Functional Analysis HW #1
Sangchul Lee
October 9, 2015
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Solutions
Solution of #1.1.
• Suppose that |X | < ∞. The only possible Hausdorff topology on X is the discrete topology, and thus every
function on X is continuous. Thus C(X ) ' C X and hence
dim C(X ) = dim C X = |X | < ∞.
• Suppose that |X | = ∞. Given any n ∈ Z+ , choose a set of n distinct points X n = {x 1, · · · , x n } ⊂ X. We
claim that
Claim 1.1. The mapping α : C(X ) → C(X n ) given by α( f ) = f | X n is a surjective linear map.
Assuming this, we know that dim C(X ) ≥ C(X n ) = n for any n. Therefore dim C(X ) = ∞.
To finish the proof we are required to show Claim 1.1. It is straightforward that α is a well-defined linear
map, thus it suffices to prove that α is surjective. Since X is compact Hausdorff, it is a normal space. Then
by the Urysohn’s lemma, for each i, j ∈ {1, · · · , n} with i , j we can choose f i j ∈ C(X ) such that
f i j (x i ) = 1
Now define gi =
the claim.
Q
j: j,i
and
f i j (x j ) = 0.
f j . Then gi ∈ C(X ) and gi (x j ) = δ i j . This proves that α is surjective and hence
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Solution of #1.2.
• Suppose that X is finite-dimensional with dim X = n. Choose a basis {x 1, · · · , x n } of X and let
α : Cn → X be the vector-space isomorphism defined by α(u1, · · · , un ) = u1 x 1 + · · · + un x n . Then we
claim that
Claim 1.2. There exists a positive constant C such that
kuk ≤ Ckα(u)k
1
∀u ∈ Cn .
Assuming this, the proof is straightforward. Let ϕ : X → C be any linear functional on X . Then writing
x = α(u) for some u = (u1, · · · , un ) we have
kϕ(x)k ≤ |u1 ||ϕ(x 1 )| + · · · + |un ||ϕ(x n )|
≤ C(|ϕ(x 1 )| + · · · + |ϕ(x n )|)k xk.
Therefore kϕk < ∞ and ϕ is continuous. So it suffices to show Claim 1.2. Notice that for any u, v ∈ Cn ,
we have
kα(u) − α(v)k = kα(u − v)k ≤ |u1 − v1 |k x 1 k + · · · |un − vn |k x n k
≤ (k x 1 k + · · · + k x n k)ku − vk
and consequently α is continuous. Moreover, since the set K = {u ∈ Cn : kuk = 1} is compact and α never
vanishes on K, the map u 7→ kα(u)k attains a positive minimum δ = minu ∈K kα(u)k > 0. Then with
C = δ−1 we have
kuk ≤ Ckuk kα(u/kuk)k = Ckα(u)k
and therefore the claim is proved.
• Assume that X is infinite-dimensional. Choose a linearly independent set {x 1, x 2, · · · } ⊂ X and let ϕ be a
linear functional on X satisfying
ϕ(x n ) = nk x n k.
(The existence of such ϕ is guaranteed by the Zorn’s lemma.) Then kϕk = ∞ and hence ϕ is not continuous.
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Sketch of Solution of #1.3. Construction of such X is a very straightforward and routine work, so we only give a
sketch of idea:
Let S be the set of Cauchy sequences in M and define an equivalence relation ∼ on S by
( f n ) ∼ (gn )
k f n − gn k → 0.
iff
Then the quotient space X = S/ ∼ gives rise to a Banach space with the norm defined as follows: for any
equivalence class [( f n )] ∈ X ,
k[( f n )]k = lim k f n k.
n→∞
It is routine to check that this is a well-defined norm on X with respect to which X is complete. Then M is
isometrically embedded into X by the mapping f 7→ ( f , f , f , · · · ).
Now if Y is any Banach space in which M is dense, then we can define an isomorphism Y → X as follows:
whenever f ∈ Y is written as f = limn f n with f n ∈ M,
f 7→ [( f n )].
Indeed, this definition is a well-defined isometric isomorphism.
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Solution of #1.5.
• c0 (Z+ ) is separable since the following set
D = { f ∈ c0 (Z+ ) : f (k) ∈ Q for all k and f (k) = 0 for all sufficiently large k}
is a countable dense subset of c0 (Z+ ). Indeed, for any f ∈ c0 (Z+ ) and ε > 0,
– Choose N such that | f (k)| < ε for k > N.
– Choose g(1), · · · , g(N ) ∈ Q such that | f (k) − g(k)| < ε for 1 ≤ k ≤ N.
