• Non ci sono risultati.

and again, up to subsequences we have TO(t) fn2

k → TO(t)(f2) a.e. in O.

TO(t) gn2

k → TO(t)(g2) a.e. in O.

Therefore

TO(t)(|f g|)(x) = lim

k→∞TO(t) (|fnkgnk|) (x)

≤ lim

k→∞

q

TO(t) fn2k (x) · TO(t) g2nk (x)

=p

TO(t)(f2)(x) · TO(t)(g2)(x) for a.e. x ∈ O.

so that

Z

uψ dγ = lim

n→∞

Z

+n |Ω dγ ≤ 0.

Now we prove the statement by contradiction: if there exists a measurable set A such that γ(A) > 0 and u|A > 0, then we take ψ(x) = χA(x) ∈ L(Ω) ⊂ L2(Ω, γ). Then

Z

uψ dγ = Z

A

u dγ > 0 that is a contradiction.

Now we use the cylindrical approximation defined in Section 1.8. Fix n ∈ N, let q ∈ N with q ≥ j(n) = dim Fn and let G = span{h1, . . . , hq}. Let πG : X → G be the finite rank projection defined by

πGx =

q

X

i=1

bhi(x)hi.

We denote by γG the induced measure γ ◦ πG−1 in G. If G is identified with Rq through the isomorphism x 7→ (bh1(x), . . . , bhq(x)) then γG is the standard Gaussian measure in Rq. Setting d = q − dim Fn, let O := On× Rd so that πG(Ωn) = O; we remark that O is an open convex subset of G. Let LG be the Ornstein-Uhlenbeck operator defined by (1.21) with Ω = O and γ = γG and let TG(t) be the associated semigroup.

Lemma 8. If w ∈ D(LG) then the map x 7→ w(πG(x)) belongs to D(L(n)) and

L(n)(w ◦ πG) = (LGw) ◦ πG.

Proof. We show that there exists f ∈ L2(Ωn, γ) such that, given ϕ ∈ W1,2(Ωn, γ) we have

Z

n

h∇Hw(πG(x)), ∇Hϕ(x)iHγ(dx) = Z

n

f (x)ϕ(x)γ(dx). (2.4) We remark that

h∇Hw(πG(x)), ∇Hϕ(x)iH =

q

X

i=1

∂w

∂ξiG(x))∂ϕ

∂hi(x)

Moreover ϕ(x) = ϕ(πG(x)+(I −πG)(x)), the space X can be split X = G× eG where eG = (I − πG)(X) and also γ = γG⊗ γ

Ge, see section 1.9. Let ξ ∈ Rq and for every y ∈ eG let

gy(ξ) := ϕ(ξ1h1+ . . . + ξqhq+ y),

then Z

n

h∇Hw(πG(x)), ∇Hϕ(x)iHγ(dx)

= Z

n

q

X

i=1

∂w

∂ξiG(x))∂ϕ

∂hiG(x) + (I − πG)(x))γ(dx)

= Z

Ge

Z

O q

X

i=1

∂w

∂ξi

(ξ)∂gy

∂ξi

(ξ)γG(dξ) γ

Ge(dy)

= Z

Ge

Z

O

LGw(ξ)gy(ξ)γG(dξ) γ

Ge(dy) = Z

n

(LGw)(πG(x))ϕ(x)γ(dx).

Then w ◦ πG ∈ D(L(n)) and L(n)(w ◦ πG) = (LGw) ◦ πG. Lemma 9. Let ev ∈ Cb(G). Then the function defined by

g(t)(x) := TG(t)ev|OG(x)), x ∈ Ωn belongs to C((0, ∞); D(L(n))).

Proof. By Lemma 8 it follows that g(t) ∈ D(L(n)) for all t > 0. Let us prove that g is continuous at t0 > 0. For t > 0 we have

Z

n

|g(t)(x) − g(t0)(x)|2γ(dx)

= Z

n

|TG(t)ev|OG(x)) − TG(t0)ev|OG(x))|2γ(dx)

= Z

O

|TG(t)ev|O(ξ) − TG(t0)ev|O(ξ)|2γG(dξ)

that goes to zero as t → t0 thanks to the strong continuity of TG(t) in L2(O, γG). Moreover

Z

n

|L(n)g(t)(x) − L(n)g(t0)(x)|2γ(dx)

= Z

n

|L(n)TG(t)ve|OG(x)) − L(n)TG(t0)ev|OG(x))|2γ(dx)

= Z

O

|LGTG(t)ev|O(ξ) − LGTG(t0)ev|O(ξ)|2γG(dξ)

that goes to zero as t → t0 since t 7→ TG(t)ev belongs to C((0, ∞), D(LG)).

