131A ASSIGNMENT 1

A. ADAM AZZAM

These solutions do not contain the detail or precision that I expect of your solu-tions. Several claims herein would require some explanation on an exam or futureassignments. Use this as a guide to write your own complete and detailed solutions.

Exercise 1. Let (X, || · ||) be a normed linear space. Define d : X ×X → R by

d(x, y) = ||x− y||.

Show that (X, d) is a metric space.

Solution.

Proof. It’s clear that d ≥ 0. If d(x, y) = 0, then since || · || is a norm it follows thatx− y = 0 or, equivalently, that x = y. Now, d is symmetric insofar as

d(x, y) = ||x− y|| = || − 1 · (y − x)|| = | − 1| · ||y − x|| = d(y, x).

Lastly, the triangle inequality for d follows from the triangle inequality for the norm:

d(x, y) = ||x− y|| = ||x− z + z − y|| ≤ ||x− z||+ ||z − y|| = d(x, z) + d(z, y).

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Exercise 2.

Solution.

Proof. Fix x = (x1, . . . , xn) ∈ Rn. Find j ∈ {1, . . . , n} with

||x||`∞(Rn) = |xj | = ||x||`∞(Rn)

Then for p ≥ 0 we see that

||x||`∞(Rn) = |xj | ≤ (|xj |p)1p ≤ ||x||`p =

[n∑

i=1

|xi|p] 1

p

≤

n∑j=1

||x||p`∞

1p

= (n||x||p`∞)1p = n

1p ||x||`∞(Rn)

By the squeeze theorem, lim ||x||p exists and equals ||x||`∞(Rn). �

Exercise 3.

Solution.

Proof. If α ∈ F and x ∈ X, then

||αx|| =√〈αx, αx〉 =

√α 〈x, αx〉 =

√αα 〈x, x〉 =

√|α|2 〈x, x〉 = |α| 〈x, x〉 = |α|·||x||.

Date: January 15, 2015.

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2 A. ADAM AZZAM

If x ∈ X then 〈x, x〉 ≥ 0 and so ||x|| ≥ 0. If ||x|| = 0, then 〈x, x〉 = 0 and so x = 0.Now, let x, y ∈ X be arbitrary. Then

||x+ y|| =√〈x+ y, x+ y〉

=√||x||2 + ||y||2 + 2 〈x, y〉

≤√||x||2 + ||y||2 + 2||x||||y||

≤√

(||x||+ ||y||)2

= ||x||+ ||y||,where the first inequality follows from Cauchy-Schwarz for a real inner productspace. �

Exercise 4. Let (X, 〈·, ·〉) be a complex inner product space. Let x, y ∈ X. Provethe Cauchy-Schwarz inequality for complex inner product spaces:

| 〈x, y〉 | ≤ 〈x, x〉1/2 〈y, y〉1/2

Solution.

Proof. There are a few proofs of this, I’ll give a couple here.

(a) Say that x, y ∈ X. The polynomial

p(t) = t2||y||2 + 2tre 〈x, y〉+ ||x||2 = ||x+ ty||2 ≥ 0

is nonnegative for all values of t, since the right most side is. Since the leadingcoefficient is positive, p(t) can have at most one real root (recall from geometrythat if a parabola has two real roots and opens upwards, then it is negativebetween the two roots). So the discriminant of the polynomial is nonpositive

b2 − 4ac = (2re 〈x, y〉)2 − 4||x||2||y||2 ≤ 0

So rearranging and taking square roots shows that

|re 〈x, y〉 | ≤ ||x||||y||(1)

Now, apply (1) to x and 〈x, y〉 y, then

| 〈x, y〉 | · ||x|| · ||y||| ≥ |re 〈x, 〈x, y〉 y〉 | = |re〈x, y〉 〈x, y〉 | = | 〈x, y〉 |2

which, up to rearranging, it precisely what we’d like to show.(b) Let λ > 0 be an parameter whose value we’ll chose later. Consider that