Extending g onto Z+ by letting g(k) = 0 for k > N. we find that g ∈ D and k f − gk < ε.
• ` 1 (Z+ ) is separable with a dense subset D as above. Indeed, for any f ∈ ` 1 (Z+ ) and ε > 0,
P
– Choose N such that k > N | f (k)| < ε/2.
– Choose g(1), · · · , g(N ) ∈ Q such that | f (k) − g(k)| < ε/2N for 1 ≤ k ≤ N.
Extending g onto Z+ by letting g(k) = 0 for k > N. we find that g ∈ D and k f − gk < ε.
• ` ∞ (Z+ ) is inseparable. Indeed, for each U ⊆ Z+ define the indicator function 1U by

 1,
1U (k) = 
 0,

k ∈U
k < U.
Then the uncountable family {1U : U ⊆ Z+ } satisfies the following condition: for any U , V ,
k1U − 1V k = 1.
In particular, there is an uncountable family of mutually disjoint open balls of radius 1/2 and thus no
countable subset of ` ∞ (Z+ ) can be dense.
• ` ∞ (Z+ ) ∗ is also inseparable by the following claim:
Claim 1.3. If X is a Banach space and X ∗ is separable, then so is X .
The following simple tweak of Hahn-Banach theorem is quite useful for our proof.
Lemma 1.4. Let M be a closed proper subspace of the Banach space X and x 0 ∈ X \ M. Then there
exists ϕ ∈ X ∗ such that
– ϕ(y) = 0 for y ∈ M,
– ϕ(x 0 ) = dist(x 0, M), and
– kϕk ≤ 1.
We defer the proof of this lemma to the next section. Assuming this, the claim is easy to verify. Indeed, let
{ϕ1, ϕ2, · · · } be a dense subset of X ∗ . (For technicality, we also assume that ϕ n , 0 for all n.) For each n,
choose x n ∈ X such that
|ϕ n (x n )|
≥ 12 kϕ n k
k xn k
Then we claim that the set D of all possible Q-linear combinations of {x 1, x 2, · · · } is dense in X . Since D
is countable, this will prove that X is separable as well.
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To this end, we assume otherwise. Then the the closure M of D in X is not equal to X . Notice also that
M is in fact a closed subspace of X . Using this, choose any x 0 ∈ X \ M and pick ϕ ∈ X ∗ as in Lemma
1.4. Clearly ϕ is not identically zero. Also we have
kϕ − ϕ n k ≥
|ϕ(x n ) − ϕ n (x n )|
|ϕ n (x n )|
=
≥ 12 kϕ n k.
k xn k
k xn k
Since {ϕ1, ϕ2, · · · } is dense in X ∗ , we can choose a subsequence (ϕ n j ) such that kϕ − ϕ n j k → 0. Then the
above inequality shows that ϕ = lim j→∞ ϕ n j = 0, a contradiction. This completes the proof.
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Sketch of Solution of #1.13. Since (X ∗ )1 is compact Hausdorff by Banach-Alaoglu theorem, a simple application
of Urysohn metrization theorem shows that (X ∗ )1 is metrizable if and only if it is second-countable.
• Assume that (X ∗ )1 is second-countable. Then X ∗ = ∪n (X ∗ )n is also second-countable and hence separable.
This implies that X is separable, by Claim 1.3.
• Assume that X is separable. Choose a dense subset D = { f 1, f 2, · · · } of X and a countable base (Uk ) of
the usual topology on C. Then we consider the set
Vn, k = {ϕ ∈ (X ∗ )1 : ϕ( f n ) ∈ Uk }.
Clearly {Vn, k }n, k ≥1 is a countable family of open sets. Now aassume that (ψα )α ∈ A is a net in (X ∗ )1 such
that limα ∈ A ψα ( f n ) = ψ( f n ) for each n. We remark that this information is completely determined by
the collection {Vn, k }n, k ≥1 . Also, by Proposition 1.21, this implies that (ψα ) w*-converges to ψ. Thus
{Vn, k }n, k ≥1 determines the topology of (X ∗ )1 and hence (X ∗ )1 is second countable.
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Solution of #1.14. Let M be the closure of N , which is a proper closed subspace of X . Utilize Lemma 1.4 to
choose ϕ ∈ X ∗ such that
• ϕ n (y) = 0 for y ∈ M,
• ϕ n (x) = 1, and
• kϕ n k ≤ 1/d.
Notice also that, for any y ∈ N , we have
kϕk ≥
|ϕ(x − y)|
1
=
.
k x − yk
k x − yk
Taking supremum over y ∈ N , it follows that kϕk ≥
1
d.
