MAST31002 Functional Analysis
Exercises 9 (week 9.11-13.11.2020)
Model solutions (suggestions)
1. Let T : `1 → `1 be the linear map T (xk) = ( kk+1xk), for x = (xk) ∈ `1.
Does there exist x = (xk) ∈ `1 for which ‖x‖1 = 1 and ‖Tx‖1 = ‖T‖?
Solution: it is clear that T is a linear map (check!). If x = (xk) ∈ `1, then
‖Tx‖1 =
∞∑
k=1
k
k + 1 |xk| ≤
∞∑
k=1
|xk| = ‖x‖1,
since 0 < k
k+1 < 1 for each k. Thus T ∈ L(`1) is a bounded linear operator
and ‖T‖ ≤ 1. Moreover, if ek = (0, . . . , 0, 1, 0, . . .) for k ∈ N (with 1 in the
k:th place), then
‖T‖ ≥ ‖Tek‖1 = ‖ k
k + 1ek‖1 =
k
k + 1 → 1
as k →∞. Thus ‖T‖ = 1.
Suppose that x = (xk) ∈ `1 satisfies ‖x‖1 = 1. Hence there is j ∈ N is
such that xj 6= 0, so that 0 < jj+1 |xj| < |xj|. Consequently
‖Tx‖1 =
∞∑
k=1
k
k + 1 |xk| <
∞∑
k=1
|xk| = ‖x‖1.
Conclusion: there is no sequence x = (xk) ∈ `1 such that ‖x‖1 = 1 and
‖Tx‖1 = ‖T‖.
2. Let E be a normed space and M ⊂ E a finite dimensional linear subspace.
Show that M is a closed subspace of E.
Suggestions: Let (x1, . . . , xn) be a normalised linear basis of M , and
(e1, . . . , en) the coordinate vector basis in Kn. Define the linear map T :
Kn →M by T (ek) = xk for k = 1, . . . , n, and show that T is a linear isomorp-
hism (Kn, ‖ · ‖1)→M . For the boundedness of T−1 verify that T is bounded
from below by using the compactness of {(c1, . . . , cn) ∈ Kn : ∑nk=1 |ck| = 1}
(Heine-Borel theorem).
Solution: define T : Kn → M by T (∑nk=1 akek) = ∑nk=1 akxk for ∑nk=1 akek ∈
Kn. By Linear Algebra T is a linear bijection Kn → M . (For instance, if
1
T (∑nk=1 akek) = ∑nk=1 akxk = 0, then ak = 0 for each k = 1, . . . , n, since
(x1, . . . , xn) be a linear basis for M .)
Moreover, T is continuous (Kn, ‖ · ‖1)→M , since by the triangle inequa-
lity in E, and ‖xk‖ = 1 for each k, we get
‖T ((
n∑
k=1
akek)‖ = ‖
n∑
k=1
akxk‖ ≤
n∑
k=1
|ak| · ‖xk‖ =
n∑
k=1
|ak| = ‖
n∑
k=1
akek‖1.
The linear map T is bounded from below so that T is a linear isomorp-
hism. Namely, the set K = {(c1, . . . , cn) ∈ Kn : ∑nk=1 |ck| = 1} is a closed and
bounded subset of (Kn, ‖ · ‖1), and hence it is compact by the Heine-Borel
theorem. (Recall that ‖ · ‖1 is equivalent to the euclidean norm in Kn.) By
metric topology the continuous function
n∑
k=1
akek 7→ ‖T (
n∑
k=1
akek)‖ = ‖
n∑
k=1
akxk‖
achieves its minimum value c > 0 onK (the minimum c > 0 since T (∑nk=1 akek) 6=
0 on K.) By homogeneity we get that
‖T (
n∑
k=1
akek)‖ ≥ c‖
n∑
k=1
akek‖1
for all ∑nk=1 akek ∈ Kn.
Finally, since completeness is preserved by linear isomorphisms, it follows
that the linear subspace M ⊂ E is complete, and hence closed in E.
Comment: we actually showed that if (M, ‖ · ‖) is an arbitrary n-dimensional
normed space, then (M, ‖·‖) is linearly isomorphic to (Kn, ‖·‖1) ≈ (Kn, ‖·‖2),
where ‖ · ‖2 is the euclidean norm. In particular, M and N are n-dimensional
normed spaces over K, then M ≈ N are linearly isomorphic. (Note that
linear isomorphism E ≈ F is an equivalence relation.)
3. Let E be a Hilbert space and (fj) ⊂ E a sequence such that
∞∑
j=1
|(f |fj)|2 <∞ for all f ∈ E.
Show that there is a constant B <∞ such that
∞∑
j=1
|(f |fj)|2 ≤ B‖f‖2 for all f ∈ E.
2
Suggestion. Define S : E → `2 by Sf = ((f |fj))j∈N for f ∈ E. Find a
sequence of bounded linear operators Sn : E → `2 such that Snf → Sf as
n → ∞ for all f ∈ E, and apply the Corollary to the Banach-Steinhaus
theorem.
Solution: define the map S : E → `2 by
Sf = ((f |fj))j∈N for f ∈ E.
By assumption the map S is defined and linear (check!). Let n ∈ N be fixed
and put
Snf = ((f |f1), (f |f2), . . . , (f |fn), 0, . . .) for f ∈ E.
Hence Snf → Sf as n→∞ for each fixed f ∈ E, since
‖Snf − Sf‖2 = ‖(0 . . . , 0, (f |fn+1), . . .)‖2 =
( ∞∑
k=n+1
|(f |fk)‖2
)1/2 → 0
as n→∞ (the remainder term of the convergent series ∑∞j=1 |(f |fj)|2). Mo-
reover, Sn is a bounded linear operator E → `2 for each n ∈ N. Hence Co-
rollary 7.5 to the Banach-Steinhaus theorem implies that S ∈ L(E, `2) is a
bounded operator, so that
‖Sf‖22 =
∞∑
j=1
|(f |fj)|2 ≤ ‖S‖2‖f‖2 for all f ∈ E.
