STUDY OF HARMONIC FUNCTIONS AND MEASURES

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1 STUDY OF HARMONIC FUNCTIONS AND MEASURES YIFEI PAN AND MEI WANG Contents. On monotonicity of positive invariant harmonic functions.. Statements of results.. Proofs of Theorem. and its corollaries 7.3. Proof of Theorem.8.4. Proofs of Theorem. and Theorem.9 7. On Harnack inequality for positive invariant harmonic functions 9.. Statement of results 9.. Proof of Theorem..3. Proof of Theorem On higher order angular derivatives an application of Faá di Bruno s formula Statements of results Proof of Theorem Proof of Proposition Proof of Theorem When is a function not flat? Introduction Proof of Theorem 4. and its corollary Positive measures on the sphere Introduction Proof of Theorem Proof of Proposition Proof of Proposition On spherical harmonic expansions of complex Borel measures on the unit sphere Introduction proofs 58 References 6

2 . On monotonicity of positive invariant harmonic functions A monotonicity property of Harnack inequality is proved for positive invariant harmonic functions in the unit ball... Statements of results. Let B n {x R n : x < }, n be the open unit ball in R n. S n B n. Consider the differential operator λ ( x x ) 4 x + λ x j + λ( n j j x j j λ), λ R. We prove a monotonicity property of invariant harmonic functions that are solutions of λ u 0 and are defined by positive Borel measures on the sphere with respect to the Poisson kernel P λ (see below). This section describes the theorems and their corollaries. The proofs are provided in the next two sections. Theorem.. Let u be a positive invariant harmonic function defined in B n by a positive Borel measure µ on S n with the Poisson kernel P λ. For ζ S n, if λ > n (if λ < n ), the function ( r) n u(rζ) ( + r) +λ is decreasing (increasing) for 0 r <, and the function ( + r) n u(rζ) ( r) +λ is increasing (decreasing) for 0 r <. Also +λ µ({ζ}), λ > n r r)n u(rζ), +λ µ({ζ}), λ < n, µ({ζ}c ) > 0 λ < n, µ({ζ}c ) 0 and Remarks. r u(rζ) Sn ( r) +λ +λ dµ(ξ). ζ ξ n+λ () Invariant harmonic functions are the solutions of λ u 0. These solutions also satisfy certain invariance property with respect to Möbius transformation. Invariant harmonic functions generally do not possess good boundary regularity, as shown in Liu and Peng []. () Let µ be a positive Borel measure on S n and P λ be the Poisson kernel P λ (x, ζ) ( x ) +λ x ζ n+λ.

3 It is known that the integral u(x) P λ (x, ζ)dµ(ζ) S n defines an invariant harmonic function in B n ([], p. 9). (3) On the completion of the current work, we learned that the it cases for n, λ 0 in Theorem. were obtained by Simon and Wolff ([8], ref. Chapter 0, p. 546 in [7]). (4) The critical value λ n yields the degenerate case with the constant Poisson kernel. The following theorem characterizes the behavior of invariant harmonic functions on the rays. Theorem.. Let u be a positive invariant harmonic function defined in B n by a positive Borel measure µ on S n with the Poisson kernel P λ. Let ζ S n and 0 r r <. If λ > n, ( ) r λ+ ( ) +r n r u(r ζ) u(rζ) +r If λ < n, ( ) + r λ+ ( ) r n + r u(r ζ) u(rζ) r For r 0, the above becomes for λ > n, and ( + r + r ( r r ) λ+ ( ) r n u(r ζ) r ) λ+ ( ) +r n u(r ζ) +r ( r) +λ ( + r)+λ u(0) u(rζ) u(0) ( + r) n ( r) n 3 for λ < n. ( + r) +λ ( r)+λ u(0) u(rζ) u(0) ( r) n ( + r) n Remark. Case λ 0 is the classical Harnack Inequality in B n. Corollary.3. Let U be the potential function defined in B n by a positive Borel measure µ on S n as follows: U(x) dµ(η). S n x η n+λ For ζ S n, if λ > n (if λ < n ), the function ( r) n+λ U(rζ)

4 4 is decreasing (increasing) for 0 r <. In Theorem., λ n corresponds to the Laplace-Beltrami operator n/ and the Poincaré metric. It is known([]) that given a positive invariant harmonic function (solutions of n/ u 0), there exists a positive Borel measure µ on S n, such that u(x) P n/ (x, ζ)dµ(ζ) S n In this case the monotonicity property in Theorem. implies the following corollary. Corollary.4. Let u be a positive solution of n/ u 0 in B n. Then ( ) r n u(rζ) decreasing in r, + r ( ) + r n u(rζ) increasing in r. r Corollary.5. Let u be a positive harmonic function with respect to the Laplace operator (λ 0) defined in B n by a positive Borel measure µ on S n with the Poisson Kernel P 0. For ζ S n, 0 r <, the function is decreasing and the function is increasing. In addition, ( r) n u(rζ) + r ( + r) n u(rζ) r ( r r)n u(rζ) µ({ζ}), u(rζ) r r S n ζ ξ n dµ(ξ). Corollary.5 is the same as a result in []. Corollary.6. Let B n (R) be the open ball of radius R. Let u(z) be an invariant harmonic function in B n (R) (a.k.a u(rz) is invariant harmonic in B n ) defined by the Poisson kernel P λ ( x R, ζ). For ζ Sn, if λ > n (if λ < n ), the function (R r) n R n λ u(rζ) (R + r) +λ is decreasing (increasing) and the function (R + r) n R n λ u(rζ) (R r) +λ

5 is increasing (decreasing) in r for 0 r < R. The case λ 0 gives the monotonicity of functions 5 ( R r R ) n R r R + r u(rζ) and which implies that, x B n (r), 0 r < R, ( R + r R ) n R + r R r u(rζ), ( R R + r ) n R r u(0) u(x) R + r the classical Harnack Inequality. ( R ) n R + r R r R r u(0) Corollary.7. Let u be a positive invariant harmonic function defined in B n by a positive Borel measure µ on S n with the Poisson kernel P λ. Let 0 r r <. If λ > n, ( r) n max ( + r) +λ u(x) ( r ) n x r ( + r ) ( + r) n min ( r) +λ u(x) ( + r ) n x r ( r ) max +λ u(x) x r min +λ u(x) x r If λ < n, ( + r) n max ( r) +λ u(x) ( + r ) n x r ( r ) max +λ u(x) x r ( r) n min ( + r) +λ u(x) ( r ) n x r ( + r ) min +λ u(x) x r Similar results are obtained in complex space C n. Let P α (z, ζ) ( z ) n+α z ζ n+α, α R be the Poisson-Szegö kernel for the operator α,β 4( z ) (δ i,j z i z j ) + α z i,j i z j j z j + β z j j z j αβ z j with α β, where z ζ n i z iζ i. Define u(z) P α (z, ζ)dµ(ζ), α R. S n

6 6 Theorem.8. Let u be a positive invariant harmonic function defined in the unit ball B n C n by a positive Borel measure µ on S n B n with the Poisson-Szegö kernel. Given ζ S n, if α > n (if α < n), the function ( r) n u(rζ) ( + r) n+α is decreasing (increasing) for 0 r <, and the function ( + r) n u(rζ) ( r) n+α is increasing (decreasing) for 0 r <. Also n+α µ({ζ}), α > n ( r r)n u(rζ), α < n, µ({ζ} c ) > 0 n+α µ({ζ}), α < n, µ({ζ} c ) 0 and r u(rζ) Sn ( r) n+α n+α dµ(η). ζ η n+α The following theorem describes invariant harmonic functions on the rays. Theorem.9. Let u be a positive invariant harmonic function defined in the unit ball B n C n by a positive Borel measure µ on S n with the Poisson-Szegö kernel. Let ζ S n and 0 r r <. if α > n, ( ) r n+α ( ) +r n r u(r ζ) u(rζ) +r If α < n, ( + r + r ) n+α ( ) r n u(r ζ) r ( ) + r n α ( ) r n+α + r r u(r ζ) u(rζ) For r 0, the above becomes for α > n, and for α < n. ( r r ( r) n+α ( + r)n+α u(0) u(rζ) ( + r) n ( r) n u(0) ( + r) n+α ( r)n+α u(0) u(rζ) ( r) n ( + r) n u(0) ) n+α ( ) +r n u(r ζ) +r

