FOR THE POISSON EQUATION ARE BOUNDED?

FOR THE POISSON EQUATION ARE BOUNDED?

DAGMAR MEDKOV ´A Received 10 September 2002

This paper deals with the problem∆u=gonGand∂u/∂n+u f =Lon∂G. Here,G⊂R^m, m >2, is a bounded domain with Lyapunov boundary, f is a bounded nonnegative func- tion on the boundary ofG,Lis a bounded linear functional onW^1,2(G) representable by a real measureµon the boundary ofG, andg∈L2(G)∩Lp(G),p > m/2. It is shown that a weak solution of this problem is bounded inGif and only if the Newtonian potential corresponding to the boundary conditionµis bounded inG.

Suppose that G⊂R^m, m >2, is a bounded domain with Lyapunov boundary (i.e., of classC^1+α). Denote byn(y) the outer unit normal ofG at y. If f,g,h∈C(∂G) and u∈C²(clG) is a classical solution of

∆u=g onG,

∂u

∂n+u f =h on∂G, (1)

then Green’s formula yields

G∇u· ∇v dᏴm+

∂Gu f v dᏴm−1=

∂Ghv dᏴm−1−

Ggv dᏴm (2) for eachv∈Ᏸ, the space of all compactly supported infinitely differentiable functions in R^m. Here,∂Gdenotes the boundary ofGand clG is the closure ofG;Ᏼk is thek- dimensional Hausdorffmeasure normalized so thatᏴk is the Lebesgue measure inR^k. Denote byᏰ(G) the set of all functions fromᏰwith the support inG.

For an open setV⊂R^m, denote byW^1,2(V) the collection of all functions f ∈L2(V), the distributional gradient of which belongs to [L2(V)]^m.

Definition 1. Let f ∈L_∞(Ᏼ), g∈L2(G) and let Lbe a bounded linear functional on W^1,2(G) such thatL(ϕ)=0 for eachϕ∈Ᏸ(G). We say thatu∈W^1,2(G) is a weak solution

Copyright©2004 Hindawi Publishing Corporation Abstract and Applied Analysis 2004:6 (2004) 501–510 2000 Mathematics Subject Classification: 35B65 URL:http://dx.doi.org/10.1155/S1085337504306196

inW^1,2(G) of the third problem for the Poisson equation

∆u=g onG,

∂u

∂n+u f₌L on∂G, (3)

if

G∇u· ∇v dᏴm+

∂Gu f v dᏴ=L(v)−

Ggv dᏴm (4)

for eachv∈W^1,2(G).

Denote byᏯ(∂G) the Banach space of all finite signed Borel measures with support in∂Gwith the total variation as a norm. We say that the bounded linear functionalLon W^1,2(G) is representable byµ∈Ꮿ(∂G) ifL(ϕ)=

ϕ dµfor eachϕ∈Ᏸ. SinceᏰis dense inW^1,2(G), the operatorLis uniquely determined by its representationµ∈Ꮿ(∂G).

Forx,y∈R^m, denote hx(y)=





(m−2)⁻¹A⁻¹|x−y|^2−m forx=y,

∞ forx=y, (5)

whereAis the area of the unit sphere inR^m. For the finite real Borel measureν, denote ᐁν(x)=

R^mhx(y)dν(y) (6)

the Newtonian potential corresponding toν, for eachxfor which this integral has sense.

We denote byᏯ_b(∂G) the set of allµ∈Ꮿ(∂G) for whichᐁµis bounded onR^m\∂G.

Remark thatᏯ_b(∂G) is the set of allµ∈Ꮿ(∂G) for which there is a polar setMsuch thatᐁµ(x) is meaningful and bounded onR^m\M, becauseR^m\∂Gis finely dense inR^m (see [1, Chapter VII, Sections 2, 6], [7, Theorems 5.10 and 5.11]) andᐁµ=ᐁµ⁺−ᐁµ⁻ is finite and fine-continuous outside of a polar set. Remark thatᏴm−1(M)₌0 for each polar setM(see [7, Theorem 3.13]). (For the definition of polar sets, see [4, Chapter 7, Section 1]; for the definition of the fine topology, see [4, Chapter 10].)

