# Nonlinear Kato Class and Unique Continuation of Eigenfunctions forp-Laplacian Operator

Authors: René Erlín Castillo; Julio C. Ramos Fernández
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

We study some basic properties of nonlinear Kato classMp(ℝn) and M~p(ℝn), respectively, for 1<p<n. Also, we study the problem -div(|∇u|p-2∇u)+V|u|p-2u=0 in Ω,  where Ω is a bounded domain in ℝn and the weight function V is assumed to be not equivalent to zero and lies in M~p(Ω), in the case where p<n. Finally, we establish the strong unique continuation property of the eigenfunction for the p-Laplacian operator in the case where V∈M~p(Ω).


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## Body

## 1. Introduction

The Kato classKn was introduced and studied by Aizenman and Simon (see [1]). For n≥3, it consists of locally integrable functions f on ℝn, such that
(1)limr→0supx∈ℝn∫B(x,r)‍|f(y)||x-y|n-2dy=0.
For 1<p<n, the following classes were defined by Zamboni (see [2]): the class M~p of functions f, such that
(2)supx∈ℝn{∫B(x,r)‍1|x-y|n-1(∫B(x,r)‍|f(z)||y-z|n-1dz)1/(p-1)dy}p-1<∞,
and the class Mp of functions f such that f∈M~p and
(3)limr→0supx∈ℝn{(∫B(x,r)‍|f(z)||y-z|n-1dz)1/(p-1)dy∫B(x,r)‍1|x-y|n-1×(∫B(x,r)‍|f(z)||y-z|n-1dz)1/(p-1)dy}p-1=0.Section2 of the present paper is devoted to the study of some basic properties of the nonlinear Kato class Mp(ℝn) and M~p(ℝn), respectively, for 1<p<n.Among other things, we show that the Lorentz spaceL(n/2,1) is embedded into Mp(ℝn) (see Lemma 10) as well as that M~p(ℝn) is a complete topological vector space (see Remark 11 and Lemma 13).Thep-Laplacian operator is a generalization of the Laplace operator, where p is allowed to range over 1<p<∞; in our case where 1<p<n, it is written as
(4)div(|∇u|p-2∇u)=∇·(|∇u|p-2∇u)=|∇u|p-4(|∇u|2Δu+(p-2)∑i,j=1n‍∂u∂xi∂u∂xj∂2u∂xi∂xj),
where
(5)∇u=(∂u∂x1,…,∂u∂xn),|∇u|=(∑i=1n‍(∂u∂xi)2)1/2,
and u∈C0∞(Ω), with Ω bounded domain in ℝn.We are concerned with the following problem:(6)-div(|∇u|p-2∇u)+V|u|p-2u=0,inΩ,
and the weight function V is assumed to be not equivalent to zero and lies in M~p(ℝn) in the case p<n.Specifically, we are interested in studying a family of functions which enjoys the strong unique continuation property, that is, functions besides the possible zero functions which have zero of infinite order.Definition 1.
We say that a functionu∈Llocp(Ω) vanishes of infinite order at point x0 if for any natural number N there exists a constant CN, such that
(7)∫B(x0,r)‍|u(x)|pdx≤CNrN,
for all N∈ℕ and for small positive number r. Here,
(8)B(x0,r)={y∈ℝn:|y-x0|<r}.Definition 2.
We say that (6) has a strong unique continuation property if and only if any solution u of (6) in Ω is identically zero in Ω provided that u vanishes of infinite order at a point in Ω.There is an extensive literature on unique continuation. We refer to the work of Zamboni on unique continuation for nonnegative solutions of quasilinear elliptic equation [3], also the work of Jerison-Kenig on the unique continuation for Schrödinger operators [4]. The same work is done by Chiarenza and Frasca, but for linear elliptic operator in the case where V∈Ln/2 when n>2 [5].Let us recall some known results concerning Fefferman's inequality as follows:(9)∫ℝn|u(x)|p|V(x)|dx≤C∫ℝn|∇u(x)|pdx∀u∈C0∞(ℝn).In [6], de Figueiredo and Gossez prove (9) in the case where p=2, assuming that V∈Lr,n-2r(ℝn) with 1<r≤n/2. Later in [7], Jerison and Kening showed the same result taking V in the Stummel-Kato class S(ℝn). We point out that it is not possible to compare the assumptions V∈Lr,n-2r(ℝn) and f∈S(ℝn). Chiarenza and Frasca [5] generalized Fefferman's result proving (9) under the assumption that V∈Lr,n-2r(ℝn) with r∈(1,n/p) and p∈(1,n). In [3], Schechter gave a new proof of (9) assuming that V∈M~p(ℝn).