Claim: Sn ∈ L(E, `2) is bounded for n ∈ N. (This is related to the proof of
Exercise 9:2 and its Comments.)
Namely, letMn = span({f1, . . . , fn}) ⊂ E be the linear subspace spanned
by f1, . . . , fn in E. Then the dimension m =: dim(Mn) ≤ n. Let Ŝn be the
restriction of Sn considered as an operator Mn → `2. Then Ŝn is bounded
Mn → `2. In fact, Mn is linearly isomorphic to (Km, ‖ · ‖1), so that the closed
unit ball BMn is compact set by the Heine-Borel theorem (compactness is
preserved by linear isomorphisms). Hence the continuous map x 7→ ‖Ŝnx‖2
is bounded on BMn , that is, Ŝn is a bounded operator. Finally, by Chapter
4 we know that E =Mn ⊕M⊥n . Let Pn =: PMn be the orthogonal projection
E → Mn. Hence Sn = ŜnPn is bounded E → `2 for each n, since Pn is
bounded.
3
Comment: Above we essentially verified the following useful fact: Let T :
M → F be any linear map, where M is an arbitrary finite-dimensional nor-
med space. Then T is a bounded linear operator, that is, T ∈ L(M,F ).
4. Let (ak) ⊂ R be a real sequence, such that the series
∞∑
k=1
akxk converges for all sequences (xk) ∈ c0. (*)
Show that (ak) ∈ `1. Suggestion: imitate the proof of Example 7.7 (see Func-
tional Analysis 2020 notes or the Finnish lecture notes).
Solution: For n ∈ N define fn : c0 → R by
fn(x) =
n∑
k=1
akxk for (xk) ∈ c0.
Clearly fn is a linear map, and
|fn(x)| = |
n∑
k=1
akxk| ≤
n∑
k=1
|ak| · |xk| ≤ ( max
k=1,...,n
|xk|)
n∑
k=1
|ak| ≤ (
n∑
k=1
|ak|)‖x‖∞
for all x = (xk) ∈ c0. Hence fn ∈ L(c0,R) for all n. Moreover, by assumption
f(x) =:
∞∑
k=1
akxk = lim
n→∞
n∑
k=1
akxk = lim
n→∞ fn(x)
exists for all x = (xk) ∈ c0. Since c0 is a Banach space, f ∈ L(c0,R) is a
bounded linear operator by Corollary 7.5 to the Banach-Steinhaus theorem.
In particular,
|f(x)| = |
∞∑
k=1
akxk| ≤ ‖f‖ · ‖x‖∞ for all x = (xk) ∈ c0.
Let n ∈ N be fixed and consider y = (yk) ∈ c0, where yk = |ak|ak if 1 ≤ k ≤ n
and ak 6= 0, and yk = 0 if ak = 0 or k > n. Then ‖y‖∞ ≤ 1, so by plugging
y ∈ c0 into the above estimate we get
n∑
k=1
|ak| =
∞∑
k=1
akyk ≤ |
∞∑
k=1
akyk| ≤ ‖f‖ · ‖y‖∞ ≤ ‖f‖.
Since n ∈ N was arbitrary, we get that ∑nk=1 |ak| <∞, that is, (ak) ∈ `1.
4
Conversely, if (ak) ∈ `1 and x = (xk) ∈ c0, then
∞∑
k=1
|ak| · |xk| ≤ ‖x‖∞
∞∑
k=1
|ak|
is finite, so that the scalar series∑∞k=1 akxk does converge for all x = (xk) ∈ c0.
Hence condition (*) holds precisely when (ak) ∈ `1.
Comment: this is part of the verification that the dual space c∗o can be iden-
tified with `1.
5. (Osgood’s uniform boundedness principle, 1897) Let (fn) ⊂ C(0, 1) be a
sequence of continuous maps [0, 1] → R, which are pointwise bounded, that
is, M(x) := supn∈N |fn(x)| <∞ for all x ∈ [0, 1]. Show that there is a (non-
trivial) open interval (a, b) ⊂ [0, 1] and and a uniform bound M < ∞, such
that |fn(x)| ≤M for all x ∈ (a, b) and all n ∈ N.
Suggestion: imitate the proof of the Banach-Steinhaus theorem!
Solution: for m ∈ N define
Fm = {x ∈ [0, 1] : |fn(x)| ≤ m for all n ∈ N}.
Then Fm =
⋂∞
n=1 f
−1
n ([−m,m]) is a closed set, since it follows from the con-
tinuity of fn that the inverse images f−1n ([−m,m]) are closed for all n ∈ N
(metric topology course). Moreover, if x ∈ [0, 1] is arbitrary, then by assump-
tion M(x) := supn∈N |fn(x)| < ∞, so that x ∈ Fm once m ≥ M(x). This
means that
[0, 1] =
∞⋃
m=1
Fm.
Since [0, 1] is complete, it follows from Corollary 7.2 to Baire’s theorem that
there is m0 ∈ N such that Fm0 has an interior point x0. Hence there is r > 0
such that the interval (a, b) =: (x0 − r, x0 + r) ⊂ Fm0 . By definition we then
have
|fn(x)| ≤M =: m0 for all n ∈ N and all x ∈ (a, b).
Comment: Osgood’s uniform boundedness principle (1897) was an early "ver-
sion" of the Uniform Boundedness principle.
5

欢迎咨询51作业君