Denote byᏴthe restriction ofᏴm−1to∂G.

Lemma2. Letµ∈Ꮿ(∂G). Then the following assertions are equivalent:

(1)µ∈Ꮿ_b(∂G), (2)ᐁµis bounded inG, (3)ᐁµ∈L_∞(Ᏼ).

Proof. (2)⇒(3). Since∂Gis a subset of the fine closure ofGby [1, Chapter VII, Sections 2, 6] and [7, Theorems 5.10 and 5.11],ᐁµ=ᐁµ⁺−ᐁµ⁻is finite and fine-continuous outside of a polar setM, andᏴm−1(M)=0 by [4, Theorem 7.33] and [7, Theorem 3.13], then we obtain thatᐁµ∈L_∞(Ᏼ).

(3)⇒(1). Letµ=µ⁺−µ⁻be the Jordan decomposition ofµ. Forz∈G, denote byµ_z the harmonic measure corresponding toGandz. Ify∈∂Gandz∈G, then

∂Ghy(x)dµz(x)=hy(z) (7)

by [7, pages 264, 299]. Using Fubini’s theorem, we get

ᐁµ⁺dµz=

∂G

∂Ghy(x)dµz(x)dµ⁺(y)=

∂Ghy(z)dµ⁺(y)=ᐁµ⁺(z). (8) Similarly,ᐁµ⁻dµ_z=ᐁµ⁻(z). Sinceᐁµ∈L_∞(Ᏼ),µ_zis a nonnegative measure with the total variation 1 (see [4, Lemma 8.12]) which is absolutely continuous with respect toᏴ by [2, Theorem 1], then we obtain that|ᐁµ(z)| ≤ ᐁµL∞(Ᏼ).

Ifz∈R^m\clG, choose a bounded domainV with smooth boundary such that clG∪ {z} ⊂V. Repeating the previous reasonings forV\clG, we get|ᐁµ(z)| ≤ ᐁµL∞(Ᏼ). Lemma3. Letf ∈L_∞(Ᏼ)andg∈L2(G)∩Lp(R^m), wherep > m/2,g=0onR^m\G. Then ᐁ(gᏴm)∈Ꮿ(R^m)∩W^1,2(G). Moreover, there is a bounded linear functionalLonW^1,2(G) representable byµ∈Ꮿ_b(∂G)such thatᐁ(gᏴm)is a weak solution inW^1,2(G)of the third problem for the Poisson equation

∆u= −g onG, ∂u

∂n+u f =L on∂G. (9) Proof. Suppose first thatgis nonnegative. Sinceᐁ(gᏴm)∈Ꮿ(R^m) by [3, Theorem A.6], the energygᐁ(gᏴm)dᏴm<∞. According to [7, Theorem 1.20], we have

∇ᐁgᏴm²dᏴm=

gᐁgᏴm

dᏴm<∞, (10) and thereforeᐁ(gᏴm)∈W^1,2(G) (see [7, Lemma 1.6] and [16, Theorem 2.1.4]).

Sinceᐁ(gᏴm)∈Ꮿ(R^m)∩W^1,2(G), f ∈L_∞(Ᏼ) and the trace operator is a bounded operator fromW^1,2(G) toL2(Ᏼ) by [8, Theorem 3.38], then the operator

L(ϕ)=

G∇ϕ· ∇ᐁgᏴm

dᏴm+

∂GᐁgᏴm

f ϕ dᏴm−1−

Ggϕ dᏴm (11) is a bounded linear functional onW^1,2(G).

