MA244 Analysis III
Solutions. Sheet 4.
Questions for credit: 2 (4 points), 8 (5 points), 4 (8 points) and 10 (8 points)
0.1
Norms
1. (a) N1 (z) is not a norm on Cn . For example, it does not separate points: take
z = (i, i, . . . , i). Then z̄ = (−i, −i, . . . , −i). So, zk + z̄k = 0 for all k and
N1 (z) = 0 even though z 6= 0.
P
(b) N2 (z) is a norm on Cn . Positivity: zk z̄k = |z|2k ≥ 0, so nk=1 zk z̄k ≥ 0.
Separation of points: if N2 (z) = 0, then zk z̄k = |zk |2 = 0 for k =
1, 2, . . . n, as each term in the sum for N2 (z)2 is non-negative. Therefore, zk = 0 for k = P
1, 2, . . . , n, meaning that z = 0. Absolute hon
mogeneity: NP
(λz)
=
(
)1/2 = |λ|N2 (z). Triangle inequality:
2
k=1 λλ̄zk z̄kP
N2 (w+z) = ( nk=1
|zk +wk |2 )1/2 = ( nk=1 ((Re(zkP
)+Re(wk ))2 +(Im(zk )+
P
n
2
2 1/2
2 1/2
+( nk=1 (Re(wk )2 +Im(wk )2 ))1/2 =
Im(w
k ) +Im(zk ) ))
Pn k )) 2))1/2 ≤ (Pnk=1 (Re(z
( k=1 |z|k ) + ( k=1 |w|2k )1/2 = N2 (z) + N2 (w). The inequality in the
middle is the triangle inequality for the Euclidean norm on R2n .
2. Non-negativity: as x − y ∈ V and || · || is a norm on V , d(x, y) = ||x − y|| ≥ 0.
Separation of points: if d(x, y) = ||x − y|| = 0, then x − y = 0, as the norm || · ||
separates points. Therefore x = y. Symmetry: d(x−y) = ||x−y|| = ||(−1)(y −
x)|| = | − 1| · ||y − x|| = d(y, x), where we used the absolute homogeneity
property of || · ||. Triangle inequality: d(x, z) = ||x − z|| = ||x − y + y − z|| ≤
||x − y|| + ||y − z|| = d(x, y) + d(y, z), where we used the triangle inequality for
|| · ||.
Note that there is a significant difference between normed and metric spaces:
the definition of the latter does not require the space to be linear!
3. Claim: || · || is a norm on Rn iff ai > 0, i = 1, 2, . . . n.
Proof. ⇒ || · ||a is a norm, therefore for every i = 1, 2, . . . , n,
||(0, 0, . . . , 0,
i-th co-ordinate
, 0, . . . 0)||a > 0.
1
Therefore, ai > 0.
⇐ If ai > 0, i = 1, 2, . . . n, the function ||·||a clearly separates points, is positive
on Rn \ {0} and is absolutely
homogeneous. Triangle inequality: ||x + y||a =
P
Pn
n
a
|x
+
y
|
≤
a
(|x
k
k | + |yk |) = ||x||a + ||y||b .
k=1 k k
k=1 k
4. (a) The most important step is to verify that SF is closed under addition and
scalar multiplication: for any a, b ∈ SF there exist N1 , N2 ∈ N: an = 0
for any n > N1 , bn = 0 for any n > N2 . Let N = max(N1 , N2 ). Then
(an + bn ) = 0 for n > N . Therefore, a + b ∈ SF . Similarly, for any λ ∈ R,
λan = for n > N1 , so λa ∈ SF .
The verification of the axioms of vector space (SF is an abelian group with
respect to addition and SF is an R-module) is elementary and students
can be excused for going through it quickly.
