myweb.ttu.edu/bban
Real Analysis
Byeong Ho BanMathematics and Statistics
Texas Tech University
Chapter 5. Elements of Functional Analysis (Last update : March 30, 2018)
1. If X is a normed vector space over K (= R or C), then addition and scalar multiplication are continuousfrom X ×X and K ×X to X . Moreover, the norm is continuous from X to [0,∞); in fact, |‖x‖− ‖y‖| ≤‖x− y‖.
Proof. (Feb 3rd 2018)
N The addition is continuous.
Let ε > 0 be given. And let δ = ε2. Then, for any (x, y) ∈ X × X , observe the below.
max(‖x− v‖, ‖y − w‖) = ‖(x− v, y − w)‖ = ‖(x, y)− (v, w)‖ < δ =ε
2
=⇒ ‖(x+ y)− (v + w)‖ = ‖(x− v) + (y − w)‖ ≤ ‖x− v‖+ ‖y − w‖ < ε
2+ε
2= ε
Therefore, the addition operation is continuous.
N The Scalar multiplication is continuous.
Let ε > 0 be given. Then, for any (α, x) ∈ K ×X , let δ = min( ε3‖x‖ ,
ε3|α| , 1,
ε3) and observe the below.
max(|α− β|, ‖x− y‖) = ‖(α− β, x− y)‖ = ‖(α, x)− (β, y)‖ < δ = min(ε
3‖x‖,ε
3|α|, 1,
ε
3)
=⇒ ‖αx− βy‖ ≤ ‖αx− βx‖+ ‖βx− βy‖ = |α− β|‖x‖+ |β|‖x− y‖
≤ |α− β|‖x‖+ |α|‖x− y‖+ |α− β|‖x− y‖ < ε
3+ε
3+ ‖x− y‖ < ε
N The norm operation is continuous.
Let ε > 0 and x ∈ X be given. And let δ = ε and, for any y ∈ X , by triangular inequality, note thebelow.
‖x‖ ≤ ‖x− y‖+ ‖y‖ =⇒ ‖x‖ − ‖y‖ ≤ ‖x− y‖‖y‖ ≤ ‖x− y‖+ ‖x‖ =⇒ ‖y‖ − ‖x‖ ≤ ‖x− y‖
Thus, |‖x‖ − ‖y‖| ≤ ‖x− y‖.
Then observe the below.1
2
‖x− y‖ < δ = ε =⇒ |‖x‖ − ‖y‖| ≤ ‖x− y‖ < ε
Therefore, the mapping ‖ · ‖ : X → K is continuous.�
2. L(X ,Y) is a vector space and the function ‖ · ‖ defined by (5.3) is a norm on it. In particular, the threeexpression on the right of (5.3) are always equal.
Proof. (Feb 4th 2018)
N L(X ,Y) is a vector space.
Note that L(X ,Y) be the space of all bounded linear maps from X to Y over a field K(= R or C). Forany T,W ∈ L(X ,Y) and α, β ∈ K, noting that there exist CT ≥ 0 and CW ≥ 0 such that ‖Tx‖ ≤ CT‖x‖and ‖Wx‖ ≤ CW‖x‖ for all x ∈ X ,observe the below.
‖(αT + βW )x‖ = ‖αTx+ βWx‖ ≤ |α|‖Tx‖+ |β|‖Wx‖ ≤ (|α|CT + |β|CW )‖x‖ ∀x ∈ XClearly, αT + βW is linear map, thus, by the relation above, L(X ,Y) is a vector space.
N The three expressions on the right of (5.3) are always equal.
For convenience, let’s define the sets for the expressions as below.
A = {‖Tx‖ : ‖x‖ = 1}
B := {‖Tx‖‖x‖
: x 6= 0}
D := {C : ‖Tx‖ ≤ C‖x‖ for all x}Clearly, A ⊂ B, so supA ≤ supB.
And observe that if x = 0, ‖Tx‖ ≤ C‖x‖ is always true for any C ∈ [0,∞) which means the below.
{C : ‖Tx‖ ≤ C‖x‖ for all x} = {C : ‖Tx‖ ≤ C‖x‖ for all x 6= 0}Let C ≥ 0 be any constant satisfying ‖Tx‖ ≤ C‖x‖ for all x ∈ X with x 6= 0. Observe the below.
‖Tx‖ ≤ C‖x‖ =⇒ ‖Tx‖‖x‖
≤ C
Since C is arbitrary in D,
‖Tx‖‖x‖
≤ inf D ∀x 6= 0 =⇒ supB = supx 6=0
‖Tx‖‖x‖
≤ inf D
Thus, supB ≤ inf D.
For any nonzero vector x ∈ X , observe the below.
‖Tx‖‖x‖
=
∥∥∥∥T x
‖x‖
∥∥∥∥ ∈ A =⇒ ‖Tx‖‖x‖
≤ supA =⇒ ‖Tx‖ ≤ supA‖x‖ =⇒ inf D ≤ supA
To sum up,
supA ≤ supB ≤ inf D ≤ supA
=⇒ supA = supB = inf D
3
Therefore, the three expressions are equivalent.
N The function ‖ · ‖ defined by (5.3) is a norm.
Observe that ‖T‖ = 0 ⇐⇒ ‖Tx‖‖x‖ = 0 ∀x ∈ X \ {0} ⇐⇒ Tx = 0 ∀x ∈ X \ {0} ⇐⇒ T = 0
For any T,W ∈ L(X ,Y),
‖T +W‖ = supx6=0
‖(T +W )x‖‖x‖
≤ supx 6=0
‖Tx‖+ ‖Wx‖‖x‖
≤ supx 6=0
‖Tx‖‖x‖
+ supx 6=0
‖Wx‖‖x‖
= ‖T‖+ ‖W‖
Lastly, for any T ∈ L(X ,Y) and α ∈ K,
‖αT‖ = supx 6=0
‖αTx‖‖x‖
= supx 6=0
|α|‖Tx‖‖x‖
= |α| supx 6=0
‖Tx‖‖x‖
= |α|‖T‖
Therefore, ‖ · ‖ defined by (5.3) is a norm. �
3. Complete the proof of Proposition 5.4.Proposition 5.4If Y is complete, so is L(X ,Y).
So prove that T ∈ L(X ,Y)(in fact, ‖T‖ = limn→∞ ‖Tn‖) and that ‖Tn − T‖ → 0.
Proof. (Feb 4th 2018)
Note that we have defined T : X → Y as below.
Tx = limn→∞
Tnx
For any x, y ∈ X and any α ∈ K, observe the below.
T (x+ y) = limn→∞
Tn(x+ y) = limn→∞
[Tnx+ Tny] = limn→∞
Tnx+ limn→∞
Tny = Tx+ Ty
T (αx) = limn→∞
Tn(αx) = limn→∞
αTnx = α limn→∞
Tnx = αTx
Therefore, T is linear.
Also, since {Tnx}∞n=1 is Cauchy, there is N ∈ N such that
‖Tnx− Tmx‖ < ‖x‖ ∀x ∈ XSince ‖ · ‖ is continuous map, letting m→∞,
‖Tnx− Tx‖ < ‖x‖ ∀x ∈ X=⇒ ‖Tx‖ < ‖x‖+ ‖Tnx‖ ≤ (1 + Cn)‖x‖
where ‖Tnx‖ ≤ Cn‖x‖ for any x ∈ X and ∀n ∈ N.Therefore, T ∈ L(X ,Y).In fact, observe the below.
‖T‖ = supx 6=0
‖Tx‖‖x‖
= supx 6=0
‖ limn→∞ Tnx‖‖x‖
= supn6=0
limn→∞ ‖Tnx‖‖x‖
= limn→∞
supx 6=0
‖Tnx‖‖x‖
= limn→∞
‖Tn‖
Lastly, observe the below.
4
limn→∞
‖Tn − T‖ = limn→∞
supx 6=0
‖(Tn − T )x‖‖x‖
= supx 6=0
limn→∞
‖Tnx− Tx‖‖x‖
= supx6=0
0
‖x‖= 0
Therefore, ‖Tn − T‖ → 0�
4. If X ,Y are normed vector spaces, the map (T, x) 7→ Tx is continuous from L(X ,Y) × X to Y. (Thatis, if Tn → T and xn → x, then Tnxn → Tx.)
Proof. (Feb 4th 2018)
Let {Tn} ⊂ L(X ,Y) and {xn} ⊂ X be sequences such that
limn→∞
Tn = T limn→∞
xn = x
Thus, for given ε > 0 there is N1, N2 ∈ N such that
‖Tn − T‖ <ε
2 ‖x‖∀n > N1 ‖xn − x‖ <
ε
2 ‖T‖∀n > N2
And, letting N = max(N1, N2), for all n ≥ N , observe the below.
