A Simple Proof

of Hunt – Muckenhoupt – Wheeden Theorem.

S. Treil and A. Volberg

July 31, 1997

First draft.

1. A weighted imbedding theorem.

Let v(λ) denotes the harmonic extension of a function v defined on a unit circle 𝕋 at the point λ ∈ 𝔻.

Theorem 1.1 Let a weight v ∈ L^{1}(𝕋) and a measure μ in 𝔻 satisfy

∫_{𝔻} \frac{1 − |λ|^{2}}{|1 − \overline{λ}z|^{2}} v(z)^{2}dμ(z) ≤ Cv(λ)

where the constant C does not depend on λ ∈ 𝔻. Then the following imbedding theorem

∫_{𝔻} |(fv^{1/2})(z)|^{2} dμ(z) ≤ K(C) ∫_{𝕋} |f|^{2},  ∀f ∈ L^{2}(T)  (1.1)

holds.

Corollary 1.2. Let w be a weight on 𝕋 such that w^{−1} ∈ L^{1}(𝕋), and μ be a measure in 𝔻. If

∫_{𝔻} \frac{1 − |λ|^{2}}{|1 − \overline{λ}z|^{2}} [w^{−1}(z)]^{2}dμ(z) ≤ Cw^{−1}(λ)

then the following weighted imbedding theorem

∫_{𝔻} |f(z)|^{2} dμ(z) ≤ K(C) ∫_{𝕋} |f|^{2}w,  ∀f ∈ L^{2}(T, v) .  (1.2)

holds.

Remark. It is easy to show that the condition w^{−1} ∈ L^{1} is necessary for the existence of a nonzero measure μ satisfying (1.2).

This Corollary follows immediately from Theorem 1 if we put v = w^{−1} and note that the imbedding (1.2) is equivalent to (1.1).

To prove Theorem 1 we need the following well known test of boundedness of an integral operator, see [1]

Theorem 1.3 (Senichkin – Vinogradov Test). Let Z be a measure space, k be a locally summable nonnegative measurable function on Z × Z. Let

∫_{Z} k(s, t)k(s, x)ds ≤ C [k(x, t) + k(t, x)] , a.e. on Z × Z .

Then for any positive g ∈ L^{2}(Z)

∫∫ _{Z×Z}k(s, t)g(s)g(t)dsdt ≤ 2C∥g∥_{L^{2}(Z)}^{2} ,

and the integral operator with kernel k(s, t) is bounded in L^{2}(Z).

We present a very simple proof of the theorem only for the sake of completeness.

Proof of Theorem 1. Let f ∈ L^{2}, ∥f∥_{2} = 1 be a nonnegative functions such that

Q =^{def} ∫∫_{Z×Z} k(s, t)f(s)f(t)dsdt < ∞ .

Then

Q^{2} = (∫_{Z} f(s) (∫_{Z} k(s, t)f(t)dt) ds)^{2} ≤ ∥f∥_{L^{2}(Z)}^{2} · ∥∫_{Z} k(., t)f(t)dt∥^{2} ≤
≤ ∫_{Z} ds (∫∫_{Z×Z} k(s, t)k(s, x)f(t)f(x)dtdx) =
= ∫∫_{Z×Z} f(t)f(x) (∫_{Z} k(s, t)k(s, x)ds) ≤
≤ 2C ∫_{Z}[k(t, x) + k(t, x)]f(t)f(x)dtdx = 2CQ .

Therefore the corresponding integral operator is bounded and its norm can be estimated above as \sqrt{2C}. □

Proof of Theorem 1. Let J be the imbedding operator that we want to prove is bounded, J : L^{2} → L^{2}(μ),

J f(z) = (fv^{1/2})(z) = ∫_{𝕋} \frac{1 − |z|^{2}}{|1 − \overline{z}t|^{2}} f(t)v^{1/2}(t)dm(t) .

Then the (formal) conjugate to J operator J^{∗}, J^{∗} : L^{2}(μ) → L^{2} can be written as

J^{∗}f(t) = v^{1/2}(t) ∫_{𝔻} \frac{1 − |z|^{2}}{|1 − \overline{z}t|^{2}} f(z)dμ(z) ,

so

∥J^{∗}f∥^{2} = ∫_{𝔻} ∫_{𝔻} (∫_{𝕋} v(t)\frac{1 − |z|^{2}}{|1 − \overline{z}t|^{2}} \frac{1 − |s|^{2}}{|1 − \overline{s}t|^{2}} dm(t)) f(z)\overline{f(s)}dμ(z)dμ(s)  (1.3)