Therefore kϕk =
1
d
and the construction is done.
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Solution of #1.15. We know from linear algebra that the mapping f 7→ fˆ is linear. So it suffices to prove that
this mapping preserves norm. To this end, let f ∈ X . Then for any ϕ ∈ X ∗ with kϕk = 1,
| fˆ(ϕ)| = |ϕ( f )| ≤ kϕk k f k = k f k.
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Taking supremum over kϕk = 1, we find that k fˆk ≤ k f k. Also, by Corollary 1.27, there exists ϕ with kϕk = 1
such that ϕ( f ) = k f k. Thus we have
k fˆk ≥ | fˆ(ϕ)| = |ϕ( f )| = k f k.
Therefore the inequality is saturated and hence k fˆk = k f k.
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Solution of #1.16.
• If X is finite-dimensional, then we know from linear algebra that dim X ∗∗ = dim X . Also, from the
previous exercise we know that X is naturally embedded into X ∗∗ . Therefore X and X ∗∗ is isometrically
isomorphic and hence reflexive.
• Inseparability of c0 (Z+ ). We know that c0 (Z+ ) ∗∗ ' ` ∞ (Z+ ), which is much larger than c0 (Z+ ). Thus
c0 (Z+ ) is not reflexive.
• Inseparability of ` 1 (Z+ ). We know that ` 1 (Z+ ) is separable. If it is reflexive, then ` 1 (Z+ ) ∗∗ ' ` 1 (Z+ )
must be also separable. But by Claim 1.3, this implies that ` 1 (Z+ ) ∗ ' ` ∞ (Z+ ) is separable, which is a
contradiction. Therefore ` 1 (Z+ ) is not reflexive.
• Inseparability of C([0, 1]). Notice that for each x ∈ [0, 1], the Dirac-delta δ x : f 7→ f (x) is a bounded
linear functional on C([0, 1]):
|δ x ( f )| = | f (x)| ≤ k f k.
Moreover, for x , y we can find f ∈ C([0, 1]) such that k f k = 1, f (x) = 1 and f (y) = 0. This implies that
kδ x − δ y k ≥
|δ x ( f ) − δ y ( f )|
= 1.
kfk
Thus C([0, 1]) ∗ is inseparable and the same is true for C([0, 1]) ∗∗ . On the other hand, C([0, 1]) is separable
by a simple application of Stone-Weierstrass theorem. Therefore C([0, 1]) is not reflexive.
• Inseparability of L 1 ([0, 1]). We know that L 1 ([0, 1]) is separable. On the other hand, it is easy to modify
the argument for inseparability of ` ∞ (Z+ ) to show that L ∞ ([0, 1]) ' L 1 ([0, 1]) ∗ is inseparable. Thus
L 1 ([0, 1]) cannot be reflexive.
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Proofs
Definition 2.1. p is called a complex sublinear functional on X if the followings are satisfied:
p(x + y) ≤ p(x) + p(y)
∀x, y ∈ X
and
p(λ x) = |λ|p(x)
∀λ ∈ C, x ∈ X,
(2.1)
Lemma 2.2. Let M be a closed proper subspace of the Banach space X and x 0 ∈ X \ M. Then there exists
ϕ ∈ X ∗ such that
• ϕ(y) = 0 for y ∈ M,
• ϕ(x 0 ) = dist(x 0, M), and
• kϕk ≤ 1.
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Proof. It is easy to check that dist(x, M) = inf y ∈M k x − yk is a complex sublinear functional on X . Now let
d = dist(x 0, M) and consider the linear functional ϕ0 : Cx 0 → C given by ϕ0 (λ x 0 ) = λd.
Since |ϕ0 (λ x 0 )| = |λ|d = dist(λ x 0, M), we can follow the proof of Theorem 1.26 mutandis mutatis to
construct a linear functional ϕ that extends ϕ0 . Indeed, consider Re ϕ0 as a R-linear functional, extend this
to a R-linear functional ψ on X̃ (X considered as a R-normed space) dominated by the sublinear functional
dist(·, M), and define φ(x) = ψ(x) − iψ(ix). Then we notice that
• ϕ(x 0 ) = ϕ0 (x 0 ) = d.
• To show that ϕ is bounded and lies in the unit ball (X ∗ )1 , write ϕ(x) = reiθ and notice that
|ϕ(x)| = e−iθ ϕ(x) = ϕ(e−iθ x) = ψ(e−iθ x) ≤ dist(e−iθ x, M) = dist(x, M) ≤ k xk.
• The inequality above also shows that |ϕ(y)| ≤ dist(y, M) = 0 for y ∈ M.
This completes the proof.
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