MATH 174A: PROBLEM SET 6

Suggested Solution

Problem 1. For f, g ∈ C(S1), let

f ∗ g =1

2π

∫S1

f(θ − ω)g(ω) dω.

(1) Show that f ∗ g ∈ C(S1) and ‖f ∗ g‖∞ ≤ ‖f‖∞‖g‖∞, where ‖f‖∞ = sup{|f(θ)| :θ ∈ S1}.

(2) Show that f ∗ g = g ∗ f .(3) Show that in fact ‖f ∗ g‖∞ ≤ ‖f‖∞‖g‖1, where ‖g‖1 = 1

2π

∫|g(θ)| dθ. Use this to

show that convolution extends to a continuous (i.e. bounded) bilinear map

∗ : C(S1)× L1(S1) → C(S1).

(4) Show that for f, g ∈ C(S1), ‖f∗g‖1 ≤ ‖f‖1 ‖g‖1. Use this to show that convolutionextends to a bilinear map

∗ : L1(S1)× L1(S1) → L1(S1).

(5) Show that if f ∈ C1(S1) and g ∈ L1(S1) then f ∗ g ∈ C1(S1), (f ∗ g)′ = f ′ ∗ g.(Hint: Assume g ∈ C(S1), and show f ∗ g ∈ C1(S1), ‖f ∗ g‖C1 ≤ ‖f‖C1‖g‖L1 , with

‖f‖Ck =∑k

j=0 ‖f (j)‖∞.)

(6) Show more generally that if f ∈ Ck(S1) and g ∈ L1(S1) then f ∗ g ∈ Ck(S1), and(f ∗ g)(k) = f (k) ∗ g.

(7) Show that F(f ∗ g) = (Ff)(Fg).

Solution:

(1) f ∗ g ∈ C(S1) Since S1 is compact, f is continuous on S1 implies that f is actually

uniformly continuous. Also, a continuous function on a compact set is bounded.So we can find an M > 0 such that |g(ω)| < M for all ω ∈ [0, 2π]. Let ε > 0, byuniform continuity of f , there exists a δ > 0 such that

|f(x)− f(y)| < ε whenever |x− y| < δ.

Suppose |θ − φ| < δ, then

|(f ∗ g)(θ)− (f ∗ g)(φ)| ≤ 1

2π

∫S1

|f(θ − ω)− f(φ− ω)||g(ω)|dω < Mε.

Therefore, f ∗ g is uniformly continuous on S1, hence f ∗ g ∈ C(S1).

‖f ∗ g‖∞ ≤ ‖f‖∞‖g‖∞ This is direct:

|(f ∗ g)(θ)| ≤ 1

2π

∫S1

|f(θ − ω)||g(ω)|dω ≤ ‖f‖∞‖g‖∞

2

for all θ. Therefore, ‖f ∗ g‖∞ ≤ ‖f‖∞‖g‖∞.

(2) f ∗ g = g ∗ f This is simply from a change of variable and periodicity:

(f ∗ g)(θ) =1

2π

∫ 2π

0

f(θ − ω)g(ω)dω =1

2π

∫ θ

θ−2π

f(φ)g(θ − φ)dφ

=1

2π

∫ 2π

0

g(θ − φ)f(φ)dφ = (g ∗ f)(θ).

(3) ‖f ∗ g‖∞ ≤ ‖f‖∞‖g‖1

|(f ∗ g)(θ)| ≤ 1

2π

∫S1

|f(θ − ω)||g(ω)|dω ≤ ‖f‖∞1

2π

∫S1

|g(ω)|dω = ‖f‖∞‖g‖1

for every θ. Taking supremum gives the inequality.