Then g ∈ C((0, ∞), D(L(n))).

Theorem 17. For all f ∈ W1,2(Ω, γ) and all t ≥ 0 we have

|∇HT(t)f |H ≤ e−tT(t)|∇Hf |H a.e. on Ω. (2.5) Proof. First we prove that (2.5) holds true for the restriction to Ω of any cylindrical regular function. Let f ∈ F Cb(X) and ϕ ∈ F Cb(X) with ϕ ≥ 0.

Then

f (x) = v(l1(x), . . . , lk(x)) ϕ(x) = w(lk+1(x), . . . , lk+m(x))

with li ∈ X, v ∈ Cb(Rk) and w ∈ Cb(Rm), w ≥ 0. We want to estimate the integral

Z

n

ϕ(x)|∇HT(n)(t)f|Ωn(x)|γ(dx).

Let G := span{Fn, Rγ(l1), . . . , Rγ(lk+m)}. Then G is a subspace of H of dimension q ≤ n + k + m; setting d = q − dim Fn let O := On × Rd. Let {hi}qi=1 ⊂ X be an orthonormal basis of G such that {h1, . . . , hj} is an orthonormal basis of Fn if dim Fn = j.

Finally we consider:

f (x) =ev(πG(x)) ϕ(x) =w(πe G(x)) where ev ∈ Cb(G), w ∈ Ce b(G).

Now we prove that

(T(n)(t)f|Ωn)(x) = TG(t)ev|OG(x)) := g(t)(x) t > 0, x ∈ Ωn. (2.6) We know that the function t 7→ TG(t)ev|O belongs to C([ 0, ∞) ; L2(O, γG)) ∩ C1((0, ∞); L2(O, γG)) ∩ C((0, ∞); D(LG)) and it is the unique classical solu-tion of

(u0(t) = LGu(t), t > 0 u(0) =ev|O

in the space L2(O, γG). Then for fixed t0 ∈ [0, +∞) we have Z

n

|g(t)(x) − g(t0)(x)|2γ(dx)

= Z

n

|TG(t)ev|OG(x)) − TG(t0)ev|OG(x))|2γ(dx)

= Z

O

|TG(t)ev|O(ξ) − TG(t0)ev|O(ξ)|2γG(dξ)

that goes to zero for t → t0 thanks to the strong continuity of TG(t) in L2(O, γG). Let us prove the differentiability of g at t0 ∈ (0, +∞). For any t such that t + t0 > 0 we have

Z

n

g(t + t0)(x) − g(t0)(x)

t − LGTG(t0)ev|OG(x))

2

γ(dx)

= Z

n

TG(t + t0)ev|OG(x)) − TG(t0)ev|OG(x))

t − LGTG(t0)ev|OG(x))

2

γ(dx)

= Z

O

TG(t + t0)ev|O(ξ) − TG(t0)ev|O(ξ)

t − LGTG(t0)ev|O(ξ)

2

γG(dξ)

which tends to zero as t → 0. Then g0(t0) = LGTG(t0)ev|OG(·)). Moreover g0 is continuous at any t0 > 0 with values in L2(Ωn, γ) since

Z

n

|g0(t)(x) − g0(t0)(x)|2γ(dx)

= Z

n

|LGTG(t)ve|OG(x)) − LGTG(t0)ve|OG(x))|2γ(dx)

= Z

O

|LGTG(t)ev|O(ξ) − LGTG(t0)ev|O(ξ)|2γG(dξ) that goes to zero as t → t0. By Lemma 9, g ∈ C((0, ∞); D(L(n))) and

g0(t) = LG(TG(t)ve|O)(πG(x)) = L(n)(TG(t)ve|OG(x))) = L(n)g(t), t > 0.

Moreover

g(0) = TG(0)ev|OG(x)) =ev|OG(x)) = f (x)|Ωn. Therefore g is a classical solution to

(g0(t) = L(n)g(t), t > 0 g(0) = f|Ωn,

so that g(t) = T(n)(t)f|Ωn and (2.6) is proved.