(b) The most economical way of proving that functions (1), (3), (4) and (5)
are norms on SF is as follows: notice that the set SF (N ) = {a ∈ SF : an =
0, n > N } is a linear subset of SF isomorphic to RN . The restriction of
functions (1), (3), (4) and (5) to SF (N ) are just the sup-norm, the taxicab
norm, the Euclidean norm and the norm from Question 4 with ai = i > 0
on RN correspondingly. As for any pair of a, b ∈ SF there exists N :
a, b ∈ SF , the check that functions (1), (3), (4) and (5) are norms on SF
is equivalent to checking that they give rise to norms on SF (N ) ' R for
any N ∈ N. This was done in the lectures for norms (1), (3), (4) and in
Question 4 for norm (5).
To check that (2) is a norm on SF , notice that it is well defined as
the max is always taken over a finite set. The separation of points
and absolute homogeneity properties are straightforward. Triangle inequality: ||a + b||w = maxn≥1 n|an + bn | ≤ maxn≥1 n(|an | + |bn |) ≤
maxn≥1 n|an | + maxn≥1 n|bn | = ||a||w + ||b||w .
(c) (i) (2)6∼(1). Let (an )n≥1 ⊂ SF be a sequence:
an = (1, 1, . . . , 1, 0, 0 . . .).
| {z }
n-times
For any n ∈ N, ||an ||∞ = 1, ||an ||w = n → ∞ for n → ∞. Therefore 6 ∃
K1 > 0: for any n ∈ N, K1 ||an ||w < ||an ||∞ , meaning that || · ||w 6∼ || · ||∞ .
(ii) (3)6∼(1). Using the sequence (an )n>1 from the previous step we see
that ||an ||1 = n → ∞ as n → ∞. Therefore, (3)6∼(1) by the argument
given in the previous step.
(iii) (4)6∼(1). Using the sequence (an )n>1 from step (i) we see that ||an ||2 =
n1/2 → ∞ as n → ∞. Therefore, (4)6∼(1) by the argument given in step
(i).
(iv) (5)6∼(1). Using the sequence (an )n>1 from step (i) we see that ||an ||u =
n(n + 1)/2 → ∞ as n → ∞. Therefore, (5)6∼(1) by the argument given
in step (i).
0.2
Completeness
5. Let (an )n≥1 ⊂ SF be a sequence such that
−k
2
if 1 ≤ k ≤ n;
(an )k =
0
if k > n.
For any pair of natural numbers k < m,
0 ≤ ||am − ak ||1 =
m
X
2−p ≤ 2−k → 0 for k → ∞.
p=k+1
Therefore, the sequence (an )n ≥1 is || · ||1 -Cauchy. However, (an )n≥1 does not
converge in SF for suppose that a ∈ SF is the limit. Then there is N ∈ N
such that aN = 0. So, for any n > N ||an − a||1 ≥ (an )N = 2−N . Hence for
= 2−N −1 and for any M ∈ N, there exists n > M such that ||an − a||1 > ,
which contradicts our assumption that a is the limit.
We have constructed a Cauchy sequence in SF which does not converge. Therefore, SF is not Banach.
6. (a) The crucial step is the proof that l1 is closed with respect to componentwise addition. This is trueP
as the sum of twoPabsolutely convergent series
∞
∞
is absolutely convergent:
n=1 |an + bn | ≤
n=1 (|an | + |bn |) ≤ ∞. The
rest is a straightforward consequence of theorems governing operations on
convergent series.
(b) Let (an )n≥1 ⊂ l1 be a Cauchy sequence. Then ∀ > 0, there exists N :
for any m > n > N ,
||am − an ||1 ≡
∞
X
|(am )k − (an )k | < /2
(1)
k=1
Therefore, for any k ∈ N, |(am )k −(an )k | < . We proved that the sequence
((an )k )n≥1 is Cauchy in (R, | · |). Therefore it converges. Let ak be the
limit. Consider the sequence a = (a1 , a2 , a3 , . . .).
We proved that an → a component-wise. Now we need to prove that (i)
an → a w. r. t. || · ||1 norm; (ii) a ∈ l1P
. Both results are a consequence of
the uniform convergence of the series ∞
k=1 (an )k with respect to n:
P
(i) Using (1): for any M ∈ N, and any m > n > N , M
k=1 |(am )k −(an )k | ≤
||am − an ||1 < /2. Taking the limit m → ∞ we find
M
X
|ak − (an )k | ≤ /2.
k=1
Now we can take the limit M → ∞:
||a − an ||1 ≤ /2.