7 .. Proofs of Theorem. and its corollaries. We need the following two lemmas for the proof of Theorem.. Lemma.0. Let x R n, x r, ζ S n. If λ > n then (.) (n+λ (n λ )r)( r ) λ x ζ n+λ ( r ) +λ (n+λ+(n λ )r)( r ) λ r x ζ n+λ x ζ n+λ If λ < n, then (.) (n+λ+(n λ )r)( r ) λ x ζ n+λ ( r ) +λ ) λ r x ζ n+λ (n+λ (n λ )r)( r x ζ Proof. then and (.3) Write x x η rη, η ζ n i η iζ i. r x ζ r ( x rη ζ + ) (r η ζ), r x ζ n+λ r ( x ζ ) n+λ ( r ) +λ r x ζ n+λ n + λ ( x ζ ) n+λ x ζ r (n + λ) x ζ n+λ (r η ζ), ( + λ)( r ) λ ( r) x ζ n+λ ( r ) +λ r x ζ n+λ x ζ (n+λ) ( + λ)( r ) λ r x ζ n+λ ( r ) +λ (n + λ) x ζ n+λ (r η ζ) x ζ (n+λ) ( + λ)( r ) λ r x ζ ( r ) +λ (n + λ)(r η ζ) x ζ n+λ+. To prove the right side inequality in (.), it suffices to show ( + λ)r x ζ ( r )(n + λ)(r η ζ) (n + λ + (n λ )r) x ζ, which is equivalent to (n + λ)( r )(r η ζ) (n + λ)( + r) x ζ. For λ > n, the above becomes or ( r )(r η ζ) ( + r) x ζ, ( r)(r η ζ) r η ζ +, 7

8 8 which, after a simple simplification, is equivalent to η ζ The inequality is true since ζ, η S n. To prove the left side inequality in (.), it suffices to show (using the result of (.3)) ( + λ)r x ζ ( r )(n + λ)(r η ζ) (n + λ (n λ )r) x ζ, which is equivalent to (n + λ)( r )(r η ζ) (n + λ)( r) x ζ. For λ > n, the inequality is equivalent to which is, after a simplification, ( r )(r η ζ) ( r) x ζ, η ζ, true since ζ, η S n. The proof of (.) for λ < n is parallel. This completes the proof of Lemma.0. Lemma.. Let u be a positive invariant harmonic function in B n defined by a positive Borel measure on S n with the Poisson kernel. If λ > n, (n + λ (n λ )r) (.4) r u(x) u(x) (n + λ + (n λ )r) r r u(x). If λ < n, (.5) (n + λ + (n λ )r) r u(x) u(x) (n + λ (n λ )r) r r u(x). Proof. By the Poisson integral representation of u in B n, Sn ( x ) +λ u(x) dµ(ζ) x ζ n+λ for a positive Borel measure µ. By (.) in Lemma.4 and µ being a positive measure, ( ( x ) +λ ) Sn (n + λ + (n λ )r)( r ) λ S n r x ζ n+λ dµ(ζ) x ζ n+λ dµ(ζ) (n + λ + (n λ )r) Sn ( x ) +λ r dµ(ζ) x ζ n+λ (n + λ + (n λ )r) r u(x)

9 when λ > n. It follows that ( u(x) ( x ) +λ ) (n + λ + (n λ )r) r r x ζ n+λ dµ(ζ) r u(x). S n The left side inequality in (.4) can be proved in the same manner. For the equality case, consider u y (x) u(x, y) ( x ) +λ x y n+λ which is invariant harmonic in R n \ {y} for y S n. A simple calculation shows that the equalities hold for u y (x) when x x y and x x y respectively. The proof of (.5) is similar. This completes the proof of Lemma.. Now we prove Theorem.. Proof. Consider ϕ(r) ( r)n ( + r)n and ψ(r) for 0 r <. ( + r) +λ ( r) +λ ϕ ϕ ψ ψ Given ω S n, consider (n + λ + (n λ )r) r, (n + λ (n λ )r) r. I(r, ω) ϕ(r)u(rω), J(r, ω) ψ(r)u(rω). To show Theorem., it suffices to show that I(r, ω) is decreasing (increasing) and J(r, ω) is increasing (decreasing) in r for 0 r < when λ > n (when λ < n ). By (.4) in Lemma., for λ > n, d ϕ (log I(r, ω)) dr ϕ + u r u (n + λ + (n λ )r) r + u r u 0. (n + λ + (n λ )r) r + (n + λ + (n λ )r) r Therefore log I(r, ω) is decreasing in r, and so is I(r, ω). Similarly, d ψ (log J(r, ω)) dr ψ + u r u (n + λ (n λ )r) r + u r u (n + λ (n λ )r) (n + λ (n λ )r) r r 0. 9

10 0 Hence, J(r, ω) is increasing in r. For λ > n and y Sn, r ( r) n ( + r) +λ P λ(rζ, y) r ( r) n ( r ) +λ ( + r) +λ rζ y n+λ ( r) n+λ δ(ζ, y) r rζ y n+λ {, ζ y 0, ζ y by applying the monotonicity properties in Theorem. to u P λ. By Lebesgue s dominated convergence theorem, ( r r)n u(rζ) ( r) n P λ (rζ, ξ)dµ(ξ) r S n ( + r) +λ ( r) n r S n r ( + r) +λ P λ(rζ, ξ)dµ(ξ) +λ µ({ζ}). ( + r) n Similarly, ( r) +λ P ( + r)n+λ λ(rζ, y) rζ ξ n+λ is increasing in r for λ > n. By Lebesgue s monotone convergence theorem, r u(rζ) ( r) +λ r ( r) +λ P λ (rζ, ξ)dµ(ξ) S n ( + r) n r ( + r) n S n r ( r) +λ P λ(rζ, ξ)dµ(ξ) ( + r) n+λ n dµ(ξ) S n r rζ ξ n+λ Sn +λ dµ(ξ). ζ ξ n+λ For λ < n, the monotonicity of I(r, ω) and J(r, ω) is proved similarly to the case λ > n using (.5) instead of (.4) in Lemma.. { ( r) n r ( + r) +λ P rζ y (n+λ) λ(rζ, y) r ( r) (n+λ), ζ y, ζ y when λ < n. Therefore, ( r r)n u(rζ) ( + r) +λ r { +λ µ({ζ}), if µ({ζ} c ) 0;, if µ({ζ} c ) > 0. S n r ( r) n ( + r) +λ P λ(rζ, ξ)dµ(ξ)

11 ( + r) n Similarly, ( r) +λ P ( + r)n+λ λ(rζ, y) rζ ξ n+λ is decreasing in r for λ < n. By Lebesgue s monotone convergence theorem, u(rζ) r ( r) +λ r ( r) +λ P λ (rζ, ξ)dµ(ξ) S n ( + r) n r ( + r) n S r n ( r) +λ P λ(rζ, ξ)dµ(ξ) ( + r) n+λ n dµ(ξ) S n r rζ ξ n+λ Sn +λ dµ(ξ) ζ ξ n+λ This completes the proof of Theorem.. The proof of Corollary. is straightforward and is omitted. Corollary.3 follows. The proof of Proof. ( r) n+λ U(rζ) Sn ( r) n+λ dµ(η) rζ η n+λ ( r)n Sn ( r ) +λ ( + r) +λ dµ(η) rζ η n+λ ( r)n u(rζ) ( + r) +λ which is decreasing (increasing) in r for λ > n (λ < n ) by Theorem.. This completes the proof of Corollary.3. Corollaries.4 and.5 are special cases of Theorem.. Corollary.6 is a straightforward generalization from B n to B n (R). The following is the proof of Corollary.7. Proof. 0 r r <. By the maximum principle, there is ζ S n such that u(rζ) max x r u(x). If λ > n, Theorem. implies ( r) n ( + r) max +λ x r ( r)n u(x) u(rζ) ( + r) +λ ( r ) n ( + r ) +λ u(r ζ) ( r ) n ( + r max ) +λ u(x) x r

12 Similarly, there is ξ S n such that u(rξ) min x r u(x). When λ > n, Theorem. yields ( + r) n min ( r) +λ x r The proof for λ < n ( + r)n u(x) u(rξ) ( r) +λ ( + r ) n ( r ) +λ u(r ξ) ( + r ) n ( r min ) +λ u(x) x r is parallel. This completes the proof of Corollary Proof of Theorem.8. In the following, B n denotes the unit ball in C n and S n B n the sphere. We need the following three lemmas for the proof of Theorem.8. Lemma.. If a C, a, then for 0 r, (.6) + r a ( + r)re(a) (.7) r a ( + r)re(a) Proof. a, so a Re(a) a and Re(a) a. If a Re(a), then + r a Re(a) + rre(a) so (.6) holds. If a < Re(a), consider f(r) +r a (+r)re(a), f (r) a Re(a) < 0. So f(r) decreases in r [0, ]. f() + a Re(a) > + Re(a) Re(a) ( Re(a)) 0. So f(r) 0 and (.6) holds. For the second inequality, +Re(a) a +Re(a) r a +rre(a), so r a ( + r)re(a) and (.7) holds. Lemma.3. Let z B n, z r, ζ S n. If α > n then (.8) (n+α + αr)( r ) n+α z ζ n+α If α < n, then ( r ) n+α r z ζ (n+α αr)( r ) n+α n+α z ζ n+α (.9) (n+α αr)( r ) n+α ( r ) n+α ) n+α z ζ n+α r z ζ n+α (n+α+αr)( r z ζ n+α Proof. Let z z η rη. r z ζ r ( rre(η ζ) + r η ζ ) (r η ζ Re(η ζ)),