## 2. Definitions and Notation

In this section, we gather definitions and notations that will be used throughout the paper. We also include several simple lemmas. ByLloc1(ℝn), we will denote the space of functions which are locally integrable on ℝn, and by Lloc,u1 the space of functions f, such that
(10)supx∈ℝn∫B(x,1)‍|f(y)|dy<∞.Definition 3.
Letf∈Lloc1(ℝn). For any 1<p<n and r>0, we set
(11)Φ(r)=supx∈ℝn(∫B(x,r)‍1|x-y|n-1(∫B(x,r)‍|f(z)|dz|z-y|n-1)1/(p-1)dy)p-1,
where B(x,r)={y:|x-y|<r}. We say that f belongs to the space M~p(ℝn) if Φ(r)<∞ for all r>0.Definition 4.
We say that a functionf∈Mp(ℝn) if
(12)limr→0Φ(r)=0.We are now ready to formulate some simple properties of the classesMpandM~p.Lemma 5 (see [3], page 152).
For1<p<n, one has (i)
Mp(ℝn)⊂M~p(ℝn),(ii)
M2(ℝn)=Kn.From Lemma5 we conclude that both Mp(ℝn) and M~p(ℝn) are generalizations of Kn.Remark 6.
The following example shows thatKnis properly contained in Mp(ℝn) for p>2. It is known that the function f(x)=|x|-2 is not in the Kato class Kn. However, f∈Mp. Indeed,
(13)limr→0supx{(∫B(x,r)‍dz|z|2|z-y|n-1)1/(p-1)∫B(x,r)‍1|x-y|n-2×(∫B(x,r)‍dz|z|2|z-y|n-1)1/(p-1)dy}p-1=0.
This can be shown by splitting the domain of integration in the interior integral into the following three parts: B(x,r)⋂‍{|z|<(1/2)|y|}, B(x,r)⋂‍{(1/2)|y|≤|z|≤(3/2)|y|}, and B(x,r)⋂‍{|z|>(3/2)|y|}.
After routine calculations, we can see that(14)∫B(x,r)‍dz|z|2|z-y|n-1
is majorized by C|y|-1. Finally, we have
(15)Csupx{∫B(x,r)‍dy|y|1/(p-1)|x-y|n-1}p-1⟶0asr⟶0.
This shows that (13) holds. Thus,f∈⋂p>2‍Mp.Definition 7.
The distribution functionDf of a measurable function f is given by
(16)Df(λ)=m({x∈ℝn:|f(x)|>λ}),
where m denotes the Lebesgue measure on ℝn. The distribution function Df provides information about the size of f but not about the behavior of f itself near any given point. For instance, a function on ℝn and each of its translates have the same distribution function. It follows from Definition 7 that Df is a decreasing function of λ (not strictly necessary).Definition 8.
Letf be a measurable function in ℝn. The decreasing rearrangement of f is the function f defined on [0,∞) by
(17)f*(t)=inf{λ:Df(λ)≤t}(t≥0).
We use here the convention that inf∅=∞.Definition 9 (Lorentz space).
Letf be a measurable function; we say that f belongs to L(n/2,1) if
(18)∥f∥(n/2,1)=∫0∞‍t2/n-1f*(t)dt<∞.
And it belongs to L(n/(n-2),∞) if
(19)∥f∥(n/(n-2),∞)=supt>1t1-2/nf*(t)<∞.Lemma 10.
ConsiderL(n/2,1)⊂Mp(ℝn).Proof.
Letf∈L(n/2,1); then
(20)∫0∞‍t2/n-1f*(t)dt<∞.
Since |f|χB(x,r)≤|f|, we have
(21)(|f|χB(x,r))*(t)≤f*(t);
then,
(22)∫0∞‍t2/n-1(|f|χB(x,r))*(t)dt≤∫0∞‍t2/n-1f*(t)dt<∞.
Thus, |f|χB(x,r)∈L(n/2,1).
On the other hand, letg(x)=|x|-(1-n); then
(23)m({x:|g(x)|>λ})=m({x:|x|-(1-n)>λ}),m({x:|x|<(1λ)1/(n-1)})=Cn(1λ)n/(n-1),
