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LANDIS-TYPE CONJECTURE FOR THE HALF-LAPLACIAN PU-ZHAO KOW AND JENN-NAN WANG Abstract.

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PU-ZHAO KOW AND JENN-NAN WANG

Abstract. In this paper, we study the Landis-type conjecture, i.e., unique continuation property from innity, of the fractional Schrödinger equation with drift and potential terms.

We show that if any solution of the equation decays at a certain exponential rate, then it must be trivial. The main ingredients of our proof are the Caarelli-Silvestre extension and Armitage's Liouville-type theorem.

1. Introduction

In this paper, we consider the following equation with the half Laplacian (1.1) (−∆)12u + b(x) · ∇u + q(x)u = 0 in Rn,

where n ≥ 1. Our aim is to investigate the minimal decay rate of nontrivial solutions of (1.1).

In other words, we consider the unique continuation property from innity of (1.1). This problem is closely related to the conjecture proposed by Landis in the 60's [KL88]. Landis conjectured that, if u is a solution to the classical Schrödinger equation

(1.2) − ∆u + q(x)u = 0 in Rn,

with a bounded potential q, satisfying the decay estimate

|u(x)| ≤ exp(−C|x|1+),

then u ≡ 0. Landis' conjecture was disproved by Meshkov [Mes91], who constructed a complex-valued potential q ∈ L(Rn)and a nontrivial solution u of (1.2) such that

|u(x)| ≤ exp(−C|x|43).

In the same work, Meshkov showed that if

|u(x)| ≤ exp(−C|x|43+),

then u ≡ 0. Based on a suitable Carleman estimate, a quantitative version of Meshkov's result was established in [BK05], see also [CS99,Dav14,DZ18,DZ19,KL19,LUW11,LW14]

for related results.

In view of Meshkov's example, Kenig modied Landis' original conjecture and asked that whether the Landis' conjecture holds true for real-valued potentials q in [Ken06]. The real version of Landis' conjecture in the plane was resolved recently in [LMNN20]. We also refer to [Dav20,DKW17,DKW19,KSW15] for the early development of the real version of Landis' conjecture.

For the fractional Schrödinger equation, the Landis-type conjecture was studied in [RW19].

The main theme of this paper is to extend the results in [RW19] to the fractional Schrödinger

2020 Mathematics Subject Classication. Primary: 35A02, 35B40, 35R11. Secondary: 35J05, 35J15.

Key words and phrases. Unique continuation property, Landis conjecture, half-Laplacian, Caarelli- Silvestre extension, Liouville-type theorem.

1

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equation with the half Laplacian (1.1). Previously, the authors in [KW19] proved some partial results for the fractional Schrödinger equation

(1.3) ((−∆)s+ b(x)x · ∇ + q(x))u = 0 in Rn,

where s ∈ (0, 1) and b, q are scalar-valued functions. The main tools used in [KW19] are the Caarelli-Silvestre extension and the Carleman estimate. The particular form of the drift coecient in (1.3) is due to the applicability of the Carleman estimate. It turns out when s = 12, i.e., the case of half Laplacian, we can treat a general vector-valued drift coecient b(x) in (1.1). The underlying reason is that the Caarelli-Silvestre extension solution of (−∆)12u = 0 in Rn is a harmonic function in Rn+1+ . Inspired by this observation, we show that if both b and q are dierentiable, then any nontrivial solution of (1.1) can not decay exponentially at innity. The detailed statement is described in the following theorem.

Theorem 1.1. Assume that there exists a constant Λ > 0 such that (1.4a) kqkL(Rn)+ k∇qkL(Rn)+ k∇bkL(Rn)≤ Λ and, furthermore, there exists an  > 0, depending only on n, such that

(1.4b) kbkL(Rn)≤ .

Let u ∈ H12(Rn) be a solution to (1.1) and u satisfy

(1.4c) |u(x)| ≤ Λe−λ|x|

for some λ > 0, then u ≡ 0.

Remark 1.2. Note that both (−∆)12u and ∇u are rst orders. In view of the Lp estimate of the Riesz transform (2.1), (2.2), the assumption (1.4b) is to ensure that the non-local operator (−∆)12u is the dominated term in (1.1).