According to [7, Theorem 4.2], there is a nonnegative ν∈Ꮿ(∂G) such that ᐁν= ᐁ(gᏴm) onR^m\clG. Choose a bounded domainV with smooth boundary such that clG⊂V. Sinceᐁνis bounded inV\clG⊂R^m\clG,Lemma 2yields thatν∈Ꮿ_b(∂(V\ clG)). Therefore,ν∈Ꮿ_b(∂G). According to [13, Lemma 4], there is ˜ν∈Ꮿ_b(∂G) such that

R^m\clG∇ϕ· ∇ᐁgᏴm

dᏴm=

R^m\clG∇ϕ· ∇ᐁνdᏴm=

∂Gϕ d˜ν (12) for eachϕ∈Ᏸ. Letµ=˜ν−fᐁ(gᏴm)Ᏼ. Sinceᐁ(fᐁ(gᏴm)Ᏼ)∈Ꮿ(R^m) by [6, Corollary 2.17 and Lemma 2.18] andᐁ(fᐁ(gᏴm)Ᏼ)(x)→0 as |x| → ∞, we have fᐁ(gᏴm)Ᏼ

∈Ꮿ_b(∂G). Therefore,µ∈Ꮿ_b(∂G).

Ifϕ∈Ᏸ, then ϕ=ᐁ((−∆ϕ)Ᏼm) by [3, Theorem A.2]. According to [7, Theorem 1.20],

R^m∇ϕ· ∇ᐁgᏴm

dᏴm=

R^m∇ᐁ(−∆ϕ)Ᏼm

· ∇ᐁgᏴm

dᏴm

=

R^mgᐁ(−∆ϕ)Ᏼm dᏴm

=

R^mgϕ dᏴm.

(13)

SinceᏴm(∂G)=0,

G∇ϕ· ∇ᐁgᏴm

dᏴm+

∂GᐁgᏴm

f ϕ dᏴm−1

=

Ggϕ dᏴm+

∂GᐁgᏴm

f ϕ dᏴm−1

−

R^m\clG∇ϕ· ∇ᐁgᏴm

dᏴm

=

Ggϕ dᏴm+

∂Gϕ dµ.

(14)

Lemma4. Let f ∈L_∞(Ᏼ)andg∈L2(G)∩L_p(R^m), wherep > m/2,g=0onR^m\G. Let Lbe a bounded linear functional onW^1,2(G)representable byµ∈Ꮿ(∂G). Ifu∈L_∞(G)∩ W^1,2(G)is a weak solution inW^1,2(G)of problem (3), thenµ∈Ꮿ_b(∂G).

Proof. Letw=u−ᐁ(gᏴm). According toLemma 3, there is a bounded linear functional L˜onW^1,2(G) representable byν∈Ꮿ_b(∂G) such thatwis a weak solution inW^1,2(G) of the problem

∆w₌0 onG,

∂w

∂n+w f =L−L˜ on∂G. (15) Fixx∈G. Choose a sequenceGjof open sets withC^∞boundary such that clGj⊂Gj+1⊂ G,x∈G1, and∪G_j=G. Fixr >0 such thatΩ2r(x)⊂G1. Choose an infinitely differ- entiable functionψsuch thatψ=0 onΩr(x) andψ=1 onR^m\Ω2r(x). According to Green’s identity,

w(x)=lim

j→∞ ∂Gj

h_x(y)∂w(y)

∂n dᏴm−1(y)−

∂Gj

w(y)n(y)· ∇h_x(y)dᏴm−1(y)

=lim

j→∞ Gj

∇w(y)· ∇

hx(y)ψ(y)dᏴm(y)

−

Gj

∇

w(y)ψ(y)· ∇h_x(y)dᏴm(y)

=

G∇w(y)· ∇

hx(y)ψ(y)dᏴm(y)−

G∇

w(y)ψ(y)· ∇hx(y)dᏴm(y)

=ᐁ(µ−ν−f wᏴ)(x)−

G∇

w(y)ψ(y)· ∇hx(y)dᏴm(y).