It is easy to estimate the inner integral. Namely,

|1 − \overline{s}z| = |1 − \overline{s}t + \overline{s}t − \overline{s}z| ≤ |1 − \overline{s}t| + |s| · |t − z| ≤ |1 − \overline{s}t| + |1 − \overline{z}t| ,

so

|1 − \overline{s}t|^{−1}|1 − \overline{z}t|^{−1} ≤ |1 − \overline{s}z|^{−1} (|1 − \overline{s}t|^{−1} + |1 − \overline{z}t|^{−1}) ,

and finally

|1 − \overline{s}t|^{−2}|1 − \overline{z}t|^{−2} ≤ 2 · |1 − \overline{s}z|^{−2} (|1 − \overline{s}t|^{−2} + |1 − \overline{z}t|^{−2}) .  (1.4)

Hence we can estimate the inner integral in (1.3):

∫_{𝕋} v(t)\frac{1 − |z|^{2}}{|1 − \overline{z}t|^{2}} \frac{1 − |s|^{2}}{|1 − \overline{s}t|^{2}} dm(t) ≤

\frac{2}{|1 − \overline{s}z|^{2}} \{∫_{𝕋} v(t)\frac{(1 − |z|^{2})(1 − |s|^{2})}{|1 − \overline{s}t|^{2}}dm(t) + ∫_{𝕋} v(t)\frac{(1 − |s|^{2})(1 − |z|^{2})}{|1 − \overline{z}t|^{2}}dm(t)\} ≤

≤ 2 \{\frac{(1 − |z|^{2})}{|1 − \overline{s}z|^{2}}v(s) + \frac{(1 − |s|^{2})}{|1 − \overline{s}z|^{2}}v(z)\}

Therefore

∥J^{∗}f∥^{2} ≤ 4 ∫_{𝔻} ∫_{𝔻} \frac{1 − |s|^{2}}{|1 − \overline{s}z|^{2}} v(z)|f(z)| · |f(s)|dμ(z)dμ(s)

Now we apply Senichkin – Vinogradov test (Theorem 1) to the integral operator in L^{2}(μ) with kernel K

K(z, s) = \frac{1 − |s|^{2}}{|1 − \overline{s}z|^{2}} v(z)

Clearly

∫_{𝔻} K(z, s)K(z, x)dμ(z) = ∫_{𝔻} \frac{1 − |s|^{2}}{|1 − \overline{s}z|^{2}} · \frac{1 − |x|^{2}}{|1 − \overline{x}z|^{2}} [v(z)]^{2} dμ(z) ≤

≤ \frac{2}{|1 − \overline{s}x|^{2}} \{∫_{𝔻} \frac{(1 − |s|^{2})(1 − |x|^{2})}{|1 − \overline{s}z|^{2}} [v(z)]^{2} dμ(z)+

+ ∫_{𝔻} \frac{(1 − |s|^{2})(1 − |x|^{2})}{|1 − \overline{x}z|^{2}} [v(z)]^{2} dμ(z)\} ≤

≤ 2C \{\frac{1 − |x|^{2}}{|1 − \overline{s}x|^{2}} v(s) + \frac{1 − |s|^{2}}{|1 − \overline{s}x|^{2}} v(x)\} ≤ 2C(K(s, x) + K(x, s))

□

2. Carleson measure.

For a function f on the circle 𝕋 let f(z), z ∈ 𝔻 always denote the harmonic extension of f at z.

Theorem 2.1. If a weight w satisfies the Muckenhoupt condition (A_{2}), then the measure \frac{|∇w(z)|^{2}}{w(z)^{2}} (1 − |z|^{2})dxdy is Carleson.

Proof. It is well known that the condition (A_{2}) can be rewritten as

sup_{z∈𝔻}(w(z))(w^{−1}(z)) < ∞ .

We can rewrite this as

sup_{z∈𝔻}{[w(z)/ exp {(log w)(z)} ] · [(w^{−1})(z)/ exp {(log w^{−1})(z)}]} < ∞

By Jensen inequality expressions in brackets are at least 1, so

sup_{z∈𝔻}[w(z)/ exp {(log w)(z)}] = C < ∞ ,

or equivalently

log {w(z)} − (log w) (z) ≤ C  (2.1)

By Green’s formula

log \{w(s)\} − (log w) (s) = −\frac{1}{2π} ∫∫_{𝔻} log |\frac{1 − \overline{s}z}{z − s} | Δ log \{w(z)\} dxdy =

\frac{1}{2π} ∫∫_{𝔻} \frac{|∇w(z)|^{2}}{w(z)^{2}} log |\frac{1 − \overline{s}z}{z − s} | dxdy = \frac{1}{4π} ∫∫_{𝔻} \frac{|∇w(z)|^{2}}{w(z)^{2}} log |\frac{1 − \overline{s}z}{z − s} |^{2} dxdy