Now using Lemma 6 we have Z

n

ϕ(x)|∇HT(n)(t)f|Ωn(x)|Hγ(dx)

= Z

n

w(πe G(x))|∇HTG(t)(ev|O)(πG(x))|Hγ(dx)

= Z

Ow(ξ)|∇Te G(t)(ev|O)(ξ)|γG(dξ)

≤ Z

Ow(ξ)ee −tTG(t)|∇(ev|O)(ξ)|γG(dξ)

= Z

n

w(πe G(x))e−tTG(t)|∇H(ev|O)(πG(x))|Hγ(dx)

= Z

n

ϕ(x)e−tT(n)(t)|∇Hf|Ωn(x)|Hγ(dx)

Therefore for every f ∈ F Cb(X) and ϕ ∈ F Cb(X), ϕ ≥ 0 and every n we have

Z

n

ϕ(x)|∇HT(n)(t)f|Ωn(x)|γ(dx) ≤ Z

n

ϕ(x)e−tT(n)(t)|∇Hf|Ωn(x)|γ(dx).

Now thanks to Proposition 32 we can take the limit as n → ∞ obtaining Z

ϕ(x)|∇HT(t)f|Ω(x)|γ(dx) ≤ Z

ϕ(x)e−tT(t)|∇Hf|Ω(x)|γ(dx) and by Lemma 7

|∇HT(t)f|Ω(x)| ≤ e−tT(t)|∇Hf|Ω(x)| for a.e. x ∈ Ω.

If f ∈ W1,2(Ω, γ) then there exists {fn} ∈ F Cb(X) such that fn|Ω → f in W1,2(Ω, γ); along a subsequence {fnk} we have

|∇HT(t)fnk(x)|H → |∇HT(t)f (x)|H a.e. in Ω and again up to a subsequence we have

T(t)|∇Hfnk(x)|H → T(t)|∇Hf (x)|H a.e. in Ω.

Therefore

|∇HT(t)f (x)|H = lim

k→∞|∇HT(t)fnk(x)|H

≤ lim

k→∞e−tT(t)|∇Hfnk(x)|H = e−tT(t)|∇Hf (x)|H

for a.e. x ∈ Ω.

Proposition 38. For all f , g ∈ L2(Ω, γ) we have

T(t)(f g)2

≤ T(t)(f2)T(t)(g2) a.e. in Ω

Proof. As before we consider f, g, ϕ ∈ F Cb(X) and using Proposition 37 we prove that

Z

n

ϕ(x)[T(n)(t)(f g)(x)]2γ(dx)

≤ Z

n

ϕ(x) q

T(n)(t)(f2)(x) · T(n)(t)(g2)(x)γ(dx).

The proof is similar to the previous one with the only difference that now there are more functions but there aren’t gradients. Then taking the limit for n → ∞ we get

Z

ϕ(x)[T(t)(f g)(x)]2γ(dx) ≤ Z

ϕ(x)p

T(t)(f2)(x) · T(t)(g2)(x)γ(dx).

Since the restrictions to Ω of the functions of F Cb(X) are dense in L2(Ω, γ) we obtain our claim.

Now we turn to the Poincaré inequality. We set m(ϕ) = 1

γ(Ω) Z

ϕ dγ, ϕ ∈ L1(Ω, γ).

We recall that if O ⊂ RN is an open convex set the then the Poincarè inequality holds (see Proposition 4.2 in [8]), that is

Z

O

|ψ − mO(ψ)|2N ≤ Z

O

|∇ψ|2N, ∀ψ ∈ W1,2(O, γN). (2.7) Proposition 39. For each ϕ ∈ W1,2(Ω, γ) we have

Z

|ϕ − m(ϕ)|2dγ ≤ Z

|∇Hϕ|2Hdγ (2.8) Proof. First we prove that (2.8) holds for every f ∈ F Cb1(X). Therefore

f (x) = v(l1(x), . . . , lk(x))

with v ∈ Cb1(Rk) and l1, . . . , lk∈ X. Let G := span{Fn, Rγ(l1), . . . , Rγ(lk)}.