‖TnxnTx‖ ≤ ‖Tnxn + Txn‖+ ‖Txn − Tx‖ ≤ ‖T − Tn‖ ‖x‖+ ‖T‖ ‖xn − x‖ < ε
Therefore,
limn→∞
Tnxn = Tx
�
5. If X is a normed vector space, the closure of any subspace of X is a subspace.
Proof. (Feb 4th 2018)
Let X be a normed vector space over a field K and Z be any subspace of X . Let x ∈ X and y ∈ accZ.Then there exists {yn} ⊂ Z such that yn → y. And for any α, β ∈ K,
αx+ βy = limn→∞
(αx+ βyn) ∈ Z ∵ {αx+ βyn}∞n=1 ⊂ Z
And if x, y ∈ accZ, there exist {xn}, {yn} ⊂ Z such that xn → x and yn → y.
αx+ βy = limn→∞
(αxn + βyn) ∈ Z {αxn + βyn}∞n=1 ⊂ Z
If x, y ∈ Z, clearly, αx+ βy ∈ Z ⊂ Z.Therefore, Z is a subspace of X . �
6. Suppose that X is a finite-dimensional vector space. Let e1, . . . , en be a basis for X , and define‖Σn
1ajej‖1 = Σn1 |aj|.
a. ‖ · ‖1 is a norm on X .b. The map (a1, . . . , an) 7→ Σn
1ajej is continuous from Kn with the usual Euclidean topology to X withthe topology defined by ‖ · ‖1.
c. {x ∈ X : ‖x‖1 = 1} is compact in the topology defined by ‖ · ‖1.d. All norms on X are equivalent. (Compare any norm to ‖ · ‖1)
5
Proof. (Feb 4th 2018)
Let X be a vector space over a field K.
a. For any Σnj=1ajej,Σ
nj=1bjej ∈ X and α ∈ K, observe the below.∥∥∥∥∥
n∑j=1
ajej +n∑j=1
bjej
∥∥∥∥∥1
=
∥∥∥∥∥n∑j=1
(aj + bj)ej
∥∥∥∥∥1
=n∑j=1
|aj + bj| ≤n∑j=1
|aj|+n∑j=1
|bj|
=
∥∥∥∥∥n∑j=1
ajej
∥∥∥∥∥1
+
∥∥∥∥∥n∑j=1
bjej
∥∥∥∥∥1∥∥∥∥∥α
n∑j=1
ajej
∥∥∥∥∥1
=
∥∥∥∥∥n∑j=1
αajej
∥∥∥∥∥1
=n∑j=1
|αaj| = |α|n∑j=1
|aj| = |α|
∥∥∥∥∥n∑j=1
ajej
∥∥∥∥∥1∥∥∥∥∥
n∑j=1
ajej
∥∥∥∥∥1
= 0 ⇐⇒n∑j=1
|aj| = 0 ⇐⇒ aj = 0 ∀j ⇐⇒n∑j=1
ajej = 0
Therefore, ‖·‖1 is a norm.
b. Let ε > 0 be given and let δ = εn. Then, for any (a1, . . . , an), (b1, . . . , bn) ∈ X observe the below.
‖(a1, . . . , an)− (b1, . . . , bn)‖ = ‖(a1 − b1, . . . , an − bn)‖ < δ
=⇒
∥∥∥∥∥n∑j=1
ajej −n∑j=1
bjej
∥∥∥∥∥1
=n∑j=1
|aj − bj| ≤ n max1≤j≤n
|aj − bj| ≤ n‖(a1 − b1, . . . , an − bn)‖ < ε
Therefore, the mapping is continuous.
c. Let A = {x ∈ X : ‖x‖1 = 1} and ‖ · ‖1 = f . Then note that ‖x‖1 ≤ 1 for all x ∈ A. Thus, Ais bounded. Also, note that f−1({1}) = A and {1} is closed since f is continuous. Thus, A is closed.Therefore, A is closed and bounded, so it is compact.
d. Let ‖ · ‖ be a given norm on X . And observe the below.∥∥∥∥∥n∑j=1
ajej
∥∥∥∥∥ ≤ max1≤j≤n
‖ej‖n∑j=1
|aj| ≤ max1≤j≤n
‖ej‖
∥∥∥∥∥n∑j=1
ajej
∥∥∥∥∥1
Conversely, due to c, A is compact. Let x ∈ X be a vector such that ‖x‖ = minz∈A ‖z‖ since ‖ · ‖ iscontinuous. Observe the below.
∥∥∥∥∥n∑j=1
ajej
∥∥∥∥∥1
‖x‖ ≤
∥∥∥∥∥n∑j=1
ajej
∥∥∥∥∥1
∥∥∥∥∥∥n∑j=1
ajej∥∥∥∑nj=1 ajej
∥∥∥1
∥∥∥∥∥∥ =
∥∥∥∥∥n∑j=1
ajej
∥∥∥∥∥Therefore, all norms on X are equivalent.
�
7. Let X be a Banach space.
6
a. If T ∈ L(X ,X ) and ‖I − T‖ < 1 where I is the identity operator, then T is invertible; in fact, theseries Σ∞0 (I − T )n converges in L(X ,X ) to T−1.
b. If T ∈ L(X ,X ) is invertible and ‖S − T‖ < ‖T−1‖−1, then S is invertible. Thus, the set of invertibleoperators is open in L(X ,X ).
Proof. (Feb 5th 2018)
a. First, note that ∃α such that ‖I − T‖ ≤ α < 1, and observe the below.∥∥∥∥∥∞∑n=0
(I − T )n
∥∥∥∥∥ ≤∞∑n=0
‖I − T‖n <∞∑n=0
αn <∞
Therefore, the series converges absolutely. Note that L(X ,X ) is complete since X is complete. Thus,the series below is convergent.
L =let
∞∑n=0
(I − T )n = limn→∞
n∑k=0
(I − T )k ∈ L(X ,X )
Now, observe the below.
LT = limn→∞
n∑k=0
(I − T )kT = limn→∞
n∑k=0
(I − T )k(I + (T − I)) = limn→∞
[n∑k=0
(I − T )k − (I − T )k+1
]= lim
n→∞
[I − (I − T )n+1
]TL = lim
n→∞
n∑k=0
T (I − T )k = limn→∞
n∑k=0
(I + (T − I))(I − T )k = limn→∞
[n∑k=0
(I − T )k − (I − T )k+1
]= lim
n→∞
[I − (I − T )n+1
]Since
limn→∞
∥∥I − [I − (I − T )n+1]∥∥ = lim
n→∞
∥∥(I − T )n+1∥∥ ≤ lim
n→∞‖I − T‖n+1 < lim
n→∞αn+1 = 0
, LT = TL = I which means T is bijection and L = T−1. Lastly, observe there is C ≥ 0 such the below.∥∥T−1x∥∥ ≤ C ‖x‖ ∀x ∈ X
Observe the below.
‖x‖ =∥∥T−1Tx
∥∥ ≤ C ‖Tx‖ ∀x ∈ XTherefore, T is invertible.
b. Suppose that T ∈ L(X ,X ) is invertible and ‖S − T‖ < ‖T−1‖−1. And observe the below.∥∥I − ST−1
∥∥ ≤ ‖T − S‖∥∥T−1∥∥ =‖S − T‖‖T−1‖−1 < 1
Since ST−1 ∈ L(X ,X ), by a, ST−1 is invertiable, so TS−1 ∈ L(X ,X ) which means S−1 = T−1(TS−1) ∈L(X ,X ). Thus, S is bijection. Also, there is CST−1 ≥ 0 such that ‖ST−1x‖ ≥ CST−1 ‖x‖ for all x ∈ X .
Since T is invertible, there is CT ≤ 0 such that ‖Tx‖ ≥ C ‖x‖ for all x ∈ X . It implies the below.
‖Sx‖ =∥∥ST−1Tx
∥∥ ≥ CST−1 ‖Tx‖ ≥ CST−1C ‖x‖ ∀x ∈ XTherefore, S is invertible. �
7
8. Let (X,M) be a measurable space, and let M(X) be the space of complex measures on (X,M). Then‖µ‖ = |µ|(X) is a norm on M(X) that makes M(X) into a Banach space. (Use Theorem 5.1.)
Proof. (Feb 5th 2018)
N ‖µ‖ = |µ|(X) is a norm.(→)For any µ, ν ∈M(X) and α ∈ C, observe the below.
