SOME PROPERTIES OF CONJUGATE HARMONIC FUNCTIONS IN A HALF-SPACE

ANATOLY RYABOGIN AND DMITRY RYABOGIN

Abstract. We prove a multi-dimensional analog of the Theorem of Hardy and Littlewood about the logarithmic bound of the L^{p}- average of the conjugate harmonic functions, 0 < p ≤ 1. We also give sufficient conditions for a harmonic vector to belong to H^{p}(R_{+}^{n+1}), 0 < p ≤ 1.

1. Introduction and statements of main results

The following result of Hardy and Littlewood [6] is classical.

Theorem 1. Let 0 < p ≤ 1, and let f(z) = u(z) + iv(z) be an analytic function in the unit disc D := {z ∈ C : |z| < 1}, such that

(1) 1) v(0) = 0,  2) M_{p}(r, u) := (\frac{1}{2π} ∫_{0}^{2π} |u(re^{iθ})|^{p}dθ)^{1/p} ≤ C, 0 ≤ r < 1.

Then

(2)  M_{p}(r, v) ≤ AC + AC( log \frac{1}{1 − r})^{1/p}.

In this paper we prove an analog of Theorem 1 for conjugate harmonic functions in R_{+}^{n+1} = R^{n} × (0, ∞). The case p < (n − 1)/n leads to additional difficulties, since |F |^{p} is subharmonic, provided p ≥ (n−1)/n, [15]. We refer the reader to the classical works [3], [15], [17], [5], [2], [4], [18] for the history and different results related to the classes S^{p}(R_{+}^{n+1}), h^{p}(R_{+}^{n+1}), H^{p}(R_{+}^{n+1}), (all definitions are given in Section 2).

We have

Theorem 2. Let 0 < p ≤ 1, and let F,

F (x, y) = (U(x, y), V_{1}(x, y), V_{2}(x, y), ..., V_{n}(x, y)), (x, y) ∈ R_{+}^{n+1},

be the harmonic vector such that

(3)  1) V_{i} ⇒_{y→∞}^{x} 0, i = 1, ..., n, 2) M_{p}(y, U) ≤ C,  3) M_{p}(1, F) ≤ C.

Then

(4)  M_{p}(y, V) ≤ AC + AC | log y|^{1/p}.

1991 Mathematics Subject Classification. Primary 30E25, secondary 42B25.

Key words and phrases. Hardy spaces, subharmonic functions.

The third condition in (3) appears after the application of the Main Theorem of calculus, see (8), (9). The logarithmic bound comes from the estimate

∫_{R^{n}} ( sup_{ξ≥t} |∇U(x, ξ)|)^{p} dx ≤ ACt^{−p}

in the integral

∫_{y}^{1} t^{p−1}dt ∫_{R^{n}} ( sup_{ξ≥t} |∇U(x, ξ)|)^{p}dx,

see Lemmata 3, 4, 5.

To control the logarithmic blow up, we use the ”Littlewood-Paley”- type condition:

I(p) := ∫_{0}^{1} t^{p−1}dt ∫_{R^{n}} ( sup_{ξ≥t} |∇U(x, ξ)|)^{p}dx < ∞.

Our second result is

Theorem 3. Let 0 < p ≤ 1, and let F = (U, V_{1}, ..., V_{n}) be the harmonic vector such that

(5)  1) F ⇒_{y→∞}^{x} 0,  2) M_{p}(1, F) ≤ C,  3) I(p) < ∞.

Then F ∈ H^{p}.

It is unlikely that F ∈ H^{p} implies I(p) < ∞.

The paper is organized as follows. In section 2 we give all necessary definitions and auxiliary results used in the sequel. In Section 3 and 4 we prove Theorems 2 and 3. For convenience of the reader we split our proofs into elementary Lemmata.