It is interesting to compare Theorem 1.1 with [RW19, Theorem 1]. Assume that u ∈ Hs(Rn) is a solution to

(1.5) (−∆)su + q(x)u = 0 in Rn

such that |q(x)| ≤ 1 and |x · ∇q(x)| ≤ 1. If

(1.6) Z

Rn

e|x|α|u|2dx < ∞ for some α > 1,

then u ≡ 0. Therefore, for s = 12, Theorem 1.1 extends their results by slightly relaxing the condition on q and also adding a drift term. Another key improvement is that the exponential decay rate e−λ|x| is sharper than (1.6).

The proof of Theorem 1.1 consists of two steps. Inspired by [RW19], we rst pass the boundary decay (1.4c) to the bulk decay of the Caarelli-Silvestre extension solution (harmonic function) in the extended space Rn× (0, ∞). In the second step, we apply the Liouville- type theorem (Theorem 6.1) to the harmonic function. It is noted that we do not use any Carleman estimate here. On the other hand, using the harmonic function in the unit ball v0(z) := <(e−1/zα), z ∈ C, 0 < α < 1 (see [Jin93]), it is not dicult to construct an example to show the optimality of the Liouville-type theorem. In view of this example, we believe that the decay assumption (1.4c) is optimal.

When b ≡ 0, the following theorem can be found in [Kow21, Theorem 1.1.9], which was obtained using similar ideas as in the proof of Theorem 1.1.

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Theorem 1.3. Let q ∈ L(Rn) (not necessarily dierentiable) satisfy kqkL(Rn)≤ Λ.

If u ∈ H12(Rn) is a solution to (1.1) with b ≡ 0 such that

(1.7) Z

Rn

e|x||u|2dx < ∞, then u ≡ 0.

It is interesting to compare this result with [RW19, Theorem 2]. There, it was proved that if u ∈ Hs(Rn) solves (1.5) with |q(x)| ≤ 1 and

(1.8) Z

Rn

e|x|α|u|2dx < ∞ for some α > 4s 4s − 1, then u ≡ 0. When s = 12, (1.8) becomes

Z

Rn

e|x|α|u|2dx < ∞

for α > 2, which is clearly stronger than (1.7). On the other hand, Theorem 1.3 holds regardless whether q is real-valued or complex-valued.

This paper is organized as follows. In Section2, we will study the decaying behavior of ∇u.

In Section 3, we localize the nonlocal operator (−∆)12 by the Caarelli-Silvestre extension.

In Section 4, we derive some useful estimates about the Caarelli-Silvestre extension ˜u of the solution u, which is harmonic. In Section 5, we obtain the decay rate of ˜u from that of u. Finally, we prove Theorem 1.1 in Section 6 by Armitage's Liouville-type theorem.

Furthermore, we provide another proof of this Liouville-type theorem in Appendix A.

2. Decay of the gradient

Let 1 < p < ∞. For each u ∈ Lp(Rn), let ψ satsify (−∆)12ψ = u and let u := ∇ψ. Using the Lp-boundedness of the Riesz transform [Ste16] (see also [BG13]), we can show that (2.1) kukLp(Rn) ≤ C(n, p)kukLp(Rn).

We remark that this estimate is also used in the proof of [CCW01, Theorem 2.1]. Note that we can formally write u = ∇(−∆)12u. Plugging (−∆)12ψ = uand u = ∇ψ into (2.1) implies (2.2) k∇ψkLp(Rn) ≤ C(n, p)k(−∆)12ψkLp(Rn).

Thanks to (2.2), we can obtain the following lemma.

Lemma 2.1. Assume that (1.4a) and (1.4b) hold. Let u ∈ H12(Rn) be a solution to (1.1).

If u satises the decay assumption (1.4c), then (−∆)12u satises

(2.3) Z

Rn

eλ2|x||(−∆)12u|2dx + Z

Rn

eλ2|x||(−∆)12u|n+1dx ≤ C for some positive constant C = C(n, λ, Λ).

