The Hilbert Transform

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The Hilbert transform

1 Deﬁnition and properties

1 Recall the distribution pv( x ), deﬁned by Z ϕ(x) pv(1/x)(ϕ) := lim dx. →0 |x|≥ x

The Hilbert transform is deﬁned via the convolution with pv(1/x), namely

1 Z f(x − t) (Hf)(x) := lim dt. π →0 |t|≥ t

The main theorem we are going to prove in this note is the following.

Theorem 1.1. For any p ∈ (1, +∞), there exists a constant Cp such that

kHfkp < Cpkfkp (1.1)

for all f ∈ S(R). Thus, H extends to a bounded linear operator on all of Lp(R).

The ﬁrst remark we make is that the above theorem is false when p = 1. To see this, note that for smooth f with compact support, when x is very large, the main contribution to the integral

Z f(x − t) dt t comes from the values t which are not far away from x. This in general gives the decay 1 rate of order |x| , unless there is a magical cancellation due to the sign changes of f, in which case one can hope for a faster decay. Indeed, it is not hard to show that if f is continuous and has compact support with R f(t)dt = a, then a (Hf)(x) = + O(1/x2) πx

1 R for all large x. As a consequence, Hf ∈ L (R) if and only if f = 0. The statement of 1 the above theorem is also not true for p = +∞. The reason is that x is not integrable at inﬁnity, so one can not bound on kHfk∞ merely by using the maximum of f without any other information (e.g., the size of the support). For p ∈ (1, +∞), the above inequality (1.1) does hold. The proof we will give below consists of the following four steps.

1 1. For p = 2, we will show kHfk2 = kfk2. This uses the Fourier transform of pv(1/x) and then Plancherel’s theorem. 2. For p = 1, the bound in (1.1) does not hold. However, one has a weak type estimate, namely there exists C > 0 independent of f such that C µx : |(Hf)(x)| > λ < kfk λ 1 for all large λ. This weak type estimate is to replace the invalid L1 inequality, and its derivation is based on the Calderón-Zygmund decomposition of a function into small and large parts in such a way that some control can be obtained for the large part. 3. Based on the result of the above two steps, we then use interpolation to conclude that the inequality (1.1) holds for all p ∈ (1, 2). The relevant result we use here is a special case of Marcinkiewicz interpolation theorem. It states that if an operator satisfies weak type (1, 1) and (q, q) estimates, then it satisfies strong (p, p) inequality for all p ∈ (1, q). 4. Finally, we will prove (1.1) for p > 2 by using the duality between Lp and Lq for 1 1 p + q = 1.

Before we proceed to the proof, we ﬁrst give a few useful properties of the Hilbert transform. As far as the Fourier transform Hfd is concerned, we need the Fourier transform of the tempered distribution pv(1/x). A formal calculation gives

pv(1\/x)(ϕ) : = pv(1/x)(ϕ ˆ) Z ϕˆ(ξ) = dξ R ξ Z 1 Z = ϕ(x)e−2πiξxdx dξ R ξ R Z Z 1 = ϕ(x) · e−2πixξdξ dx. R R ξ If we let I = R 1 · e−2πixξdξ, then by symmetry, we have R ξ Z sin(2πxξ) Z +∞ sin(2πxξ) I = −i dξ = −2i dξ. R ξ 0 ξ R +∞ sin ξ π Now, recall the Dirichlet integral 0 ξ dξ = 2 ; this gives the value of I by I = −πi · sgn(x), where sgn(x) = 1 if x > 0 and −1 for x < 0. Substituting back into the expression of pv(1\/x), we get Z pv(1\/x)(ϕ) = −πi ϕ(x) · sgn(x)dx.

We have thus obtained the following proposition.

2 Proposition 1.2. The Fourier transform of pv(1/x) is pv\(1/x)(ξ) = −πi · sgn(ξ).

The above calculations are only formal in the sense that when changing the order of integration, the hypothesis of Fubini’s theorem is not satisﬁed. One needs the range of integration for ξ to be ﬁnite in order for the integrand to be L1, jointly in (ξ, x). So,

Exercise 1.3. Justify rigorously the above calculations.