‖µ+ ν‖ = |µ+ ν|(X) ≤ |µ|(X) + |ν|(X) = ‖µ‖+ ‖ν‖ ∵ Prop. 3.14
‖αµ‖ = |αµ|(X) = |α||µ|(X) = |α| ‖µ‖‖µ‖ = 0 ⇐⇒ |µ|(X) = 0 ⇐⇒ µ ≡ 0
Thus, the operation ‖·‖ is a norm on M(X), so M(X) is a normed vector space.
N M(X) is a complete space.(→)Let {µn}∞n=1 ⊂M(X) is a sequence such the below.
∞∑n=1
‖µn‖ <∞
Let
µ =∞∑n=1
µn
Then, for a sequence of disjoint sets {Xn}∞n=1 ⊂M, observe the below.
µ(∅) =∞∑n=1
µn(∅) =∞∑n=1
0 = 0
µ(∞⋃k=1
Xk) =∞∑n=1
µn(∞⋃k=1
Xk) =∞∑n=1
∞∑k=1
µn(Xk) =∞∑k=1
∞∑n=1
µn(Xk) =∞∑k=1
µ(Xk)
Therefore, µ ∈ M(X), so the series converges in M(X). Thus, by Proposition 5.1, M(X) is complete,so is Banach space.
�
9. Let Ck([0, 1]) be the space of functions on [0, 1] possessing continuous derivatives up to order k on [0, 1],including one-sided derivatives at the endpoints.
a. If f ∈ C([0, 1]), then f ∈ Ck([0, 1]) iff f is k times continuously differentiable on (0, 1) and limx↘0 f(j)(x)
and limx↗1 f(j)(x) exist for j ≤ k. (The mean value theorem is useful.)
b. ‖f‖ =∑k
0
∥∥f (j)∥∥u
is a norm on Ck([0, 1]) that makes Ck([0, 1]) into a Banach space. (Use induction
on k. The essential point is that if {fn} ⊂ C1([0, 1]), fn → f uniformly, and f ′n → g uniformly, thenf ∈ C1([0, 1]) and f ′ = g. The easy way to prove this is to show that f(x)− f(0) =
∫ x0g(t)dt.)
Proof. (Feb 13rd 2018)
a. If f ∈ Ck([0, 1]), then clearly, by definition, f is k times continuously differentiable on [0, 1], so on(0, 1). Also, it is clear that limx↘0 f
(j)(x) and limx↗1 f(j)(x) exist for j ≤ k from the definition.
8
Conversely, suppose that f is k times continuously differentiable on (0, 1) and limx↘0 f(j)(x) and limx↗1 f
(j)(x)exist for j ≤ k. We need to verify if f is differentiable on the end points. Let’s define two values as below.
L := limx↘0
f (j)(x) R := limx↗1
f (j)(x) ∀j
Let ε > 0 be given. Then there exist δL > 0 and δR > 0 such that
x− 0 < δL =⇒ |f (j)(x)− L| < ε
1− x < δR =⇒ |f (j)(x)− L| < ε
By mean value theorem, ∀x ∈ (0, δL) and ∀x ∈ (1− δR, 1) there exist xL ∈ (0, δL) and xR ∈ (1− δR, 1)such that
f (j−1)(x)− f (j−1)(0)
x− 0= f (j)(xL)
f (j−1)(x)− f (j−1)(1)
x− 1= f (j)(xR)
Therefore, ∀x ∈ (0, δL) and ∀x ∈ (1− δR, 1)∣∣∣∣f (j−1)(x)− f (j−1)(0)
x− 0− L
∣∣∣∣ =∣∣f (j)(xL)− L
∣∣ < ε
and
∣∣∣∣f (j−1)(x)− f (j−1)(1)
x− 1−R
∣∣∣∣ =∣∣f (j)(xR)−R
∣∣ < ε
Thus, each f (j) is differentiable on end points of [0, 1].
b. The given ‖·‖ is clearly a norm as can be observed as below.
For any f, g ∈ Ck([0, 1]) and λ ∈ C,
‖λf‖ =k∑j=0
∥∥λf (j)∥∥u
=k∑j=0
|λ|∥∥f (j)
∥∥u
= |λ| ‖f‖
‖f + g‖ =k∑j=0
∥∥f (j) + g(j)∥∥u≤
k∑j=0
∥∥f (j)∥∥u
+k∑j=0
∥∥g(j)∥∥u
= ‖f‖+ ‖g‖
‖f‖ = 0 ⇐⇒∥∥f (j)
∥∥u
= 0 ∀j ⇐⇒ f = 0
Thus, ‖·‖ is a norm.
Now we want to check if Ck([0, 1]) is a Banach Space with respect to the norm ‖·‖. So, let {fn}∞n=1 ⊂Ck([0, 1]) is a Cauchy sequence with the given norm. �
10. Let L1k([0, 1]) be the space of all f ∈ Ck−1([0, 1]) such that f (k−1) is absolutely continuous on [0, 1]
(and hence f (k) exists a.e. and is in L1([0, 1])). Then ‖f‖ =∑k
0
∫ 1
0|f (j)(x)|dx is a norm on L1
k([0, 1]) thatmakes L1
k([0, 1]) into a Banach space.(See Exercise 9 and its hint.)
Proof. (March 8th 2018)
(1) ‖f‖ =∑k
0
∫ 1
0|f (j)(x)|dx is a norm on L1
k([0, 1])( =⇒ )
9
‖λf‖ =k∑j=0
∫ 1
0
|λf (j)(x)|dx =k∑j=0
∫ 1
0
|λ||f (j)(x)|dx = |λ|k∑j=0
∫ 1
0
|f (j)(x)|dx = |λ| ‖f‖ ∀λ ∈ C
‖f + g‖ =k∑0
∫ 1
0
|f (j)(x) + g(j)(x)|dx ≤k∑j=0
{∫ 1
0
|f (j)(x)|dx+
∫ 1
0
|g(j)(x)|dx}
=k∑j=0
∫ 1
0
|f (j)(x)|dx+k∑j=0
∫ 1
0
|g(j)(x)|dx
= ‖f‖+ ‖g‖ ∀f, g ∈ L1k([0, 1])
‖f‖ = 0 ⇐⇒k∑0
∫ 1
0
|f (j)(x)|dx = 0
⇐⇒ f (j) ≡ 0 ∀j ≤ k
⇐⇒ f ≡ 0
(2) L1k([0, 1]) is a Banach Space with the given norm.
( =⇒ )First of all, note that
‖f‖ =k∑k=0
∫ 1
0
|f (j)(x)|dx =k∑j=0
∥∥f (j)∥∥L1
Suppose that {fn} ⊂ L11([0, 1]) is a Cauchy sequence. And observe the below.
‖fn − fm‖ = ‖f ′n − f ′m‖L1([0,1]) + ‖fn − fm‖L1([0,1])
Note that each fn is in L1 since fn is continuous on compact interval.Thus, {fn} and {f ′n} are also Cauchy in L1([0, 1]). Since L1([0, 1]) is a Banach space, ∃f, g ∈ L1([0, 1])
such that fn → f and f ′n → g in L1([0, 1]).Since each fn is absolutely continuous,
fn(x)− fn(0) =
∫ x
0
f ′ndm =
∫ x
0
f ′n(y)dy
Let ε > 0 be given and N = max(N1, N2) where N1, N2 ∈ N satisfies the following condition .
‖fn − f‖L1 <ε
4∀n ≥ N1 ‖f ′n − g‖L1 <
ε
4∀n ≥ N2
Then observe the below.
∥∥∥∥f(x)− f(0)−∫ 1
0
g(y)dy
∥∥∥∥L1
≤ ‖f(x)− f(0)− fN(x) + fN(0)‖L1 +
∥∥∥∥∫ 1
0
f ′N(y)dy −∫ 1
0
g(y)dy
∥∥∥∥L1
<ε
4+ ‖‖f ′N − g‖L1‖L1 =
ε
4+ ‖f ′N − g‖L1 <
ε
2
10
Since ε > 0 is arbitrary,∥∥∥∥f(x)− f(0)−∫ x
0
g(y)dy
∥∥∥∥L1
= 0 =⇒∫ x
0
f ′(y)dy = f(x)− f(0) =
∫ x
0
g(y)dy a.e.
Therefore, f ′ = g a.e., so
‖f − fn‖ = ‖f − fn‖L1 + ‖f ′ − f ′n‖L1 ≤ε
4+ ‖f ′ − g‖L1 + ‖g − f ′n‖L1 < ε ∀n ≥ N
Therefore, L11([0, 1]) is a Banach space.
