Homework 4

1. Let X and Y be normed spaces, T ∈ B(X, Y ) and {xn} a sequence in X. If xn → xweakly, show that Txn → Tx weakly.

Solution: We need to show that

g(Tnx)→ g(Tx) ∀g ∈ Y ∗.

It suffices to do this when ‖g‖Y ∗ = 1. Observe that

|g(Txn)− g(Tx)| = |g(Txn − Tx)|≤ ‖g‖Y ∗ ‖Txn − Tx‖Y≤ ‖g‖Y ∗ ‖T‖ ‖xn − x‖X= ‖T‖ ‖xn − x‖X .

Note that then we have

‖xn − x‖X = sup06=f∈X∗

|f(xn − x)|‖f‖X∗

So, let ε > 0 be given, and select f ∈ X∗ with ‖f‖∗X = 1 such that

‖xn − x‖X < |f(xn − x)|+ ε

2 ‖T‖.

Since xn → x weakly, there exists an integer N such that if n > N we have

|f(xn)− f(x)| = |f(xn − x)| < ε

2 ‖T‖.

This then gives,

|g(Txn)− g(Tx)| ≤ ‖T‖ ‖xn − n‖X ≤ ‖T‖(|f(xn − x)|+ ε

2 ‖T‖

)< ε.

2. If {xn} is a weakly convergent sequence in a real normed space X with xn → x0. Showthat there is a sequence {ym} of linear combinations of elements of {xn} which convergestrongly to x0.

Hint: Prove that the weak closure of a convex set is the same as the strong closure.

Solution: Note that if suffices to prove that the following statement. If E ⊂ X is aconvex subset of a normed space, then the weak closure of E, denoted by Ew is the sameas the strong closure of E, denoted by E. Once we have this statement, we can apply it to

E = span{xn} in the following way. We have that x0 ∈ Ew by hypothesis. But, Ew = Eby the claim. So we have that x ∈ E, which when we unpack the definitions gives theexistence of {ym} a sequence of linear combinations of {xn} that converges strongly to x0.

So it suffices to prove the claim. Since strong convergence implies weak convergence,we have that E ⊂ Ew. Suppose that x0 6∈ E. By your exam, there exists a functionalf ∈ X∗ such that

f(x0) < δ ≤ f(x) ∀x ∈ E.

Consider the set {x : f(x) < δ}, this is a weak open set of x0 that does not intersect E,and so x0 /∈ Ew. Which gives Ew ⊂ E.

3. Let Tn ∈ B(X, Y ). Show that Tn → T uniformly if and only if for every ε > 0 there is anN , depending only on ε such that for all n > N and all x ∈ X of norm 1 we have

‖Tnx− Tx‖Y < ε.

Solution: Suppose first that Tn → T uniformly. Then we have that for any ε > 0 thereexists N such that

‖Tn − T‖X→Y < ε.

Then we clearly have that

‖Tnx− Tx‖Y ≤ ‖Tn − T‖X→Y ‖x‖X = ‖Tn − T‖X→Y < ε

when n > N and x ∈ X with ‖x‖X = 1.For the other direction, suppose that ε > 0 is given, then we need to show that

‖Tn − T‖X→Y < ε

when n > N(ε). Note that

‖Tn − T‖X→Y = sup06=x∈X

‖Tnx− Tx‖Y‖x‖X

But, by the hypothesis, we have that

‖Tnx− Tx‖Y < ε ‖x‖X

when n > N(ε). This then clearly implies that

‖Tn − T‖X→Y = sup0 6=x∈X

‖Tnx− Tx‖Y‖x‖X

≤ ε

which proves the result.

4. If a normed space X is reflexive, show that X∗ is reflexive.

Solution: Let h ∈ X∗∗∗, then for every g ∈ X∗∗ there is an x ∈ X such that g = Cxsince X is reflexive. Here C is the canonical map identifying X and X∗∗. Hence,

h(g) = h(Cx) := f(x)

defines a bounded linear functional f on X, and C̃f = h, where C̃ : X∗ → X∗∗∗ is thecanonical mapping. Thus, C̃ is surjective and so X∗ is reflexive.

5. Let X be a normed space and {fn} a sequence of linear functionals on X. Show

(a) If fn → f weakly, then fn → f weak* too;

(b) Show that if X is reflexive, then the converse holds.

Solution: Part (a) is immediate since we have that X ⊂ X∗∗ where we have usedthe canonical mapping. Namely, each x ∈ X defines a functional x̂ : X∗ → C byx̂(f) := f(x). So, if we have weak convergence of the sequence of functionals, then wehave weak-* convergence as well.

If X is reflexive, then we have that X∗∗ = X, and so every linear functional on X∗ canbe identified with a functional of the form x̂.

6. A weak Cauchy sequence in a normed space is a sequence {xn} in X such that for everyf ∈ X∗ the sequence {f(xn)} is Cauchy. Show that a weak Cauchy sequence is bounded.

Solution: Define x̂n : X∗ → C by x̂n(f) = f(xn). Since {xn} is a weak Cauchy sequence,then we have that {f(xn)} is a Cauchy sequence in C and hence bounded. Thus, wehave that {x̂n(f)} is a bounded sequence, and since X∗ is a Banach space we can applythe Uniform Boundedness Principle to conclude that

supn‖x̂n‖ <∞

But, this then gives that supn ‖xn‖X <∞, which is what we wanted.

7. A normed space X is said to be weakly complete if each weak Cauchy sequence in Xconverges weakly in X. If X is reflexive, show that X is weakly complete.

