1. Hw 7 part 4 (5) (d) Since a(n) = 1/nα for n ≥ 1,
∞
X
n=1
|a(n)|p =
∞
X
n=1
1 npα.
By p-test, the series is convergent if and only if pα > 1. Hence a ∈ lp(N) if and only if α > 1/p.
(a) The statement is false if p > q. Take p = 2 and q = 1. Consider the function a : N → R defined by n 7→ 1/n. By p-test,
∞
X
n=1
|a(n)|2=
∞
X
n=1
1 n2 is convergent while
∞
X
n=1
|a(n)| =
∞
X
n=1
1 n
is divergent. In other words, a ∈ l2(N) while a 6∈ l1(N). In fact, the statement is true if p < q.
Proof. Let a ∈ lp(N). Then P∞
n=1|a(n)|p is convergent. By n-th term test, limn→∞|a(n)|p= 0 and hence by the property of limit, lim
n→∞|a(n)| = 0. Taking
= 1, we can choose N ∈ N such that |a(n)| < = 1 for any n ≥ 1. When p < q,
|a(n)|q = |a(n)|p|a(n)|q−p< |a(n)|p· 1 = |a(n)|p, for any n ≥ N. By Comparison test, P∞
n=1|a(n)|q is convergent and hence
a ∈ lq(N).
(f) Let a ∈ lp(N). As we have seen in (d), limn→∞|a(n)| = 0 and hence limn→∞a(n) = 0. We found that (a(n)) is a convergent sequence in R. Therefore (a(n)) is bounded in R, i.e. a ∈ l∞(N).
(6) (a) When p = 2, the H¨older inequality for convergent series of real numbers is exactly the Cauchy-Schwarz inequality for convergent series of real numbers:
∞
X
n=1
|a(n)b(n)| ≤
∞
X
n=1
|a(n)|2
!1/2 ∞
X
n=1
|b(n)|2
!1/2
. This provides the boundedness of the sequence (un) where un=Pn
i=1|a(n)b(n)|.
Since (un) is nondecreasing and bounded above, it is convergent in R by mono- tone sequence property.
You can also use the inequality:
(1.1) |a(n)b(n)| ≤ 1
2(|a(n)|2+ |b(n)|2), n ≥ 1.
Since a, b ∈ l2(N), P∞
n=1|a(n)|2 and P∞
n=1|b(n)|2 are both convergent in R.
By the property of convergence of infinite series,
(1.2) X
n=1
1
2(|a(n)|2+ |b(n)|2) = 1 2
∞
X
n=1
|a(n)|2+1 2
∞
X
n=1
|b(n)|2
1
2
is convergent in R. By (1.1) and (1.2) and the comparison test,P∞
n=1|a(n)b(n)|
is convergent.
(b) It is routine to check that h·, ·i is an inner product on l2(N).
(c) For any i, j ∈ N, hei, eji =
∞
X
n=1
ei(n)ej(n) =
∞
X
n=1
δinδnj = δij.
Hence {ei : i ∈ N} is an orthonormal family of vectors in l2(N).
(d) It follows from the definition that the set S∞ is bounded. To show that S∞ is closed, we prove that S∞contains all of its adherent points. Let a be an adherent point of S∞. There exists a sequence (ai) in S∞such that limi→∞ai = a in l2(N).
By triangle inequality, for any i ≥ 1,
0 ≤ |1 − kak2| = |kaik2− kak2| ≤ kai− ak2.
Since limn→∞ai = a in l2(N), limn→∞kai− ak2 = 0. By Sandwich principle,
|kak2− 1| = 0. Therefore kak2 = 1. This implies that a ∈ S∞. (e) For any i 6= j,
kei− ejk2= hei− ej, ei− eji = keik22− 2hei, eji + kejk2 = 2.
Here we use the orthonormality of {ei}. This shows that kei− ejk2 = √ 2 if i 6= j. Assume that (ei) has a convergent subsequence (eik) in S∞. Since (eik) is convergent, it is a Cauchy sequence in l2(N). Choose < √
2. We can find N ∈ N so that keik − eilk <√
2 for any k, l ≥ N. When we choose k, l ≥ N so that k > l, we find
√
2 = keik− eilk <√ 2
which is impossible. We conclude that (ei) does not have any convergent sub- sequence in S∞.
(f) We leave it to the reader to verify that V is a vector subspace of l2(N). Let a ∈ V. Then ha, fii = 0 for any i ≥ 1. By definition,
ha, fii = ha, ei− 2ei+1i
= ha, eii − 2ha, ei+1i
=
∞
X
n=1
a(n)ei(n) − 2
∞
X
n=1
a(n)ei+1(n)
=
∞
X
n=1
a(n)δin− 2
∞
X
n=1
a(n)δi+1,n
= a(i) − 2a(i + 1).