2 Equality in L2

The assertion kHfk2 = kfk2 is an immediate consequence of the Fourier transform of 1 pv(1/x). Since Hf is the convolution of f with π pv(1/x), we easily get

Hfd(ξ) = −i · sgn(ξ)fˆ(ξ), which has the same L2 norm of fˆ. Thus, an application of Plancherel’s theorem gives ˆ kHfk2 = kHfdk2 = kfk2 = kfk2.

Furthermore, if we neglect the multiplication of −i for a moment, we see what the operator H does on f is simply changing the sign of its Fourier transform for negative ξ’s. If we do it twice, then we just change the sign back as if nothing had happened. Thus, taking into account the multiplication of −i, we have the relation

H2 = − id .

This says H is a unitary operator on L2 with H−1 = −H.

3 Weak L1 estimate

We now establish the weak-type (1, 1) estimate for the convolution operator H. For this, we need the Calder´on-Zygmund decomposition of an integrable function, which in turn depends on the following lemma.

1 Lemma 3.1. Let f ∈ L . Then, for any λ, there exists a subset Eλ ⊂ R such that

c 1. |f(x)| ≤ λ for almost every x ∈ Eλ;

1 2. µ(Eλ) < λ kfk1;

3. Eλ is a union of intervals Ik’s whose interiors are disjoint, and that 1 Z λ ≤ |f(x)|dx ≤ 2λ |Ik| Ik for each k.

3 Proof. Since f ∈ L1, we can have a family of disjoint intervals with equals size whose union is the whole real line and that for each I in this family, we have 1 Z |f|dx ≤ λ. |I| I This can be achieved as long as the interval length is large enough. Now, for each such interval I, we decompose it into two equal sub-intervals. If the average of |f| of one sub-interval is bigger than λ, then we label it as Ik and put it as one of the sub-intervals that consist of our set Eλ. We do this process for each of the initial intervals, and stop splitting when the average of |f| of one children is bigger than λ. Thus, we have obtained a sequence of disjoint intervals {Ik}. If Jk denotes the immediate parent of Ik, then the fact that one splits Jk into two sub-intervals (one of them being Ik) implies 1 Z |f| < λ. |Jk| Jk

Since |Jk| = 2|Ik|, we easily have 1 Z λ < |f|dx < 2λ. |Ik| Ik Also, we have X X 1 Z 1 |E | = |I | < |f|dx ≤ kfk . λ k λ λ 1 k k Ik

It remains to show f is almost everywhere bounded by λ outside Eλ. First, we note that c every x ∈ Eλ belongs to a nested sequence of intervals whose lengths shrink to 0, and that each of the interval in this sequence satisﬁes 1 Z |f(y)|dy ≤ λ. |I| I

c By Lebesgue’s diﬀerentiation theorem, for almost every x ∈ Eλ, f(x) is the limit of the c above average as |I| → 0, which then implies |f(x)| ≤ λ a.e. in Eλ. This completes the proof of the lemma.

Now, for f ∈ L1, we let

c f(x), x ∈ Eλ g(x) = 1 R , f(y)dy, x ∈ Ik. |Ik| Ik R Let bk = f · 1{Ik} − g, then bk is supported in the interval Ik with I bkdx = 0, and the P k function b = k bk satisﬁes f = g + b.

This is the Calder´on-Zygmund decomposition of f. The tail measure of Hf can be bounded by λ λ µx : |(Hf)(x)| > λ ≤ µx : |(Hg)(x)| > + µx : |(Hb)(x)| > , 2 2

4 and it suffices to bound each of the two terms on the right hand side. For the first one, we first note that our construction of g satisfies

2 g ∈ L , sup |g(x)| ≤ 2λ, kgk1 = kfk1. x Thus, we get λ C C C C µx : |(Hg)(x)| > ≤ kHgk ≤ kgk ≤ kgk kgk < kgk . 2 λ2 2 λ2 2 λ2 1 ∞ λ 1

