- If x ∈ X then there exists a sequence (an) in D such that an → x as n → ∞, since D is dense in X. The sequence (an) is Cauchy in D, and the uniform continuity of f implies that (f (an)) is Cauchy in Y. Since Y is complete, there exists y ∈ Y such that f (an) → y as n → ∞. We define F (x) = y.
- This function F : X → Y is well-defined. If (an), (a′ n) are two sequences in D that converge to x ∈ X, then dX (an, a′ n) → 0 as n → ∞, and the uniform continuity of f implies that
dY (f (an), f (a′ n)) → 0 as n → ∞.
It follows that lim n→∞ f (an) = lim n→∞ f (a′ n),
so the value F (x) is independent of the choice of sequence converging to x.
- If a ∈ D, then we may choose an = a for every n ∈ N, which gives F (a) = f (a), so that the restriction of F to D is equal to f.
- To prove the continuity of F , suppose that x, x′^ ∈ X. Choose sequences (an), (a′ n) in D such that an → x, a′ n → x′. Given > 0, the uniform continuity of f implies that there exists δ > 0 such that a, a′^ ∈ D and dX (a, a′) < δ implies that dY (f (a), f (a′)) < /3. Choose N such that
d(aN , x) <
δ 3
, dY (f (aN ), F (x)) <
d(a′ N , x′) <
δ 3
, dY (f (a′ N ), F (x′)) <
Then
dX (aN , a′ N ) ≤ dX (aN , x) + dX (x, x′) + dX (x′, a′ N ).
Hence, if dX (x, x′) < δ/3, then dX (aN , a′ N ) < δ, and
dY (F (x), F (x′)) ≤ dY (F (x), f (aN )) + dY (f (aN ), f (a′ N )) +dY (f (a′ N ), F (x′)) < .
It follows that F is uniformly continuous on X
- We only give a brief outline of a proof that (Q, d) is not complete. (Perhaps there is a simpler one.) We use a long-division algorithm to prove that every rational number r ∈ Q has a |·|p-convergent expansion of the form
r =
∑^ ∞
i=k
ripi^ with k, ri ∈ Z and 0 ≤ ri ≤ p − 1 (1)
in which the ri are periodic functions of i.^1 For example,
1 1 − p
∑^ ∞
i=
pi^ ∈ Q.
It then follows that the partial sums of a series with non-periodic co- efficients cannot converge to any rational number. For example, the series (rn) defined by
rn =
∑^ n
i=
pi!^ = p + p^2 + p^6 +... + pn!
is Cauchy in Qp but does not converge to any r ∈ Q.
- The completion Qp may be thought of concretely as the space of all sequences of the form (1).
- (b) We equip Q × Q with the product metric dQ×Q = d × d,
dQ×Q ((r, s), (r′, s′)) = d(r, r′) + d(s, s′).
We will temporarily use the notation +(r, s) = r + s for the addition function. (^1) To obtain this algorithm, write
m =
∑^ M
i=
mipi, n =
∑^ N
i=
nipi, r =
∑
i≥k
ripi
where 0 ≤ mi, ni, ri ≤ p − 1 and m 0 , n 0 6 = 0, multiply the series in the equation nr = mpk, carry multiples of powers of p to the succeeding terms so that all coefficients of pi^ are between 0 and p − 1, and equate coefficients of pi. One finds that ri is determined by the previous N coefficients {ri− 1 ,... , ri−N }. Since there are only finitely many such sequences, a sequence must eventually repeat, and then the coefficients ri will repeat.
- If (r, s), (r′, s′) ∈ Q × Q, then by the ultrametric property of | · |p,
d (+(r, s), +(r′, s′)) = |r + s − (r′^ + s′)|p = |r − r′^ + s − s′)|p
≤ max
|r − r′|p , |s − s′)|p
≤ |r − r′|p + |s − s′)|p ≤ dQ×Q ((r, s), (r′, s′)) ,
which proves that + : Q × Q → R is uniformly continuous.
- If (X, dX ) and (Y, dY ) are metric spaces with completions ( X, d˜ (^) X˜ ) and ( Y , d˜ (^) Y˜ ), respectively, then the completion of (X × Y, dX × dY ) is ( X˜ × Y , d^ ˜ (^) X˜ × d (^) Y˜ ). (The proof is left as a exercise.) Thus, the completion of (Q × Q, d × d) is (Qp × Qp, dp × dp), and it follows from the result of Problem 2 that + : Q × Q → R extends to a unique uniformly continuous map +|p : Qp × Qp → R.
- The sum of p-adic numbers r, s ∈ Qp may be computed by adding their series expansions and carrying any multiples of p. For example, if
r = 1 + 0p + 0p^2 +... , s = (p − 1) + (p − 1)p + (p − 1)p^2 +... ,
then r + s = 0, so s is the additive inverse of 1 in Qp, meaning that the sum of the series for s is −1.
Remark. Elements of Qp are called p-adic numbers, important in algebraic number theory. Multiplication of rational numbers also extends continuously from Q to Qp, so Qp is a complete field. Analysis on (Qp, |·|p) is an ultrametric analog of the more familiar analysis on the Euclidean real line (R, | · |).
is dense in X.
Solution.