13 3 and r z ζ n+α r ( z ζ ) n+α We have (.0) (n + α) ( z ζ ) n+α z ζ r (n + α) z ζ n+α (r η ζ Re(η ζ)). ( r ) n+α r z ζ (n + α)( r ) n+α ( r) z ζ n+α n+α z ζ 4n+4α ( r ) n+α r z ζ n+α z ζ 4n+4α (n + α)( r ) n+α r z ζ n+α z ζ 4n+4α ( r ) n+α (n + α) z ζ n+α (r η ζ Re(η ζ)) z ζ 4n+4α (n + α)( r ) n+α r z ζ z ζ n+α+ ( r ) n+α (n + α)(r η ζ Re(η ζ)) z ζ n+α+ To prove the right side inequality of (.8), it suffices to prove (n+α)r z ζ ( r )(n+α)(r η ζ Re(η ζ)) (n+α αr) z ζ which is equivalent to ( r )(n + α)(r η ζ Re(η ζ)) (n + α)( + r) z ζ. For α > n, the above inequality is equivalent to which is, after a simple simplification, ( r)((r η ζ Re(η ζ)) z ζ ( + r)re(η ζ) + r η ζ. The inequality is true by (.6) in Lemma.. To prove the left side inequality of (.8), it suffices to show (n+α)r z ζ ( r )(n+α)(r η ζ Re(η ζ)) (n+α+αr) z ζ which is equivalent to ( r )(n + α)(r η ζ Re(η ζ)) (n + α)( r) z ζ. For α > n, the above inequality becomes which is, after a simple simplification, ( + r)((r η ζ Re(η ζ)) z ζ ( + r)re(η ζ) r η ζ.

14 4 The inequality is true by (.7) in Lemma.. The proof of (.9) is parallel to that of (.8), using the same inequalities in Lemma.. This completes the proof of Lemma.3. Lemma.4. Let u(z) be a positive invariant harmonic function in B n defined by a positive Borel measure on S n with the Poisson-Szegö kernel, z r. If α > n, (n + α + αr) (.) r u(z) u(z) (n + α αr) r r u(z). If α < n, (.) (n + α αr) r u(z) u(z) (n + α + αr) r r u(z). Proof. By the Poisson-Szegö integral representation of u in B n, S n Sn ( z ) n+α u(z) dµ(ζ) z ζ n+α for a positive Borel measure µ on S n. By (.8) in Lemma.3 and µ being a positive measure, ( ( z ) n+α ) Sn (n + α αr)( r ) n+α dµ(ζ) dµ(ζ) r z ζ n+α z ζ n+α S n (n + α αr) r (n + α αr) r u(z) Sn ( z ) n+α dµ(ζ) z ζ n+α when α > n. It follows that ( u(z) ( z ) n+α ) (n + α αr) dµ(ζ) r r z ζ n+α r u(z). The left side inequality in (.) is proved similarly. For the equality case, consider u w (z) P α (z, w) ( z ) n+α z w n+α. It is known that u w(z) is invariant harmonic in C n \ {w} for w S n. A simple calculation shows that the equalities in (.) hold for u w (z) when z z w and z z w respectively. The proof of (.) is parallel to that of (.), using (.9) instead of (.8) in Lemma.8. This completes the proof of Lemma.4. Now we prove Theorem.8.

15 ( r)n ( + r)n Proof. Consider ϕ(r), ψ(r) for 0 r <. ( + r) n+α ( r) n+α Given ω S n, consider ϕ ϕ ψ ψ n + α αr r, n + α + αr r. I(r, ω) ϕ(r)u(rω), J(r, ω) ψ(r)u(rω). To show Theorem.8, it suffices to show that I(r, ω) is decreasing (increasing) and J(r, ω) is increasing (decreasing) in r when α > n (when α < n). By (.) in Lemma.4, when α > n, ϕ log I(r, ω) r ϕ + u r u n + α αr r + u r u 0. n + α αr r + n + α αr r Therefore log I(r, ω) is decreasing in r, and so is I(r, ω). Similarly, ψ log J(r, ω) r ψ + u r u n + α + αr r + u r u n + α + αr n + α + αr r r 0. Hence, J(r, ω) is increasing in r. For α > n and ζ, w S n, r ( r) n ( + r) n+α P α(rζ, w) r ( r) n ( r ) n+α ( + r) n+α rζ w n+α ( r) n+α δ(ζ, w) r rζ w n+α by applying the monotonicity results in Theorem.8 to u P α. By Lebesgue s dominated convergence theorem, ( r r)n u(rζ) ( r) n P α (rζ, ξ)dµ(ξ) r S n ( + r) n+α r n+α µ({ζ}). S n r ( r) n ( + r) n+α P λ(rζ, ξ)dµ(ξ) 5

16 6 ( + r) n Similarly, ( r) n+α P ( + r)n+α α(rζ, w) is increasing in r for α > n. rζ w n+α By Lebesgue s monotone convergence theorem, u(rζ) r ( r) n+α r ( r) n+α P α (rζ, ξ)dµ(ξ) S n ( + r) n r ( + r) n S n r ( r) n+α P α(rζ, ξ)dµ(ξ) ( + r) n+α n dµ(ξ) S n r rζ ξ n+α Sn n+α dµ(ξ). ζ ξ n+α For α < n, the monotonicity of I(r, ω) and J(r, ω) is proved similarly by applying (.) in Lemma.4. { ( r) n r ( + r) n+α P rζ w (n+α), ζ w α(rζ, w) r ( r) (n+α), ζ w when α < n. Therefore, ( r r)n u(rζ) ( + r) n+α r S n r { n+α µ({ζ}), if µ({ζ} c ) 0;, if µ({ζ} c ) > 0. ( r) n ( + r) n+α P α(rζ, ξ)dµ(ξ) ( + r) n Similarly, ( r) n+α P ( + r)n+α α(rζ, w) is decreasing in r for α < n. rζ w n+α By Lebesgue s monotone convergence theorem, u(rζ) r ( r) n+α r ( r) n+α P α (rζ, ξ)dµ(ξ) S n ( + r) n r ( + r) n S r n ( r) n+α P α(rζ, ξ)dµ(ξ) ( + r) n+α n dµ(ξ) S n r rζ ξ n+α Sn n+α This completes the proof of Theorem.8. dµ(ξ) ζ ξ n+α Remark. Notice that the monotonicity of the auxiliary functions ϕ and ψ in the proofs of Theorem. and Theorem.8 may vary depending on the values of the parameter λ (or α) and the dimension n. When λ > n ( or α > n), we have ϕ < 0 and ψ > 0, i.e. ϕ increases and ψ decreases in r for 0 < r <. For λ < n

17 (or α < n), the monotonicity does not necessarily hold. For example, in the real case in Theorem., for λ < n, ( ) λ n ( > 0, r 0, r)n ϕ(r) ( + r) +λ, λ + (n ) ϕ (r) ( ) λ n < 0, r λ + (n ), i.e. the monotonicity may change for certain combinations of n and λ. However, the monotonicity of ϕ(r)u(rζ) and ψ(r)u(rζ) holds Proofs of Theorem. and Theorem.9. The proofs for the two theorems are based on the following lemma. Lemma.5. Let f(r) be a positive function on r [0, ). If for a, b R, (.3) a + br r f(r) f (r) a br r f(r), then for 0 r r <, (.4) ( + r + r ) a ( r r )b+a f(r ) f(r) ( ) + r a ( )b a r + r r f(r ). Proof. a br r dr a ln( + r) + (b a) ln( r ) + C Thus for 0 r r <, by (.3), ln f(r ) ln f(r ) r r f (r) f(r) dr r r ( a br + r dr ln r + r ) a ( r r )b a i.e. the right side inequality in (.4) holds. Similarly, by the left side of (.3), r ln f(r ) ln f(r ) a + br ( ) + r a ( )b+a r dr ln r + r r. i.e. the left side inequality in (.4) holds. Now we prove Theorem.. r Proof. If λ > n, u(rζ) satisfies (.4) in Lemma.. Therefore (.3) holds with f(r) u(rζ), a n + λ, b n + λ +. Let 0 r r <. (.4) in Lemma.5 implies ( ) + r n λ ( ) r λ+ ( ) + r n+λ ( ) r + r r u(r n+ ζ) u(rζ) + r r u(r ζ).

18 8 If λ < n, u(rζ) satisfies (.5) in Lemma., thus (.3) holds with f(r) u(rζ), a n λ, b n + λ +. Applying (.4), ( ) + r n+λ ( ) r n+ ( ) + r n λ ( ) r + r r u(r λ+ ζ) u(rζ) + r r u(r ζ). This completes the proof of Theorem.. The proof of Theorem.9 is similar to that of Theorem.. Proof. If α > n, u(rζ) satisfies (.) in Lemma.4. Therefore (.3) holds with f(r) u(rζ), a n + α, b α. Let 0 r r <. (.4) in Lemma.5 implies ( + r + r ) n α ( ) r n+α u(r ζ) u(rζ) r ( ) + r n+α ( ) r n + r r u(r ζ). If α < n, u(rζ) satisfies (.) in Lemma.9, thus (.3) holds with f(r) u(rζ), a n α, b α. From (.4), ( ) + r n+α ( ) r n ( ) + r n α ( ) r + r r u(r n+α ζ) u(rζ) + r r u(r ζ). This completes the proof of Theorem.9. Most results in this paper are on the function values at two points in B n on the same ray. Similar results can be obtained for any two points in B n (ref. [5] and Section ).