where Cn=m(B(0,1)).
Next, we sett=Cn(1/λ)n/(n-1), and then λ=Cnt1/n-1. Thus, g*(t)=Cnt1/n-1. From this, we obtain
(24)∥g∥(n/(n-2),∞)=∥1|·|n-1∥(n/(n-2),∞)=supt>1Cnt1-2/nt1/n-1=supt>1Cnt-(1/n)≤Cn<∞,
which means that g∈L(n/(n-2),∞). Finally, by Fubini's theorem and Hölder's inequality, we have
(25)ϕ(r)=supx∈ℝn((∫B(x,r)‍|f(z)||y-z|n-1dz)1/(p-1)dy∫B(x,r)‍1|x-y|n-1×(∫B(x,r)‍|f(z)||y-z|n-1dz)1/(p-1)dy)p-1≤supx∈ℝn((∫B(x,2r)‍|f(z)||y-z|n-1dz)1/(p-1)dy∫B(x,r)‍1|x-y|n-1×(∫B(x,2r)‍|f(z)||y-z|n-1dz)1/(p-1)dy)p-1=supx∈ℝn((∫ℝn‍|f(z)|χB(0,2r)(y-z)dz|y-z|n-1)1/(p-1)dy∫B(x,r)‍1|x-y|n-1×(∫ℝn‍|f(z)|χB(0,2r)(y-z)dz|y-z|n-1)1/(p-1)dy)p-1≤supx∈ℝn(∫B(x,r)‍dy|x-y|n-1)p-1×∥fχB(0,2r)∥(n/2,1)∥1|·|n-1∥(n/(n-2),∞)=Cnrp-1∥fχB(0,2r)∥(n/2,1)⟶0asr⟶0,
which means that f∈Mp(ℝn) and the proof is complete.Remark 11.
(i) For0<r<1, it is not hard to check that for 1<p≤2, the expression
(26)∥f∥M~p(ℝn)=supx∈ℝn((∫B(x,1)|f(z)||z-y|n-1dz‍)1/(p-1)dy∫B(x,1)1|x-y|n-1‍×(∫B(x,1)|f(z)||z-y|n-1dz‍)1/(p-1)dy)p-1
defines a norm on M~p(ℝn).
(ii) Forp>2, the expression (26) satisfies the following inequality:
(27)∥f+g∥M~p(ℝn)≤2p-2(∥f∥M~p(ℝn)+∥g∥M~p(ℝn))
for all f and g in M~p(ℝn).
IfU is a neighborhood of 0 from (27), we have
(28)2p-1U+2p-1U⊂U;
then, M~p(ℝn) is a topological vector space.Lemma 12.
ConsiderM~p(ℝn)⊂Lloc,u1(ℝn) for 1<p<n.Proof.
Letf∈M~p(ℝn), and fix r0>0. Then, there exists a positive constant C such that Φ(r0)≤C. It follows that
(29)supx∈ℝn(∫B(x,r0)‍1|x-y|n-1(∫B(x,r0)‍|f(z)||z-y|n-1dz)1/(p-1)dy)p-1≥supx∈ℝn(∫B(x,r0)‍dyr0n-1(∫B(x,r0)‍|f(z)|(2r0)n-1dz)1/(p-1))p-1≥supx∈ℝn(12r0)n-1(m(B(x,r0))r0n-1)p-1∫B(x,r0)‍|f(z)|dz.
Therefore,
(30)supx∈ℝn∫B(x,r0)‍|f(z)|dz<BC,
where
(31)B=(2r0)n-1(r0m(B(0,1)))p-1.
Finally, let B(x,1)⊆⋃k=1n‍B(xk,r0); then
(32)supx∈ℝn∫B(x,1)‍|f(z)|dz≤∑k=1nsupx∈ℝn∫B(xk,r0)‍|f(z)|dz,
so
(33)supx∈ℝn∫B(x,1)‍|f(z)|dz<∞.
Therefore,
(34)M~p(ℝn)⊂Lloc,u1(ℝn).Lemma 13.
For1<p<n, M~p(ℝn) is a complete space.Proof.
Let{fn}n∈𝒩 be a Cauchy sequence in
(35)B¯(0,r)={f∈M~p(ℝn):∥f∥M~p(ℝn)≤r}.
By Lemma 10, {fn}n∈𝒩 is a Cauchy sequence in Lloc,u1(ℝn). Since this space is complete, there exists a function f∈Lloc,u1(ℝn) such that fn→f in Lloc,u1(ℝn). By Fatous's lemma, we have
(36)∥f∥M~p(ℝn)≤liminf∥fn∥M~p(ℝn)≤r.
Thus, f∈B¯(0,r), which means that B¯(0,r) is complete with respect to the topology generated by Lloc,u1(ℝn)-norm. By Corollary 2 of Proposition 9 in [8, Chapter III, Section 3, no. 5] we obtain the assertion.Lemma 14.
If1<p<n, then Mp(ℝn) is closed in M~p(ℝn).Proof.
Let us define the mapφ:M~p(ℝn)→[0,∞) by φ(f)=limr→0ϕf(r) (see Definition 3).
It is not hard to prove that the family{φr}r>0 where φr(f)=ϕf(r) is equicontinuous and φr→φ pointwise as r→0. Since Mp(ℝn)=φ-1(0), we obtain the result.For more details on nonlinear Kato class, we refer the readers to [9].