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Proof. Let p = n + 1 in (2.2) and denote C(n, p) = C(n). We rst estimate the Ln+1-norm of ∇u. Taking Ln+1-norm on (1.1) and using (2.2), (1.4c), we have

k∇ukLn+1(Rn)≤ C(n)k(−∆)12ukLn+1(Rn)

≤ C(n)



nkbkL(Rn)k∇ukLn+1(Rn)+ kqkL(Rn)kukLn+1(Rn)



≤ C(n)k∇ukLn+1(Rn)+ C(n, λ, Λ).

Choosing  = (2C(n))−1 in the estimate above gives

(2.4) k∇ukLn+1(Rn)≤ C(n, λ, Λ).

Next, we estimate the Ln+1-norm of ∇2u. Dierentiating (1.1) yields

(2.5) (−∆)12ju + b(x) · ∇(∂ju) + ∂jb(x) · ∇u + q(x)∂ju + ∂jq(x)u = 0 for each j = 1, · · · , n. Taking the Ln+1-norm of (2.5), we have

k∇(∂ju)kLn+1(Rn) ≤ C(n)k(−∆)12jukLn+1(Rn)

≤ C(n)



nkbkL(Rn)k∇(∂ju)kLn+1(Rn)+ n2k∇bkL(Rn)k∇ukLn+1(Rn) + kqkL(Rn)k∇ukLn+1(Rn)+ nk∇qkL(Rn)kukLn+1(Rn)



≤ C(n)



nk∇(∂ju)kLn+1(Rn)+ Λ(n2+ 1)k∇ukLn+1(Rn)+ ΛnkukLn+1(Rn)



≤ 1

2k∇(∂ju)kLn+1(Rn)+ C(n, λ, Λ), and hence

(2.6) k∇2ukLn+1(Rn) ≤ C(n, λ, Λ).

Hence, it follows from the Sobolev embedding that k∇ukL(Rn)≤ C(n, λ, Λ).

Now we would like to derive the L2-decay of ∇u (for the case of n = 1). Combining (2.4) and (2.6), it is easy to see that

Z

Rn

eλ2|x||∂ju|2dx = Z

Rn

eλ2|x|(∂ju)(∂ju) dx

= −λ 2

Z

Rn

xj

|x|eλ2|x|u∂ju dx − Z

Rn

eλ2|x|(∂j2u)u dx

≤ λ 2

Z

Rn

eλ2|x||u||∂ju| dx + Z

Rn

eλ2|x||u||∂j2u| dx

≤ λΛ 2

Z

Rn

eλ2|x||∂ju| dx + Λ Z

Rn

eλ2|x||∂j2u| dx (by (1.4c))

≤ λΛ 2

 Z

Rn

eλ2p0|x|dx

p01

k∂jukLn+1(Rn)+ Λ

 Z

Rn

eλ2p0|x|dx

p01

k∂j2ukLn+1(Rn)

≤ C(n, λ, Λ) (where p0 = (n + 1)/n),

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that is, we obtain

(2.7) Z

Rn

eλ2|x||∇u|2dx ≤ C(n, λ, Λ).

We now continue to obtain the L2-decay of (−∆)12u. In view of (1.1) and using (1.4a), (1.4b), (1.4c), (2.7), we have

Z

Rn

eλ2|x||(−∆)12u|2dx

≤ Z

Rn

eλ2|x||b(x) · ∇u|2dx + Z

Rn

eλ2|x||q(x)u|2dx

≤ n2kbk2L(Rn)

Z

Rn

eλ2|x||∇u|2dx + kqk2L(Rn)

Z

Rn

eλ2|x||u|2dx

≤ C(n, λ, Λ).

(2.8)

Here we may choose a smaller  if necessary. The estimate (2.8) implies (2.3) for n = 1.

Our next task is to derive the Ln+1-decay of ∇u for n ≥ 2. First of all, let n be even. We then have

Z

Rn

eλ2|x||∂ju|n+1dx = Z

Rn

eλ2|x||∂ju|(∂ju)ndx

= Z

{∂ju6=0}

eλ2|x||∂ju|(∂ju)n−1(∂ju) dx

= −λ 2

Z

Rn

xj

|x|eλ2|x||∂ju|(∂ju)n−1u dx − Z

{∂ju6=0}

eλ2|x|ju

|∂ju|(∂j2u)(∂ju)n−1u dx

− (n − 1) Z

Rn

eλ2|x||∂ju|(∂ju)n−2(∂j2u)u dx

≤ λΛ 2

Z

Rn

eλ2|x||∂ju|ndx + n Z

Rn

eλ2|x||∂ju|n−1|∂j2u| dx (by (1.4c))

≤ C(n, λ, Λ) + n Z

Rn

eλ2|x||∂ju|n−1|∂j2u| dx (using (2.4)).