(16) According to [16, Theorem 2.3.2], there is a sequence of infinitely differentiable func- tionswnsuch thatwn→wψinW^1,2(G). According to [6, Section 2],

w(x)=ᐁ(µ−ν−f wᏴ)(x)−lim

n→∞

G∇w_n(y)· ∇h_x(y)dᏴm(y)

=ᐁ(µ−ν−f wᏴ)(x)−lim

n→∞

∂Gw_n(y)n(y)· ∇h_x(y)dᏴm−1(y).

(17)

Since the trace operator is a bounded operator fromW^1,2(G) toL2(Ᏼ) by [8, Theorem 3.38], we obtain

w(x)=ᐁ(µ−ν−f wᏴ)(x)−

∂Gw(y)n(y)· ∇hx(y)dᏴm−1(y). (18) Sincew∈L_∞(G) byLemma 3, the trace ofwis an element ofL_∞(Ᏼ). Since

∂Gw(y)n(y)· ∇hx(y)dᏴm−1(y)

≤ wL∞(Ᏼ)

∂G

n(y)· ∇h_x(y)dᏴm−1(y)

≤ wL∞(Ᏼ) sup

z∈∂G

∂G

n(y)· ∇h_z(y)dᏴm−1(y) +1 2

<∞

(19)

by [6, Lemma 2.15 and Theorem 2.16] and the fact that∂Gis of classC^1+α, the function x−→

∂Gw(y)n(y)· ∇hx(y)dᏴm−1(y) (20) is bounded inG. Sinceᐁνis bounded inGandᐁ(f wᏴ) is bounded inGby [6, Corollary 2.17 and Lemma 2.18], the functionᐁµis bounded inGby (18). Thus,µ∈Ꮿ_b(∂G) by

Lemma 2.

Notation 5. LetXbe a complex Banach space andTa bounded linear operator onX. We denote by KerTthe kernel ofT, byσ(T) the spectrum ofT, byr(T) the spectral radius of T, byXthe dual space ofX, and byTthe adjoint operator ofT. Denote byIthe identity operator.

Theorem6. LetXbe a complex Banach space andKa compact linear operator onX. LetY be a subspace ofXandTa closed linear operator fromYtoXsuch thaty(Tx)=x(T y)for eachx,y∈Y. Suppose thatK(Y)⊂Y andKT y=TKyfor eachy∈Y. Letα∈C\ {0}, Ker(K₋αI)²₌Ker(K₋αI)⊂Y, and _{β_∈σ(K);(β₋α)_·α_≤0} ⊂ {α_}. Ifx,y_∈X, (K−αI)x=y, thenx∈Y if and only ify∈Y.

Proof. Ifx∈Y, then y∈Y. Suppose that y∈Y. SinceK is a compact operator, the operatorKis a compact operator by [14, Chapter IV, Theorem 4.1]. Suppose first that α∈σ(K). SinceKis compact, thenαis a pol of the resolvent by [5, Satz 50.4]. Since

Ker(K−αI)²=Ker(K−αI), (21)

the ascent of (K−αI) is equal to 1. Sinceαis a pol of the resolvent and the ascent of (K−αI) is equal to 1, [5, Satz 50.2] yields that the spaceXis the direct sum of Ker(K− αI) and (K−αI)(X) and the descent of (K−αI) is equal to 1. Since the descent of (K−αI) is equal to 1, we have

(K−αI)²(X)=(K−αI)(X). (22) Since the spaceXis the direct sum of Ker(K−αI) and (K−αI)(X)=(K−αI)²(X), the operator (K−αI) is invertible on (K−αI)(X). Ifα∈σ(K), then the spaceXis the direct sum of Ker(K−αI) and (K−αI)(X), and the operator (K−αI) is invertible on (K−αI)(X). Therefore, there arex1∈Ker(K−αI)⊂Y andx2∈(K−αI)(X) such thatx1+x2=x. We have (K−αI)x2=y.

