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.

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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.

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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.

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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.

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