Solution: Let {xn} be a weak Cauchy sequence in X. Again, we will identify xn with x̂nvia the canonical mapping. Note that by Problem 6 above, we have that supn ‖xn‖X <∞.

Since {f(xn)} is Cauchy in C, it has a limit, and we can define

ϕ(f) = limn→∞

f(xn) = limn→∞

x̂n(f)

which exists for each f ∈ X∗. This defines a map ϕ : X∗ → C. This map is clearly linearsince each x̂n is linear. Also, it is bounded since

|x̂n(f)| = |f(xn)| ≤ ‖f‖X∗ ‖xn‖X ≤ supn‖xn‖ ‖f‖X∗ .

Passing to the limit we have

|ϕ(f)| ≤ supn‖xn‖ ‖f‖X∗ .

So we have that ϕ ∈ X∗∗. But, since X is reflexive, there exists x ∈ X such that x̂ = ϕ.But, then we have

f(x) = x̂(f) = ϕ(f) = limn→∞

x̂n(f) = limn→∞

f(xn)

so that xn → x weakly since f(xn)→ f(x) for all f ∈ X∗.

8. Put fn(t) = eint, t ∈ [−π, π]. Let Lp = Lp(−π, π) with respect to Lebesgue measure. If1 ≤ p <∞, show that fn → 0 weakly in Lp, but not strongly.

Solution: Note that if p is a polynomial on [−π, π] then we clearly have that

〈p, fn〉L2 → 0.

The polynomials are dense in Lp when 1 ≤ p <∞, so for any f ∈ Lp, select a polynomialp such that ‖f − p‖Lp < ε. Then we have that

〈f, fn〉L2 = 〈f − p, fn〉L2 + 〈p, fn〉L2 .

The second term goes to zero as n→∞, in particular if n > N we have |〈p, fn〉L2| < ε .For the first term, note that we have

|〈f − p, fn〉L2| ≤ ‖f − p‖p ‖fn‖ q = ‖f − p‖p < ε

where 1p

+ 1q

= 1. Thus for n > N we have that

|〈f, fn〉L2| < 2ε.

It is immediate that fn does not converge strongly to 0 in Lp when 1 ≤ p <∞ since eachelement has norm 1.

9. A subset of a Banach space X is weakly bounded if for all f ∈ X∗, supx∈S |f(x)| < ∞.Prove that S is weakly bounded if and only if it is strongly bounded, i.e., supx∈S ‖x‖X <∞.

Solution: It is clear that if the set S is strongly bounded, then it must be weaklybounded since

supx∈S|f(x)| ≤ sup

x∈S‖f‖X∗ ‖x‖X = ‖f‖X∗ sup

x∈S‖x‖X <∞.

Suppose now that S is weakly bounded. Then we have that

supx∈S|f(x)| <∞

for all f ∈ X∗. Define x̂ : X∗ → C by x̂(f) = f(x). Then, we have that

supx∈S|x̂(f)| = sup

x∈S|f(x)| <∞.

So, we have that for all x ∈ S that x̂ is a uniformly bounded family of operators fromX∗ → C. Hence by the Uniform Boundedness Principle, we have that

supx∈S‖x̂‖X∗∗ <∞

But ‖x‖X = ‖x̂‖X∗∗ , and so we have the result.

10. If {xn} ⊂ `1 show that∑∞

j=1 y(j)xn(j)→ 0 for every y ∈ c0 if and only if supn ‖xn‖`1 <∞ and xn(j)→ 0 for all j ≥ 1.

Solution: Note that we have c∗0 = `1. For notational simplicity, let 〈xn, y〉 denote theinfinite sum

∞∑j=1

xn(j)y(j).

The easy direction then follows from testing on an obvious element in c0 and the UniformBoundedness Principle. If we have

∑∞j=1 y(j)xn(j)→ 0 for every y ∈ c0, then taking

y = (0, . . . , 0, 1, 0, . . .)

with the one occurring in the jth position, we have that

xn(j) =∞∑n=0

y(j)xn(j)→ 0.

Since xn ∈ `1, then it is a bounded linear functional on so we have that |〈xn, y〉| ≤‖xn‖`1 ‖y‖`∞ . The Uniform Boundedness Principle gives that

supn‖xn‖`1 <∞.

Conversely, if we have supn ‖xn‖`1 < ∞ and xn(j) → 0 for all j ≥ 1, we need to show∑∞j=1 y(j)xn(j)→ 0. Fix y ∈ c0 and note that for any ε > 0 there exists J such that

|y(j)| < ε.

Now we have that

∞∑j=1

y(j)xn(j) =J−1∑j=1

y(j)xn(j) +∞∑j=J

y(j)xn(j).

For the second term, we have that∣∣∣∣∣∞∑j=J

y(j)xn(j)

∣∣∣∣∣ ≤∞∑j=J

|y(j)| |xn(j)| ≤ ε ‖xn‖`1 ≤ ε supn‖xn‖`1 .

This can be made as small as we wish by choosing ε appropriately. Now for the firstterm, we have that xn(j) → 0 for j = 1, . . . , J − 1. Select N so that |xn(j)| ≤ ε

J. Then

we have that ∣∣∣∣∣J−1∑j=1

y(j)xn(j)

∣∣∣∣∣ ≤ ‖y‖`∞J−1∑j=1

|xn(j)| ≤ ‖y‖`∞ ε.

Combining these two estimates gives that 〈xn, y〉 → 0 as n→∞.