Extending ∗ to a bilinear map ∗ : C(S1)× L1(S1) → C(S1) Note that ∗ : C(S1)×C(S1) → C(S1) is a bilinear map. Fix any f ∈ C(S1), the map f ∗ · : C(S1) →C(S1) is linear. The inequality we have just proved tells us that f ∗ · is continuousconsidered as a map (C(S1), ‖.‖1) → (C(S1), ‖.‖∞). By B.L.T. theorem(note that(C(S1), ‖.‖∞) is complete), we can extend this map to a bounded linear mapon its completion L1(S1). In this way, we have extended the convolution to acontinuous bilinear map ∗ : C(S1)× L1(S1) → C(S1). Linearity and continuity inthe second slot comes from the extension automatically. For the first slot, fix anyg ∈ L1(S1), pick a Cauchy sequence gn ∈ C(S1) converging to g in L1. Then f ∗ gn

converges uniformly to f ∗ g for all f ∈ C(S1). Hence (c1f1 + c2f2) ∗ gn convergesto c1(f1 ∗ g) + c2(f2 ∗ g) uniformly, proving linearity. Also ‖f ∗ gn‖∞ ≤ ‖f‖∞‖gn‖1

implies that ‖f ∗ g‖∞ ≤ ‖f‖∞‖g‖1 hence ∗ is continuous in the first slot. Sinceboth (C(S1), ‖.‖∞) and (L1(S1), ‖.‖1) are Banach spaces, separate continuity ofa bilinear map implies joint continuity. Thus, ∗ : C(S1) × L1(S1) → C(S1) is ajointly continuous bilinear map.

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(4) For f, g ∈ C(S1), ‖f ∗ g‖1 ≤ ‖f‖1‖g‖1

‖f ∗ g‖1 =1

2π

∫ 2π

0

|(f ∗ g)(θ)|dθ

≤ 1

(2π)2

∫ 2π

0

∫ 2π

0

|f(θ − ω)g(ω)|dωdθ

=1

(2π)2

∫ 2π

0

∫ 2π

0

|f(θ − ω)g(ω)|dθdω

=1

(2π)2

∫ 2π

0

(

∫ 2π

0

|f(θ − ω)|dθ)|g(ω)|dω

=1

(2π)2

∫ 2π

0

(

∫ 2π

0

|f(θ)|dθ)|g(ω)|dω

=1

2π

∫ 2π

0

|f(θ)|dθ · 1

2π

∫ 2π

0

|g(ω)|dω

= ‖f‖1‖g‖1

Next, we will show that this inequality actually holds for any g ∈ L1(S1). Letgn ∈ C(S1) be a Cauchy sequence converging to g in L1. Then, ‖f ∗ gn‖1 ≤‖f‖1‖gn‖1 for all n. When n tends to infinity, ‖gn‖1 tends to ‖g‖1. Also, as f isbounded, ‖f ∗gn‖1 tends to ‖f ∗g‖1. Hence ‖f ∗g‖1 ≤ ‖f‖1‖g‖1 for any f ∈ C(S1),g ∈ L1(S1).

The convolution extend to a bilinear map ∗ : L1(S1)× L1(S1) → L1(S1) Fix any

g ∈ L1(S1), the inequality above shows that we have a map · ∗g : C(S1) → L1(S1).This is actually a bounded linear map. By B.L.T. theorem, this map extends toa continuous linear map from L1(S1) to L1(S1). Similar argument as in part (3)proves that this is a jointly continuous bilinear map.

(5) For f ∈ C1(S1), g ∈ L1(S1), f ∗ g ∈ C1(S1) and (f ∗ g)′ = f ′ ∗ g. First, we assume

that g ∈ C(S1). We want to show that f ∗ g ∈ C1(S1). Note that

(f ∗ g)(θ + h)− (f ∗ g)(θ)

h=

1

2π

∫ 2π

0

f(θ + h− ω)− f(θ − ω)

hg(ω)dω.

Take h → 0, and we can actually take the limit inside the integral by DominatedConvergence Theorem, so f ∗ g is differentiable and (f ∗ g)′ = f ′ ∗ g ∈ C(S1) sinceboth f ′, g ∈ C(S1). Therefore,

‖f ∗ g‖C1 = ‖f ∗ g‖∞ + ‖(f ∗ g)′‖∞ = ‖f ∗ g‖∞ + ‖f ′ ∗ g‖∞≤ ‖f‖∞‖g‖1 + ‖f ′‖∞‖g‖1 = ‖f‖C1‖g‖1.