Using an elementary inequality log(1/a) ≥ 1 − a for 0 < a ≤ 1 we can estimate the last integral as

∫∫_{𝔻} \frac{|∇w(z)|^{2}}{w(z)^{2}} log |\frac{1 − \overline{s}z}{z − s} |^{2} dxdy ≥ ∫∫_{𝔻} \frac{|∇w(z)|^{2}}{w(z)^{2}} (1 − |\frac{1 − \overline{s}z}{z − s} |^{2}) dxdy =

= ∫∫_{𝔻} \frac{|∇w(z)|^{2}}{w(z)^{2}} · \frac{(1 − |s|^{2})(1 − |z|^{2})}{|1 − \overline{s}z|^{2}} dxdy

Together with (2.1) this imply

∫∫_{𝔻} \frac{(1 − |s|^{2})}{|1 − \overline{s}z|^{2}} · \frac{|∇w(z)|^{2}}{w(z)^{2}} (1 − |z|^{2}) dxdy ≤ C , ∀s ∈ 𝔻 ,

that yields that the measure \frac{|∇w(z)|^{2}}{w(z)^{2}} (1 − |z|^{2}) dxdy is Carleson. □

3. Equivalence of norms.

Let as usual given a function f on the unit circle 𝕋, f(z), z ∈ 𝔻 denotes its harmonic extension at point z.

Lemma 3.1 Let a weight w satisfies the Muckenhoupt (A_{2}) condition, and μ be a Carleson measure on 𝔻. Then for any function f ∈ L^{2}(𝕋, w), the following inequality holds,

∫∫_{𝔻} |f(z)|^{2}w(z) dμ(z) ≤ C ∫_{𝕋} |f|^{2}w ,

where the constant C depends on Carleson norm of μ and Mackenhoupt constant of w.

Proof. By Corollary 1 it is sufficient to show that

∫_{𝔻} \frac{1 − |s|^{2}}{|1 − \overline{s}z|^{2}} [w^{−1}(z)]^{2}w(z)dμ(z) ≤ C ∫_{𝔻} \frac{1 − |s|^{2}}{|1 − \overline{s}z|^{2}} [w^{−1}(z)]dμ(z) ≤ C′w^{−1}(s)  (3.1)

(the first inequality trivially holds because of Muckenhoupt condition).

If we introduce a new variable z̃ = \frac{z−s}{1−\overline{s}z} (so z = \frac{z̃+s}{1+\overline{s}z̃}) and a new measure μ̃

μ̃(z̃(Ω)) = ∫∫_{Ω} |(\frac{z − s}{1 − \overline{s}z} )′| dμ(z) = ∫∫_{𝔻} \frac{1 − |s|^{2}}{|1 − \overline{s}z|^{2}} dμ(z)

then (3.1) can be rewritten as

∫∫_{𝔻} w^{−1} (\frac{z̃ + s}{1 + \overline{s}z̃}) dμ̃(z̃) == ∫∫_{𝔻} w̃^{−1}(z̃)dμ̃(z̃)

It is well known that if μ is Carleson, then μ̃ is also Carleson, and Carleson norms of μ and μ̃ are equivalent. It is also well known that a Möbius transform of a Muckenhoupt weight is again a Muckenhoupt weight with equivalent constant. So it is sufficient to show that (3.1) holds for s = 0.

By Cauchy and Jensen inequalities

exp{[log w^{−1}](z)} ≤ [w^{−1/2}(z)]^{2} ≤ w^{−1}(z) ,

and since the weight w^{−1} satisfies the Muckenhoupt (A_{2}) condition,

w^{−1}(z)/ exp{[log w^{−1}](z)} ≤ C.

Therefore

w^{−1}(z) ≤ C[w^{−1/2}(z)]^{2} .

Since the measure μ is Carleson,

∫_{𝔻}[w^{−1}(z)]dμ(z) ≤ ∫_{𝔻}[w^{−1/2}(z)]^{2}dμ(z) ≤ ∫_{𝕋}[w^{−1/2}(t)]^{2}dm(t) = ∫_{𝕋}[w^{−1}(t)]dm(t) = w^{−1}(0)

and (3.1) for s = 0 and hence the lemma is proved. □

Theorem 3.2 let w be a Muckenhoupt (A_{2}) weight, f be a function in L^{2}(w), such that f(0) = 0. Then there exists a constant C such that

\frac{1}{C} ∫_{𝕋} |f|^{2}w ≤ ∫∫_{𝔻} |∇f(z)|^{2}w(z)(1 − |z|^{2}) dxdy ≤ C ∫_{𝕋} |f|^{2}w