2. Auxiliary results

Let U(x, y) be a harmonic function in R_{+}^{n+1} ≡ R^{n} × (0, ∞). We say that the vector-function V (x, y) = (V_{1}(x, y), ..., V_{n}(x, y)) is the conjugate of U(x, y) in the sense of M. Riesz [14], [16], if V_{k}(x, y), k = 1, ..., n are harmonic functions, satisfying the generalized Cauchy-Riemann conditions:

\frac{∂U}{∂y} + ∑_{k=1}^{n} \frac{∂V_{k}}{∂x_{k}} = 0,  \frac{∂V_{i}}{∂x_{k}} = \frac{∂V_{k}}{∂x_{i}} ,  \frac{∂U}{∂x_{i}} = \frac{∂V_{i}}{∂y} ,  i ≠ k, k = 1, ..., n.

If U(x, y) and V (x, y) are conjugate in R_{+}^{n+1} in the above sense, then the vectorfunction

F (x, y) = (U(x, y), V (x, y)) = (U(x, y), V_{1}(x, y), ..., V_{n}(x, y))

is called a harmonic vector.

Define

M_{p}(y, F) = ( ∫_{R^{n}} (U^{2}(x, y) + ∑_{i=1}^{n} V_{i}^{2}(x, y))^{p/2}dx)^{1/p}, p > 0.

H^{p} SPACES OF CONJUGATE HARMONIC FUNCTIONS 3

Now we define the space H^{p}(R_{+}^{n+1}). We follow the work of Fefferman and Stein[4]. Let U(x, y) be a harmonic function in R_{+}^{n+1}, and let U_{j_{1}j_{2}j_{3}...j_{k}} denote a component of a symmetric tensor of rank k, 0 ≤ j_{i} ≤ n, i = 1, ..., n. Suppose also that the trace of our tensor is zero, meaning

∑_{j=0}^{n} U_{jjj_{3}...j_{k}}(x, y) = 0, ∀j_{3}, ..., j_{k}.

The tensor of rank k + 1 can be obtained from the above tensor of rank k by passing to its gradient:

U_{j_{1}j_{2}...j_{k}j_{k+1}}(x, y) = \frac{∂}{∂x_{j_{k+1}}} (U_{j_{1}j_{2}j_{3}...j_{k}}(x, y)), x_{0} = y, 0 ≤ j_{k+1} ≤ n.

Definition ([4]).We say that U ∈ H^{p}(R_{+}^{n+1}), p > 0, if there exists a tensor of rank k of the above type with the properties:

U_{0...0}(x, y) = U(x, y),  sup_{y>0} ∫_{R^{n}} ( ∑_{(j)} U_{(j)}^{2}(x, y))^{p/2}dx < ∞,  (j) = (j_{1}, ...j_{k}).

It is well-known that the function ( ∑_{(j)} U_{(j)}^{2}(x, y))^{p/2} is subharmonic for p ≥ p_{k} =

(n − k)/(n + k − 1), see [3],[4],[16].

We remind that the radial and the non-tangential maximal functions are defined as follows:

F^{+}(x) = sup_{y>0} |F (x, y)|, N_{α}(F)(x^{0}) = sup_{(x,y)∈Γ_{α}(x^{0})} |F (x, y)|.

Here

Γ_{α}(x^{0}) = {(x, y) ∈ R_{+}^{n+1} : |x − x^{0}| < αy}, α > 0,

is an infinite cone with the vertex at x^{0}. It is well-known [4] that

F ∈ H^{p}(R_{+}^{n+1}) ⟺ N_{α}(F) ∈ L^{p} ⟺ F^{+} ∈ L^{p}, p > 0.

We also define the weak maximal function

W F (x, y) = sup_{ζ≥y} |F (x, ζ)|, y > 0.

The above expression is understood as folows: we fix x, and for fixed y we find the supremum over all ζ ≥ y.

We will use the following results.

Lemma 1. ([4], p.173). Suppose w is harmonic in R_{+}^{n+1}, and M_{p}(y, u) ≤ C for some p, 0 < p < ∞. Then

(6)  sup_{x∈R^{n}} |u(x, y)| ≤ Ay^{−n/p},  0 < y < ∞.

Theorem 4. ([5], p.267). Let 0 < p ≤ 1, a > 0, let w : R_{+}^{n+1} → [0, ∞) be a function such that w^{p} is subharmonic and satisfies

J_{a,p} := ∫_{R_{+}^{n+1}} t^{ap−1}w(x, t)^{p}dxdt < +∞,

and for each (x, t) ∈ R^{n} × [0, +∞) let

w_{a}(x, t) := \frac{1}{Γ(a)} ∫_{0}^{+∞}s^{a−1}w(x, s + t) ds.