We proved that for any > 0, there is N such that for any n > N ,
||a − an ||1 < . Therefore, (an )n≥1 converges to a with respect to || · ||1
norm.
(ii) It also follows from (1) that
0≤
M
X
|(am )k | ≤ ||am ||1 ≤ /2 + ||an ||1 .
k=1
Taking the
PMlimit m → ∞ in the above equation we conclude
P∞ that for any
M ∈ N, k=1 |ak | ≤ /2+||an ||1 . Therefore the series k=1 |ak | converges
and a ∈ SF .
We proved that any || · ||1 -Cauchy series in l1 converges. Therefore, (l1 , || ·
||1 ) is Banach.
7. Let (xk )k≥1 ⊂ X be a Cauchy sequence, (xkp )p≥1 ⊂ (xk )k≥1 - its convergent
subsequence, which exists by the BW property of X. Let x ∈ X be the limit
of (xkp )p≥1 . Thus for any > 0, there exists N : (i) For any p : kp > N ,
||xkp −x|| < /2 (convergence of (xkp )); (ii) For any m, n > N , ||xm −xn || < /2
(Cauchy property of (xk )). Therefore, for any > 0, there exists N : for any
n > N , ||xn − x|| ≤ ||xn − xkp || + ||xkp − x|| < /2 + /2 < . So xn → x by
definition. We proved that every Cauchy sequence in X converges. In other
words, X is complete.
8. Let n > 0. ||xn −w|| = ||T xn−1 −T w|| ≤ K||xn−2 −w||, where we used T w = w.
We proved that for any n > 0,
||xn − w|| ≤ K||xn−1 − w||.
Iterating the above inequality n times we find
n→∞
||xn − w|| ≤ K n ||x0 − w|| −→ 0,
as 0 < K < 1. So xn → w by definition.
0.3
Closed and opened sets, continuity
9. (a) Noticing that E ∼
= SF , we can immediately conclude (Question 4(a)) that
E ∈ l∞ is a linear subspace. It is not closed: Let (an )n≥1 ⊂ E be a
sequence such that
−k
2
if 1 ≤ k ≤ n;
(an )k =
0
if k > n.
This sequence converges to a = (2−k )k≥1 ∈ l∞ :
n→∞
||an − a||∞ = 2−n−1 −→ 0.
Notice that a 6∈ E as (a)k = 2−k > 0 for any k ∈ N. We found a convergent
sequence contained in E which has a limit in l∞ \ E. Therefore, E is not
closed by our Proposition 45.
(b) The check that F is closed with respect component-wise addition and
scalar multiplication is elementary. Let (an )n≥1 ⊂ F be a convergent
subsequence. Let a ∈ l∞ be the limit. Then for any p ∈ N,
n→∞
0 ≤ |(a)2p | = |(a)2p − (an )2p | ≤ supk≥1 |(a)k − (an )k | = ||a − an ||∞ −→ 0.
Therefore, (a)2p = 0 for all p ∈ N and a ∈ F . We proved that any
convergent sequence in F converges to an element of F . Therefore F is
closed by Proposition 45.
(c) Let (U, || · || be a normed linear space, V ⊂ U - a finite dimensional
linear subspace. Then (V, || · ||) is a finite dimensional normed vector
space with respect to the restriction of the norm || · || to V. We know
that (V, || · ||) is Banach (Theorem 37). Let (xn )n≥1 ⊂ V be a convergent
sequence in V . Any convergent sequence is Cauchy, therefore the limit
must belong to V due to the completeness of the latter. We proved that
every convergent subsequence in V converges to an element of V , so V is
closed by Proposition 45.
PN
10. (a) For any a, b ∈ SF and λ, µ ∈ R, T (λa + µb) =
k=1 (µak + λbk ) =
PN
PN
µ k=1 ak + λ k=1 bk = µT (a) + λT (b), so the map is linear. In the
derivation, N ∈ N : ak = 0, bk = 0 for any k > N . It exists as a, b ∈ SF .