Now suppose that L1k([0, 1]) is a Banach space. And we need to show L1
k+1([0, 1]) is a Banach space.Suppose {fn} is a Cauchy sequence with respect to the norm
‖f‖ =k+1∑j=0
∥∥f (j)∥∥L1
Again, since L1([0, 1]) is a Banach Space and {f (j)n } ⊂ L1
k+1([0, 1]) ⊂ L1k([0, 1]) is a Cauchy sequence in
L1([0, 1]),
∀j ∃gj such that f (j)n → gj
Therefore, by inductive assumption, gj = f (j) ∀j ≤ k.Then with the same procedure with the above, since f (k+1) is absolutely continuous, gk+1 = f (k+1) a.e.,
so ∃M ∈ N such that ‖fn − f‖ < ε ∀n ≥M . In other words, fn → f in L1k+1([0, 1]), so it is a Banach Space.
In conclusion, ∀k ∈ N, L1k([0, 1]) is a Banach space. �
11. If 0 < α ≤ 1, let Λα([0, 1]) be the space of Holder continuous functions of exponent α on [0, 1]. Thatis, f ∈ Λα([0, 1]) iff ‖f‖Λα
<∞, where
‖f‖Λα= |f(0)|+ sup
x,y∈[0,1],x 6=y
|f(x)− f(y)||x− y|α
a. ‖·‖Λαis a norm that makes Λα([0, 1]) into a Banach space.
b. Let λα([0, 1]) be the set of all f ∈ Λα([0, 1]) such that
|f(x)− f(y)||x− y|α
→ 0 as x→ y ∀y ∈ [0, 1]
If α < 1, λα([0, 1]) is an infinite-dimensional closed subspace of Λα([0, 1]). If α = 1, λα([0, 1]) containsonly constant functions.
Proof. (Feb 24th 2018)
a.N ‖·‖Λα
is a norm.
11
( =⇒ )
‖f‖Λα= 0 ⇐⇒ |f(0)|+ sup
x,y∈[0,1],x 6=y
|f(x)− f(y)||x− y|α
= 0
⇐⇒ f(0) = 0 ∧ |f(x)− f(y)||x− y|α
= 0 ∀x, y ∈ [0, 1] with x 6= y
⇐⇒ |f(0)| = 0 ∧ |f(x)− f(y)| = 0 ∀x, y ∈ [0, 1]
⇐⇒ f(x) = 0 ∀x ∈ [0, 1]
⇐⇒ f ≡ 0 on [0, 1]
For any f, g ∈ Λα([0, 1]),
‖f + g‖Λα= |f(0) + g(0)|+ sup
x,y∈[0,1],x 6=y
|f + g(x)− f + g(y)||x− y|α
≤ |f(0)|+ |g(0)|+ supx,y∈[0,1],x 6=y
|f(x)− f(y)||x− y|α
+ supx,y∈[0,1],x 6=y
|g(x)− g(y)||x− y|α
= ‖f‖Λα+ ‖g‖Λα
For any f ∈ Λα([0, 1]) and β ∈ C,
‖βf‖Λα= |βf(0)|+ sup
x,y∈[0,1],x 6=y
|βf(x)− βf(y)||x− y|α
= |β||f(0)|+ |β| supx,y∈[0,1],x 6=y
|f(x)− f(y)||x− y|α
= |β|
(|f(0)|+ sup
x,y∈[0,1],x 6=y
|f(x)− f(y)||x− y|α
)= |β| ‖f‖Λα
N(Λα([0, 1]), ‖‖Λα
)is a Banach space.
( =⇒ ) Let {fn}∞n=1 ⊂ Λα be a Cauchy Sequence. Then, for given ε > 0, ∃N ∈ N such that
n,m > N =⇒ ‖fn − fm‖Λα= |fn(0)− fm(0)|+ sup
x,y∈[0,1],x 6=y
|fn(x) + fm(x)− fn(y)− fm(y)||x− y|α
< ε
Which means
n,m > N =⇒ |fn(x)− fm(0)| < ε
Thus, {fn(0)}∞n=1 is a Cauchy sequence in C, so there is f(0) such that
limn→∞
fn(0) = f(0)
12
Also, noting that for any x ∈ [0, 1] there is M such that
n,m > M =⇒ |fn(0)− fm(0)|+ supx,y∈[0,1],x 6=y
|fn(x) + fm(x)− fn(y)− fm(y)||x− y|α
<ε
2and |fn(0)− fm(0)| < ε
2
=⇒ |fn(x)− fm(x)| − |fn(0)− fm(0)| ≤ |fn(x) + fm(x)− fn(0)− fm(0)| ≤ |x|α ε2≤ ε
2=⇒ |fn(x)− fm(x)| < ε
Thus, there is f(x) such that
limn→∞
fn(x) = f(x) ∀x ∈ [0, 1]
Clearly, f ∈ Λα([0, 1]) since for any x, y ∈ [0, 1] there is N ∈ N such that
|f(x) + fN(x)− f(y)− fN(y)||x− y|α
≤ |f(x)− fN(x)|+ |fN(y)− f(y)||x− y|α
<|x− y|α + |x− y|α
|x− y|α= 1
Thus,
|f(x)− f(y)||x− y|α
≤ |fN(x)− fN(y) + f(x)− f(y)|+ |fN(x)− fN(y)||x− y|α
<1 + |fN(x)− fN(y)|
|x− y|α<∞
Similarly,
|f(0)| ≤ |fN(0)− f(0)|+ |fN(0)| ≤ 1 + |fN(0)| <∞
Therefore, f ∈ Λα([0, 1]), so Λα([0, 1]) is a Banach Space.
b.
N When α < 1
Let f ∈ acc(λα)([0, 1]), then there is a sequence {fn} ⊂ λα([0, 1]) such that
limn→‖fn − f‖Λalpha
= 0
And observe the below.
limx→y
|f(x)− f(y)||x− y|α
= limx→y
limn→∞ |fn(x)− fn(y)||x− y|α
= limn→∞
limx→y
|fn(x)− fn(y)||x− y|α
= limn→0
0
= 0
Thus, f ∈ λα([0, 1]) so λα([0, 1]) is closed.
N When α = 1
13
Note that for any f ∈ λ1([0, 1]) and any y ∈ [0, 1]
limx→y
|f(x)− f(y)||x− y|
= f ′(y) = 0
And it means that f is constant function. Thus, λ1([0, 1]) consists only of constant functions. �
12. Let X be a normed vector space and M a proper closed subspace of X .
a. ‖x+M‖ = infy∈M ‖x+ y‖ is a norm on X
b. For any ε > 0 there exists x ∈X such that ‖x‖ = 1 and ‖x+M‖ ≥ 1− ε
c. The projection map π(x) = x+M from X to X /M has norm 1.
d. If X is complete, so is X /M. (Use Theorem 5.1)
e. The topology defined by the quotient norm is the quotient topology as defined in Exercise 28 in §4.2.
Proof. (Feb 25th 2018)
a.
‖x+M‖ = 0 ⇐⇒ infy∈M‖x+ y‖ = 0
⇐⇒ ∃y ∈M such that y = −x⇐⇒ x ∈M⇐⇒ x+M = 0 +M
For ∀x ∈X /M and ∀a ∈ C
‖ax+M‖ = infy∈M‖ax+ y‖ = |a| inf
y∈M
∥∥∥x+y
a
∥∥∥ = |a| infy∈M‖x+ y‖ = |a| ‖x+M‖
∀x, y ∈X /M,
‖x+ y +M‖ = infz∈M‖x+ y + z‖
≤ infz∈M
∥∥∥x+z
2
∥∥∥+ infz∈M
∥∥∥y +z
2
∥∥∥= inf
z∈M‖x+ z‖+ inf
z∈M‖y + z‖
= ‖x+M‖+ ‖y +M‖
Therefore, ‖x+M‖ is a norm on X /M.
b.
14
If ε ≥ 1, it is trivial. Thus, let’s assume that 1 > ε > 0 be given. Then, for x ∈ X /M observe thebelow.
‖x+M‖ ≤ ‖x+M‖1− ε
=⇒ ∃y ∈M such that ‖x+ y‖ ≤ ‖x+M‖1− ε
=‖(x+ y) +M‖
1− ε
=⇒ 1− ε ≤∥∥∥∥ x+ y
‖x+ y‖+M
∥∥∥∥And
∥∥∥ x+y‖x+y‖
∥∥∥ = 1.
c.
Observe that π is linear.∀a, b ∈ C ∀x, y ∈X
π(ax+ by) = ax+ by +M = ax+M+ by +M = a(x+M) + b(y +M) = π(ax) + π(by)
And it is bounded, for any x ∈X , due to below.
‖x+M‖ = infy∈M‖x+ y‖ ≤ ‖x+ 0‖ = ‖x‖ =⇒ ‖π‖ = sup
x 6=0
‖x+M‖‖x‖
≤ 1
Also, from b.