Then w_{a} is subharmonic on R_{+}^{n+1} and is finite a.e. on R^{n}, and for all t ≥ 0,

∫_{R^{n}} w(x, t)^{p} dx ≤ AC(a, n, p) J_{a,p}.

Theorem 5. ([5], p.269). Let m ∈ N, p ≥ (n − 1)/(m + n − 1) (if n = 1 we suppose p > 0), and let u : R_{+}^{n+1} → R be harmonic. Then, for all t > 0,

∫_{R^{n}} |∇^{m}u(x, t)|^{p}dx ≤ A(m, n, p)t^{−mp−1} ∫_{t/2}^{3t/2}ds ∫_{R^{n}} |u(x, s)|^{p}dx.

Lemma 2. ([13], p.2464). Let p > 0 and let F = (U, V_{1}, ..., V_{n}) be such that V_{i} ⇒_{y→∞}^{x} 0, i = 1, ..., n, M_{p}(y, U) ≤ C. Then

M_{p}(y, ∇^{k}F) ≤ ACy^{−k}, k ∈ N.

Notation. We denote by D_{i}^{k}f(x, y) the partial derivative of the function f of the order k with respect to x_{i}, i = 1, 2, ..., n + 1. The notation f(x, y) ⇒_{y→∞}^{x} 0 means that f(x, y) converges to 0 uniformly with respect to x, provided y → ∞, ∇^{k}f(x) = (\frac{∂^{k}f(x)}{∂x_{1}^{k}}, ..., \frac{∂^{k}f(x)}{∂x_{n}^{k}}). Everywhere below the constants A(k, n), C, K depend only on the parameters pointed in parentheses, and may be different from time to time.

3. Proof of Theorem 2.

Lemma 3. Let p > 0, and let F = (U, V_{1}, ..., V_{n}) satisfy M_{p}(1, F) ≤ C. Then

(7) M_{p}(y, V_{i}) ≤ AC + ∫_{R^{n}} ( ∫_{y}^{1} sup_{ξ≥t} |∇U(x, ξ)|dt)^{p}dx,  i = 0, 1, ..., n, V_{0} = U.

Proof. By the Main Theorem of Calculus, and the Cauchy-Riemann equations, we have

(8)  V_{i}(x, y) − V_{i}(x, 1) = − ∫_{y}^{1} \frac{∂V_{i}(x, t)}{∂t} dt = − ∫_{y}^{1} \frac{∂U(x, t)}{∂x_{i}} dt,

i = 1, 2, ..., n,

(9)  U(x, y) − U(x, 1) = − ∫_{y}^{1} \frac{∂U(x, t)}{∂t} dt.

Then,

M_{p}(y, V_{i}) ≤ M_{p}(1, V_{i}) + ∫_{R^{n}} ( ∫_{y}^{1} sup_{ξ≥t} |∇U(x, ξ)|dt)^{p}dx,

and the result follows. □

H^{p} SPACES OF CONJUGATE HARMONIC FUNCTIONS 5

Lemma 4. Let p > 0 and let 0 < y < 1. Then

∫_{R^{n}} ( ∫_{y}^{1} sup_{ξ≥t} |∇U(x, ξ)| dt)^{p}dx ≤

(10)  y^{p} ∫_{R^{n}} ( sup_{ξ≥y} |∇U(x, ξ)|)^{p}dx + 2 p ∫_{y}^{1} t^{p−1}dt ∫_{R^{n}} ( sup_{ξ≥t} |∇U(x, ξ)|)^{p}dx.

Proof. Denote

Ψ(x, y) := ∫_{y}^{1} sup_{ξ≥t} |∇U(x, ξ)| dt.

Following [6] consider

Φ(x, y) := Ψ(x, y)^{p} − y^{p}( − \frac{∂Ψ(x, y)}{∂y})^{p}, Ω(y) := \{x ∈ R^{n} : Φ(x, y) > 0\}.