(2.9) Note that

Z

Rn

eλ2|x||∂ju|n−1|∂j2u| dx

 Z

Rn

eλ2r1|x|dx

r11  Z

Rn

|∂ju|r2(n−1)dx

r21  Z

Rn

|∂j2u|r3dx

r31 , where 1 < r1, r2, r3 < ∞ satisfy

1 r1 + 1

r2 + 1 r3 = 1.

Choosing r2 = n+1n−1 and r3 = n + 1, we have r1 = n + 1. Hence, we obtain from (2.4) and (2.6) that

(2.10) Z

Rn

eλ2|x||∂ju|n−1|∂j2u| dx ≤ C(n, λ, Λ).

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Combining (2.9) and (2.10) gives

(2.11) Z

Rn

eλ2|x||∇u|n+1dx ≤ C(n, λ, Λ).

When n ≥ 2 is odd, estimate (2.11) follows from the same argument above by noting

|∂ju|n+1 = (∂ju)n+1.

Finally, we estimate the Ln+1-decay of (−∆)12u for n ≥ 2. Using the equation (1.1) and by (1.4a), (1.4b), (1.4c), (2.11)), we have

Z

Rn

eλ2|x||(−∆)12u|n+1dx

≤ C

 Z

Rn

eλ2|x||b(x) · ∇u|n+1dx + Z

Rn

eλ2|x||q(x)u|n+1dx



≤ C



nn+1kbkn+1L(Rn)

Z

Rn

eλ2|x||∇u|n+1dx + kqkn+1L(Rn)

Z

Rn

eλ2|x||u|n+1dx



≤ C(n, λ, Λ).

(2.12)

Finally, (2.3) is a direct consequence of (2.8) and (2.12).  3. Caffarelli-Silvestre extension

In this section, we shall restate some notations in [RW19]. Let Rn+1+ := Rn × R+ =

 x = (x0, xn+1) x0 ∈ Rn, xn+1 > 0

and x0 = (x0, 0) ∈ Rn× {0}. For R > 0, we denote BR+(x0) := x ∈ Rn+1+ |x − x0| ≤ R ,

BR0 (x0) := x ∈ Rn× {0} |x − x0| ≤ R .

To simplify the notations, we also denote BR+ := BR+(0) and BR0 := BR0 (0). We dene two Sobolev spaces

1(Rn+1+ ) := n

v : Rn+1+ → R R

Rn+1+ |∇v|2dx < ∞ o , Hloc1 (Rn+1+ ) := n

v ∈ ˙H1(Rn+1+ ) R

Rn×(0,r)|v|2dx < ∞ for some constant r > 0 o . For u ∈ H12(Rn), let ˜u ∈ ˙H1(Rn+1+ ) be the solution to the Dirichlet problem

(3.1)

(∆˜u = 0 in Rn+1+ ,

˜

u(x0, 0) = u(x0) on Rn,

where ∇ = (∇0, ∂n+1) = (∂1, · · · , ∂n, ∂n+1). From [Sti10, page 4849], we have that ˜u ∈ Hloc1 (Rn+1+ ) and

(3.2) k˜u(•, xn+1)kL2(Rn) ≤ kukL2(Rn).

In [CS07], we know that the half Laplacian is equivalent to the Dirichlet-to-Neumann map of the extension problem (3.1), precisely,

(3.3) − (−∆)12u(x0) = lim

xn+1→0n+1u(x).˜

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Using this observation, we can reformulate (1.1) as the following local elliptic equation:

(3.4)





∆˜u = 0 in Rn+1+ ,

˜

u(x0, 0) = u(x0) on Rn,

xn+1lim→0n+1u(x) = b(x˜ 0) · ∇0u + q(x0)u on Rn.

4. Some estimates related to the extension problem

The following lemma is a special case of [RW19, Equation (19)] (see also [KW19, Lemma 3.2]).

