Math 641 Lecture #45 ¶9.5,9.6 the Codomain of the Fourier Transform

Math 641 Lecture #45 ¶9.5,9.6 The Codomain of the Fourier Transform

Question. Into what space does the Fourier transform send L1(R) functions?

Deﬁnition. Let f be any function on R.

For each y ∈ R, the translate of f by y is the function fy deﬁned by fy(x) = f(x − y).

p Theorem (9.5). Let 1 ≤ p < ∞. For each f ∈ L (R), the mapping y → fy from R into Lp(R) is uniformly continuous. Proof. Let f ∈ Lp(R), and ﬁx > 0. p Since Cc(R) is dense in L (R), there is a g ∈ Cc(R) such that supp(g) ⊂ [−A, A] for A > 0, and kf − gkp < . Since a continuous function with a compact support is uniformly continuous, there is δ ∈ (0,A)[δ = min{δ∗, A/2} where δ∗ corresponds to ∗] such that

|s − t| < δ ⇒ |g(s) − g(t)| < (3A)−1/p [(3A)−1/p = ∗].

min{−A + s, −A + t} < x < max{A + s, A + t}.

The diﬀerence max{A + s, A + t} − min{−A + s, −A + t} is at most 2A + δ. Thus

Z ∞ Z ∞ p p |gs(x) − gt(x)| dx = |g(x − s) − g(x − t)| dx −∞ −∞ Z max{A+s,A+t} = |g(x − s) − g(x − t)|p dx < (3A)−1p(2A + δ) < p, min{−A+s,−A+t} so that kgs − gtkp < . Since the Lp-norms are translation invariant (with respect to Lesbegue measure),

kfs − ftkp = kfs − gskp + kgs − gtkp + kgt − ftkp

= k(f − g)skp + kgs − gtkp + k(f − g)tkp < 3 whenever |s − t| < δ.

Therefore y → fy is uniformly continuous on R. 1 ˆ ˆ Theorem (9.6). If f ∈ L (R), then f ∈ C0(R) and kfku ≤ kfk1. Proof. The Fourier transform of f ∈ L1(R) is in L∞(R): Z ∞ Z ∞ ˆ −ixt |f(t)| = f(x)e dm(x) ≤ |f(x)| dm(x) = kfk1, −∞ −∞ ˆ so that kfk∞ ≤ kfk1. ∞ ˆ ˆ ˆ Recall that C0(R) ⊂ L (R) where kfku = kfk∞ when f ∈ C0(R).

It remains to show that f ∈ C0(R).

−itnx −itx Fix t ∈ R, let tn → t, and set gn(x) = |f(x)| |e − e |. 1 The sequence {gn} converges pointwise to 0, and is dominated by 2|f| ∈ L (R). By LDCT, Z ∞ ˆ ˆ −itnx itx lim |f(tn) − f(t)| ≤ lim |f(x)| |e − e | dm(x) n→∞ n→∞ −∞ Z ∞ = lim gn(x) dm(x) = 0, −∞ n→∞ so that fˆ is continuous. Now consider t with |t| large. Since eiπ = −1, then Z ∞ fˆ(t) = f(x)e−itxe−iπeiπ dm(x) −∞ Z ∞ = eiπ f(x)e−it(x+π/t) dm(x) −∞ Z ∞ = − f(x − π/t)e−itxdm(x). −∞ This implies that

2fˆ(t) = fˆ(t) + fˆ(t) Z ∞ Z ∞ = f(x)e−ixt dm(x) − f(x − t/π)e−itx dm(x) −∞ −∞ Z ∞ π = f(x) − f x − e−ixt dm(x) −∞ t so that ˆ 2|f(t)| ≤ kf0 − fπ/tk1. ˆ By Theorem 9.5, kf0 − fπ/tk1 → 0 as |0 − π/t| → 0, so that |f(t)| → 0 as |t| → ∞. ˆ ˆ ˆ Therefore, f ∈ C0(R) and kfku = kfk∞ ≤ kfk1. Remarks. The Banach space C0(R) is a commutative Banach algebra when equipped with the pointwise product: kfgku ≤ kfkukgku.

There is no unit in this C0(R), i.e. no element u ∈ C0(R) such that uf = f for all f ∈ C0(R). [The only candidate for a unit is the constant 1 function, which is continuous, but does not vanish at inﬁnity].

Deﬁnition. Let A and B be (complex) Banach algebras. An algebra homomorphism is a linear transformation Λ : A → B such that

Λ(fg) = Λ(f)Λ(g) for all f, g ∈ A.

Notation. Let F denote the Fourier transform: F(f) = fˆ.

1 Proposition. F : L (R) → C0(R) is an algebra homomorphism for which (a) kFk ≤ 1 [the Fourier transform is a bounded linear transformation], and

(b) kF(f ∗ g)ku ≤ kfk1kgk1 for all f, g. Proof. Linearity of F is clear, as is F(f ∗ g) = F(f)F(g). 1 By Theorem 9.6, kF(f)ku ≤ kfk1 for all f ∈ L (R), which implies that

kFk = sup{kF(f)ku : kfk1 = 1} ≤ 1.

This and the Convolution Theorem imply that kF(f ∗ g)ku ≤ kFk kf ∗ gk1 ≤ kf ∗ gk1 ≤ kfk1kgk1.////

Proposition. The commutative Banach algebra L1(R) has no unit. Proof. Suppose that there is a unit v ∈ L1(R), i.e. f ∗ v = f for all f ∈ L1(R). Since F is an algebra homomorphism, it follows that F(f) = F(f)F(v); hence fˆ(t) vˆ(t) − 1 = 0 for all t ∈ R.

1 Consider the L (R) function f(x) = χ[−1,1](x). For this, ( 2t−1 sin t if t 6= 0, fˆ(t) = F(f)(t) = 2 if t = 0.

[see Homework Problem Ch.9 #2]. iαx ˆ ˆ ˆ ˆ Since F f(x)e = f(t − α) = fα(t) for α ∈ R, then fα(α) = f(0) = 2. ˆ Thus fα(α) vˆ(α) − 1 = 0 implies thatv ˆ(α) = 1 for each α ∈ R.

But the constant 1 function is not in C0(R).