19 . On Harnack inequality for positive invariant harmonic functions A refined estimate of Harnack inequality is proved for positive invariant harmonic functions... Statement of results. Let B n {x R n : x < }, n be the unit ball in R n, S n B n. We prove a refined estimate of Harnack inequality for positive invariant harmonic functions defined by positive Borel measures on the sphere with respect to the Poisson kernel P λ (defined below). First we consider positive harmonic functions, and the more general case follows. 9 Theorem.. Let u be a positive harmonic function in B n. ξ, ξ S n, 0 r r <. Then (.) f( r, r ) exp{ g(r )} u(r ξ ) u(r ξ ) f(r, r ) exp{g(r )}. where ( ) ( + r r f(r, r ) + r r g(r ) g(r, ξ, ξ ) π ξ nr ξ ( r ) ) n Remark. When r r x, r 0, (.) becomes r u(x) ( + r) n u(0) the classical Harnack inequality in B n. Denote the differential operator λ ( x x ) 4 j x j + λ j + r ( r) n x j + λ( n x j λ), λ R. Invariant harmonic functions are solutions of λ u 0 and are of certain invariant property with respect to Möbius transformations. Let µ be a positive Borel measure on S n and P λ be the Poisson kernel P λ ( x ) +λ x η n+λ, λ R. It is known that u(x) P λ (x, η)dµ(η) S n is an invariant harmonic function in B n ([], p. 9).

20 0 Theorem.. Let u be a positive invariant harmonic function in B n defined by a positive Borel measure µ on S n with the Poisson kernel P λ. Let ξ, ξ S n and 0 r r <. If λ > n, (.) f λ ( r, r ) exp{ g λ (r )} u(r ξ ) u(r ξ ) f λ(r, r ) exp{g λ (r )}. If λ < n, (.3) f λ (r, r ) exp{ g λ (r )} u(r ξ ) u(r ξ ) f λ( r, r ) exp{g λ (r )} where f λ (r, r ) ( ) + λ+ ( r r + r r g λ (r ) g λ (r, ξ, ξ ) π ξ ξ n + λ r ( r ) ) n Case λ n corresponds to the Laplace-Beltrami operator n and the Poincaré metric. It is known ([]) that, for positive u, n u 0, there exists a positive Borel measure µ on S n such that u(x) P n S n (x, η)dµ(η). In this case, Theorem. has the following form. Corollary.3. Let u be a positive solution of n u 0 in B n. Let ξ, ξ S n and 0 r r <. Then C u(r ξ ) u(r ξ ) C, where C C(r, r, ξ, ξ ) ( + r r ) n { exp π ξ ξ (n )r } + r r ( r ). The proofs will need an earlier result in Part (also [4]) on the monotonicity of positive invariant harmonic functions. Here we state the result as a proposition. Proposition.4. (Theorem. in Par and in [4]) Let u be a positive invariant harmonic function defined in B n by a positive Borel measure µ on S n with the Poisson kernel P λ. Let ζ S n and 0 r r <. If λ > n, ( ) r λ+ ( ) +r n r u(r ζ) u(rζ) +r ( + r + r ) λ+ ( ) r n u(r ζ). r

21 If λ < n, ( ) + r λ+ ( ) r n + r u(r ζ) u(rζ) r ( r r ) λ+ ( ) +r n u(r ζ). +r.. Proof of Theorem.. We need five lemmas before proving Theorem.. Let x y n k x ky k denote the inner product in R n. Lemma.5. Let η, ξ, ξ S n. Let ϕ(t), t [0, ] be the shortest arc on the great circle connecting ξ and ξ. Then (.4) Proof. d dt rϕ(t) η n nrϕ (t) η rϕ(t) η n+ d dt rϕ(t) η d ( r + rϕ(t) η ) rφ (t) η dt d dt rϕ(t) η n d ( rϕ(t) η ) n/ dt n ( rϕ(t) η ) n d rϕ(t) η dt nrϕ (t) η rϕ(t) η n+ Lemma.6. Let u be a positive harmonic function in B n. ξ, ξ S n. Let ϕ(t), t [0, ] be the shortest arc on the great circle connecting ξ and ξ. Then d dt u(rϕ(t)) nr ϕ (t) u(rϕ(t)), r [0, ]. ( r) Proof. By (.4), d S n dt rϕ(t) η n S dµ(η) nrϕ (t) η n rϕ(t) η n+ dµ(η) ϕ (t) η nr rϕ(t) η n ( r) dµ(η) S n nr ϕ (t) ( r) ( r ) nr ϕ (t) ( r) ( r ) u(rϕ(t)) Sn r rϕ(t) η n dµ(η)

22 where we applied the inequality rϕ(t) η rφ(t) η r. Therefore by the Lebesgue s Dominant Convergence Theorem, d dt u(rϕ(t)) d Sn rϕ(t) dt rϕ(t) η n dµ(η) ( r ) d dt rϕ(t) η n dµ(η) S n nr ϕ (t) ( r) u(rϕ(t)). Lemma.7. Let ξ, ξ S n. Then there exists a Möbius transformation T in R n such that T (S n ) S n and for all r [0, ], and T (rξ i ) (0,, 0, r cos θ i, r sin θ i ), i,, θ θ π (.5) det T (x), T (rξ ) T (rξ ) rξ rξ, where T (x) denotes the Jacobian matrix of the Möbius transformation T. Proof. The transformations involves a rotation in R n with respect to the origin such that ξ, ξ are in the R plane of the last two coordinates. If θ θ π is not yet satisfied, it can be achieved by a reflection with respect to the origin. Both rotation and reflection preserve the Euclidean norm and distance, and the absolute value of the determinant of the Jacobian matrix det T (x). Lemma.8. Let and Then ξ i (0,, 0, r cos θ i, r sin θ i ) S n, i,, θ θ π ϕ(t) (0,, 0, r cos θ t, r sin θ t ), θ t tθ + ( t)θ, t [0, ]. (.6) ϕ (t) π ξ ξ. Proof. It suffices to prove for n. For computation convenience, we use the complex plane notations in R. Denote ξ i e θ i. ξ ξ e iθ e iθ ( exp i θ ) ( + θ exp ( exp i θ + θ ( i θ ) ( θ exp i θ )) θ ) (i) sin θ θ

23 3 Notice that sin x x for x π, π so θ θ π implies ( ) e iθ e iθ sin θ θ π θ θ. Furthermore, Thus ϕ (t) d ) (e itθ +i( t)θ ϕ(t)i(θ θ ). dt ϕ (t) θ θ π e iθ e iθ π ξ ξ. Lemma.9. Let u be a positive invariant harmonic function. ξ, ξ S n. Then for r [0, ), { exp π } ξ nr ξ ( r) u(rξ { } ) π u(rξ ) exp ξ nr ξ ( r). Proof. Let T be the Möbius transformation in R n such that for r [0, ], rζ i T (rξ i ) (0,, 0, r cos θ i, r sin θ i ), i,, θ θ π. By (.5) in Lemma.3 and [], U(x) u(t (x)) is still a positive harmonic function in B n with respect to the measure µ(t (x)), det T (x). Furthermore, and Let U(rζ i ) u(t (rζ i )) u(rξ i ), i, ξ ξ T (ξ ) T (ξ ) ζ ζ e iθ e iθ. ϕ(t) (0,, 0, r cos θ t, r sin θ t ), θ t tθ + ( t)θ, t [0, ] be the shortest arc on the great circle connecting ζ and ζ, ϕ(0) ζ, ϕ() ζ. By (.6) in Lemma.8, d dt U(rϕ(t)) d U(rϕ(t)) dt dt U(rϕ(t)) U(rϕ(t)) dt 0 0 nr ( r) ϕ (t) dt 0 π ζ nr ζ ( r)

24 4 Since we have Therefore ln u(rξ ) u(rξ ) ln U(rζ ) U(rϕ()) ln U(rζ ) U(rϕ(0)) 0 d dt U(rϕ(t)) U(rϕ(t)) dt, ln u(rξ ) u(rξ ) π ζ nr ζ ( r) π ξ nr ξ ( r). π ξ nr ξ ( r) ln u(rξ ) u(rξ ) π ξ nr ξ ( r). This completes the proof of Lemma.9. Now we prove Theorem.. Proof. u(r ξ ) u(r ξ ) u(r ξ ) u(r ξ ) u(r ξ ) u(r ξ ) Proposition.4 implies ( ) r + n r u(r ξ ) r + r u(r ξ ) + r ( ) n r. + r r Combine the above with the results in Lemma.9. Theorem. follows..3. Proof of Theorem.. We need the following three lemmas for the proof of Theorem.. Lemma.0. Let η, ξ, ξ S n. Let ϕ(t), t [0, ] be the shortest arc on the great circle connecting ξ and ξ. Then for λ R, (.7) d dt rϕ(t) η n+λ (n + λ)rϕ (t) η rϕ(t) η n+λ+ Proof. The proof is similar to that of Lemma.5. d dt rϕ(t) η n+λ d ( rϕ(t) η ) n+λ dt n + λ (n + λ)rϕ (t) η rϕ(t) η n+λ+ using the result from the proof of (.4) in Lemma.5. ( rϕ(t) η ) n+λ d rϕ(t) η dt