Since (C1(S1), ‖.‖C1) is complete, so f ∗ g ∈ C1(S1) even for g ∈ L1(S1). The onlything left is the equality (f ∗ g)′ = f ′ ∗ g. Suppose g ∈ L1(S1) and gn ∈ C(S1) is aCauchy sequence converging to g in L1. Then (f ∗ gn)′ = f ′ ∗ gn for all n. Since gn

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converges to g in L1 and ‖f ∗ g‖C1 ≤ ‖f‖C1‖g‖1 holds for any g ∈ L1, f ∈ C1(S1),we have (f ∗gn)′ and f ′ ∗gn converging uniformly to (f ∗g)′ and f ′ ∗g respectively.Hence (f ∗ g)′ = f ′ ∗ g.

(6) If f ∈ Ck(S1) and g ∈ L1(S1), then f ∗ g ∈ Ck(S1) and (f ∗ g)(k) = f (k) ∗ g. We will

do it by induction on k. The cases k = 0, 1 are proved above. Assume the state-ment has shown to be true for k−1. Then f (k−1) ∈ C1(S1), hence apply the resultof (5) again, we conclude that f (k−1)∗g ∈ C1(S1) and (f (k−1)∗g)′ = f (k)∗g. Applyinduction hypothesis, we know that f ∗ g ∈ C(k)(S1) and (f ∗ g)(k) = f (k) ∗ g.

(7) F(f ∗ g) = (Ff)(Fg) We have to show that f ∗ g(k) = f(k)g(k) for every integer

k. Recall

f ∗ g(k) =1

2π

∫ 2π

0

(f ∗ g)(θ)e−ikθdθ

=1

(2π)2

∫ 2π

0

∫ 2π

0

f(θ − ω)g(ω)e−ikθdωdθ

=1

(2π)2

∫ 2π

0

∫ 2π

0

f(θ − ω)g(ω)e−ikθdθdω

=1

(2π)2

∫ 2π

0

(∫ 2π

0

f(θ − ω)e−ik(θ−ω)dθ

)g(ω)e−ikωdω

=1

(2π)2

∫ 2π

0

(∫ 2π

0

f(θ)e−ik(θ)dθ

)g(ω)e−ikωdω

=

(1

2π

∫ 2π

0

f(θ)e−ik(θ)dθ

)(1

2π

∫ 2π

0

g(ω)e−ikωdω

)= f(k)g(k).

Problem 2. Show that L2(S1) ⊂ L1(S1). That is, show that the identity map i : C(S1) →C(S1) extends to an injective continuous linear map ι : L2(S1) → L1(S1).

Hint: Show that there is C > 0 such that for all f ∈ C(S1), ‖f‖1 ≤ C‖f‖2, and use thisto conclude that we get a bounded linear extension ι : L2(S1) → L1(S1). Here ‖.‖2 is theL2-norm. Now suppose ι(f) = 0, f ∈ L2(S1). Let {fn}, fn ∈ C(S1) be a Cauchy sequencein L2, fn → f . Show that there is φ ∈ C(S1) such that (f, φ)L2 6= 0 (hint: C(S1) is densein L2(S1)), hence limn→∞(fn, φ)L2 6= 0. But now show that |(fn, φ)| ≤ C‖fn‖1.

Solution: ‖f‖1 ≤ C‖f‖2 Let f ∈ C(S1), we have, using Holder inequality,

‖f‖1 =

∫S1

|f | ≤(∫

S1

|f |2)1/2

·(∫

S1

12

)1/2

=√

2π‖f‖2.

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Hence, the inclusion ι : C(S1) → L1(S1) can be extended to a bounded linear mapι : L2(S1) → L1(S1).