By definition of Ω(y),

(11)  ∫_{R^{n}\Ω(y)} Ψ(x, y)^{p}dx ≤ y^{p} ∫_{R^{n}\Ω(y)} ( − \frac{∂Ψ(x, y)}{∂y})^{p}dx.

Next, the reasons which are similar to those in [6], imply

(12) ∫_{Ω(y)} Φ(x, y) dx − ∫_{Ω(a)} Φ(x, a) dx = ∫_{y}^{a} dξ ∫_{Ω(ξ)} −\frac{∂Φ(x, ξ)}{∂ξ} dx, 0 < y < a ≤ 1.

Moreover, using ∂^{2}Ψ/∂ξ^{2} ≥ 0 for almost every 0 < ξ < 1, we have

−\frac{∂Φ}{∂ξ} = −p Ψ^{p−1} \frac{∂Ψ}{∂ξ} + p ξ^{p−1}( − \frac{∂Ψ}{∂ξ})^{p} − p ξ^{p}( − \frac{∂Ψ}{∂ξ})^{p−1} \frac{∂^{2}Ψ}{∂ξ^{2}} ≤

p ( − \frac{∂Ψ}{∂ξ})(Ψ^{p−1} + ξ^{p−1}( − \frac{∂Ψ}{∂ξ})^{p−1}) ≤ 2p ξ^{p−1}( − \frac{∂Ψ}{∂ξ})^{p}.

Here the last inequality follows from the definition of Ω(ξ) and 0 < p < 1. Since

Ψ(x, y) ≤ (1 − y) |\frac{∂Ψ(x, y)}{∂y}|,

the function Φ(x, y) is negative, provided y is sufficiently close to 1, and we can take a such that Ω(a) = ∅. Hence, (12) yields

∫_{Ω(y)} Φ(x, y) dx ≤ 2p ∫_{y}^{a} ξ^{p−1} dξ ∫_{Ω(ξ)} ( − \frac{∂Ψ(x, ξ)}{∂ξ}(x, ξ))^{p} dx ≤

(13)  2p ∫_{y}^{1} ξ^{p−1} dξ ∫_{R^{n}} ( − \frac{∂Ψ(x, ξ)}{∂ξ}(x, ξ))^{p} dx.

Adding

∫_{Ω(y)} Ψ(x, y)^{p} dx ≤ ∫_{Ω(y)} Φ(x, y) dx + y^{p} ∫_{Ω(y)} ( − \frac{∂Ψ(x, y)}{∂y})^{p}dx

with (11), and using (13), we obtain (10). □

The next result is crucial.

Lemma 5. Let p > 0 and let F = (U, V_{1}, ..., V_{n}) be such that

(14)  1) V_{i} ⇒_{y→∞}^{x} 0, i = 1, ..., n,  2) M_{p}(y, U) ≤ C.

Then

(15)  ( ∫_{R^{n}} ( sup_{ξ≥y} |ϕ_{ij}(x, ξ)|)^{p}dx)^{1/p} ≤ ACy^{−1},

where ϕ_{ij}(x, y) is a coordinate of ∇V_{i}(x, y), j = 1, ..., n + 1, x_{n+1} = y, i = 0, ..., n, V_{0} = U.

Proof. Fix p > 0 and let l = inf{j ∈ N : p > p_{j} := (n − 1)/(j + n − 1)}. Since ∇V_{i}(x, y) ⇒_{y→∞}^{x} 0, we may use the following relation (see [5] or [4])

ϕ_{ij}(x, y) = \frac{1}{(2l − 2)!} ∫_{y}^{∞}(s − y)^{2l−2}D_{n+1}^{2l−1}ϕ_{ij}(x, s)ds =

\frac{1}{(2l − 2)!} ∫_{0}^{∞}s^{2l−2}D_{n+1}^{2l−1}ϕ_{ij}(x, s + y)ds.

We have

sup_{ξ≥y} |ϕ_{ij}(x, ξ)| ≤ R(x, y),

where

R(x, y) := \frac{1}{(2l − 2)!} ∫_{0}^{∞}s^{2l−2}( sup_{ξ≥y} |∇^{l}D_{n+1}^{l−1}ϕ_{ij}(x, s + ξ)|)ds.