‖π‖ = sup‖x‖=1
‖x+M‖ ≥ 1
Therefore, ‖π‖ = 1.
d.
Suppose X is complete and that∞∑n=1
(xn +M)
is a absolutely convergent series in X /M. Then, there exists y ∈M such that
∞∑n=1
‖xn + y‖ <∞∑n=1
(‖xn +M‖+
1
2n
)<∞
Thus, {xn + y}∞n=1 ⊂ X is a sequence such that the series corresponding to the sequence is absolutelyconvergent. Since X is complete, the series converges to (let)x ∈X .
∞∑n=1
(xn +M) =
(∞∑n=1
xn
)+M =
(∞∑n=1
(xn + y)−∞∑n=1
y
)+M = x+M
Thus, the absolutely convergent series converges, so X /M is complete.
15
e.
I will come back after solving the corresponding problem. �
13. If ‖·‖ is a seminorm on the vector space X , let M = {x ∈X : ‖x‖ = 0}.Then M is a subspace, andthe map x+M 7→ ‖x‖ is a norm on X /M.
Proof. (Feb 25th 2018)
NM is a subspace.
Clearly, 0 ∈M.For any a ∈ C and any x ∈M,
‖ax‖ = |a| ‖x‖ = 0
Thus, ax ∈M
And for any x, y ∈X ,
0 ≤ ‖x+ y‖ ≤ ‖x‖+ ‖y‖ = 0 + 0 = 0
Thus, x+ y ∈M.
Therefore, M is a subspace of X .
N The map x+M 7→ ‖x‖ is a norm on X /M.
Since ‖·‖ is already seminorm, and
‖x‖ = 0 ⇐⇒ x ∈M ⇐⇒ x+M = 0 +M
, we can conclude that ‖·‖ is a norm on X /M �
14. If X is a normed vector space and M is a nonclosed subspace, then ‖x+M‖, as defined in Exercise12, is a seminorm on X /M. If one divides by its nullspace as is Exercise 13, the resulting quotient spaceis isometrically isomorphic to X /M. (Cf. Exercise 5)
Proof. (Feb 25th 2018)
N ‖x+M‖ is a seminorm on X /M
∀a ∈ C and ∀x+M∈X /M
‖ax+M‖ = infy∈M‖ax+ y‖ = |a| inf
y∈M
∥∥∥x+y
a
∥∥∥ = |a| infy∈M‖x+ y‖ = |a| ‖x+M‖
∀x, y ∈X
16
‖x+ y +M‖ = infz∈M‖x+ y + z‖ ≤ inf
z∈M
∥∥∥x+z
2
∥∥∥+ infz∈M
∥∥∥y +z
2
∥∥∥= inf
z∈M‖x+ y‖+ inf
z∈M‖y + z‖ = ‖x+M‖+ ‖y +M‖
N
Let N = {x ∈X : ‖x+M‖ = 0}, then, by Exercise 13, ‖·‖ is a norm on X /N .Let L(x+N ) = x+M, then it is well defined and injection.
�
15. Suppose that X and Y are normed vector spaces and T ∈ L(X ,Y). Let N (T ) = {x ∈ X : Tx = 0}a. N (T ) is a closed subspace of X .b. There is a unique S ∈ L(X/N (T ),Y) such that T = S ◦ π where π : X → X/N (T ) is the projection
π(x) = x+N (T ). Moreover, ‖S‖ = ‖T‖.
Proof. (Feb 21st 2018)
a.Observe the below.
T0 = 0 =⇒ 0 ∈ N (T )
T (αx+ βy) = αTx+ βTy = 0 + 0 = 0 ∀x, y ∈ N (T )∀α, β ∈ C.
Thus, N (T ) is a subspace. And it is closed since T is bounded so continuous, {0} is closed in Y , and
N (T ) = T−1({0})
b.Let’s define the S : X/N (T )→ Y as below.
S(x+N (T )) = Tx ∀x ∈ X
It is well defined since the below.
x+N (T ) = y +N (T ) =⇒ (x− y) ∈ N (T )
=⇒ S(x+N (T )) = Tx = Ty + T (x− y) = Ty = S(y +N (T ))
Then note the below.
S ◦ π(x) = S(x+N (T )) = Tx ∀x ∈ XDue to the following reasoning, S is linear.
∀α, β ∈ C∀x, y ∈ XS((αx+ βy) +N (T )) = T (αx+ βy) = αTx+ βTy = αS(x+N (T )) + βS(y +N (T ))
Also, it is bounded because T is bounded and of the below.
17
∀x ∈ X‖S(x+N (T ))‖ = ‖Tx‖ ≤ ‖x‖
Therefore, S ∈ L(X/N (T ),Y).
Moreover,
‖T‖ = supx 6=0
‖Tx‖‖x‖
= supx 6∈N (T )
‖Tx‖‖x‖
= supx 6∈N (T )
‖S(π(x))‖‖x‖
= supx 6∈N (T )
‖S(π(x))‖‖π(x)‖
‖π(x)‖‖x‖
= supx 6∈N (T )
‖S(π(x))‖‖π(x)‖
supx 6∈N (T )
‖π(x)‖‖x‖
= ‖S‖ ‖π‖ = ‖S‖
Also, the S is unique. If there are S1 and S2 that satisfy the given conditions,
‖S1 − S2‖ = supx 6∈N (T )
‖(S1 − S2)(π(x))‖‖π(x)‖
= supx 6∈N (T )
‖S1(π(x))− S2(π(x))‖‖π(x)‖
= supx 6∈N (T )
‖Tx− Tx‖‖π(x)‖
= 0
=⇒ S1 = S2
�
16.
Proof. �
17. A linear functional f on a normed vector space X is bounded iff f−1({0}) is closed. (Use Exercise12.b)
Proof. (Feb 26th 2018)
( =⇒ )
If a linear functional f is bounded, f is continuous. Thus, the inverse image of closed set {0}, f−1({0})is closed (C is Normal space).
(⇐=)
Note that f−1({0})(=let M) is a closed subspace of X . By Exercise 12 (b), there exists x ∈ X suchthat ‖x‖ = 1 and
‖x+M‖ ≥ 1− 1
2=
1
2
Note that x ∈X \M and let Mx = Cx+MFor any y ∈X \Mx, observe the below.
y =f(y)
f(x)x+
(y − f(y)
f(x)x
)∈ Cx+M =Mx
Therefore, X =Mx. Also, for any x ∈X there exists y ∈M, observe the below.
18
|f(z)| = |f(λx+ y)| = |λ||f(x)| ≤ 2|λ|12|f(x)| ≤ 2|λ| ‖x+M‖|f(x)| ≤ 2|λ|
∥∥∥x+y
λ
∥∥∥ |f(x)| = 2|f(x)| ‖λx+ y‖
Therefore, f is bounded.�
18. Let X be a normed vector space.
a. If M is a closed subspace and x ∈X \M then M+ Cx is closed. (Use Theorem 5.8 a.)
b. Every finite-dimensional subspace of X is closed.
Proof. (Feb 26th 2018)
a.
If X =M, then it is trivially true. Thus, let’s assume that M is a closed proper subspace of X .
Let z ∈ acc(M+ Cx), then there exists a sequence {yn + λnx}∞n=1 ⊂M+ Cx such that {yn}∞n=1 ⊂M,{λn}∞n=1 ⊂ C, and
limn→∞
yn + λnx = z
Note that, by Theorem 5.8a., there is a function f ∈X ∗ such that
f(x) 6= 0 f |M = 0
And note the below.
limn→∞
λn = limn→∞
f(z)
f(x)=f(z)
f(x)∵ f(z) = f( lim
n→∞(yn + λnx)) = lim
n→∞f(yn + λnx) = lim
n→∞λnf(x)
The exchanging the function and limit is valid here since f is bounded, thus continuous.Now, observe the below.
z = limn→∞
(yn + λnx) = limn→∞
(yn + λnx)− limn→∞
λnx+ limn→∞
λnx = limn→∞
yn + limn→∞
λnx = limn→∞
yn +f(z)
f(x)x
=⇒ limn→∞
ynz −f(z)
f(x)x =let y exists
Since M is closed, z − f(z)f(x)
x = y ∈M. And consequently, z = y + f(z)f(x)
x ∈M+ Cx.
Therefore, M+ Cx is closed.
b.