Lemma 4.1. Let ˜u ∈ Hloc1 (Rn+1+ ) be a solution to (3.1). Then the following estimate holds for any x0 ∈ Rn× {0}:

k˜ukL2(B+cR(x0))≤C



k˜ukL2(B+16R(x0))+ R12kukL2(B0

16R(x0))

α

×

 R32

xn+1lim→0n+1

L2(B016R(x0))

+ R12kukL2(B0

16R(x0))

1−α

(4.1)

for some positive constants C = C(n), α = α(n) ∈ (0, 1) and c = c(n) ∈ (0, 1), all of them are independent of R and x0.

By choosing σ = 12, ν = 2 and a(x0) ≡ 0 in [JLX14, Proposition 2.6(i)], we obtain the following version of De Giorgi-Nash-Moser type theorem.

Lemma 4.2. Let ˜u ∈ Hloc1 (Rn+1+ ) be a solution to (3.1) and p > n. There exists a constant C = C(n, p) > 0 such that

(4.2) k˜ukL(B+1

4

) ≤ C



k˜ukL2(B1+)+

xn+1lim→0n+1Lp(B10)

 .

Combining (4.1) and (4.2), together with some suitable scaling, we can obtain the following lemma.

Lemma 4.3. Let ˜u ∈ Hloc1 (Rn+1+ ) be a solution to (3.1), and let p > n. The following inequality holds for all x0 ∈ Rn× {0} and R ≥ 1:

k˜ukL(B+cR(x0))≤ C



k˜ukL2(B+16R(x0))+ R12kukL2(B0

16R(x0))

α

×



R32k(−∆)12ukL2(B0

16R(x0))+ R12kukL2(B0

16R(x0))

1−α

+ R32k(−∆)12ukLp(B0

R(x0))

(4.3)

for some positive constants C = C(n, p), α = α(n) ∈ (0, 1) and c = c(n) ∈ (0, 1), all of them are independent of R and x0.

Proof. Without loss of generality, it suces to take x0 = 0. Let ˜v(x) = ˜u(Rx) and let v(x0) = u(Rx0), we observe that

(∆˜v = 0 in Rn+1+ ,

˜

v(x0, 0) = v(x0) on x0 ∈ Rn.

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From (4.1) and (4.2), it follows that k˜vkL(B+c)≤ C



k˜vkL2(B+16)+ kvkL2(B0

16)

α

xn+1lim→0n+1L2(B016)

+ kvkL2(B0

16)

1−α

+

xn+1lim→0n+1Lp(B01)

(4.4) . Note that

k˜vkL(B+c) = k˜ukL(BcR+), k˜vkL2(B+16) = Rn+12 k˜ukL2(B16R+ ), kvkL2(B0

16) = Rn2kukL2(B0

16R),

xn+1lim→0n+1Lp(B10)

= R1−np

xn+1lim→0n+1Lp(B0R)

, p ≥ 2.

Hence, (4.4) becomes

k˜ukL(BcR+) ≤ CRn2Rα2



k˜ukL2(B+16R)+ R12kukL2(B0

16R)

α

× R12(1−α)

 R32

xn+1lim→0n+1

L2(B16R0 )

+ R12kukL2(B0

16R)

1−α

+ R12npR32

xn+1lim→0n+1Lp(BR0)

.

Since R ≥ 1, (4.3) follows immediately. 

5. Boundary decay to bulk decay

In this section, we will establish that the boundary decay implies the bulk decay.

Proposition 5.1. Assume that (1.4a) and (1.4b) are satised. Let u ∈ H12(Rn)be a solution to (1.1). If u satises the decay assumption (1.4c), then

(5.1) |˜u(x)| ≤ Ce−c|x| for x = (x0, xn+1) ∈ Rn+1+ . Proof. Setting p = n + 1 in (4.3) gives

k˜ukL(BcR+(x0)) ≤ C



k˜ukL2(B16R+ (x0))+ R12kukL2(B0

16R(x0))

α

×



R32k(−∆)12ukL2(B0

16R(x0))+ R12kukL2(B0

16R(x0))

1−α

+ R32k(−∆)12ukLn+1(B0

R(x0)). (5.2)

Given any R ≥ 1, choosing x0 ∈ Rn× {0}with |x0| = 32R. By (1.4c), we have

(5.3) kukL2(B0

16R(x0))≤ Ce−cR. Furthermore, (3.2) yields

(5.4) k˜ukL2(B16R+ (x0)) ≤ k˜ukL2(Rn×(0,16R))≤ 4R12kukL2(Rn) ≤ C(n, λ, Λ)R12.