To prove (15) it is enough to show that

(16)  M_{p}(y, R) ≤ AC y^{−1}.

Since

(|∇^{l}D_{n+1}^{l−1}ϕ_{ij}(x, ξ)|)^{p}

is subharmonic [3], the function

w^{p}(x, s + y) := ( sup_{ξ≥y} |∇^{l}D_{n+1}^{l−1}ϕ_{ij}(x, s + ξ)|)^{p}

is also subharmonic, and we may apply Theorem 4 (take a = 2l − 1, A = A(l, n, p)) to obtain

∫_{R^{n}} |R(x, y)|^{p}dx ≤ A ∫_{0}^{∞}s^{(2l−1)p−1}ds ∫_{R^{n}} ( sup_{ξ≥y} |∇^{l}D_{n+1}^{l−1}ϕ_{ij}(x, s + ξ)|)^{p}dx =

A ∫_{0}^{∞}s^{(2l−1)p−1}ds ∫_{R^{n}} ( sup_{ξ≥y} |∇^{l}D_{n+1}^{l−1}ϕ_{ij}(x, s + ξ)|^{p_{j}})^{p/p_{j}}dx.

By the choice of p_{j}, we have p/p_{j} > 1 and we may use the well-known [4] L^{p/p_{j}}- boundedness of the maximal operator:

∫_{R^{n}} |R(x, y)|^{p}dx ≤ A ∫_{0}^{∞}s^{(2l−1)p−1}ds ∫_{R^{n}} |∇^{l}D_{n+1}^{l−1}ϕ_{ij}(x, s + y)|^{p}dx.

Since D_{n+1}^{l−1}ϕ_{ij}(x, y) is the l-th derivative of V_{i}, we use Lemma 2 to get

(17)  ∫_{R^{n}} |∇^{l}D_{n+1}^{l−1}ϕ_{ij}(x, y)|^{p}dx ≤ ∫_{R^{n}} |∇^{2l}F (x, y)|^{p}dx ≤ C y^{−2lp}.

This gives

∫_{R^{n}} |R(x, y)|^{p}dx ≤ A(l, n, p)C ∫_{0}^{∞}s^{(2l−1)p−1}(s + y)^{−2lp}ds = A(l, n, p)Cy^{−p},

and (16) is proved. □

Proof of Theorem 2. The proof follows from Lemmata 3, 4, 5.

4. Proof of Theorem 3.

Lemma 6. Let p > 0. Then F = (U, V_{1}, ..., V_{n}) ∈ H^{p} iff

1) F ⇒_{y→∞}^{x} 0, 2) ∫_{R^{n}} ( sup_{η≥y} |U(x, η)|)^{p}dx < C.

Proof. Let F ∈ H^{p}, then both 1) and 2) are well-known, [4]. We prove the converse in two steps. At first we show that

(18)  ∫_{R^{n}} ( sup_{y>0} |U(x, y)|)^{p}dx ≤ C.

Then we prove that (18) implies

(19)  (sup_{y>0} |V_{i}(·, y)|)^{p} ∈ L^{1}(R^{n}),  i = 1, ..., n.

To prove (18), we observe that

sup_{y>0} |U(x, y)| = sup_{y>0} sup_{η≥y} |U(x, η)| = lim_{y→0} sup_{η≥y} |U(x, η)|.

Hence, using 2) and Fatou’s Lemma, we obtain

∫_{R^{n}} ( sup_{y>0} |U(x, y)|)^{p}dx ≤ lim_{y→0} ∫_{R^{n}} ( sup_{η≥y} |U(x, η)|)^{p}dx ≤ C.