Let {hi}qi=1 ⊂ X be an orthonormal basis of G such that {h1, . . . , hj} is an orthonormal basis of Fn if dim Fn = j. Then G is a subspace of H of

dimension q ≤ n + k; setting d = q − dim Fn let O := On× Rd. Let {bhi}qi=1 be an orthonormal basis of G such that {bh1, . . . , bhj} is an orthonormal basis of Fn if dim Fn= j. Then we have

f (x) = ϕ(πG(x)), ϕ ∈ Cb1(G), mn(f ) = 1

γ(Ωn) Z

n

f (x) γ(dx) = 1 γG(O)

Z

O

ϕ(ξ) γG(dξ) =: mO,γG(ϕ),

n→∞lim mn(f ) = m(f ).

Applying (2.7) we obtain Z

n

|f (x) − mn(f )|2γ(dx)

= Z

n

|f (x) − mO,γG(ϕ)|2γ(dx) = Z

O

|ϕ(ξ) − mO,γG(ϕ)|2γG(dξ)

≤ Z

O

|∇ϕ(ξ)|2γG(dξ) = Z

n

|∇Hf (x)|2Hγ(dx).

Taking the limit as n → ∞ we get Z

|f (x) − m(f )|2γ(dx) ≤ Z

|∇Hf (x)|2Hγ(dx).

Let now f ∈ W1,2(Ω, γ). There exists a sequence {fn} ⊂ F Cb1(X) such that fn|Ω → f in W1,2(Ω, γ). It follows that

Z

|f − m(f )|2dγ = lim

n→∞

Z

|fn− m(f )|2

≤ lim

n→∞

Z

|∇Hfn|2Hdγ = Z

|∇Hf |2Hdγ.

The Poincaré inequality (2.8) implies that if ∇Hϕ vanishes a.e. in Ω, then ϕ is constant a.e. in Ω. By the definition of L, this implies that the kernel of L consists of constant functions. As a consequence of the Poincaré inequality other spectral properties of L follow.

Proposition 40. For all ϕ ∈ L2(Ω, γ) we have

kT(t)ϕ − m(ϕ)kL2(Ω,γ) ≤ e−tkϕkL2(Ω,γ) (2.9) and consequently

σ(L)\{0} ⊂ {λ ∈ C : Re λ ≤ −1}. (2.10)

Proof. Let f ∈ D(L) be such that m(f ) = 0. By using (2.8) we have Z

Lf · f dγ = − Z

h∇Hf, ∇Hf iHdγ = − Z

|∇Hf |2Hdγ ≤ −kf k2L2(Ω,γ). For every ϕ ∈ L2(Ω, γ), m(ϕ) = 0, T(t)ϕ ∈ D(L) for t > 0. Therefore

d

dtkT(t)ϕk2L2(Ω,γ) = 2 Z

LT(t)ϕ T(t)ϕ dγ ≤ −2kT(t)ϕk2L2(Ω,γ), t > 0.

It follows that

kT(t)ϕk2L2(Ω,γ) ≤ e−2tkϕk2L2(Ω,γ), t > 0. (2.11) Let now ϕ ∈ L2(Ω, γ). Then

Z

|T(t)ϕ − m(ϕ)|2dγ = Z

|T(t)(ϕ − m(ϕ))|2dγ ≤ e−2t Z

|ϕ − m(ϕ)|2

= e−2t

Z

|ϕ|2dγ − γ(Ω)[m(ϕ)]2



≤ e−2t Z

|ϕ|2dγ.

Note that ϕ 7→ m(ϕ) is the orthogonal projection on Ker(L). Splitting L2(Ω, γ) = Ker(L) ⊕ (Ker(L)), T(t) maps (Ker(L)) = Ker(m) into itself and the infinitesimal generator of the restriction of T(t) to (Ker(L)) is the part L0 of Lin (Ker(L)). By (2.11), the spectrum of L0 is contained in {λ ∈ C : Re(λ) ≤ −1}. Since the spectrum of L consists of the spectrum of L0 plus the eigenvalue 0, (2.10) follows.

Lemma 10. If ϕ ∈ F Cb1(X), ϕ ≥ 0, then T(t)ϕ|Ω≥ 0 a.e. in Ω.

Proof. Let ϕ ∈ F Cb1(X), ϕ ≥ 0, then

ϕ(x) = v(l1(x), . . . , lk(x))

with l1, . . . , lk ∈ X and v ∈ Cb1(Rk) with v ≥ 0. First we prove that T(n)(t)ϕ ≥ 0 a.e. in Ωn. Let G := span{Fn, Rγ(l1), . . . , Rγ(lk)}. Then G is a subspace of H of dimension q ≤ n + k; setting d = q − dim Fn let O := On× Rd. Let {hi}qi=1 ⊂ X be an orthonormal basis of G such that {h1, . . . , hj} is an orthonormal basis of Fn if dim Fn = j. Then we have

ϕ(x) =ev(πG(x)) where ev ∈ Cb1(Rq) andv ≥ 0 in Re q. Therefore, by

(T(n)(t)ϕ|Ωn)(x) = TG(t)evOG(x)), for t > 0, and x ∈ Ωn.