25 Lemma.. Let u be a positive invariant harmonic function in B n defined by a positive Borel measure µ on S n with the Poisson kernel P λ. ξ, ξ S n. Let ϕ(t), t [0, ] be the shortest arc on the great circle connecting ξ and ξ. Then for λ R, λ n, (.8) d dt u(rϕ(t)) r (n + λ)ϕ (t) ( r) u(rϕ(t)), r [0, ]. Proof. By (.7) and rϕ(t) η rφ(t) η r, d S n dt rϕ(t) η n+λ dµ(η) (n + λ)rϕ (t) η S n rϕ(t) η n+λ+ dµ(η) ϕ (t) η n + λ r rϕ(t) η n+λ ( r) dµ(η) S n r (n + λ)ϕ (t) ( r) ( r ) +λ r (n + λ)ϕ (t) ( r) ( r u(rϕ(t)). ) +λ By the Lebesgue s Dominant Convergence Theorem, d dt u(rϕ(t)) d Sn ( rϕ(t) ) +λ dt rϕ(t) η n+λ dµ(η) ( r ) +λ d dµ(η) S n dt rϕ(t) η n+λ r (n + λ)ϕ (t) ( r) u(rϕ(t)). 5 Sn ( r ) +λ dµ(η) rϕ(t) η n+λ Lemma.. Let u be a positive invariant harmonic function in B n defined by a positive Borel measure µ on S n with the Poisson kernel P λ. ξ, ξ S n. Then for r [0, ), (.9) exp { π ξ ξ } n + λ r ( r) u(rξ { ) π u(rξ ) exp ξ ξ } n + λ r ( r) Proof. Let T be the Möbius transformation in R n such that rζ i T (rξ i ) (0,, 0, r cos θ i, r sin θ i ), i,, θ θ π. By (.5) in Lemma.7 and [], U(x) u(t (x)) is also a positive invariant harmonic function in B n with respect to the measure µ(t (x)), det T (x). U(rζ i ) u(t (rζ i )) u(rξ i ), i,

26 6 and ξ ξ T (ξ ) T (ξ ) ζ ζ e iθ e iθ. Let ϕ(t) (0,, 0, r cos θ t, r sin θ t ), θ t tθ + ( t)θ, t [0, ]. Then ϕ(0) ζ, ϕ() ζ. By (.8) in Lemma., d dt U(rϕ(t)) d U(rϕ(t)) dt dt U(rϕ(t)) U(rϕ(t)) dt Since we have Therefore 0 0 n + λ r ( r) π ζ ζ ln u(rξ ) u(rξ ) ln U(rζ ) U(rϕ()) ln U(rζ ) U(rϕ(0)) 0 0 ϕ (t) dt n + λ r ( r). d dt U(rϕ(t)) U(rϕ(t)) dt, ln u(rξ ) u(rξ ) π ζ n + λ r ζ ( r) π ξ n + λ r ξ ( r). π ξ n + λ r ξ ( r) ln u(rξ ) u(rξ ) π ξ n + λ r ξ ( r). This completes the proof of Lemma.. The proof Theorem. is similar to that of Theorem.. Proof. u(r ξ ) u(r ξ ) u(r ξ ) u(r ξ ) u(r ξ ) u(r ξ ) Apply (.9) in Lemma. and Proposition.4. Theorem. follows.

27 3. On higher order angular derivatives an application of Faá di Bruno s formula 3.. Statements of results. We study the singular behavior of kth angular derivatives of analytic functions in the unit disk in the complex plane C and positive harmonic functions in the unit ball in R n. Faá di Bruno s formula plays an important role in our proofs. Let D {z : z < } C and T D. Let ϕ : D D be analytic and ζ T. ζ is a fixed point of ϕ if r ϕ(rζ) ζ. The angular derivative at ζ is defined as ϕ (ζ) r ϕ (rζ). It is a consequence of the Julia Lemma [6] that the angular derivative at the fixed point exists and that ϕ (ζ) (0, ]. When the angular derivative of a fixed point is finite, what could be the iting behavior of the higher order angular derivatives? We describe an asymptotic property of the higher order derivatives of the fixed point in the following theorem. 7 Theorem 3.. Let ϕ(z) : D D be analytic. Let ζ T be a fixed point of ϕ with angular derivative ϕ (ζ) <. Then l, the l-th angular derivative ( ) (3.) ϕ (l) (rζ) o ( r) l as r. The above is equivalent to r ( r) l ϕ (l) (rζ) 0 by the definition of the little o notation. The order l in Theorem 3. is sharp in the sense illustrated in the following proposition and its proof. Proposition 3.. For any ε (0, ) and ζ T, there exists an analytic function ψ(z) : D D such that ζ is a fixed point of ψ, ψ (ζ) <, and (3.) r ( r) l ε ψ (l) (rζ) > 0, l. Furthermore, for any integer m, there exists an analytic function ψ(z) : D D such that ζ is a fixed point of ψ, (3.3) ψ (m j) (ζ) <, j 0, ψ (m+k) (ζ), k, ( r r)m+k ψ (n+k) (rζ) 0, ( r r)m+k ε ψ (m+k) (rζ) > 0. Results analogous to Theorem 3. can be obtained for positive harmonic functions, as stated in the following theorem.

28 8 Theorem 3.3. Let u be a positive harmonic function in the unit ball B n R n, n. Let ζ S n B n. Then for k, } n+k dk (3.4) {( r) r dr k u(rζ) (n + k )! ( r) n u(rζ). (n )! r + r Consequently, } n+k dk (3.5) {( r) r dr k u(rζ) 0 except possibly on a countable set of points on the sphere. From the proof of Theorem 3.3 we can see that the results can be extended to harmonic functions defined by complex measures. We may restate Theorem 3.3 as the following. Theorem 3.3. Let u be a harmonic function in the unit ball B n, n defined by a complex measure µ on S n (with the Poisson kernel). Let ζ S n. Then for k, r } n+k dk {( r) dr k u(rζ) (n + k )! (n )! r ( r) n u(rζ). + r 3.. Proof of Theorem 3.. First we prove a lemma needed for the proof of Theorem 3.. Lemma 3.4. Let f(z) be analytic and Ref(z) > 0 for z D. Let ζ T. Then ( r r)k+ f (k) (rζ) ζ k r k! r + r f(rζ). Proof. The proof follows the steps similar to the proof of Theorem.3 in [4]. First consider the case f(0). Since Ref(z) > 0, there exists a unique positive Borel measure µ such that (ref. [5]) f(z) Direct calculation yields f (k) (z) k! Consider z rζ. Since T r + ηz dµ(η), µ(t). ηz T η k dµ(η), k. ( ηz) k+ r rζη {, η ζ; 0, η ζ;

29 we have T r r + r + rζη dµ(η) µ({ζ}), rζη T r T r By the Lebesgue s dominated convergence theorem, { } { } r r r + r f(rζ) + rζη r + r T rζη dµ(η) { r + rζη + r rζη and r { } ( r) k+ f (k) (rζ) Therefore, r k! { ( r) k+ k! T r 9 η k ( r) k+ ( rζη) k+ dµ(η) ζk µ({ζ}), k. T } dµ(η) µ({ζ}), η k } dµ(η) ( ηz) k+ η k ( r) k+ ( rζη) k+ dµ(η) ζk µ({ζ}), k. ( r r)k+ f (k) (rζ) ζ k r k! r + r f(rζ). If f(0), consider f(z) i Imf(0) g(z), Ref(0) we have Thus g (k) (z) f (k) (z) Ref(z), g(0), Re(g(z)) > 0 for z D. Ref(0) Ref(0) Consequently, ( f (k) (rζ) r r)k+ Ref(0) { } ζ k r f(rζ) i Imf(0) k! r + r Ref(0) { } ζ k r f(rζ) k!. r + r Ref(0) ( r r)k+ f (k) (rζ) ζ k r k! r + r f(rζ). The following is the proof of Theorem 3.. Proof. By considering the analytic function ζϕ(ζz) : D D, we only need to prove Theorem 3. for the case ζ without loss of generality. Let f(z) + ϕ(z) ϕ(z), z D.