## 3. Some Useful Inequalities

For the sake of completeness and convenience of the reader, we include the proof of the next result which is due to Schechter [3].Theorem 15.
Assume thatV∈M~p(ℝn). Then, for anyr>0 there exists a positive constantC(n,p), such that (37)∫ℝn|V(x)||u(x)|pdx≤C(n,p)Φ(2r)∫ℝn|∇u(x)|pdx,
for anyu∈C0∞(ℝn)supported inB(x0,r).Proof.
For anyu∈C0∞(ℝn) supported in B(x0,r), using the well-known inequality
(38)|u(x)|≤C(n,p)∫B(x0,r)‍|∇u(y)||x-y|n-1dy,
Fubini's theorem, and Hölder's inequality, we obtain
(39)∫ℝn|V(x)||u(x)|pdx=∫B(x0,r)|V(x)||u(x)|pdx≤C(n,p)∫B(x0,r)|V(x)||u(x)|p-1×(∫B(x0,r)|∇u(y)||x-y|n-1dy)dx≤C(n,p)∫B(x0,r)|∇u(y)|×(∫B(x0,r)|V(x)||u(x)|p-11|x-y|n-1dx)dy≤C(n,p)(∫B(x0,r)|∇u(y)|pdy)1/p×[(∫B(x0,r)|V(x)||u(x)|p-11|x-y|n-1dx)p/(p-1)×1|x-y|n-1dx×1|x-y|n-1dx∫B(x0,r)|V(x)||u(x)|p-1)p/(p-1)∫B(x0,r)|V(x)||u(x)|p-1)p/(p-1)∫B(x0,r)(×1|x-y|n-1dx∫B(x0,r)|V(x)||u(x)|p-1)p/(p-1)∫B(x0,r)|V(x)||u(x)|p-1×1|x-y|n-1dx×1|x-y|n-1dx∫B(x0,r)|V(x)||u(x)|p-1)p/(p-1)∫B(x0,r)|V(x)||u(x)|p-1)p/(p-1)dy].
On the other hand, using Hölder's inequality one more time, we have(40)∫B(x0,r)(∫B(x0,r)|V(x)||u(x)|p-11|x-y|n-1dx)p/(p-1)dy≤∫B(x0,r)(∫B(x0,r)|V(z)||z-y|n-1dz)1/(p-1)×∫B(x0,r)|V(x)||u(x)|p|x-y|n-1dxdy=∫B(x0,r)|V(x)||u(x)|p×∫B(x0,r)1|x-y|n-1(∫B(x0,r)|V(z)||z-y|n-1dz)1/(p-1)dydx≤[Φ(2r)]1/(p-1)∫B(x0,r)|V(x)||u(x)|pdx.
By (39) and (40), we obtain
(41)∫B(x0,r)|V(x)||u(x)|pdx≤C(n,p)[Φ(2r)]1/(p-1)(∫B(x0,r)|∇u(y)|pdy)1/p×(∫B(x0,r)|V(x)||u(x)|pdx)1-1/p.
Hence(42)∫B(x0,r)|V(x)||u(x)|pdx≤C(n,p)[Φ(2r)]q/p∫B(x0,r)|∇u(x)|pdx.The next corollary is an easy consequence of the previous theorem. It can be obtained via a standard partition of unity.Corollary 16.
LetV∈M~p(ℝn) and let Ω be a bounded subset of ℝn, suppV⊆Ω. Then, for any σ>0 there exists a positive constant K depending on σ, such that