By Proposition 34 we have

(T(n)(t)ϕ|Ωn)(x) ≥ 0 for all x ∈ Ωn. Thanks to Proposition 32, up to a subsequence,

(T(t)ϕ|Ω)(x) = lim

n→∞(T(n)(t)ϕ|Ωn)(x) ≥ 0 a.e. in Ω.

Now we have all the ingredients to prove the logarithmic Sobolev inequal-ity by the Deuschel-Strook’s method (see [12]).

Proposition 41. For all f ∈ W1,2(Ω, γ) we have Z

f2log(f )dγ ≤ Z

|∇Hf |2Hdγ + kf k2L2(Ω,γ)log(kf kL2(Ω,γ)). (2.12) Proof. First we suppose that f ∈ F Cb1(Ω) and f ≥ c > 0 in Ω. Setting ϕ = f2 we have

Hf = 1 2

Hϕ

√ϕ so (2.12) is equivalent to

Z

ϕ log(ϕ)dγ − Z

ϕdγ log

Z

ϕdγ



≤ 1 2

Z

1

ϕ|∇Hϕ|2Hdγ.

We remark that d

dt Z

T(t)(ϕ) log(T(t)(ϕ))dγ = Z

LT(t)(ϕ) log(T(t)(ϕ))dγ +

Z

LT(t)(ϕ)dγ.

The second term vanishes for the invariance of γ while for the first we recall

that Z

(Lψ)g(ψ)dγ = − Z

g0(ψ)|∇Hψ|2dγ with g(ξ) = log ξ and ψ = T(t)ϕ. Hence

d dt

Z

T(t)(ϕ) log(T(t)(ϕ))dγ = − Z

1

T(t)(ϕ)|∇HT(t)(ϕ)|2dγ.

Since

(|∇HT(t)(ϕ)|H)2 ≤ e−tT(t)|∇H(ϕ)|H2

and

T(t)|∇H(ϕ)|H2

=



T(t)√

ϕ|∇Hϕ|H

√ϕ

2

≤ T(t)(ϕ)T(t) |∇Hϕ|2H ϕ

 , we have

d dt

Z

T(t)(ϕ) log(T(t)(ϕ))dγ ≥ −e−2t Z

T(t) |∇Hϕ|2H ϕ

 dγ

= −e−2t Z

|∇Hϕ|2H ϕ dγ.

(2.13)

We recall that, by (2.9),

T(t)ϕ−−−→ mt→∞ (ϕ), in L2(Ω, γ).

Moreover

t→∞lim Z

| log(T(t)ϕ) − log(mϕ)|2dγ = 0

indeed by assumption, we have ϕ ≥ c2 and, by Lemma 10, it follows that T(t)ϕ ≥ c2. Moreover m(ϕ) ≥ c2, and

log(T(t)ϕ) − log(m(ϕ)) =

Z m(ϕ) T(t)ϕ

1 t dt

≤ 1

min{T(t)ϕ, m(ϕ)}|T(t)ϕ − m(ϕ)|

≤ 1

c2|T(t)ϕ − m(ϕ)|.

Then Z

T(t)(ϕ) log(T(t)(ϕ))dγ −−−→ mt→∞ (ϕ) log (m(ϕ)) Then integrating (2.13) with respect to t between 0 and ∞ we get

m(ϕ) log (m(ϕ)) − Z

ϕ log |ϕ|dγ ≥ −1 2

Z

|∇Hϕ|2H ϕ dγ.

Now let f ∈ W1,2(Ω, γ) and f ≥ 0 a.e. in Ω. Then there exists a sequence {fn} ⊂ F Cb1(Ω) such that fn ≥ 1/n and fn → f in W1,2(Ω, γ) and almost everywhere. Since t2log(t) > −1 for all t > 0, we can apply the Fatou’s

lemma and obtain Z

f2log(f )dγ ≤ lim

n→∞

Z

fn2log(fn)dγ

≤ lim

n→∞

Z

|∇Hfn|2Hdγ + kfnk2L2(Ω,γ)log(kfnkL2(Ω,γ))



= Z

|∇Hf |2Hdγ + kf k2L2(Ω,γ)log(kf kL2(Ω,γ)).