Note that {0} =let M0 is closed subspace of X . And from a. we know that {0} + Cx1 =let M1 is aclosed set for any x1 ∈X . Assuming we have selected xn ∈X \Mn−1 whereMn−1 is a closed subspace,we can create Mn = Mn−1 + Cxn. From a. we know Mn is a closed subspace. Therefore, any finitedimensional subspace of X is closed. �
19
19. Let X be an infinite-dimensional normed vector space.
a. There is a sequence {xj} in X such that ‖xj‖ = 1 for all j and ‖xj − xk‖ ≥ 12
for all k 6= j. (Con-struct xj inductively, using Exercise 12b and 18.)
b. X is not locally compact.
Proof. (Feb 27th 2018)
a.
Let M0 = {0}, then M0 is a closed subspace of X .
Then, by Exercise 12, we can choose a vector x1 ∈X \M0 such that ‖x1‖ = 1 and ‖x1 +M0‖ ≥ 12.
Suppose that we have chosen {xk}n−1k=1 ⊂ X such that ‖xj‖ = 1 and ‖xj − xi‖ ≥ 1
2for all i, j ∈
{1, 2, . . . , n− 1}.
Let Mn =M0 +∑n
k=1 Cxk, and it is a closed proper subspace of X from exercise 18. Thus, again byExercise 12, we can choose xn ∈X \Mn such that ‖xn‖ = 1 and ‖xn +Mn‖ ≥ 1
2. Noting that −xi ∈Mn
∀i ∈ {1, 2, . . . , n− 1}, observe the below.
‖xn − xi‖ ≥ ‖xn +Mn‖ ≥1
2
Thus, inductively, we can construct a sequence {xn}∞n=1 as above.
b.
Assume that X is compact. Then there exists a compact neighborhood K0 of 0. Then we have r > 0such that B(0, r) ⊂ K0. Since B(0, r) is closed subset of a compact set, it is compact.
Firstly, note that {rxn}∞n=1 where {xn}∞n=1 is from part a satisfy the following properties.
‖xj‖ = r ∀j ∈ N
‖xj − xi‖ ≥r
2∀i, j ∈ N
And let’s construct a open cover as below.
C = {B(xk,r
2) : k ∈ N} ∪ U
where U =
(⋃k∈N
B(xk,
r
4
))c
Note that C is an open cover of B(0, r). However, there exists no finite subcover since if you miss B(xk,r2)
for some k ∈ N, we cannot cover xk. Therefore, B(0, r) is not a compact ans it is a contradiction. �
20
20. If M is a finite-dimensional subspace of a normed vector space X , there is a closed subspace N suchthat M∩N = {0} and M+N = X .
Proof. (Mar 11st 2018)
If X =M, then N = {0} satisfies the conditions.
Since M is finite dimensional subspace of X , let
M =⊕
1≤i≤n
Cui
Note that M is closed (by Exercise 18). Thus, if
Bj =⊕
1≤i≤ni 6=j
{ui}
by a theorem,
∀j ∈ {1, . . . , n} ∃fj ∈X ∗ such that fj(uj) 6= 0 and fj|Bj = 0
Now let
N =n⋂i=1
f−1i ({0})
Suppose that x ∈M∩N . Since x ∈M, x should be a linear combination of ui’s. However, since x ∈ Nand ui 6∈ N for each i, we have x = 0. Therefore,
M∩N = {0}Also, if x ∈X , then
x =
(n∑i=1
fi(x)
fi(ui)ui
)+
(x−
n∑i=1
fi(x)
fi(ui)ui
)∈M∩N
Since
fj
(x−
n∑i=1
fi(x)
fi(ui)ui
)= fj(x)−
n∑i=1
fi(x)
fi(ui)fj(ui) = fj(x)− fj(x)
fj(uj)f(uj) = 0
Therefore,
X =M+N�
21. If X and Y are normed vector spaces, define α : X ∗ ×Y ∗ → (X ×Y )∗ by
α(f, g)(x, y) = f(x) + g(y).
Then α is an isomorphism which is isometric if we use the norm ‖(x, y)‖ = max(‖x‖ , ‖y‖) on X ×Y ,the corresponding operator norm on (X ×Y )∗, and the norm ‖(f, g)‖ = ‖f‖+ ‖g‖ on X ∗ ×Y ∗.
Proof. (March 11st 2018)
Let β : (X ×Y )∗ → X ∗ ×Y ∗ be defined as below.
β(F (x, y)) = (F (x, 0), F (0, y))
21
And observe the below.
∀(x, y) ∈ X ×Y ∀F : (X ×Y )∗ → X ∗ ×Y ∗ ∀f ∈ X ∗, g ∈ Y ∗
(α ◦ β)(F (x, y)) = α(β(F (x, y))) = α(F (x, 0), F (0, y)) = F (x, 0) + F (0, y) = F (x, y)
(β ◦ α)(f(x), g(y)) = β(α(f(x), g(y))) = β(f(x) + g(y)) = (f(x) + g(0), f(0) + g(y)) = (f(x), g(y))
Therefore, α is a bijection. Also,
‖α(f, g)(x, y)‖ = ‖f(x) + g(y)‖ ≤ ‖f(x)‖+ ‖g(y)‖ ≤ ‖f‖ ‖x‖+ ‖g‖ ‖y‖= (‖f‖+ ‖g‖) max(‖x‖ , ‖y‖) = (‖f‖+ ‖g‖) ‖(x, y)‖=⇒ ‖α(f, g)‖ ≤ ‖(f, g)‖
so, α is bounded.Conversely, let ε > 0 be given. And note that there exists x ∈ X and y ∈ Y with ‖x‖ = ‖y‖ = 1 such
that
‖f‖ < ‖f(x)‖+ ε
‖g‖ < ‖g(y)‖+ ε
Now, observe the below.
‖(f, g)‖ = ‖f‖+ ‖g‖ < ‖f(x)‖+ ‖g(y)‖+ 2ε = ‖|f(x)|+ |g(y)|‖+ 2ε
= ‖f(Sfx) + g(Sgy)‖+ 2ε = ‖α(f, g)(Sfx, Sgy)‖+ 2ε
where
Sf =
{f(x)|f(x)| f(x) 6= 0
0 f(x) = 0Sg =
{g(y)|g(y)| g(y) 6= 0
0 g(y) = 0
Since ε is arbitrary,
‖(f, g)‖ < ‖α(f, g)(Sfx, Sgy)‖ ≤ ‖α(f, g)‖max(‖Sfx‖ , ‖Sgy‖) = ‖α(f, g)‖
Therefore,
‖α(f, g)‖ = ‖(f, g)‖
So, α is an isometry and so is an isomorphism. �
22. Suppose that X and Y are normed vector spaces and T ∈ L(X,Y).
a. Define T † : Y ∗ → X ∗ by T †f = f ◦ T . Then T † ∈ L(Y∗,X∗) and∥∥T †∥∥ = ‖T‖.
T † is called the adjoint or transpose of T .
b. Applying the construction in (a) twice, one obtain T †† ∈ L(X∗∗,Y∗∗). If X and Y are identified with
their natural image X and Y in X ∗∗ and Y ∗∗, then T ††|X = T .
c. T † is injective iff the range of T is dense in Y .
d. If the range of T † is dense in X ∗, then T is injective; the converse is true if X is reflective.
Proof. (March 11st 2018)
22
a. For any f, g ∈ Y ∗, λ ∈ C and x ∈ X , observe the linearity of T † as below.
T †(f + g)(x) = ((f + g) ◦ T )(x) = (f ◦ T )(x) + (g ◦ T )(x) = T †(f)(x) + T †(g)(x)
=⇒ T †(f + g) = T †(f) + T †(g)
T †(λf)(x) = (λf ◦ T )(x) = λ(f ◦ T )(x) = λT †(f)(x)
=⇒ T †(λf) = λT †(f)
Also, it is bounded as we can observe from below.
‖(f ◦ T )(x)‖ = ‖f‖ ‖Tx‖ ≤ ‖f‖ ‖T‖ ‖x‖=⇒
∥∥T †(f)∥∥ = ‖(f ◦ T )‖ ≤ ‖f‖ ‖T‖
Therefore, T † ∈ L(Y∗,X∗).Also, ∀x ∈ X with ‖x‖ = 1∥∥T †∥∥ = sup
‖f‖=1
∥∥T †(f)∥∥ ≤ sup
‖f‖=1
‖f ◦ T‖ ≤ sup‖f‖=1
‖f‖ ‖T‖ = ‖T‖∥∥T †∥∥ ≥ ∥∥T †(‖·‖)∥∥ = ‖‖·‖ ◦ T‖ = ‖T‖
since ‖·‖ ∈ Y ∗ , ‖‖·‖‖ = 1, and
‖(‖·‖ ◦ T )(x)‖ = ‖Tx‖=⇒ ‖‖·‖ ◦ T‖ = sup
‖x‖=1
‖(‖·‖ ◦ T )(x)‖ = sup‖x‖=1
‖Tx‖ = ‖T‖
Therefore,∥∥T †∥∥ = ‖T‖.
b.Note that
T ††(x)(f) = (x ◦ T †)(f) = x ◦ (f ◦ T ) = f(Tx) = T x(f) ∀f ∈ Y ∗
Thus,
T ††|X = T ††|X = TX = T
c.Suppose that ∃y ∈ Y \ T (X ). Note that T (X ) is a subspace, soM =let T (X ) is also a subspace (by
Exercise 5). Then observe the below.