30 30 Then Furthermore, r Subsequently, By Lemma 3.4, Ref(z) > 0 and ϕ(z) f(z), z D. f(z) + r r + ϕ(r) f(r) + r r + r ϕ(r) r ( r)(f(r) + ) r ϕ (). ( f (k) (r) r r)k+ k! k! r r + r f(r) ϕ () r + ϕ(r) ϕ(r) + r ϕ (). for k 0. Let h(z) z, then ϕ(z) h(f(z)). By Faà di Bruno s formula [7], z + ( ϕ (l) (r) dl dz l h(f(r)) l! m! m! m l! h(m + +m l ) (f(r)) j where the sum is over all l-tuples (m, m,, m l ) satisfying Since we have m + m + + lm l l. h (k) (z) ( )k+ k!, k, (z + ) k+ ϕ (l) (r) l! ( ) (m + +m l +) (m + + m l )! m! m! m l! (f(r) + ) m + +m l + ( l! ( ) (m + +m l +) (m + + m l )! m! m! m l! f(r) + j ) f (j) mj (r) j! ( ) f (j) mj (r) j! j ) f (j) mj (r). (f(r) + )j! Notice that for each term of the sum, m + m + + lm l l, therefore ( ) r ( f (j) mj (r) r)l f(r) + (f(r) + )j! j ( ) ( r) j+ f (j) mj (r)/j! r ( r)(f(r) + ) ( r)(f(r) + ) j ( /ϕ ) () mj /ϕ () /ϕ ϕ (). () j

31 3 Consequently, { } ( r) l ϕ (l) (r) r ϕ ()l! ( ) (m + +m l +) (m + + m l )!. m! m! m l! To see that the above sum is zero, consider the function g(x) x, g (k) (x) k!( ) k x (k+), g (k) (g(x)) k!( x) k+, x (0, ]. Applying Faà di Bruno s formula to x g(g(x)), we have d l dl (x) dxl dr l g(g(x)) Hence, l! m! m! m l! g(m + +m l ) (g(r)) j l! (m + + m l )!( x) m + +m l + m! m! m l! ( g (j) (r) j! ( ( ) j l! (m + + m l )!( ) m + +m l + x ( ) l m! m! m l! x l ( ) l l! ( ) m + +m l + (m + + m l )! x m! m! m l! j ) mj x j+ ) mj 0, x (0, ], l. { } ( r) l ϕ (l) (r) r ϕ ()l! ( ) (m + +m l +) (m + + m l )! m! m! m l! ( ) d ϕ ()( ) l l dx l (x) 0, x therefore ( ) ϕ (l) (r) o ( r) l as r, l Proof of Proposition 3.. The following lemma is needed for the proof of Proposition 3..

32 3 Lemma 3.5. Let α (0, ), f (x) a n+ x n+, a n+ f (x) n0 b n x n, b n n0 h(x) f (x) f (x) c n x n. n0 ( ) α n ( ) α n + α(α ) (α n), (n + )! α(α ) (α n + ), (n)! Then c n 0, c n+ > 0, n 0. Proof. By the symmetry of f and f, Furthermore, h( x) f ( x) f ( x) f (x) f (x) h(x) c n 0, n 0. f (x) h(x)f (x) a n+ c n+ b 0 + c n b + + c b n, n 0. Since b 0, we have For n 0,, c a α > 0, c 3 a 3 c b c n+ a n+ (c n b + + c b n ), n 0. α(α )(α ) 3! ( α(α ) α α(α ) ) (α + ) > 0.! 3 Now assume c j > 0 for j,, k for some k. Notice that for α (0, ), a k+ > 0, k 0 and b k < 0, k. Therefore c k+ a k+ c k b c b k > 0, because every term is positive. By induction, c n+ > 0, n 0. The following is the proof of Proposition 3.. Proof. We prove (3.) in Proposition 3. by constructing a function ψ such that ψ (or its rotation) satisfies (3.). First consider ϕ is its own inverse: ϕ(z) z z, Reϕ(z) > 0, z D. + z + z ϕ ϕ : ϕ(ϕ(z)) z + z + z + z z.

33 For α (0, ), let g(z) z α, and define ( ) z α f ϕ + z g ϕ ( ) z α, D H D. + + z Then f(0) 0, f(), f (z) 0. Therefore f : D D is univalent and z is a fixed point of f. Considering the Taylor expansions ( ) α ( ) α ( + z) α z n, ( z) α ( ) n z n, n n we have n0 f(z) ( + z)α ( z) α ( + z) α + ( z) α Define F (z) n0 z 0 f(w)dw z 0 ( ) α z n+ n + ( α )z n n n0 n0 c n+ w n+ dw n0 n0 n0 c n z n n0 c n+ z n+. n0 c n+ n + zn+, z D. By Lemma 3.5, c n+ > 0 for n 0. Therefore F (z) achieves its maximum on the boundary at z : c n+ c n+ c n+ F (z) n + z n+ n + z n+ F () max F (z). n + z n0 By the maximal principle, the function ψ(z) F (z) F (), n0 z D maps D into D. Furthermore, z is a fixed point of ψ, and Notice that ψ(), ψ () F () F () f() F (), ψ () f () F (), d dz ( ) z α α + z ψ (k) (r) f (k ) (r), k. F () ( ) z α + z ( + z) α ( ) z α ϕ (z), + z 33

34 34 and d dz ( ) z α α(α ) + z ( z + z ) α ( ϕ (z) ) ( ) z α + α ϕ (z), + z etc. Using the big O notation for z near, we may write d k ( ) z α ( ) ( ) α z α k ( dz k k! ϕ (z) ) ( k + O ( z) α k+). + z k + z Applying the little o notation, we have d k ( ) dk z α g(ϕ(r)) dzk dz k + z zr ( ) ( ) α r α k ( ) k k! k + r ( + r) + O (( r) α k+) ( α k! )( r) α k ( ) k k ( k ( + r) α+k + o ( r) α k) By Faà di Bruno s formula [7], f (l) (r) dl dz l ϕ(g(ϕ(r))) l! m! m! m l! ϕ(m + +m l ) (g(ϕ(r)) j d k dz k g(ϕ(r)) j! m j where the sum is over all l-tuples (m, m,, m l ) satisfying Notice that m + m + + lm l l. ϕ (k) (z) ( ) k k!(+z) (k+), ϕ (k) (g(ϕ(r)) ( ) k k! and j d k dz k g(ϕ(r)) j! m j j ( + ( ) r α ) (k+), + r {( α )( r) α j ( ) j j j ( + r) α+j + o ( ( r) α j) } mj ( ) ( ) m + +m l l ( r) α l + r j ( + o ( r) α l), where each term in the product ( ) α ( ) j α(α ) (α j + )( )j j ( + r) α j! ( + r) α {( α j ) } ( ) j ( + r) α α( α) (j α) j! ( + r) α > 0

35 35 for α (0, ), and the little o term is obtained by the fact that r j ( r)α j ( r) (m+ +ml)α l ( r) α l r ( r) α l ( r) (m + +m l )α 0. r Since for given α (0, ) and l, l! (m + + m l )! m! m! m l! is bounded and > 0 as r, we have f (l) (r) l!(m + + m l )! m! m! m l! + o ( ( r) α l). ( + ( + ( ) r α ) (m + +m l +) + r ( ) α ) (m + +m r l +) ( ( r) α l + r + r ) l j ( ) α ( ) j j ( + r) α Consequently ( r r)l α f (l) (r) l! (m + + m l )! m! m! m l! j ( ) α ( ) j j α C l,α > 0, where C l,α is a constant for any given α (0, ) and l. Therefore we have ( r r)n α ψ (n) (r) ( r) n α f (n ) (r) C n,α r F () F () > 0, n. We have shown that (3.) in Proposition 3. holds for ψ with ζ. arbitrary ζ T, (3.) is satisfied by ζψ(ζz). For an To prove (3.3), we show that for any m, there exists a function ψ m such that ψ m (or its rotation) satisfies the conditions in (3.3). Let where ψ m (z) F z m(z) F m (), F j(z) 0 F F, F 0 f, ψ ψ F j (w) dw, j m, F j ()

36 36 are the functions used in the above proof of (3.). By the construction, c n+ F (z) n + zn+ F (z) F (z) F (z) n0 n0 z etc. For j,,, m, F j (z) F j (z) z 0 0 F () F () c n+ n + z n+ F (w) F () dw F () n0 n0 F j (w) F j () dw F j () F j () F j () n0 n0 By the maximal principle, the functions map D into D. Furthermore, ψ (k) n0 c n+ n + F (), n0 c n+ (n + )(n + 3) z n+3 c n+ (n + )(n + 3) F (), n0 c n+ (n + )(n + 3) zn+3, c n+ (n + )(n + 3) (n + + j) zn++j, c n+ (n + )(n + 3) (n + + j) z n++j c n+ (n + )(n + 3) (n + + j) F j(). ψ m (z) F m(z) F m (), ψ m() m () F m (k) () F m () F (k ) m () F m ()F m () F (k ) m () F m ()F m ()F m () F m k () k j0 F <, k,,, m, m j() especially, ψ (m) m () f() m j F j() <. Notice that ψ (m) m (z) ψ(z) m j F j() <.

