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myweb.ttu.edu/bban [email protected] Real Analysis Byeong Ho Ban Mathematics 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 continuous from X×X and K ×X to X . Moreover, the norm is continuous from X to [0, ); in fact, |kxk-kyk| ≤ kx - yk. Proof. (Feb 3rd 2018) N The addition is continuous. Let > 0 be given. And let δ = 2 . Then, for any (x, y) ∈X×X , observe the below. max(kx - vk, ky - wk)= k(x - v,y - w)k = k(x, y) - (v,w)k = 2 = ⇒k(x + y) - (v + w)k = k(x - v)+(y - w)k≤kx - vk + ky - wk < 2 + 2 = Therefore, the addition operation is continuous. N The Scalar multiplication is continuous. Let > 0 be given. Then, for any (α, x) K ×X , let δ = min( 3kxk , 3|α| , 1, 3 ) and observe the below. max(|α - β |, kx - yk)= k(α - β,x - y)k = k(α, x) - (β,y)k = min( 3kxk , 3|α| , 1, 3 ) = ⇒kαx - βyk≤kαx - βxk + kβx - βyk = |α - β |kxk + |β |kx - yk ≤|α - β |kxk + |α|kx - yk + |α - β |kx - yk < 3 + 3 + kx - yk < N The norm operation is continuous. Let > 0 and x ∈X be given. And let δ = and, for any y ∈X , by triangular inequality, note the below. kxk≤kx - yk + kyk = ⇒kxk-kyk≤kx - yk kyk≤kx - yk + kxk = ⇒kyk-kxk≤kx - yk Thus, |kxk-kyk| ≤ kx - yk. Then observe the below. 1

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Page 1: myweb.ttu.edu/bban lev.ban@ttu.edu Real Analysis · 2018. 3. 30. · myweb.ttu.edu/bban lev.ban@ttu.edu Real Analysis Byeong Ho Ban Mathematics and Statistics Texas Tech University

myweb.ttu.edu/bban

[email protected]

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

Page 2: myweb.ttu.edu/bban lev.ban@ttu.edu Real Analysis · 2018. 3. 30. · myweb.ttu.edu/bban lev.ban@ttu.edu Real Analysis Byeong Ho Ban Mathematics and Statistics Texas Tech University

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

Page 3: myweb.ttu.edu/bban lev.ban@ttu.edu Real Analysis · 2018. 3. 30. · myweb.ttu.edu/bban lev.ban@ttu.edu Real Analysis Byeong Ho Ban Mathematics and Statistics Texas Tech University

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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.

Page 4: myweb.ttu.edu/bban lev.ban@ttu.edu Real Analysis · 2018. 3. 30. · myweb.ttu.edu/bban lev.ban@ttu.edu Real Analysis Byeong Ho Ban Mathematics and Statistics Texas Tech University

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)

Page 5: myweb.ttu.edu/bban lev.ban@ttu.edu Real Analysis · 2018. 3. 30. · myweb.ttu.edu/bban lev.ban@ttu.edu Real Analysis Byeong Ho Ban Mathematics and Statistics Texas Tech University

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.

Page 6: myweb.ttu.edu/bban lev.ban@ttu.edu Real Analysis · 2018. 3. 30. · myweb.ttu.edu/bban lev.ban@ttu.edu Real Analysis Byeong Ho Ban Mathematics and Statistics Texas Tech University

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. �

Page 7: myweb.ttu.edu/bban lev.ban@ttu.edu Real Analysis · 2018. 3. 30. · myweb.ttu.edu/bban lev.ban@ttu.edu Real Analysis Byeong Ho Ban Mathematics and Statistics Texas Tech University

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.

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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])( =⇒ )

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‖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

4+ ‖‖f ′N − g‖L1‖L1 =

ε

4+ ‖f ′N − g‖L1 <

ε

2

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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.

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( =⇒ )

‖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)

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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

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

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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.

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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.

Page 15: myweb.ttu.edu/bban lev.ban@ttu.edu Real Analysis · 2018. 3. 30. · myweb.ttu.edu/bban lev.ban@ttu.edu Real Analysis Byeong Ho Ban Mathematics and Statistics Texas Tech University

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

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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.

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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.

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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. �

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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. �

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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))

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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)

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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

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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.)

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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.

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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+

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.�

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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.

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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. �

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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.

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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

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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.

Proof. �

41.

Proof. �

42.

Proof. �

43.

Proof. �

44.

Proof. �

45.

Proof. �

46.

Proof. �

47.

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31

Proof. �

48.

Proof. �

49.

Proof. �

50.

Proof. �

51.

Proof. �

52.

Proof. �

53.

Proof. �

54.

Proof. �

55.

Proof. �

56.

Proof. �

57.

Proof. �

58.

Proof. �

59.

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32

Proof. �

60.

Proof. �

61.

Proof. �

62.

Proof. �

63.

Proof. �

64.

Proof. �

65.

Proof. �

66.

Proof. �

67.

Proof. �

68.

Proof. �

69.

Proof. �

70.

Proof. �

71.

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33

Proof. �

72.

Proof. �

73.

Proof. �

74.

Proof. �

75.

Proof. �

76.

Proof. �

77.

Proof. �

78.

Proof. �

79.

Proof. �

80.

Proof. �

81.

Proof. �

82.

Proof. �

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34

83.

Proof. �

84.

Proof. �

85.

Proof. �

86.

Proof. �

87.

Proof. �

88.

Proof. �

89.

Proof. �

90.

Proof. �

91.

Proof. �

92.

Proof. �

93.

Proof. �

94.

Proof. �

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35

95.

Proof. �

96.

Proof. �