It remains to show (19). Using Cauchy-Riemann equations, we have

V_{i}(x, y) = [V_{i}]_{0...0}(x, y) = \frac{(−1)^{k}}{(k − 1)!} ∫_{0}^{∞}s^{k−1}D_{n+1}^{k}V_{i}(x, s + y)ds =

(20)  \frac{(−1)^{k}}{(k − 1)!} ∫_{0}^{∞}s^{k−1}D_{n+1}^{k−1}D_{i}U(x, s + y)ds,

where k is chosen such that the function ( ∑_{(j)} U_{(j)}^{2}(x, y))^{p/2} is subharmonic, (p ≥ p_{k} =

(n − k)/(n + k − 1), see [3],[4],[16]). Since the expression in (20) is one of the tensor coordinates of U_{(j)}, see ([4], page 169), we have

∫_{R^{n}} ( sup_{y>0} |V_{i}(x, y)|)^{p}dx ≤ ∫_{R^{n}} ( sup_{y>0} ∑_{(j)} U_{(j)}^{2}(x, y))^{p/2}dx =

sup_{y>0} ∫_{R^{n}} ( ∑_{(j)} U_{(j)}^{2}(x, y))^{p/2}dx ≤ C ∫_{R^{n}} ( sup_{y>0} |U(x, y)|)^{p}dx.

The last estimate is proved in [4], page 170. □

Lemma 7. Let p > 0 and let F = (U, V_{1}, ..., V_{n}) be such that V_{i} ⇒_{y→∞}^{x} 0, i = 1, ..., n. Then F ∈ H^{p}, provided

(21)  1) M_{p}(1, F) ≤ C,  2) M_{p}(y, U) ≤ C,  3) I(p) < ∞.

Proof. At first we prove that

(22) 2^{−p} ∫_{G} ( sup_{y>0} |V_{i}(x, y)|)^{p}dx ≤ AC + ∫_{G} ( sup_{y>0} |∇U(x, y)|)^{p}dx,  ∀G ⊂ R^{n}.

This follows from (8), (9) the inequality

2^{−p}( sup_{y>0} |V_{i}(x, y)|)^{p} ≤ (|V_{i}(x, 1)|)^{p} + ( sup_{y>0} |∇U(x, y)|)^{p},

j = 0, 1, 2, ..., n, V_{0} = U, and the first condition in (21).

Now we finish the proof. Assume that the lemma is not true. Then ∀N > 0 there exists a set E ⊂ R^{n}, 0 < m(E) < ∞, such that

∫_{E} ( sup_{y>0} |V_{i}(x, y)|)^{p}dx ≥ 4N,

for some i = 0, 1, ..., n, V_{0} = U. Hence, taking G = E in (22) we have

∫_{E} ( sup_{y>0} |∇U(x, y)|)^{p}dx ≥ 2N.

Then

sup_{y>0} |∇U(x, y)| = sup_{y>0} sup_{ξ≥y} |∇U(x, ξ)| = lim_{y→0} sup_{ξ≥y} |∇U(x, ξ)|

and Lemma Fatou imply

lim_{y→0} ∫_{E} ( sup_{ξ≥y} |∇U(x, ξ)|)^{p}dx ≥ ∫_{E} ( sup_{y>0} |∇U(x, y)|)^{p}dx ≥ 2N.

But this contradicts the third condition of the lemma, since

I(p) ≥ ∫_{0}^{1} dy ∫_{R^{n}} ( sup_{ξ≥y} |∇U(x, ξ)|)^{p}dx ≥ ∫_{0}^{1} dy ∫_{E} ( sup_{ξ≥y} |∇U(x, ξ)|)^{p}dx ≥ N.

□

Lemma 8. Let p > 0 and let F = (U, V_{1}, ..., V_{n}) be the harmonic vector satisfying conditions of Theorem 3. Then (18) holds.

Proof. We argue as in the previous Lemma. We have (9), and

2^{−p} ∫_{R^{n}} ( sup_{y>0} |U(x, y)|)^{p}dx ≤ ∫_{R^{n}} |U(x, 1)|^{p}dx + ∫_{R^{n}} ( sup_{y>0} |∇U(x, y)|)^{p}dx

leads to a contradiction. □

Proof of Theorem 3. The proof follows from Lemmata 8, 7 and 6.

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Anatoly Ryabogin, Department of Physics and Mathematics, Stavropol State University, Pushkin’s Street 1, Stavropol, Russia

Dmitry Ryabogin, Department of Mathematical Sciences, Kent State University, Kent, OH 44242, USA

E-mail address: ryabogin@math.kent.edu