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Plugging (2.3), (5.3) and (5.4) into (5.2) implies

k˜ukL(BcR+(x0)) ≤ Ce−cR.

Following the chain of balls argument described in [RW19, Proposition 2.2, Step 2], we nally

conclude our result. 

6. Proof of Theorem 1.1

Recall the following Liouville-type theorem in [Arm85, Theorem B].

Theorem 6.1. Suppose that ∆˜u = 0 in Rn+1+ . If ˜u satises the decay property (5.1), then

˜ u ≡ 0.

It is obvious that Theorem 1.1 is an easy consequence of Proposition5.1 and Theorem 6.1.

We now say a few words about the proof of Theorem 1.3. As Proposition 5.1, the boundary decay (1.7) implies the bulk decay (5.1). In the case of b ≡ 0, the proof of Proposition 5.1 remains true when q is bounded.

To make the paper self-contained, we will give another proof of Theorem6.1in AppendixA.

Appendix A. Proof of Theorem 6.1

First of all, we introduce a mapping from the ball to the upper half-space, and back, which preserves the Laplacian. For convenience, we dene

x := x

|x|2 for x ∈ Rn+1\ {0},

see e.g. [ABR01]. Let s = (0, · · · , 0, −1) be the south pole of the unit sphere Sn, and we dene

Φ(z) := 2(z − s)+ s = (2z0, 1 − |z|2)

|z0|2+ (1 + zn+1)2 for all z = (z0, zn+1) ∈ Rn+1\ {s}. It is easy to see that Φ2 = Id.

Let B1(0) be the unit ball in Rn+1. The following lemma can be found in [ABR01].

Lemma A.1. The mapping Φ : Rn+1\ {s} → Rn+1\ {s} is injective. Furthermore, it maps B1(0) onto Rn+1+ , and maps Rn+1+ onto B1(0). It also maps Sn\ {s} onto Rn and maps Rn onto Sn\ {s}.

Given any function w dened on a domain Ω in Rn+1\ {s}. The Kelvin transform K[w] of w is dened by

(A.1) K[w](z) := 2n−12 |z − s|1−nw(Φ(z)) for all z ∈ Φ(Ω).

The following lemma can be found in [ABR01], which exhibits a crucial property of the Kelvin transform.

Lemma A.2. Let Ω be any domain in Rn+1\ {s}. Then u is harmonic on Ω if and only if K[u] is harmonic on Φ(Ω).

Now, we are ready to prove Theorem 6.1.

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Proof of Theorem 6.1. To begin, it is not hard to compute

|Φ(z)| = p4|z0|2+ (|z|2− 1)2

|z0|2 + (1 + zn+1)2 =

2|z0| + i((−zn+1)2+ |z0|2− 1) (−zn+1− 1)2+ |z0|2

=

(−zn+1+ 1) + i|z0| (−zn+1− 1) + i|z0| .

The decay assumption (5.1) implies that for z near the south pole s,

|K[˜u](z)| = 2n−12 |z − s|1−n|˜u(Φ(z))| ≤ C|z − s|1−ne−c|Φ(z)|

= C|z − s|1−nexp



− c

(−zn+1+ 1) + i|z0| (−zn+1− 1) + i|z0|



≤ C|z − s|1−nexp



− c 1

|z − s|



≈ C exp



− c 1

|z − s|

 .

From Lemma A.2, we know that K[˜u] is harmonic on B1(0). By [Jin93, Theorem 1], we obtain that K[˜u] ≡ 0. In view of (A.1) and Lemma A.1, we then conclude that ˜u ≡ 0 in

Φ(B1) = Rn+1+ . 

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Email address: d07221005@ntu.edu.tw

Institute of Applied Mathematical Sciences, NCTS, National Taiwan University, Taipei 106, Taiwan.

Email address: jnwang@math.ntu.edu.tw

參考文獻

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