ι : L2(S1) → L1(S1) is injective. Suppose ι(f) = 0, f ∈ L2(S1), f 6= 0. Let {fn}, fn ∈C(S1) be a Cauchy sequence in L2, fn → f in L2. Next, we claim that there is φ ∈ C(S1)such that (f, φ)L2 6= 0. Suppose not, then (f, φ) = 0 for all φ ∈ C(S1). We would have(f, fn)L2 = 0 for all n. Since fn → f in L2, this implies that (f, f)L2 = 0, so f = 0. As aresult, we can pick a φ ∈ C(S1) such that (f, φ)L2 6= 0, hence lim(fn, φ)L2 = (f, φ)L2 6= 0.On the other hand,

|(fn, φ)L2 | ≤∫

S1

|fn||φ| ≤ ‖φ‖∞∫

S1

|fn| ≤ C‖fn‖1,

since φ ∈ C(S1). Since ι(f) = 0, i.e. ‖f‖1 = 0, ‖f‖1 ≤ C‖f‖2, fn → f in L2 implies thatfn → f in L1, so lim ‖fn‖1 = ‖f‖1 = 0. Hence lim(fn, φ)L2 = 0, cntradicting our choiceof φ such that lim(fn, φ)L2 6= 0.

Problem 3. Show that if x ∈ Rn and δ > 0 then there exists φ ∈ C∞(Rn) such thatφ ≥ 0, φ(x) > 0 and if ‖y − x‖ ≥ δ then φ(y) = 0. (Hint: Show first that the functionχ : R → R defined by χ(t) = 0 for t ≤ 0, χ(t) = e−1/t, t > 0, is C∞.)

Solution: Define χ : R → R as

χ(t) =

{0 , t ≤ 0

e−1/t , t > 0.

χ is C∞ This is clear that χ is smooth everywhere except possibly at 0. Note that by

successive differentiation, over t > 0,

χ′(t) = e−1/tP (1/t),

where P is a polynomial. Using the fact that limx→+∞xn

ex = 0 for all n, we know that

limt→0+ χ(n)(t) = 0 for all n. Therefore, χ is smooth everywhere.

Construction of φ Define φ : Rn → R as

φ(y) = χ(δ2/4− ‖y − x‖2).

Then φ ≥ 0 since χ ≥ 0. φ(x) = χ(δ2/4) > 0 and if ‖y − x‖ ≥ δ, then φ(y) = χ(t) forsome t < 0, so φ(y) = 0.

Problem 4. (Taylor 3.2.1.) Suppose u satisfies the following Neumann boundary problemin the disk D′:

∆u = 0 in D′,∂u

∂r= g on S1.

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If u = PI(f), show that f and g must be related by g(k) = |k|f(k), for all k ∈ Z, that is,

g = Nf,

with N defined by (2.13).

Solution: Recall that

u = PI(f)(r, θ) =∑

f(k)r|k|eikθ.

Hence,∂u

∂r=∑

f(k)|k|r|k|−1eikθ.

On S1,

g =∂u

∂r(1, θ) =

∑f(k)|k|eikθ.

In other words, g(k) = f(k)|k|.

Problem 5. (Taylor 3.2.2.) Define the function kN by

kN(θ) =∑k 6=0

|k|−1eikθ.

Show that kN ∈ L2(S1) ⊂ L1(S1). Also show that, provided g ∈ L2(S1) and∫S1

g(θ)dθ = 0,

a solution to (2.45) is given by

f(θ) = (2π)−1

∫S1

kN(θ − ϕ)g(ϕ)dϕ = Tg(θ).

Solution: kN ∈ L2(S1) ⊂ L1(S1) The inclusion L2(S1) ⊂ L1(S1) was proved in Problem

2. To see that kN ∈ L2, note that kN is the Fourier series of the smooth functionf(θ) = −iθ, so kN is a uniformly convergent series. This allows us to permute summationand integral signs. Using orthogonality of eikθ,

‖kN‖22 =

∫S1

(∑k 6=0

|k|−1eikθ

)·

(∑l 6=0

|l|−1e−ilθ

)=

∑k 6=0,l 6=0

|kl|−1

∫S1

ei(k−l)θ =∑k 6=0

1

k2< ∞.

Therefore, kN ∈ L2.

g(k) = |k|f(k) Note that both kN , g ∈ L1(S1), observe that (see Problem 1) f = kN ∗ g.

By Problem 1.(7), for k 6= 0,

f(k) = kN(k)g(k) = |k|−1g(k),

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in other words, g(k) = |k|f(k). For k = 0, since∫S1

g(θ)dθ = 0,

we have g(0) = 0. Hence g(k) = |k|f(k) for all integers k.