37 Consequently the proven result (3.) implies (3.3) for ζ. For an arbitrary ζ T, (3.3) is satisfied by ζψ m (ζz). 37 This completes the proof of Proposition Proof of Theorem 3.3. We need several lemmas to prove Theorem 3.3. Lemma 3.6. If ϕ(x) x + ax + b, then for any m 0, m d l (3.6) dx l h(ϕ(x)) j0 m j0 (m)! (j)!(m j)! h(m+j) (ϕ)(ϕ ) j, l m; (m + )! (j + )!(m j)! h(m+j+) (ϕ)(ϕ ) j+, l m +. Proof. Again by Faà di Bruno s formula [7], d l dx l h(ϕ(x)) l! m! m! m l! h(m + +m l ) (ϕ(x)) j ( ) ϕ (j) mj (x) j! where the sum is over all l-tuples (m, m,, m l ) satisfying m + m + + lm l l. Since ϕ, ϕ (j) 0 for j 3, the product in Faà di Bruno s formula simplifies to ( ) ϕ (j) mj (x) ϕ (x) m (with 0 0 ), j! which implies (3.7) j d l dx l h(ϕ(x)) m +m l l! m! m! h(m +m ) (ϕ)(ϕ ) m, where { m m m, l m; m m, l m +. Relabeling the summation index by j, { { m j, l m; m + j, l m; m then m + m, l m +, m + j +, l m +. Replacing m and m by j and m, (3.7) becomes (3.6).

38 38 Lemma 3.7. Let f(r) rζ η n, r [0, ), ζ, η Sn, n. Let θ r [0, π] denote the angle between the n-vectors ζ and rζ η for any r [0, ]. Then m (cos θ r ) j (m)! rζ η n+l (j)!(m j)! f (l) j0 (r) r ζ η m rζ η n+l+ j0 ( ) m+j n(n + ) (n + m + j ) m j, l m; (cos θ r ) j (m + )! ( ) m+j+ n(n + ) (n + m + j) (j + )!(m j)! m j, l m +. Proof. Let h(x) x n/, ϕ(r) rζ η. Then ϕ (r) d n (rζ j η j ) (r ζ η), ϕ (r), ϕ (j) (r) 0 for j 3, dr j ( h (α) (x) n ) ( n ) ( n ) (α ) x n α α n(n + ) (n + (α )) ( ) α x (n+α)/. Denote ζ (ζ,, ζ n ), η (η,, η n ) and ζ η j ζ jη j the Euclidean inner product of ζ and η. Notice that therefore (r ζ η) rζ ζ ζ η ζ (rζ η) rζ η cos θ r, h (m+j) (ϕ(r))(ϕ (r)) j m+j n(n + ) (n + (m + j )) j (r ζ η) j ( ) m+j rζ η n+m+j and (cos θ r ) j n(n + ) (n + m + j ) ( )m+j rζ η n+m m j, h (m+j+) (ϕ(r))(ϕ (r)) j+ m+j+ n(n + ) (n + (m + j)) j+ (r ζ η) j+ ( ) m+j+ rζ η n+m+j+ (rζ η)(cos θ r) j n(n + ) (n + m + j) ( )m+j+ rζ η n+m+ m j. Applying Lemma 3.6 to f(r) h(ϕ(r)), the result of Lemma 3.7 follows. Notation. Denote where (3.8) C n,k ( ) k (C(n, k) + kc(n, k )) for k, n, m (m)! (j)!(m j)! j0 C(n, l) m (m + )! (j + )!(m j)! j0 ( ) m+j n(n + ) (n + m + j ) m j, lm; ( ) m+j+ n(n + ) (n + m + j) m j, lm+;

39 39 for m 0, with C(n, 0). Lemma 3.8. Let ζ, η S n, n. Then for k, ( )} { n+k dk r 0, ζ η; {( r) r dr k rζ η n C n,k, ζ η. Proof. Let g(r) r, f(r) rζ η n. Then For k, d k dr k ( r rζ η n ) g (r) r, g (r) and g (j) (r) 0, j 3. k ( ) k dr k (f(r)g(r)) f (k j) (r)g (j) (r) j j0 ( ) ( ) ( ) k k k f (k) (r)g(r) + f (k ) (r)g (r) + f (k ) (r)g (r) 0 dk ( r )f (k) (r) rkf (k ) (r) k(k )f (k ) (r) with the convention that f (α) (r) g (α) (r) 0 for α < 0. Let m j0 C(n, l, θ) m Then j0 From Lemma 3.7, f (l) (r) (cos θ) j (m)! ( ) m+j n(n + ) (n + m + j ) (j)!(m j)! m j, lm; (cos θ) j (m + )! ( ) m+j+ n(n + ) (n + m + j) (j + )!(m j)! m j, lm+; C(n, l) C(n, l, 0), l, n. A(l, r, ζ, η) rζ η n+l C(n, l, θ r), where A(l, r, ζ, η), l even; r ζ η rζ η, l odd. Thus ( )} n+k dk r {( r) r dr k rζ η { [ n ]} ( r) n+k ( r )f (k) (r) rkf (k ) (r) k(k )f (k ) (r) r { ( r )( r) n+k A(k, r, ζ, η) r rζ η n+k C(n, k, θ r ) rk( r)n+k A(k, r, ζ, η) rζ η n+k C(n, k, θ r ) k(k )( r)n+k A(k, r, ζ, η) rζ η n+k C(n, k, θ r ) }.

40 40 Notice that A(k, r, ζ, η) is bounded, θ r 0 as r, and C(n, l, 0) is bounded, hence the last term 0 as r. In addition, r { r rζ η 0, ζ η;, ζ η; so we have r r and n+k dk {( r) dr k { ( r)a(k, r, ζ, η) 0, ζ η; r rζ η ( ) k, ζ η; ( r )} rζ η n { ( + r)( r) n+k A(k, r, ζ, η) rζ η n+k C(n, k, θ r ) rk( } r)n+k A(k, r, ζ, η) rζ η n+k C(n, k, θ r ) { 0, ζ η; ( ) k C(n, k, 0) ( ) k kc(n, k, 0), { ζ η; 0, ζ η; ( ) k C(n, k) ( ) k kc(n, k ) C n,k, ζ η. (n + k )! Lemma 3.9. C n,k (n )! for k, n. We postpone the proof of Lemma 3.9 until after the proof of Theorem 3.3. Now we prove Theorem 3.3. Proof. For any x B n, x rζ, ζ S n, n, we may write By Lemma 3.8, for any k, S n r Sn r u(rζ) rζ η n dµ(η). ( )} { n+k dk r 0, µ({ζ}) 0; {( r) dr k rζ η n dµ(η) C n,k µ({ζ}), µ({ζ}) > 0. By the interchangeability of differentiation and integration when the integral of the derivative converges and the Lebesgue s dominated convergence theorem, we

41 have r n+k dk {( r) dr k u(rζ) } r r 4 n+k dk {( Sn r } r) dr k rζ η n dµ(η) {( Sn r) n+k d k ( ) } r dµ(η) S n r C n,k µ({ζ}). n+k dk {( r) dr k dr k rζ η n ( r rζ η n )} dµ(η) By Theorem. in [4] (or by going through the proof of Lemma 3.8 with k 0), { ( r) n u(rζ) } µ({ζ}). Thus and r { } ( r) n u(rζ) µ({ζ}) r + r } { } n+k dk ( r) n {( r) r dr k u(rζ) C n,k u(rζ). r + r So (3.4) holds by Lemma 3.9. Consequently (3.5) holds because each positive harmonic function in the unit ball corresponds to a positive measure on the sphere (ref. []), and that the set of non-zero point mass of the measure is countable. This completes the proof of Theorem 3.3. The following is the proof of Lemma 3.9. Proof. Lemma 3.9 is proved by induction. By the definition, C(n, 0), C(n, ) ( ) n 0 n, C(n, )! ( ) n +! ( ) n(n + )! 0 n + n(n + ) n(n + ). For k and, C n, ( ) (n + )! (C(n, ) + C(n, 0)) ( n + ), (n )! C n, ( ) (n + )! (C(n, ) + C(n, )) (n(n + ) n). (n )! For any k, assuming C n,k ( ) k (C(n, k) + kc(n, k )) (n + k )!, (n )!

42 4 we will prove C n,k+ ( ) k+ (C(n, k + ) + (k + )C(n, k)) (n + k )!. (n )! Using the induction assumption, the above equation can be written as ( ) k+ (C(n, k + ) + (k + )C(n, k)) so it is sufficient to show (n + k )(n + k )! (n )! (n + k )C n,k (n + k )( ) k (C(n, k) + kc(n, k )), (3.9) (C(n, k + ) + (k + )C(n, k)) (n + k ) (C(n, k) + kc(n, k )) Write C(n, k) j C(n, k, j). Notice that (m + )(j + ) C(n, m +, j) C(n, m +, j), 0 j m; m + j (m + )(n + m + j + ) C(n, m +, j + ) C(n, m +, j), 0 j m. j + Therefore, C(n, m + ) m+ j0 m j0 j0 ( m + j m + + j ) C(n, m +, j) m + m+ m + j m + C(n, m +, j) + j C(n, m +, j) m + j m m + j m m + C(n, m +, j) + j + C(n, m +, j + ) m + j0 m m (j + )C(n, m +, j) (n + m + j + )C(n, m +, j) j0 j0 m (n + m + )C(n, m +, j) j0 (n + m + )C(n, m + ) for m 0.