∃f ∈ Y ∗ such that f(y) 6= 0 and f |M = 0
Let g ∈ Y ∗ be defined as g ≡ 0. Then observe the below.
T †(f)(x) = f(Tx) = 0 = g(Tx) = T †(g)(x) ∀x ∈ X
Thus, T †(f) = T †(g) but f 6= g. Therefore, T † is not an injection.
Conversely, suppose that T † is not an injection. Then ∃f 6= g(∈ Y ∗) such that T †(f) = T †(g). Since
f 6= g, ∃y ∈ Y such that f(y) 6= g(y). If T (X ) = Y , ∃{xn} such that
limn→∞
Txn = y
23
However,
f(y) = limn→∞
f(T (xn)) = limn→∞
T †(f)(xn) = limn→∞
T †(g)(xn) = limn→∞
g(Txn) = g(y)
It shows a contradiction.
d. (Need to be fixed) If T †(Y ∗) = X ∗, by c, T †† is injective. And since T †† = T ††|X = T . Thus,T is an injection. If X is reflexive and T is injective, then T †† is also injective and, by c, T † has denseimage. �
23. Suppose X is a Banach space. If M is a closed subspace of X and N is a closed subspace of X ∗, letM0 = {f ∈ X ∗ : f |M = 0} and N⊥ = {x ∈ X : f(x) = 0 ∀f ∈ N}.(Thus, if we identify X with itsimage in X ∗∗, N⊥ = N 0 ∩X )
a. M0 and N⊥ are closed subspaces of X ∗ and X , respectively.
b. (M0)⊥ =M and (N⊥)0 ⊃ N . If X is reflective, (N⊥)0 = N .
c. Let π : X → X /M be the natural projection, and define α : (X /M)→ X ∗ by α(f) = f ◦π. Thenα is an isometric isomorphism from (X /M)∗ onto M0, where X /M has a quotient norm.
d. Define β : X ∗ →M∗ by β(f) = f |M; then β induces a map β : X ∗/M0 →M∗ as in Exercise 15,and β is an isometric isomorphism.
Proof. �
24. Suppose that X is a Banach space.
a. Let X , (X ∗) be the natural images of X , X ∗ in X ∗∗, X ∗∗∗, and let X 0 = {F ∈ X ∗∗∗ : F |X = 0}.Then (X ∗) ∩ X 0 = {0} and (X ∗) + X 0 = X ∗∗∗.
b. X is reflexive iff X ∗ is reflexive.
Proof. �
25.
Proof. �
26.
Proof. �
27. There exists meager subsets of R whose complements have Lebesgue measure zero.
Proof. �
28. The Baire category theorem remains true if X is assumed to be LCH space rather than a completemetric space. (The proof is similar, the substitute for completeness is Proposition 4.21.)
24
Proof. �
29. Let Y = L1(µ) where µ is counting measure on N, and let X ={f ∈ Y :∑∞
1 n|f(n)| <∞}, equippedwith the L1 norm.
a. X is a proper dense subspace of Y ; hence X is not complete.
b. Define T : X → Y by
Tf(n) = nf(n)
Then T is closed but not bounded.
c. Let S = T−1. Then S : Y → X is bounded and surjective but not open.
Proof. (March 12nd 2018)
a. [f ∈ X =⇒
∞∑n=1
|f(n)| ≤∞∑n=1
n|f(n)| <∞ =⇒ f ∈ Y
]=⇒ X ⊂ Y
[f, g ∈ X =⇒
∞∑n=1
|f(n)| <∞∧∞∑n=1
|g(n)| <∞ =⇒∞∑n=1
|f(n) + g(n)| ≤∞∑n=1
|f(n)|+∞∑n=1
|g(n)| <∞
]=⇒ f + g ∈ X
[λ ∈ C ∧ f ∈ X =⇒
∞∑n=1
|λf(n)| = |λ|∞∑n=1
|f(n)| <∞
]=⇒ λf ∈ X
Therefore, X is a proper subspace of Y since f(n) = 1n2 ∈ Y but f 6∈ X .
Now, we will show that X is dense in Y .Let ε > 0 and f ∈ Y be given. Note that
∃N ∈ N such that∞∑n=N
|f(n)| < ε
If we define g as below,
g(n) =
{f(n) n < Nf(n)n
n ≥ N
then
∞∑n=1
|f(n)− g(n)| =∞∑n=N
(1− 1
n)|f(n)| ≤
∞∑n=1
|f(n)| < ε
Therefore, X is dense in Y .
b.
25
Suppose that fm → f and Tfm(n) → g(n) where g ∈ Y . Let ε > 0 be given. Then there existN1, N2 ∈ N such that
∞∑n=1
|fm(n)− f(n)| < ε
3∀m ≥ N1
∞∑n=1
|Tfm(n)− g(n)| < ε
3∀m ≥ N2
Now, let N = max(N1, N2). Note that it is possible that f ∈ Y because of a. However, since if f 6∈ X ,we cannot define Tf , we need to assume that f ∈ X . Then ∃N3 ∈ N such that
∞∑n=N3
n|f(n)− fN(n)| < ε
3
Now, for all m ≥ N , observe the below.
‖Tf − g‖ =∞∑n=1
|nf(n)− g(n)| ≤∞∑n=1
n|f(n)− fN(n)|+∞∑n=1
|TfN(n)− g(n)|
<
N3−1∑n=1
n|f(n)− fN(n)|+∞∑
n=N3
n|f(n)− fN(n)|+ ε
3
< N3ε
3+
2ε
3
Since ε is arbitrary, Tf = g.Therefore, T is closed. However, it is not bounded since, if we define f as below,
fN(n) =
{1 n = N
0 Otherwise
for some N ∈ N, then fN ∈ X , ‖f‖ = 1, and
‖TfN‖ =∞∑n=1
n|f(n)| = N ‖fN‖
Thus, for any constant C, there exists M ∈ N such that M > C, so that
‖TfM‖ = M ‖fM‖ > C ‖fM‖Therefore, T is unbounded.
c.Note that Sf(n) = f(n)
n, and, for any f ∈ Y , observe the below.
‖Sf‖ =∞∑n=1
|Sf(n)| =∞∑n=1
|f(n)|n≤
∞∑n=1
|f(n)| = ‖f‖
Thus, S is bounded.Also, observe that, for any g(n) ∈ X , ng(n) ∈ Y , and
S(ng(n)) = g(n)
Therefore, S is a surjection. However, if S is open, then T is bounded. It is a contradiction to b.�
26
30. Let Y = C([0, 1]) and X = C1([0, 1]), both equipped with the uniform norm.
a. X is not complete.
b. The map (d
dx
): X → Y
is closed (see exercise 9) but not bounded.
Proof. (March 13rd 2018)
a.
Consider {fn} ={√
x+ 1n
}∞n=1⊂ X . Then fn →
√x uniformly due to below.
supx∈[0,1]
∣∣∣∣∣√x+
1
n−√x
∣∣∣∣∣ < 1√n→ 0
Clearly,√x is a continuous function. However, the derivative is not defined at x = 0.
Thus,√x 6∈ Y , so X is not complete.
b.If {fn} ⊂ X and fn → f pointwisely and f ′n → g uniformly, then f ′ is continuous and g = f ′ from basic
analysis course. Thus, the map is closed.
On the other hands, assume the map is bounded so there exists C > 0 such that∥∥∥∥ dfdx∥∥∥∥ < C ‖f‖ ∀f ∈ X
But we have fn(x) = xn which satisfies∥∥∥∥ dfdx∥∥∥∥ =
∥∥nxn−1∥∥ = sup
x∈[0,1]
nxn−1 = n
Since ∃N ∈ N such that N > C, fN is an example that makes the bounded condition fail. �
31. Let X , Y be Banach spaces and let S : X → Y be an unbounded linear map (for the existence ofwhich, see §5.6). Let Γ(S) be the graph of S, a subspace of X ×Y .
a. Γ(S) is not complete.
b. Define T : X → Γ(S) by Tx = (x, Sx). Then T is closed but not bounded.
c. T−1 : Γ(S)→ X is bounded and surjective but not open.