- (a) The series defining d converges, since the terms are bounded by the terms 2−n^ of a convergent geometric series. The function d is clearly symmetric and nonnegative. If d(s, t) = 0, then δn = 0 and sn = tn for every n ∈ N, so s = t. Finally, if s = (s 1 , s 2 , s 3 ,.. .), t = (t 1 , t 2 , t 3 ,.. .), r = (r 1 , r 2 , r 3 ,.. .), and
δn =
0 if sn = tn, 1 if sn 6 = tn, n =
0 if tn = rn, 1 if tn 6 = rn, ηn =
0 if sn = rn, 1 if sn 6 = rn,
then ηn ≤ δn + n, so d(s, r) ≤ d(s, t) + d(t, r).
- (b) To show that X is compact, we prove that it is complete and totally bounded. We use the following properties of the metric: 1. if d(s, t) < 1 / 2 N^ then sn = tn for 1 ≤ n ≤ N ; 2. if > 1 / 2 N^ and sn = tn for 1 ≤ n ≤ N then d(s, t) < .
- Suppose that (sk)∞ k=1 is a Cauchy sequence in X. We write
sk = (s 1 ,k, s 2 ,k,... , sn,k,.. .).
For any N ∈ N, there exists K ∈ N such that d(sj , sk) < 1 / 2 N^ for all j, k > K, and hence sn,j = sn,k for all n ≤ N. Thus, the terms in the sequences are eventually the same. We denote the eventual common value of (sn,k)∞ k=1 by sn, and define
s = (s 1 , s 2 ,... , sn,.. .).
Then for k > K we have
d(sk, s) ≤
∑^ ∞
n=N +
2 n
≤
2 N^
It follows that sk → s as k → ∞, so X is complete.
- Given > 0, choose N ∈ N such that > 1 / 2 N^. If
s = (s 1 , s 2 , s 3 ,... , sN , sN +1, sN +2,.. .)
then the open ball B(s) contains all sequences of the form
t = (s 1 ,... , sN , tN +1, tN +2.. .) where tn ∈ { 0 , 1 } for n > N.
There are 2N^ initial sequences (s 1 ,... , sN ) ∈ { 0 , 1 }N^ , so X is covered by 2N^ balls of radius , and X is totally bounded.
- Next we show that X is totally disconnected. Given N ∈ N, let
UN = {s ∈ X | sN = 0} , VN = {s ∈ X | sN = 1}.
Then UN and VN are open. For example, suppose s ∈ UN. If d(s, t) < 1 / 2 N^ then tN = sN = 0, so t ∈ UN , meaning that B 1 / 2 N (s) ⊂ UN , so UN is open. (It follows that X = UN ∪ VN is the disjoint union of non-empty open sets so it is not connected.)
- Suppose that A ⊂ X contains at least two distinct points s, t. There exists N ∈ N such that sN 6 = tN. Let U = A∩UN and V = A∩VN. Then U , V are disjoint, non-empty sets that are open in A, and A = U ∪ V. Hence A is not connected and X is totally disconnected.
- Suppose that s = (s 1 , s 2 ,... , sn,.. .) ∈ X and define
sk = (s 1 ,k, s 2 ,k,... , sn,k,.. .) ∈ X
by sn,k = sn if n 6 = k and sn,k 6 = sn if n = k. Thus, sk differs from s only in the kth term. Then sk 6 = s and d(sk, s) = 1/ 2 k. Hence sk → s as k → ∞, and X is perfect.
- Compare the Cantor set with the closed interval [0, 1] — a compact space that is perfect but connected — and the finite set { 1 , 2 ,... , N } with the discrete metric — a compact space that is totally disconnected but not perfect.
- (c) If a ∈ C has the base-three expansion a = 0.a 1 a 2 a 3... with no 1’s, we define φ(a) = s ∈ X where s = (s 1 , s 2 , s 3 ,.. .) is given by
sn =
0 if an = 0, 1 if an = 2.
tn 6 = sn for n ≥ N + 1. Then d(s, t) = 1/ 2 N^ < , so t ∈ U. On the other hand,
σN^ (s) = (sN +1, sN +2,.. .), σN^ (t) = (tN +1, tN +2,.. .)
differ in every term, so
d
σN^ (s), σN^ (s
= 1 > δ.
- (f) Define s ∈ X by listing all one-term sequences, followed by all two- term sequences, followed by all three-term sequences, and so on. For example, we could define
s = (0, 1 , 0 , 0 , 0 , 1 , 1 , 0 , 1 , 1 , 0 , 0 , 0 , 0 , 0 , 1 , 0 , 1 , 0 , 0 , 1 , 1 , 1 , 0 , 0 ,.. .).
Then for every t ∈ X and every N ∈ N there exists an n ∈ N such that σn(s) and t have the same first N terms. Hence, d(σn(s), t) ≤ 1 / 2 N^ , which proves that the orbit of s under σ is dense in X.
Remark. The metric space X is an example of a symbol space. Such symbol spaces arise in the study of chaotic dynamical systems, and the representa- tion of chaotic dynamical systems by shift maps on symbol spaces is called symbolic dynamics. In this context, the property in (e) is called ‘sensitive dependence on initial conditions.’ For example, consider the logistic map Fμ : Λ ⊂ [0, 1] → [0, 1] defined by Fμ(x) = μx(1 − x). For μ > 4 the map Fμ has an invariant Cantor set Λ of points that remain in [0, 1] under all iterates F (^) μn. One can prove that there is a homeomorphism φ : Λ → X such that Fμ = φ−^1 ◦ σ ◦ φ is topologically conjugate to the shift map σ.