We now turn to the functions bk. Each of them has support in the interval Ik with R R bkdx = 0. We claim that for any function u supported in an interval I with udx = 0, Ik I we have Z |Hu|dx < Ckuk1 (I∗)c for some universal constant C, where I∗ is the interval with the same center as I but R ∗ c with doubled size. To see this, ﬁrst note that since I udx = 0, then if x ∈ (I ) , we will have 1 Z u(t) u(t) (Hu)(x) = − dt. π I x − t x − c In particular, if x ∈ (I∗)c and t ∈ I, we will have

1 1 t − c |I| − = ≤ , x − t x − c (x − t)(x − c) (x − c)2 and as a consequence, we obtain Z Z ∗ c 1 (I ) |(Hu)(x)|dx ≤ C|I|kuk1 2 dx ≤ Ckuk1. (I∗)c (x − c)

5 ∗ c ∗ c P where in the middle line we have used (Eλ) ⊂ (Ik ) as well as |Hb| ≤ k |Hbk|. This ﬁnishes the proof of the weak (1, 1) estimate.

4 Interpolation for p ∈ (1, 2)

Given that the transform operator H satisﬁes both weak tpye (1, 1) and (2, 2) estimates, the strong (p, p) inequality then becomes a simple consequence of the following interpolation theorem.

Theorem 4.1. Suppose T is a sub-additive operator that is both weak (p, p) and weak (q, q). Then, T is of strong type (r, r) for all r ∈ (p, q).

To prove this theorem, we need the concept of a distribution function. For any function f, its distribution function ρ : R+ → R+ is deﬁned by ρf (λ) := µ x : |f(x)| > λ .

We will omit the subscript f if there is no confusion of the underlying function. The importance of the distribution function lies in the fact that knowing ρ(λ) for all λ would give complete knowledge of kfkp. In fact, for any p > 1, we have the identity

Z Z +∞ |f(x)|pdx = p λp−1ρ(λ)dλ. R 0 Now, let f ∈ Lr and write f = g + h with

g = 1{|f|>λ}, h = 1{|f|≤λ}.

In other words, g is the large part of f and h is the small part. By the assumption r ∈ (p, q), it is easy to see that g ∈ Lp and h ∈ Lq. Now let T be an operator satisfying the assumptions in the above theorem, then

|(T f)(x)| ≤ |(T g)(x)| + |(T h)(x)|, and as a consequence, we have λ λ x : |(T f)(x)| > λ ⊂ x : |(T g)(x)| > ∪ x : |(T h)(x)| > . 2 2 If ρ denotes the distribution function of T f, then the above inclusion together with the assumption T being both weak (p, p) and weak (q, q) gives

C C ρ(λ) ≤ p · kgkp + q · khkq. λp p λq q By the deﬁnition of g and h, we get Z Z C C q ρ(λ) ≤ p |f(x)|dx + q |f(x)| dx. (4.1) λ |f|>λ λ |f|≤λ

6 Now, by the relation

Z Z +∞ |T f|rdx = r λr−1ρ(λ)dλ, 0

r−1 we see that to get a bound on kT fkr, it suﬃces to multiply λ on both sides of (4.1) and integrate λ. For the ﬁrst term, we have

Z +∞ Z Z Z |f(x)| λr−p−1 |f|dx dλ = |f(x)| λr−p−1dλ dx 0 |f|>λ R 0 1 Z = |f|rdx. r − p R For the second one, we similarly have

Z +∞ Z Z Z +∞ λr−q−1 |f|rdx dλ = |f(x)|r λr−1−1dλ dx 0 |f|≤λ R |f(x)| 1 Z = |f(x)|rdx. q − r R Since p < r < q, we immediately get

kT fkr < Crkfkr.

r C C We can see from the above calculations that Cr ≈ r−p or q−r as r → p or q.

5 p > 2 case

Finally, we deal with the case p > 2 by duality. Let q be the conjugate of p, so the Lp norm of a function f could be represented by Z

kfkp = sup fgdx . kgkq=1

Now, let f ∈ Lp, we then have Z

kHfkp = sup (Hf)(x)g(x)dx kgkq=1 Z

= sup f(x)(H ∗ g)(x)dx kgkq=1

≤ sup kfkpkHgkq kgkq=1

≤ Ckfkp, which establishes the theorem. Note that here H∗ acts on g as a complex conjugate of H, so it does not change the norm anyway.