(b) Here we apply Theorem 39 of the lecture notes:
(i) Norm (1). Let (an )n≥1 ⊂ SF be a sequence:
an = (1, 1, . . . , 1, 0, 0 . . .).
| {z }
n-times
For any n, ||an ||∞ = 1, but T (an ) = n → ∞ as n → ∞. Therefore, T is
not bounded, hence NOT continuous.
(ii) Norm (2). Let (an )n≥1 ⊂ SF be a sequence:
an = (1, 1/2, . . . , 1/n, 0, 0 . . .).
P
For any n, ||an ||w = 1, but T (an ) = nk=1 (1/k) → ∞ as n → ∞. Therefore, T is not bounded, hence NOT continuous.
(iii) Norm (3). Consider any a ∈ SF P: ||a||1 ≤ 1.
P
P∞This means that
∞
∞
k=1 |ak | ≤ 1. Therefore, |T (a)| = |
k=1 ak | ≤
k=1 |ak | ≤ 1. We
proved that T is bounded, so it IS continuous.
(iv) Norm (4). Let (an )n≥1 ⊂ SF be a sequence:
an =
where Z =
1 1/2
k=1 k2
P∞
1
(1, 1/2, . . . , 1/n, 0, 0 . . .),
Z
< ∞. For any n,
1 1/2
k=1 k2
P∞ 1 1/2
k=1 k2
Pn
||an ||2 =
≤ 1,
P
but T (an ) = Z1 nk=1 (1/k) → ∞ as n → ∞. Therefore, T is not bounded,
hence NOT continuous.
(v)
This means
F : ||a||u ≤
P∞ that
P1.
P∞Norm (5). Consider any a ∈ SP
∞
∞
k=1 k|ak | ≤
k=1 |ak | ≤
k=1 ak | ≤
k=1 k|ak | ≤ 1. Therefore, |T (a)| = |
1. We proved that T is bounded, so it IS continuous.
11. (a) Linearity: For any j ∈ N, T (µa) + λb)j = (1 − j −1 )(µaj + λbj ) =
µT (a)j + λT (b)j .
P
Continuity: Consider any aP
∈ SF : ||a||U ≤ 1. This
means that ∞
k=1 |ak | ≤
P
∞
∞
−1
1. Therefore, ||T (a)||V = | k=1 (1 − k )ak | ≤ k=1 |ak | ≤ 1. We proved
that T : U → V is bounded, so it is continuous by our Thm 39.
A byproduct of the continuity proof is that ||T || ≤ 1. Let (an )n≥1 ⊂ SF
be a sequence:
an = (1, 1, . . . , 1, 0, 0 . . .).
| {z }
n-times
Then
||T an ||1 /||an ||1 =
(n −
Pn
k=1 (1/k))
→ 1, n → ∞.
n
Therefore, ||T || = 1 by the definition of sup. On other hand, for any
b ∈ SF \ {0},
∞
X
||T (b)||1 =
(1 − k −1 )|bk | < ||b||1 .
k=1
So there exists no non-zero element a of SF such that ||T a||V = ||T ||||a||U .
(b) Yes. Here is the sketch of the argument: using the definition of sup in the
operator norm, we can construct a sequence (xn )n≥1 ⊂ U , such that for
any n ∈ N, ||xn ||U = 1 and ||T (xn )||V → ||T || as n → ∞. As we proved in
the lecture, any bounded sequence in a finite dimensional normed space
has a convergent subsequence. Let (xnk )k≥1 be such a subsequence. Let
x ∈ U be its limit. Then,
0 ≤ |||T xnk ||V −||T x||V | ≤ ||T (xnk −x)||V ≤ ||T ||||xnk −x||U → 0, k → ∞.
Therefore, ||T x||V = ||T || = ||T ||||x||U .
November the 17th, 2014
Daniel Ueltschi and Oleg Zaboronski.