Denote byZthe closure ofY. SinceK(Y)⊂Y, we obtainK(Z)⊂Z. Denote byK_Z the restriction ofKtoZ. ThenK_Z is a compact operator inZ. Since Ker(K−αI)²⊂Y, we have

KerK_Z−αI²=Ker(K−αI)²=Ker(K−αI)=KerK_Z−αI. (23) Ifα∈σ(K_Z), then the spaceZis the direct sum of Ker(K_Z−αI) and (K_Z−αI)(Z), and the operator (K_Z−αI) is invertible onZ. Suppose thatα∈σ(K_Z). SinceK_Z is compact, thenαis a pol of the resolvent by [5, Satz 50.4]. Since

KerK_Z−αI²=KerK_Z−αI, (24) the ascent of (K_Z−αI) is equal to 1. Sinceαis a pol of the resolvent and the ascent of (K_Z−αI) is equal to 1, [5, Satz 50.2] yields that the spaceZis the direct sum of Ker(K_Z− αI) and (K_Z−αI)(Z) and the descent of (K_Z−αI) is equal to 1. Since the descent of (K_Z−αI) is equal to 1, we have

K_Z−αI²(Z)=(K−αI)(Z). (25) Since the spaceZis the direct sum of Ker(K_Z−αI) and (K_Z−αI)(Z)=(K_Z−αI)²(Z), the operator (K_Z−αI) is invertible on (K_Z−αI)(Z). Since y∈Y ⊂Z, there are y1∈ Ker(K_Z−αI) andy2∈(K_Z−αI)(Z) such that y=y1+y2. SinceXis the direct sum of Ker(K−αI)=Ker(K_Z−αI) and (K−αI)(X)⊃(K_Z−αI)(Z) andy∈(K−αI)(X), we obtain thaty1=0 and y2=y. Thus,y∈(K_Z−αI)(Z). Since (K_Z−αI) is invertible on (K_Z−αI)(Z), there isz∈(K_Z−αI)(Z) such that (K_Z−αI)(z)=y. Since (K−αI) is invertible on (K−αI)(X), we deduce thatx2=z∈(K_Z−αI)(Z)⊂Z.

Now, letw∈Ker(K−αI). Fix a sequence{z_k} ⊂Y such thatz_k→z=x2. Then w(T y)=y(Tw)=

(K−αI)x2

(Tw)=lim

k→∞

(K−αI)z_k(Tw)

=lim

k→∞z_k(K−αI)Tw=lim

k→∞z_kT(K−αI)w=lim

k→∞z_k(0)=0. (26) Sincew(T y)=0 for eachw∈Ker(K−αI), [15, Chapter 10, Theorem 3] yields T y∈ (K−αI)(X).

Denote by ˜K the restriction ofK to (K−αI)(X). If we denote byP the spectral projection corresponding to the spectral set{α}and the operatorK, thenP(X)=(K− αI)(X) by [5, Satz 50.2] andσ( ˜K)=σ(K)\ {α} by [14, Chapter VI, Theorem 4.1].

Therefore,

σK˜=σ(K)\ {α} ⊂

β; (β−α)·α >0⊂ ∪

t>0

β;|β−α−tα|<|tα|

. (27) Since{β;|β−α−t1α|<|t1α|} ⊂ {β;|β−α−t2α|<|t2α|}for 0< t1< t2 andσ( ˜K) is a compact set (see [14, Chapter VI, Theorem 1.3, and Lemma 1.5], there ist >0 such thatσ( ˜K)⊂ {β;|β−α−tα|<|tα|}. Therefore,r( ˜K−αI−tαI)<|tα|. Since we have r(t⁻¹α⁻¹( ˜K−αI−tαI))<1, the series

V= ∞ k=0

(−1)^kt⁻¹α⁻¹K˜−αI−tαI^k (28) converges. Easy calculation yields that V is the inverse operator of the operator I+ t⁻¹α⁻¹( ˜K−αI−tαI)=t⁻¹α⁻¹( ˜K−αI). Since t⁻¹α⁻¹y=t⁻¹α⁻¹( ˜K−αI)x2, we have x2=t⁻¹α⁻¹V y. Denotezk=t⁻¹α⁻¹[−t⁻¹α⁻¹( ˜K−αI−tαI)]^ky. Then

x2=^∞

k=0

z_k. (29)

SinceK(Y)⊂Y,zk∈Yfor eachk. SinceKT=TKonY, we haveTzk=t⁻¹α⁻¹[−t⁻¹α⁻¹(K

−αI−tαI)]^kT y.