43 43 Similarly, (m + )(n + m + j) C(n, m +, j) C(n, m, j), 0 j m; j + (m + )(j + ) C(n, m +, j) C(n, m, j + ), 0 j m. m j Consequently, C(n, m + ) m j0 m j + m + C(n, m +, j) + j0 m j C(n, m +, j) m + m m (n + m + j)c(n, m, j) + (j + )C(n, m, j + ) j0 m (n + m + j)c(n, m, j) + j0 m (n + m)c(n, m, j) j0 j0 m (j)c(n, m, j) j0 (n + m)c(n, m) for m 0. The above relations between the adjacent C(n, l) s can be summarized as C(n, l) (n + l )C(n, l ) for l. Applying the above equation repeatedly, for any k we have and L.H.S. of (3.9) C(n, k + ) (k + )C(n, k) (n + k)c(n, k) (k + )C(n, k) (n )C(n, k), R.H.S. of (3.9) (n + k ) (C(n, k) + kc(n, k )) (n + k )C(n, k) + k(n + k )C(n, k ) (n + k )C(n, k) kc(n, k) (n )C(n, k). Therefore (3.9) holds for all k. This completes the proof of Lemma 3.9. Acknowledgments. We thank Pietro Poggi-Corradini for his valuable idea for Proposition 3..

44 Introduction. The function is well known for its property 4. When is a function not flat? f(x) {e x, x 0 0, x 0 f (k) (0) 0, k 0 but f 0. Such a function is called flat at the origin. On the other hand, if f is a real analytic function with Taylor expansion on an open interval containing 0, then f (k) (0) 0, k 0 implies f 0. The unique continuation problem in PDE is to find conditions such that the solutions of PDE enjoy the same property. There are a large amount of literature in this area originated from the ideas by Carleman [3], called Carleman s method. In this paper we consider the simplest case of one variable. Theorem 4.. Let f(x) C ([a, b]), 0 [a, b], and (4.) f (n) (x) C n k0 for some constant C and n. Then f (k) (x), x [a, b] x n k f (k) (0) 0, k 0 implies f 0 on [a, b]. From Theorem 4. we obtain the following corollary. Corollary 4.. Let f(x) C ([a, b]), 0 [a, b], and (4.) holds for some constant C and n. Then f 0 implies the zero set {f (0)} [a, b] is finite. An example. We use an example in [6] to show that the order of singularity in (4.) is best possible, i.e., there exists a function f(x) C ([ a, a]), a > 0, f (n) (x) C n k0 f (k) (x) x n k+ε for x [ a, a] and f (k) (0) 0, k 0 for some constant C and ε > 0, but f 0 on [ a, a]. For m >, ε > 0, the following equation is considered in [6]: or equivalently, x u (x) + mxu (x) cx ε u(x) 0, x (0, ), (4.) u (x) + m x u (x) c u(x) 0, x (0, ), x+ε

45 a Bessel differential equation. The general Bessel differential equation takes the form (4.3) z u (z) + ( α)zu (z) + { β γ z γ + (α ν γ ) } u(z) 0, z C. It is well known that for (non-integer) ν Z, the solution for (4.3) is u(z) z α [ C J ν (βz γ ) + C J ν (βz γ ) ], where C, C are arbitrary complex numbers, and J ν is the Bessel function of order ν, i.e., a solution of equation (4.3) with α 0, γ. Notice that equation (4.) with c > 0 is equation (4.3) with α m, β i c ε, γ ε, ν m. ε By choosing m >, ε (0, ) such that C C πe iνπ sin(νπ), ν m Z, ε the solution of equation (4.) can be written as ( ) c (4.4) u(x) x (m )/ K (m )/ε ε x ε/, x (0, ) where K ν (z) π e iνπ (J ν (iz) J ν (iz)), arg z ( π, π/) sin(νπ) is the modified Bessel function of the third kind [9], with the asymptotic property K ν (x) π x / e x as x +. Therefore in (4.4), the function u(x) π ( ) / { c x m + ε4 exp } c ε ε x ε/ as x 0 is a nontrivial solution of (4.) vanishing at x 0 of infinite order. Hence f(x) u( x ), x [ a, a], a (0, ) is well defined and f C ([ a, a]). Taking derivative of equation (4.) n times, d n { dx n u (x) + m x u (x) c } x +ε u(x) 0, x (0, ), we obtain u (n) (x) + a n (x)u (n ) (x) + + a 0 (x)u(x) 0, x (0, ). For given m, c and ε, the coefficients ( ) a j (x) C o, j 0,,, n, x (0, ) x n j+ε for some constant C o > 0. By the property of u(x) near x 0, we have f (n) (x) + a n (x)f (n ) (x) + + a 0 (x)f(x) 0, x [ a, a] 45

46 46 with f (k) (0) 0, k 0, for some constant C >. Thus f (k) (0) 0, k 0, ( a j (x) C and f 0 from the non-triviality of u. f (n) (x) C x n j+ε n k0 ), j 0,,, n. f (k) (x) x n k+ε, x [ a, a], Theorem 4. leads to applications in ODE as stated in the following two propositions. Based on the above example, the order of the singularity of the coefficients in the assumption of the propositions is sharp. Proposition 4.3. Let f(x) C be a solution of with Then y (n) + a n (x)y (n ) + + a 0 (x)y 0, x [ a, a], a > 0 ( ) a k (x) O x n k as x 0, k 0,,, n. f (k) (0) 0, k 0 f 0 on [ a, a]. Proposition 4.4. Let f(x), g(x) C be solutions of with Then y (n) + a n (x)y (n ) + + a 0 (x)y b(x), x [ a, a], a > 0 ( ) a k (x) O x n k as x 0, k 0,,, n. f (k) (0) g (k) (0), k 0 f g on [ a, a]. 4.. Proof of Theorem 4. and its corollary. Several lemmas are needed for the proof of Theorem 4.. The basic idea of the following lemma was considered in [3]. Lemma 4.5. Let v(x) C ([0, b]). Assume v (k) (0) 0, k 0. Then for α, (4.5) b [v(x)] x α+ dx 4 b [v (x)] (α + ) 0 x α dx 0

47 Proof. Write v v(x), v v (x) and so on. 0 d dx ( x (α+) v ) (α + )x (α+) v + x (α+) vv Since v (k) (0) 0 for k [α/] +, we have [v(x)] O(x (k+) ) o(x α+ ), thus b d (x (α+) v ) dx [v(b)] dx b α+ [v(x)] x 0+ x α+ [v(b)] b α+ 0. Therefore, (α + ) b 0 v dx xα+ b 0 b 0 b 0 vv dx xα+ { (α ) + / v α + v dx + xα+ x (α+)/ b 0 α + 47 } { ( ) } / v α + x α/ dx (v ) x α dx by applying ab a + b to the last inequality. Consequently, α + b which is equivalent to (4.5). 0 v dx xα+ α + b 0 (v ) x α dx, Lemma 4.6. Let u(x) C ([0, b]). Assume u (k) (0) 0, k 0. Then for β, n, b [ u (k) (x) ] x β+(n k) dx 4 b [u (n) (x)] (β + ) x β dx, for k 0,, n. 0 0 Proof. Applying Lemma 4.5 to v u (n j), α β +j, β, j 0,,, n, we obtain b [ u (n j) (x) ] 4 b [u (n j) (x)] 0 x β+j+ dx (β + j + ) 0 x β+j dx, j 0,,, n, or equivalently, b [ u (k) (x) ] x β+(n k) dx 4 b [u (k+) (x)] (β + (n k ) + ) dx, k 0,,, n. xβ+(n k ) 0 Applying the above inequality repeatedly, we have b [ u (k) (x) ] { n } dx 0 xβ+(n k) 4 4 b [u (n) (x)] (β + (n m) + ) (β + ) 0 x β dx mk 4 (β + ) b 0 0 [u (n) (x)] dx, x β

48 48 because β, β + (n m ) + for m 0,..., n. Lemma 4.7. Let u(x) C ([0, b]). Assume u (k) (0) 0, k 0. Then for β, n, ( b n 0 x β k0 ) u (k) (x) x n k dx 4n b (β + ) 0 x β [u(n) (x)] dx. Proof. Apply Lemma 4.6 to k 0,, n and sum up both sides. Lemma 4.8. Let f(x) C ([a, b]), 0 [a, b]. Assume f (k) (0) 0, k 0. Then for β, n, ( ) b n f (k) (x) 4n b a x β x n k dx (β + ) a x β [f (n) (x)] dx. k0 Proof. The function u(x) f( x) satisfies the assumptions for Lemma 4.7 on [0, a], so ( ) a n f (k) ( x) 4n a 0 x β x n k dx (β + ) 0 x β [f (n) ( x)] dx. a k0 Substitute the variable x by x. Since the terms in the sum are of even powers and both sides have the same x β term, the above inequality can be written as ( ) 0 a n f (k) (x) 4n 0 x β x n k dx (β + ) a x β [f (n) (x)] dx. k0 From Lemma 4.7 the desired inequality is already true for f(x) on [0, b]. Combining the results on [a, 0] and [a, b], Lemma 4.8 follows. The following is the proof of Theorem 4.. Proof. f(x) satisfies the assumptions in Lemma 4.8 on [a, b], so for any β, (β + ) ( ) b n f (k) (x) b (4.6) 4n x β x n k dx x β [f (n) (x)] dx. From (4.), a k0 ( ) n f (n) (x) C f (k) (x) x n k k0 a