For a general f ∈ W1,2(Ω, γ), (2.12) follows from the fact that |f | ∈ W1,2(Ω, γ) and |∇|f ||H = |∇f |H almost everywhere.

Remark 14. The Log-Sob inequality allows to prove an interesting property of the space W1,2(Ω, γ), namely if f ∈ W1,2(Ω, γ) and v is a measurable function with |v(x)| ≤ k(kxkX + 1) for some k > 0 and for a.e. x ∈ Ω, then f v ∈ L2(Ω, γ). To this aim we recall that, by Theorem 6, there exists a constant α > 0 such that

Z

X

eαkxk2Xdγ < ∞.

Fix c < α/4. Then we have Z

(f (x)v(x))2

≤ k2 Z

f (x)2(kxkX + 1)2dγ ≤ 2k2 Z

f (x)2kxk2Xdγ + 2k2 Z

f (x)2

= k2 Z

{x∈Ω: ckxk2X>log |f (x)|}

f (x)2kxk2Xdγ + k2

Z

{x∈Ω: ckxk2X≤log |f (x)|}

f (x)2kxk2Xdγ + 2k2kf k2L2(Ω,γ)

≤ k2 Z

X

kxk2X e2ckxk2Xdγ + k2 c

Z

|f (x)|2log |f (x)|dγ + 2k2kf k2L2(Ω,γ)

≤ k2

Z

X

kxk4X

1/2Z

X

e4ckxk2X

1/2

+ k2 c

Z

|f (x)|2log |f (x)|dγ + 2k2kf k2L2(Ω,γ)

≤ C + k2 c

Z

|∇Hf |2Hdγ + kf k2L2(Ω,γ)log(kf kL2(Ω,γ))



+ 2k2kf k2L2(Ω,γ), that is f v ∈ L2(Ω, γ). The previous estimates shows that the linear function Λv : f 7→ f v, maps bounded subsets of W1,2(Ω, γ) into bounded subsets of L2(Ω, γ); this implies that Λv is continuous.

Chapter 3

Maximal Sobolev Regularity for Ornstein-Uhlenbeck equation

In this chapter we study the elliptic equation λu−Lu = f in an open convex subset Ω of an infinite dimensional separable Banach space X endowed with a centered non-degenerate Gaussian measure γ, where L is the Ornstein-Uhlenbeck operator. We prove that for λ > 0 and f ∈ L2(Ω, γ) the weak solution u belongs to the Sobolev space W2,2(Ω, γ). Moreover, under some regularity assumption on the boundary of Ω, we prove that u satisfies a gen-eralized Neumann boundary condition in the sense of traces at the boundary of Ω.

3.1 Finite-dimensional estimates

Let O be an open smooth convex subset of Rn, with fixed n. We assume that O = {x ∈ Rn : g(x) < 0}

where g is a smooth convex function whose gradient does not vanish at the boundary ∂O. We denote by ν(x) the exterior normal vector to ∂O at x, ν(x) = |∇g(x)|∇g(x). Let µ be the standard Gaussian measure in Rn and let LO be the associated Ornstein-Uhlenbeck operator, that is

LOψ(x) =

n

X

i=1

Diiψ(x) −

n

X

i=1

xiDiψ(x) for ψ ∈ D(LO).

In this section we consider the following problem

λψ − LOψ = f in O ⊂ Rn,

∂u

∂ν = 0 on ∂O (3.1)

where f ∈ L2(O, µ) and λ > 0.

Let us introduce a weighted surface measure on ∂O:

dσ = N (x)dHn−1

where N (x) = (2π)−n/2exp(−|x|2/2) is the Gaussian weight and dHn−1 is the usual Hausdorff (n − 1) dimensional measure. Using the surface measure dσ the integration by parts formula reads as:

Z

O

Dkϕ ψ dµ = − Z

O

ϕDkψ dµ + Z

O

xkϕψ dµ + Z

∂O

Dkg

|∇g|ϕψ dσ, (3.2) for each ϕ, ψ ∈ W1,2(O, µ) one of which with bounded support, so the bound-ary integral is meaningful. Indeed W1,2(O, µ) ⊂ Wloc1,2(O, dx) and the trace at the boundary of any function in W1,2(O, µ) belongs to L2loc(∂O, dHn−1) = L2loc(∂O, dσ).