0 ≤ ||x− λy||2 = ||x||2 + λ2||y||2 − 2λre(〈x, y〉)Then rearranging shows that

re 〈x, y〉 ≤ ||x||2

2λ+λ||y||2

2(2)

Let’s minimize the right hand side as a function of λ. The derivative of thefunction

p(λ) =||x||2

2λ+λ||y||2

2is

p′(λ) =||y||2

2− ||x|||

2

2λ2

131A ASSIGNMENT 1 3

which vanishes at

λ =||x||||y||

(recall that λ > 0, so we ignore the negative value). The second derivative testverifies that this is a minimum (this point is more or less unnecessary, actually).Since (2) holds for every positive value of λ, we’ll choose λ = ||x||/||y||, whichgives us that

re 〈x, y〉 ≤ ||x||||y||Now since this holds for every choice of x, y, it holds for x and 〈x, y〉 y, whichgives the same conclusion as above.

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Exercise 5.

Solution.

Proof. We’ll show that A = {(x, y) ∈ R2 : x ≥ 0}. Let (x, y) ∈ R2. If x > 0, thenx ∈ A and so x is an adherent point of A vacuously. Suppose then that x = 0. Ifε > 0, then (ε/2, y) ∈ A and

||(x, y)− (ε.2, y)|| = ||(−ε/2, 0)|| = ε/2 < ε

and so (x, y) is an adherent point of A. If x < 0, then choosing ε = −x/2, we seethat the ball of radius ε about (x, y) does not intersect A, and so (x, y) is not anadherent point of A. �

Exercise 6.

Solution.

Proof. We’ll prove a more general result. From the proof in Exercise 2, we see thatif 1 ≤ p, q ≤ ∞, then

||x||`p(Rn) ≤ n1p ||x||`∞(Rn) ≤ n

1p ||x||`q(Rn) ≤ n

1p+ 1

q ||x||`∞(Rn) ≤ n1p+ 1

q ||x||`p(Rn)

or, in short,

n− 1

p ||x||`p(Rn) ≤ ||x||`q(Rn) ≤ n1q ||x||`p(Rn)

with the convention that 1/∞ = 0. Now, say {xj}∞j=1 is a sequence in Rn and

x0 ∈ Rn with xj converging to x0 with respect to d`p . Then d`p(Rn)(x0, xj) → 0 as

j →∞, and so

d`q(Rn)(xj , x0) = ||xj − x0||`q(Rn) ≤ n1/q||xj − x0||`p(Rn) = n1/qd`p(Rn)(x

j , x0)→ 0

as j →∞, and so xj → x0 with respect to d`q as j →∞. �

Exercise 7.

Solution.

Proof.

(i) Suppose x0 is an adherent point of E. Then B(x0, r) ∩E 6= ∅ for every r > 0.In particular, x0 is not an exterior point. If x0 is an interior point, we’re done.If x0 is not an interior point, then since x0 is not an exterior point by definitionit is a boundary point.

4 A. ADAM AZZAM

(ii) Suppose x0 is an interior point. For n ≥ 1 let xn = x0, then {xn}∞n=1 is asequence of elements in E which converges to x0 with respect to the metric d.Suppose then that x0 is a boundary point. Then B(x0, 2

−n)∩E 6= ∅ for n ≥ 1and so we may find some xn ∈ B(x0, 2

−n)∩E. Then {xn}∞n=1 is a sequence ofelements in E with

d(x0, xn) < 2−n,

so xn → x0 with respect to d.(iii) Suppose {xn}∞n=1 is a sequence of elements in E with xn → x0 with respect to

d. Let r > 0. Since xn → x0, we may find N ∈ N so that d(xn, x0) < r for alln ≥ N . But then

xN ∈ B(x0, r) ∩ E,and so B(x0, r) ∩ E 6= ∅. Since r was chosen arbitrarily, it follows that x0 isan accumulation point.

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Exercise 8.

Solution.