Proof. Applying Green’s formula and using the notation ⟨., .⟩ for standard linear duality, ⟨a, b⟩ = ∑ a_{k}b_{k}, one can write

∫_{𝕋} |f|^{2}w = \frac{1}{2π} ∫∫_{𝔻} Δ (|f(z)|^{2}w(z)) log \frac{1}{|z|} dxdy =

= \frac{1}{2π} ∫∫_{𝔻} |∇f(z)|^{2}w(z) log \frac{1}{|z|} dxdy +

+\frac{1}{2π} ∫∫_{𝔻} f(z)⟨∇\overline{f}(z), ∇w(z)⟩ log \frac{1}{|z|} dxdy + \frac{1}{2π} ∫∫_{𝔻} \overline{f}(z)⟨∇f(z), ∇w(z)⟩ log \frac{1}{|z|} dxdy =

= \frac{1}{2π} ∫∫_{𝔻} |∇f(z)|^{2}w(z) log \frac{1}{|z|} dxdy + \frac{1}{π} Re \{∫∫_{𝔻} \overline{f}(z)⟨∇f(z), ∇w(z)⟩ log \frac{1}{|z|} dxdy\}

Since f(0) = 0, the logarithmic singularity log \frac{1}{|z|} does not influence the integrals, so

∫∫_{𝔻} Δ (|f(z)|^{2}w(z)) log \frac{1}{|z|} dxdy ≍ ∫∫_{𝔻} Δ (|f(z)|^{2}w(z)) (1 − |z|^{2}) dxdy

where ≍ means equivalence in a sense of two-sided estimate.

Let us estimate

|∫∫_{𝔻} \overline{f}(z)⟨∇f(z), ∇w(z)⟩ log \frac{1}{|z|} dxdy| ≤ C ∫∫_{𝔻} |f(z)| |∇f(z)| |∇w(z)|(1 − |z|^{2}) dxdy ≤

≤ (∫∫_{𝔻} |∇f(z)|^{2}w(z)(1 − |z|^{2}) dxdy)^{1/2} (∫∫_{𝔻} |f(z)|^{2}w(z)\frac{|∇w(z)|^{2}}{w(z)^{2}} (1 − |z|^{2}) dxdy)^{1/2} ≤

≤ (∫∫_{𝔻} |∇f(z)|^{2}w(z)(1 − |z|^{2}) dxdy)^{1/2} (∫_{𝕋} |f|^{2}w)^{1/2}

(the last inequality holds by Lemma 3) If we denote

X = ∫_{𝕋} |f|^{2}w, A = ∫∫_{𝔻} |∇f(z)|^{2}w(z)(1 − |z|^{2}) dxdy ,

then the above estimates give us X ≤ C(A + (AX)^{1/2}) and A ≤ C(X + (AX)^{1/2}). Therefore X and A are equivalent. □

4. Boundedness of the Hilbert Transform.

Now we can prove that if w satisfies the Muckenhoupt (A_{2}) condition then the Hilbert transform (or, equivalently, the Riesz Projection P_{+}) bounded operator in L^{2}(w).

Consider a weighted vector-valued L^{2}-space H = L^{2}(𝔻 → ℂ^{2}, w(z)(1 − |z|^{2}) dxdy). Let L_{0}^{2}(w) = \{f ∈ L^{2}(w) : f(0) = 0\}. Since |∇f|^{2} = 2(|∂f|^{2} + |\overline{∂}f|^{2}), Theorem 3 asserts that the mapping

f ↦ ( \begin{matrix}∂f(z) \\ \overline{∂}f(z) \end{matrix} )

gives a representation of L_{0}^{2}(w) as a subspace of H (with equivalent norm). Here again f(z), z ∈ 𝔻 denotes the harmonic extension of f.

Consider an operator ℙ_{+} on H,

ℙ_{+} ( \begin{matrix}f \\ g \end{matrix} ) = ( \begin{matrix}f \\ 0 \end{matrix} )

The operator ℙ_{+} is obviously bounded, and and its restriction onto L_{0}^{2}(w) is nothing but the Riesz Projection P_{+}. Therefore P_{+} is bounded on L_{0}^{2}(w). To complete the proof note that for f ∈ L^{2}(w)

P_{+}f = P_{+}(f − f(0)) + f(0) .

The operator

f ↦ f(0) · 1

is a bounded operator in L^{2}(w); it is an easy exercise on the Cauchy inequality that its norm is exactly (w(0)w^{−1}(0))^{1/2}. Since f − f(0) ∈ L_{0}^{2}(w) the operator P_{+} is bounded on L^{2}(w). □

References

[1] N. K. Nikolskii, “Treatise on the shift operator,” Springer-Verlag, NY etc. 1985.