(43)∫Ω|V(x)||u(x)|pdx≤σ∫Ω|∇u(x)|pdx+K(σ)∫Ω|u(x)|pdx,
for all u∈C0∞(Ω).Proof.
Letσ>0. Let r be a positive number that will be chosen later. Let {αkp}, k=1,2,…,N(r), be a finite partition of the unity of Ω¯, such that suppV⊆B(xk,r) with xk∈Ω¯. We apply Theorem 15 to the functions αk and we get
(44)∫Ω|V(x)||u(x)|pdx=∫Ω|V(x)||u(x)|p∑k=1N(r)αkp(x)dx=∑k=1N(r)‍∫Ω|V(x)||u(x)αkp(x)|pdx≤∑k=1N(r)‍CΦV(2r)(∫Ω|∇u(x)|pαnp(x)dx+∫Ω|∇u(x)|p|u(x)|pdx)≤CΦV(2r)(∫Ω|∇u(x)|pdx+N(r)rp∫Ω|u(x)|pdx).
Finally, to obtain the result, it is sufficient to choose r such that CΦV(2r)=σ. After that, we note that N(r)≈r-n and the corollary follows.Lemma 17.
LetBr and B2r be two concentric balls contained in Ω. Then,
(45)∫Br|∇u(x)|pdx≤Crp∫B2r|u|p,
where the constant C does not depend on r and u∈C0∞(Br).Proof.
Takeφ∈C0∞(Ω), with suppφ⊂B2r, φ(x)=1 for x∈Br and |∇φ|≤C/r using φp as a test function in (6); we get
(46)∫B2r-div(|∇u|p-2∇u)φpu+∫B2rV|u|p-2uφpu=0.
Thus,
(47)∫B2r|∇u|pφp=-p∫B2r|∇u|p-2φp-2∇u·∇φ(φu)-∫B2rV|φu|p.
Using Young's inequalities for(p-1)/p+1/p=1, we can estimate the first integral in the right-hand side of (47) by
(48)(p-1)ɛp/(p-1)∫B2r|∇u|pφp+ɛ-p∫B2r|∇φ|p|u|p.
Also by result of Corollary16, we can estimate the second integral in the right-hand side of (47) by
(49)ɛ∫B2r|∇(φu)|p+Cɛ∫B2r|φu|p.
Using these estimates in (47), we have
(50)∫B2r|∇u|pφp≤((p-1)ɛp/(p-1)+ɛ)∫B2r|∇u|pφp+(ɛp+ɛ)∫B2r|u|p|∇u|p+Cɛ∫B2r|∇u|p|φ|p.
Using the fact that|∇φ|≤C/r, |φ|≤C/r, and φ=1 in Br, we immediately have inequality (45).Lemma 18.
Letu∈C0∞(Br) where Br is the ball of radius r in ℝn and let E={x∈Br:u(x)=0}. Then, there exists a constant β depending only on n, such that
(51)∫A|u|≤βrnm(E)[m(A)]1/n∫B2r|∇u|,
for all ball Br, u as above, and all measurable sets A⊂Br.To prove this lemma see [5]. Note that m(A) and m(E) denote the Lebesgue measure of the sets A and E.