Since (K−αI), (K−αI)², (K−αI), and (K−αI)²are Fredholm operators with in- dex 0 (see [14, Chapter V, Theorem 3.1]), [14, Chapter VII, Theorem 3.2] yields

dim Ker(K−αI)²=dim Ker(K−αI)²=dim Ker(K−αI)=dim Ker(K−αI), (30) and thus Ker(K−αI)²=Ker(K−αI). Ifα∈σ(K), then the spaceXis the direct sum of Ker(K−αI) and (K−αI)(X), and the operator (K−αI) is invertible onX. Suppose that α∈σ(K). SinceKis compact, thenαis a pol of the resolvent by [5, Satz 50.4]. Since

Ker(K−αI)²=Ker(K−αI), (31)

the ascent of (K−αI) is equal to 1. Sinceα is a pol of the resolvent and the ascent of (K−αI) is equal to 1, [5, Satz 50.2] yields that the spaceXis the direct sum of Ker(K−αI) and (K−αI)(X) and the descent of (K−αI) is equal to 1. Since the descent of (K−αI) is equal to 1, we have (K−αI)²(X)=(K−αI)(X). Since the spaceX is the direct sum

of Ker(K−αI) and (K−αI)(X)=(K−αI)²(X), the operator (K−αI) is invertible on (K−αI)(X). Denote byK the restriction ofK to (K−αI)(X). If we denote byQ the spectral projection corresponding to the spectral set{α}and the operatorK, thenQ(X)= (K−αI)(X) by [5, Satz 50.2] andσ(K)=σ(K)\ {α}by [14, Chapter VI, Theorem 4.1].

Sinceσ(K)=σ(K) by [14, Chapter VI, Theorem 4.6], we obtainσ(K)⊂ {β;|β−α− tα|<|tα|}. Therefore,r(K−αI−tαI)<|tα|. SinceT y∈(K−αX) andr(t⁻¹α⁻¹(K− αI−tαI))<1, the series

∞ k=0

Tzk= ∞ k=0

t⁻¹α⁻¹−t⁻¹α⁻¹ K−αI−tαI^kT y (32) converges. SinceTis closed,x2=

z_k, andTz_kconverges, then the vectorx2lies inY,

the domain ofT.

Theorem7. Let f ∈L_∞(Ᏼ), f ≥0, andg∈L2(G)∩Lp(R^m), wherep > m/2,g=0on R^m\G. LetLbe a bounded linear functional onW^1,2(G)representable byµ∈Ꮿ(∂G). Ifu is a weak solution inW^1,2(G)of problem (3), thenu∈L_∞(G)if and only ifµ∈Ꮿ_b(∂G).

Proof. Ifu∈L_∞(G), thenµ∈Ꮿ_b(∂G) byLemma 4.

Suppose now thatµ∈Ꮿb(∂G). Letw=u−ᐁ(gᏴm). According toLemma 3, there is a bounded linear functional ˜LonW^1,2(G) representable by ˜µ∈Ꮿ_b(∂G) such thatwis a weak solution inW^1,2(G) of the problem

∆w=0 onG,

∂w

∂n+w f =L˜ on∂G. (33)

Define forϕ∈L_∞(Ᏼ) andx∈∂G, Tϕ(x)₌1

2ϕ(x) +

∂Gϕ(y) ∂

n(y)h_x(y)dᏴ(y) +ᐁ(f ϕᏴ). (34) Since ᐁ(fᏴ)∈Ꮿ(R^m) by [6, Corollary 2.17 and Lemma 2.18], the operator T is a bounded linear operator on L_∞(Ᏼ) by [11, Proposition 8] and [6, Lemma 2.15]. The operatorT−(1/2)Iis compact by [12, Theorem 20] and [6, Theorem 4.1 and Corollary 1.11]. According to [10, Theorem 1], there isν∈Ꮿ(∂G)⊂(L_∞(Ᏼ))such thatTν=µ˜

and

G∇ᐁν· ∇v dᏴm+

∂Gᐁνf v dᏴ=

v d˜µ, (35)

for eachv∈Ᏸ.