Applying (3.2) with ϕ replaced by Dkϕ and summing up, we find Z

O

LOϕψ dµ = − Z

O

h∇ϕ, ∇ψidµ + Z

∂O

h∇ϕ, ∇gi

|∇g| ψ dσ (3.3) for every ϕ ∈ W2,2(O, µ) , ψ ∈ W1,2(O, µ) one of which with bounded support.

Now we give dimension free estimates for the weak solution u ∈ W1,2(O, µ) to (3.1) with λ > 0 and f ∈ L2(O, µ). We can suppose f ∈ Cc(O) because Cc(O) is dense in L2(O, µ). In this case, thanks to classical results about elliptic equations with smooth coefficients we know that the weak solution u of (3.1) belongs to C(O) ⊂ Wloc2,2(O, µ). Since O can be unbounded, we introduce a smooth cutoff function θ : Rn → R such that

0 ≤ θ(x) ≤ 1, |∇θ(x)| ≤ 2 ∀x ∈ Rn, θ ≡ 1 in B(0, 1), θ ≡ 0 outside B(0, 2) and we set, for R > 0

θR(x) = θ(x/R), x ∈ Rn.

For the W1,2 estimates we take u as a test function in the definition of weak solution and we get

λ Z

O

u2dµ + Z

O

|∇u|2dµ = Z

O

f u dµ, (3.4)

then Z

O

u2dµ ≤ 1

λ2kf k2L2(O,µ), Z

O

|∇u|2dµ ≤ 1

λkf k2L2(O,µ). (3.5) The following lemma takes into the account the convexity of O.

Lemma 11. If u ∈ C2(O) satisfies h∇u, νi = 0 on ∂O then hD2u · ∇u, νi ≤ 0 on ∂O.

Proof. We recall that ∂O = g−1(0) where g : Rn → R is a smooth convex function. Let τ ∈ Rn such that hτ, ν(x)i = 0 for x ∈ ∂O, then we have

h∂ν

∂τ(x), τ i ≥ 0 ∀x ∈ ∂O. (3.6)

Indeed

∂ν

∂τ = ∂

∂τ

 ∇g

|∇g|



= 1

|∇g|

∂τ(∇g) + ∂

∂τ

 1

|∇g|



∇g

= 1

|∇g|D2g · τ + h∇

 1

|∇g|



, τ i∇g,

then for x ∈ ∂O we have h∂ν

∂τ(x), τ i = 1

|∇g(x)|hD2g(x) · τ, τ i + h∇

 1

|∇g(x)|



, τ ih∇g(x), τ i

= 1

|∇g(x)|hD2g(x) · τ, τ i ≥ 0

since D2g is a positive semi-definite symmetric matrix. Now we recall that h∇u, νi = 0 on ∂O therefore

∂τ(h∇u(x), ν(x)i) = hD2u(x) τ, ν(x)i + h∇u(x), ∂ν

∂τ(x)i = 0, x ∈ ∂O for each τ ∈ Rn such that hτ, νi = 0 on ∂O. If we take τ = ∇u(x) then we get

hD2u(x) · ∇u(x), ν(x)i = −hτ,∂ν

∂τ(x)i ≤ 0, x ∈ ∂O.

Now we can give an estimate of the second order derivatives of u.

Proposition 42. For every f ∈ Cc(O) and ε > 0 there exists R0 > 0 such that for R > R0 the solution u to (3.1) satisfies

(1 − ε) Z

O

θR2Tr[(D2u)2]dµ ≤  2 + ε

λ

kf k2L2(O,µ). (3.7)

Proof. Recall that u ∈ C(O); differentiating (3.1) with respect to xh yields λDhu − ∆Dhu − hx, ∇(Dhu)i + Dhu = Dhf.

Multiplying by DhR2 we obtain

(λ + 1) (Dhu)2θ2R− ∆Dhu · DhR2 − hx, ∇(Dhu)iDhR2 = Dhf DhR2. Integrating over O and using (3.3) yields

Z

O

(λ + 1)(Dhu)2θR2dµ + Z

O

|∇(Dhu)|2θ2Rdµ + 2 Z

O

θRh∇(Dhu), ∇θRiDhu dµ

= Z

∂O

h∇(Dhu), ∇giDhu

|∇g| θ2Rdσ + Z

O

Dhf Dh2Rdµ.