Proof. (March 13rd 2018)
a.If Γ(S) is complete, S is closed. Then, by the Closed Graph Theorem, S should be bounded. It is a
contradiction.
b.
27
Suppose xn → x in X and Txn → y in Γ(S). Since X is complete, x ∈ X , therefore, y = (x, Sx) ∈ Γ(S),so T is closed.
Also, observe the below.
‖Tx‖ = max(‖x‖ , ‖Sx‖) ≥ ‖Sx‖
Since S is unbounded, T is also unbounded.
c. For any y = (x, Sx) ∈ Γ(S),∥∥T−1y∥∥ = ‖x‖ ≤ max(‖x‖ , ‖Sx‖) = ‖y‖
Thus, T−1 is bounded. Also, since, for any x ∈ X , we have (x, Sx) ∈ Γ(S), T−1 is surjective. However,if T−1 is open, then T is continuous, so it is bounded, it is a contradiction to b. �
32. Let ‖·‖1 and ‖·‖2 be norms on the vector space X such that ‖·‖1 ≤ ‖·‖2. If X is complete with respectto both norms, then the norms are equivalent.
Proof. (March 13rd 2018)
Note that X 1 = (X , ‖·‖1) and X 2 = (X , ‖·‖2) are Banach spaces. Further note that T : X 2 → X 1
defined by Tx = x is clearly a bijective linear map. Now, for any x ∈ X , with the given condition, observethe below.
‖Tx‖1 = ‖x‖1 ≤ ‖x‖2 = ‖x‖2
Therefore, T is bounded. Then, by the open mapping theorem, T is open, so is an isomorphism.Therefore, there exists C > 0 such that for any x ∈ X ,
‖x‖2 =∥∥T−1x
∥∥2≤ C ‖x‖1
Thus, the two norms are equivalent. �
33. There is no slowest rate of decay of the terms of an absolutely convergent series;That is, there is no sequence {an} of positive numbers such that
∞∑n=1
an|cn| <∞ ⇐⇒ {cn} is bounded.
(The set of bounded sequences is the space B(N) of bounded functions on N, and the set of absolutelysummable sequences is L1(µ) where µ is counting measure on N. If such an {an} exists, consider T :B(N)→ L1(µ) defined by Tf(n) = anf(n). The set of f such that f(n) = 0 for all but finitely many n isdense in L1(µ) but not in B(N).)
Proof. (March 13rd 2018)
Assume that such {an} exists. And let’s define a map T : B(N)→ L1([0, 1]) as below.
Tf(n) = anf(n)
Clearly, T is a linear map. �
28
34. With reference to Exercise 9 and 10, show that the inclusion map of L1k([0, 1]) into Ck−1([0, 1]) is
continuous (a) by using the closed graph theorem, and (b) by direct calculation. (This is to illustrate theuse of the closed graph theorem as a labor-saving device.)
Proof. �
35. Let X and Y be Banach spaces, T ∈ L(X ,Y ), N (T ) = {x : Tx = 0}, and M = range(T ). ThenX /N (T ) is isomorphic to M iff M is closed.(See Exercise 15.)
Proof. (March 13rd 2018)
From exercise 15, N (T ) is closed and there exists unique S ∈ L(X /N (T ),Y ) such that T = S ◦ πdefined as below.
S(x+N (T )) = T (x)
We now show that X /N (T ) is complete so is Banach space.Let {xn +N (T )}∞n=1 =let {yn} ⊂ X /N (T ) be a given sequence satisfying the following condition.
∞∑n=1
‖yn‖ <∞
Note that for any n ∈ N we can choose x′n ∈ X satisfying the condition below.
‖x′n‖ < ‖yn‖+ ‖yn‖ = 2 ‖yn‖Therefore,
∞∑n=1
‖xn‖ ≤∞∑n=1
2 ‖yn‖ <∞
Since X is a Banach space, ∃x ∈ X such that∞∑n=1
xn = x
Now, observe that∥∥∥∥∥(∞∑n=1
yn
)− (x+N (T ))
∥∥∥∥∥ =
∥∥∥∥∥((
∞∑n=1
xn
)− x
)+N (T )
∥∥∥∥∥ ≤∥∥∥∥∥(∞∑n=1
xn
)− x
∥∥∥∥∥ = 0
Therefore,∞∑n=1
yn = x+N (T )
so X /N (T ) is a Banach space.
( =⇒ ) Now, suppose that X /N (T ) is isomorphic to M.Then there is an isomorphism J : X /N (T ) → M. Let x ∈ M, then ∃{xn} ⊂ M such that xn → x.
Then, since J is an isomorphism, and∥∥J−1xn − J−1(x)∥∥ ≤ ∥∥J−1
∥∥ ‖xn − x‖J−1xn → J−1x. Now, since X /N (T ) is a Banach space, J−1x ∈ X , so JJ−1x = x ∈ M. Therefore, Mis closed.
(⇐=) Suppose that M is closed. Thus, M is a Banach space since Y is a Banach space.
29
Consider the map S ∈ L(X /N (T ),Y ) as beginning of this proof. Clearly, S is an injection as we canobserve from below.
S(x+N (T )) = S(y +N (T )) ⇐⇒ T (x) = T (y)
⇐⇒ T (x− y) = 0
⇐⇒ x− y ∈ N (T )
⇐⇒ x+N (T ) = y +N (T )
Also, if y ∈M, then ∃x ∈ X such that Tx = y. Thus, from below, S is a surjection.
S(x+N (T )) = Tx = y
Now, since X /N (T ) and M are Banach spaces and S ∈ L(X /N (T ),M), S is the isomorphism.Therefore, X /N (T ) and M are isomorphic. �
36. Let X be a separable Banach
Proof. �
37. Let X and Y be Banach spaces. If T : X → Y is a linear map such that f ◦ T ∈ X ∗ for everyf ∈ Y ∗, then T is bounded.
Proof. (March 14th 2018)
In order to prove it, due to the Closed Graph Theorem, it is enough to prove that T is closed.Let {xn} ⊂ X be a sequence such that xn → x. Since X is a Banach space, x ∈ X . Now, we need to
prove that Txn → Tx. For any f ∈ Y ∗, observe the below.
limn→∞
f ◦ T (xn) = f ◦ T ( limn→∞
xn) = f ◦ T (x) ∵ f ◦ T ∈ X ∗
limn→∞
f ◦ T (xn) = f( limn→∞
Txn) ∵ f ∈ Y ∗
Let I be identity mapping and note that I ∈ Y ∗. Thus,
Tx = I ◦ T (x) = limn→∞
I ◦ T (xn) = I( limn→∞
Txn) = limn→∞
Txn
Therefore, T is closed, so is bounded. �
38. Let X and Y be Banach spaces, and let {Tn} be a sequence in L(X ,Y ) such that limTnx exists forevery x ∈ X . Let Tx = limTnx, then T ∈ L(X ,Y ).
Proof. (March 14th 2018)
Clearly, T is linear as below.
T (x+ y) = limn→∞
Tn(x+ y) = limn→∞
Tnx+ limn→∞
Tny = Tx+ Ty ∀x, y ∈ X
T (αx) = limn→∞
Tn(αx) = α limn→∞
Tn(x) = αT (x) ∀x ∈ X , α ∈ C
Also, note that
limn→∞
Tnx exists ∀x ∈ X
Thus, for any x ∈ X , {Tnx}∞n=1 is bounded. In other words,
supn∈N‖Tnx‖ <∞ ∀x ∈ X
30
Then, the Uniform Boundedness principle,
lim supn→∞
‖Tn‖ ≤ supn∈N‖Tn‖ <∞
Now, there exists C > 0 such that for any x ∈ X
‖Tx‖ =∥∥∥ limn→∞
Tnx∥∥∥ = lim
n→∞‖Tnx‖ = lim sup
n→∞‖Tnx‖ ≤ lim sup
n→∞‖Tn‖ ‖x‖ ≤ C ‖x‖
Therefore, T ∈ L(X ,Y ). �
39. Let X ,Y ,Z be Banach spaces and let B : X × Y → Z be a separately continuous bilinearmap; that is, B(x, ·) ∈ L(Y ,Z ) for each x ∈ X and B(·, y) ∈ L(Y ,Z ) for each y ∈ Y . ThenB is jointly continuous, that is, continuous from X × Y to Z . (Reduce the problem to proving that‖B(x, y)‖ ≤ C ‖x‖ ‖y‖ for some C > 0.)
Proof. (March 14th 2018)
Let B1(x) = B(x, ·) and B2(y) = B(·, y), then B(x, y) = B1◦ �
40.
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41.
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42.
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44.
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45.
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46.
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47.
31
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