Remark thatᏯ(∂G) is a closed subspace of (L_∞(Ᏼ)). According to [11, Proposition 8], we haveT(Ꮿ(∂G))⊂Ꮿ(∂G). Denote byτthe restriction ofTtoᏯ(∂G). Accord- ing to [10, Lemma 11] and [14, Chapter VI, Theorem 1.2], we haveσ(τ)⊂ {β;β≥0}. Sinceσ(τ)=σ(τ) (see [15, Chapter VIII, Section 6, Theorem 2]), eachβ∈σ(T) is an eigenvalue (see [14, Chapter VI, Theorem 1.2]), andTis the restriction ofτtoL_∞(Ᏼ), we obtain thatσ(T)=σ(T)⊂ {β;β≥0}by [15, Chapter VIII, Section 6, Theorem 2].

According to [9, Theorem 1.11], we have KerT⊂Ꮿ_b(∂G). According to [9, Lemma 1.10]

and [10, Lemmas 12 and 13], KerT=Ker(T)². Denote, forρ∈Ꮿ_b(∂G), byV ρthe re- striction ofᐁρto∂G. ThenVis a closed operator fromᏯ_b(∂G) toL_∞(Ᏼ) by [13, Lemma 5]. Ifρ∈Ꮿ_b(∂G), thenV Tρ=TV ρby [13, Lemma 4]. Ifρ1,ρ2∈Ꮿ_b(∂G), thenρ1and ρ2have finite energy by [13, Proposition 23], [7, Theorem 1.20], and

ᐁρ1dρ2=

R^m∇ᐁρ1· ∇ᐁρ2dᏴm=

ᐁρ2dρ1. (36) Since Tν=µ˜∈Ꮿ_b(∂G), Theorem 6yields that ν∈Ꮿ_b(∂G). Since ν has finite energy ᐁνdνandᐁνdν=

|∇ᐁν|²dᏴmby [7, Theorem 1.20], we obtain thatᐁν∈W^1,2(G) (see [7, Lemma 1.6] and [16, Theorem 2.14]). SinceᏰis dense inW^1,2(G) by [16, The- orem 2.3.2], relation (35) yields that the functionᐁνis a weak solution inW^1,2(G) of (33). Sincev=ᐁν−wis a weak solution inW^1,2(G) of the problem

∆v=0 onG,

∂v

∂n+v f =0 on∂G, (37)

and f ≥0, we obtain 0=

G∇v· ∇v dᏴm+

∂Gv f v dᏴ≥

G|∇v|²dᏴm≥0. (38) Therefore,∇v=0 on G and there is a constant csuch that v(x)=cfor Ᏼm-a.a. x∈ G by [16, Corollary 2.1.9]. Sinceν∈Ꮿ_b(∂G), the functionᐁνis bounded in G. Since u(x)=ᐁ(gᏴm)(x) +ᐁν(x)−cforᏴm-a.a.x∈Gandᐁ(gᏴm)∈Ꮿ(R^m) byLemma 3,

we obtainu∈L_∞(G).

Acknowledgment

The author was supported by GA ˇCR Grant no. 201/00/1515.

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Dagmar Medkov´a: Mathematical Institute, Academy of Sciences of the Czech Republic, ˇZitn´a 25, 115 67 Praha 1, Czech Republic

Current address: Department of Technical Mathematics, Faculty of Mechanical Engineering, Czech Technical University, Karlovo n´am. 13, 121 35 Praha 2, Czech Republic

E-mail address:[email protected]