Proof. This solution is long, straightforward, and omitted. If you find that you’dlike a clarification, I invite you to attend my office hours. �

Exercise 9.

Solution.

Proof. Say {x(j)}∞j=k converges to some x ∈ X and x(jn) is a subsequence of x(j).

Let ε > 0. Find N ∈ N so that d(x(j), x0) < ε for all j ≥ N . Then for all n ≥ N we

see that jn ≥ N and so d(x(jn), x0) < ε. So x(jn) → x0, as desired. �

Exercise 10.

Solution.

Proof. Say (x(j))j=k∞ converges to x ∈ X with respect to d. Let ε > 0. Since

x(j) → x, there exists N so that d(x, x(j)) < ε/2 for all j ≥ N . But then for alln,m ≥ N we have

d(x(n), x(m)) ≤ d(x(n), x0) + d(x0, x(m)) < ε/2 + ε/2 = ε

and so x(j) is a Cauchy sequence. �

Exercise 11.

Solution.

Proof.

(i) Say (Y, d |Y×Y ) is complete. Suppose xn is a sequence in Y converging tox0 ∈ X with respect to d. Then {xn}∞n=1 is a Cauchy sequence with respect tod |Y×Y . Since (Y, d |Y×Y ) is complete, there exists some y0 ∈ Y with xn → y0with respect to d |Y×Y . Now,

0 ≤ d(x0, y0) ≤ d(x0, xn) + d(xn, y0) = d(x0, xn) + d |Y×Y (xn, y0)→ 0

and so by the squeeze theorem d(x0, y0) = 0. So x0 = y0 ∈ Y . So Y containsits limit points, and hence Y is closed.

131A ASSIGNMENT 1 5

(ii) Say Y is a closed subset of X. Say xn is a Cauchy sequence in Y with respectto d |Y×Y . Then xn is a Cauchy sequence in X, and since X is complete, thereexists some x0 ∈ X with xn → x0 with respect to d as n → ∞. Since Y isclosed, x0 ∈ Y and

d(x0, xn)→ 0 as n→∞.So every Cauchy sequence in Y has a limit in Y with respect to d, and so Y iscomplete.

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Exercise 12.

Solution.

Proof. Let X be an open subset of R. Define an equivalence relation on X bydeclaring that x ∼ y if the closed interval

[min{x, y},max{x, y}] ⊆ X.Fix x ∈ X. I claim that the equivalence class [x] = {y ∈ X : y ∼ x} is an openinterval. Let’s first prove it is open. Say y ∈ [x]. Since y ∈ X, there exists someε > 0 so that B(x, ε) ⊆ X. But then

[min{y, y + r},max{y, y + r}], [min{y, y − r},max{y, y − r}] ⊆ Xfor every 0 < r < ε/2. So y + r ∼ y − r ∼ y ∼ x for every 0 < r < ε/2. HenceB(y, ε/2) ⊆ [x], and so [x] is open. Now, let

ax = inf{y ∈ X : (y, x) ⊆ X} and bx = inf{y ∈ X : (y, x) ⊆ X}It’s not difficult to prove that (ax, bx) ⊆ [x]. If y ∈ [x] and x ≤ y. Then

[x, y] ⊆ XSince X is open, there exists some ε > 0 so that

(x, y + ε) ⊆ XBut then ax ≤ x ≤ y < y + ε ≤ bx. So y ∈ (ax, bx). This holds similarly if y ≤ x,which proves the reverse conclusion. Since equivalence classes are disjoint, andpartition X, this proves that X is a disjoint union of open intervals. To see why theremust be a countable number of intervals, define a map Φ : Q ∩X → {[x] : x ∈ X}by

Φ(q) = [q]

I claim this map is surjective. Indeed, say that x ∈ X. Then [x] is an interval. SinceQ is dense in R, there exists some q ∈ [x] ⊆ X. So q ∈ X ∩Q and Φ(q) = [x]. SinceQ is countable and Φ is surjective, we know that {[x] : x ∈ X} is at most countable,i.e. countable infinite or finite. �