Hence ha, fii = 0 if and only if a(i) − 2a(i + 1) = 0 for any i ≥ 1. By induction, we see
a(i) = 1
2i−1a(1), for any i ≥ 1.
Let v : N → R be the function v(i) = 1/2i−1 for any i ∈ N. Then v ∈ F(N ).
Let us check that v ∈ l2(N). By definition, kvk22 =
∞
X
n=1
1 2i−1
2
=
n
X
i=1
1
4i−1 = 1
1 − (1/4) = 4 3 < ∞.
3
Hence v ∈ l2(N). Let us check that v ∈ V. One can see that v(i) − 2v(i + 1) = 1
2i−1 − 2 · 1 2i = 0.
Let us show that V = span{v} = {av : a ∈ R}. Let V0 = span{v}. Since v ∈ V and V is a vector space, V contains av for any a ∈ R. Hence V contains V0. For any a ∈ V, we have shown that a(i) = a(1)v(i) for any i ≥ 1. This implies that a = av for any a ∈ V with a = a(1). In other words, a ∈ V0 for any a ∈ V.
Hence V is contained in V0. We conclude that V = V0.
(g) Similarly, you can check that W is a vector subspace of l2(N) and for any a ∈ W, a(i) − 3a(i + 1) = 0 for any i ≥ 1.
Let w : N → R be the function i 7→ 1/3i−1. Then w ∈ l2(N) and W = span{w}.
We leave it to the reader to fill in all the details.
(h) Prove that Z is a vector subspace of l2(N) and a ∈ Z, if and only if a(i) − 5a(i + 1) + 6a(i + 2) = 0 for any i ≥ 1.
Check v, w ∈ Z by verifying that v and w satisfy the above equation. Since Z is a vector space and Z contains both v and w, Z contains V + W. It is also not difficult to verify that V ∩ W = {0}. In fact, {v, w} is linearly independent.
To show that Z is contained in V + W, we need to use the technique learned in Calculus 2 HW 4 (22). For any a ∈ Z, there exist A, B ∈ R so that
a(i) = A
2i−1+ B
3i−1, n ≥ 1.
To do this, we construct a power series f (X) =
∞
X
i=0
a(i + 1)Xi
and show that f (X) is a rational function using the recursive relation. Then use partial fraction expansion to find A, B so that
f (X) = A
1 −X2 + B 1 −X3 .
The details of this method can be found in Calculus 2 HW 4 (22). Hence a = Av + Bw. This shows that a is a linear combination of {v, w}. Hence a ∈ V + W. This proves that Z is contained in V + W. We conclude that Z = V + W. Furthermore, Z = V ⊕ W by the fact that V ∩ W = {0}.
(i) For any n ≥ 1, we need to compute
ksn− ak22 = ksnk22− 2hsn, ai + kak22. One can show that ksnk22 = hsn, ai =Pn
i=1|a(i)|2. Hence for any n ≥ 1, ksn− ak22=
∞
X
i=n+1
|a(i)|2. Let xn =Pn
i=1|a(i)|2 for any n ≥ 1. Since a ∈ l2(N), (xn) is convergent in R;
the limit of (xn) kak22 =P∞
i=1|a(i)|2. Furthermore, 0 ≤ x − xn=
∞
X
i=n+1
|a(i)|2.
4
Since (xn) is convergent to x, for any > 0, there exists N∈ N such that 0 ≤ x − xn< 2 whenever n ≥ N
This shows that P∞
i=n+1|a(i)|2 < 2 whenever n ≥ N. Thus ksn− ak2 < whenever n ≥ N. This shows that
a = lim
n→∞sn=
∞
X
n=1
a(n)en.
(7) (a) It is routine to check R∞ is a vector subspace of lp(N).
(b) To show that R∞ is not a closed subset of lp(N), let us find an adherent point of R∞ which does not belong to R∞.
Let a : N → R be the function n 7→ 1/n2. By p-test,P∞ n=1 1
n2p is convergent for any p ≥ 1. Hence a ∈ lp(N) for any p ≥ 1. Let sn: N → R be the function
sn= e1+ 1
22e2+ · · · + 1
n2en, n ≥ 1.
Then sn∈ R∞ for any n ≥ 1. Moreover, ksn− akpp =
∞
X
i=n+1
1 i2p.
Using the same argument as that of (6-i), we can see that limn→∞sn= a while a 6∈ R∞. This proves that a is an adherent point of R∞ and does not belong to R∞.