## 4. Strong Unique Continuation

In this section, we proceed to establish the strong unique continuation property of the eigenfunction for thep-Laplacian operator in the case V∈M~p(Ω).Theorem 19.
Letu∈C0∞(Ω) be a solution of (6). If u=0 on a set E of positive measures, then u has zero of infinite order in p-mean.Proof.
We know that almost every point ofE is a point of density of E. Let x0∈E be such point. This means that
(52)limr→0m(E∩Br)m(Br)=1,
where Br denotes the ball of radius r centered at x0, and thus, given that ɛ>0, there is an r0=r0(ɛ), such that
(53)m(Ec∩Br)m(Br)<ɛ,m(E∩Br)m(Br)>1-ɛforr≤r0,
where Ec denotes the complement of the set E. Taking r0, smaller if necessary, we can assume that Br0⊂Ω. Sinceu=0 onE, by Lemmas 18 and 11, we have
(54)∫Br|u|p=∫Br∩Ec|u|p≤βrnm(E∩Br)[m(Ec∩Br)]1/n×∫Br|∇(u)|p,pβrnɛ1/n[m(Br)]1-1/n(1-ɛ)∫Br|u|p-1|∇(u)|.
By Hölder inequality
(55)∫Br|u|p≤Cɛ1/n1-ɛ(∫Br|∇(u)|p)1/p(∫Br|u|p)(p-1)/p,
and by using the Young inequality, we get
(56)∫Br|u|p≤Cɛ1/n1-ɛr(rp-1∫Br|∇(u)|p+p-1r∫Br|up|).
Finally, by Lemma 17, we have
(57)∫Br|u|p≤Cɛ1/n1-ɛ∫B2r|up|,
where C is independent of ɛ and of r, as r→0. Now, let us introduce the following function:
(58)f(r)=∫B2r|u|p.
And let us fixn∈ℕand choose ɛ>0 such that (Cɛ1/n)(1-ɛ)≤2-n. Observe that consequently r0 depends on n. Then, (57) can be written as
(59)f(r)≤2-nf(2r),forr≤r0.
Iterating (59), we get
(60)f(ρ)≤2-knf(2kρ),if2k-1ρ≤r0.
Now, given that0<r<r0(n), choose k∈ℕ, such that
(61)2-kr0≤r≤2-k+1r0.
From (60), we obtain
(62)f(r)≤2-knf(2kr)≤2-knf(2r).
Since2-k≤r/r0, we finally obtain
(63)f(r)≤(rr0)nf(2r0).
And thus, we have
(64)∫Br(x0)‍|u(y)|pdy≤(rr0)nf(2r0),
and this shows that (7) holds, which means that u has a zero of infinite order in p-mean at x0.Corollary 20.
Equation (6) has a strong unique continuation property.



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