Summing over h we obtain Z

O

(λ + 1)|∇u|2θ2Rdµ + Z

O

Tr[(D2u)2R2dµ + 2 Z

O

hD2u · ∇u, ∇θRR

= Z

∂O

hD2u · ∇u, ∇gi

|∇g| θ2Rdσ + Z

O

h∇f, ∇uiθR2dµ.

Since f has compact support, for R large enough θR ≡ 1 on the support of f . For such R we obtain

Z

O

h∇f, ∇uiθ2R

=

− Z

O

LOuf dµ

= Z

O

(λu − f )f dµ

≤ 2kf k2L2(O,µ). Moreover

Z

O

hD2u∇u, ∇θRR

≤ Z

O n

X

i,j=1

|Diju Dju DiθRR

≤ 1 2

Z

O n

X

i,j=1

|Diju|2|DiθRR2 dµ + 1 2

Z

O n

X

i,j=1

|Dju|2|DiθR| dµ

≤ 1 2

k|∇θ|k

R Z

O

θR2Tr[(D2u)2]dµ + 1 2

k|∇θ|k

R Z

O

|∇u|2

≤ 1 2

Z

O

Tr[(D2u)22Rdµ + 1

2λkf k2L2(O,µ)

 k|∇θ|k

R .

Taking R large enough, such that k|∇θ|k/R ≤ ε, we get (1 − ε)

Z

O

θ2RTr[(D2u)2]dµ ≤  2 + ε

λ

kf k2L2(O,µ)+ Z

∂O

θR2hD2u · ∇u, ∇gi

|∇g| dσ.

By using Lemma 11, the statement follows.

Theorem 18. For each λ > 0 there exists C = C(λ) > 0, independent of n and O, such that for each f ∈ L2(O, µ) the weak solution u of problem (3.1) belongs to W2,2(O, µ), and satisfies

kukW2,2(O,µ) ≤ Ckf kL2(O,µ). (3.8) Proof. Let f ∈ Cc(O). Taking the limit as R → ∞ in (3.7) and using the monotone convergence theorem, we get

(1 − ε) Z

O

Tr[(D2u)2]dµ ≤ 2 + ε

λ

kf k2L2(O,µ).

Now taking the limit as ε → 0 we get Z

O

Tr[(D2u)2]dµ ≤ 2kf k2L2(O,µ). (3.9) Taking into account (3.4), (3.5), and (3.9) we obtain

kuk2W2,2(O,µ) = kukL2(O,µ)+ k|∇u|kL2(O,µ) + kTr[(D2u)2]kL2(O,µ)

≤ 1 λ2 + 1

λ + 2



kf k2L2(O,µ)

which is (3.8) with C(λ) = λ12+λ1+2. For f ∈ L2(O, µ) the statement follows approaching it by a sequence of functions belonging to Cc(O).

We recall that (1.26) holds, and we prove that the condition h·, ∇f (·)i ∈ L2(O, µ) can be omitted.

Proposition 43. If f ∈ W2,2(O, µ) then the map x 7→ hx, ∇f (x)i belongs to L2(O, µ), moreover the map

f 7→ h·, ∇f (·)i is continuous from W2,2(O, µ) to L2(O, µ).

Proof. Let f ∈ W2,2(O, µ), then Z

O

|h∇f, xi|2dµ = Z

O n

X

i=1

(Dif xi)2

by assumption Dif ∈ W1,2(O, µ) and if c < 1/4, by using (1.28), we have Z

O

(Dif (x))2x2ie−|x|2/2dx

= Z

{x∈O: cx2i>log |Dif (x)|}

(Dif (x))2x2ie−|x|2/2dx +

Z

{x∈O: cx2i≤log |Dif (x)|}

(Dif (x))2x2ie−|x|2/2dx

≤ Z

O

e2cx2ix2ie−|x|2/2dx + Z

O

1

c|Dif |2log |Dif |e−|x|2/2dx

≤ C1+ 1 c

Z

O

|∇Dif |dµ + 1 2

Z

O

(Dif )2dµ log

Z

O

(Dif )2



. Summing over i from 1 to n we have h∇f, xi ∈ L2(O, µ).

3.2 Maximal Sobolev regularity for infinite

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