Restriction of Gaussian Sobolev p-Space

Chapter 4 Restriction of Gaussian Sobolev p-Space Based on Chapters 1–3, in this chapter we settle the restriction/trace question asked in the preface of this monograph. 4.1 Gaussian p-Capacitary-Strong-Type Inequality Lemma 4.1.1. Let 1 ≤ p < ∞ and f ∈ W 1,p (Gn ) be continuous. For any t ∈ (0, ∞) set   Et (f ) := x ∈ Rn : | f (x)| > t . Then ∫ (4.1) 0 ∞ p Capp (Et (f ); Gn ) dt p   f W 1,p (Gn ) . Proof. Let f ∈ W 1,p (Gn ) be a nonzero continuous function. In what follows, for any t ∈ (0, ∞), we simply write Et (f ) as Et . By Proposition 3.1.4(ii), (4.2) ∫ 0 ∞ Capp (Et ; Gn ) dt p = ∫ k ∈Z 2k +1 2k ≤ (2p − 1) Capp (Et ; Gn ) dt p  2kp Capp (E 2k ; Gn ). k ∈Z As in [41, p. 155, Remark 1], we choose a function τ : R → R such that ⎧ ⎪ τ ∈ C 1 (R) ⎪ ⎪ ⎪ ⎪ ⎪ 0 ≤ τ(t) ≤ 1 ⎪ ⎨ ⎪ τ(t) = 0 ⎪ ⎪ ⎪ τ(t) = 1 ⎪ ⎪ ⎪ ⎪ ⎪ 0 ≤ τ′(t) ≤ 3 ⎩ is even; ∀ t ≥ 0; ∀ t ∈ [0, 2−1 ]; ∀ t ≥ 1; ∀ t ≥ 0. © Springer Nature Switzerland AG 2018 L. Liu et al., Gaussian Capacity Analysis, Lecture Notes in Mathematics 2225, https://doi.org/10.1007/978-3-319-95040-2_4 55 56 4 Restriction of Gaussian Sobolev p-Space Indeed, on the interval [2−1 , 1], we can define τ by smoothing the line passing over the points (2−1 , 0) and (1, 1) so that 0 ≤ τ′(t) ≤ 2 + ǫ for some small ǫ ∈ (0, 1). For all k ∈ Z and x ∈ Rn , we define fk (x) := τ f (x) 2k which is also a continuous function owing to f ∈ C 0 (Rn ). For all k ∈ Z, observe that | f (x)| f (x) 0 =τ = k k 2 2 1 fk (x) = τ ∀ x  E 2k −1 ; ∀ x ∈ E 2k . and 0 ≤ fk (x) ≤ 1 ∀ x ∈ E 2k −1 \ E 2k . Thus ∫ Rn and ∫ Rn p |∇fk | dVγ = ∫ Rn τ ′ | fk |p dVγ < Vγ (Rn ) = 1 f (x) ∇f (x) 2k 2k p p dVγ ≤ 3 ∫ |∇f (x)|p 2−kp dVγ , E 2k −1 \E 2k which implies fk ∈ W 1,p (Gn ). Also, since fk = 1 on the open set E 2k , it is easy to verify  ◦ E 2k ⊆ x ∈ Rn : fk (x) ≥ 1 . Hence, fk ∈ Ap (E 2k ) and p Capp (E 2k ; Gn ) ≤  fk W 1,p (Gn ) ∫ ∫ p | fk |p dVγ |∇fk | dVγ + ≤ Rn Rn ∫ ≤ |∇f (x)|p 2−kp dVγ (x) + Vγ (E 2k −1 ). E 2k −1 \E 2k 4.2 Trace Inequality for W 1,p (Gn ) Under 1 ≤ p ≤ q < ∞ 57 Combining this with (4.2) yields ∫ ∞ Capp (Et ; Gn ) dt p 0 ∫   (4.3) |∇f (x)|p dVγ + 2kp Vγ (E 2k −1 )   p 2kp Vγ (E 2k −1 ).   |∇f | Lp (Gn ) + k ∈Z E 2k −1 \E 2k k ∈Z k ∈Z By the Hölder inequality, we have  2kp Vγ (E 2k −1 ) =  2kp = Vγ (E 2i \ E 2i +1 ) 2p  ip 2 Vγ (E 2i \ E 2i +1 ) 1 − 2−p i ∈Z k ∈Z k ∈Z ∞  i=k−1 p   f Lp (Gn ) . From this and (4.3), it follows that (4.1) holds.  4.2 Trace Inequality for W 1,p (Gn ) Under 1≤p ≤q <∞ Now we use Lemma 4.1.1 to establish the first restriction/trace result for W 1,p (Gn ). Theorem 4.2.1. Let 1 ≤ p ≤ q < ∞ and μ be a nonnegative Radon measure on Rn . Then the following two assertions are equivalent. (i) There exists a positive constant C 1 such that for all compact sets K ⊆ Rn ,  q μ(K) ≤ C 1 Capp (K; Gn ) p . (ii) There exists a positive constant C 2 such that (4.4) ∫ 1 | f |q dμ Rn q ≤ C 2  f W 1,p (Gn ) ∀ q f ∈ C 0 (Rn ) ∩ W 1,p (Gn ). Moreover, C 1 ≃ C 2 with the implicit constants depending only on p and q. Proof. Assuming (i) we prove (ii). Fix f ∈ C 0 (Rn ) ∩ W 1,p (Gn ). 58 4 Restriction of Gaussian Sobolev p-Space   Et := x ∈ Rn : | f (x)| > t , For t ∈ (0, ∞), let which is open. Consider the set E 2k with k ∈ Z. Since μ is a nonnegative Radon measure on Rn , we can find a compact set Kk ⊆ E 2k such that μ(E 2k ) ≤ 2μ(Kk ). By this last inequality and Proposition 3.1.4(ii), we have ∫ Rn | f |q dμ = ∫ 2k +1 μ(Et ) dt q 2k k ∈Z ≤ (2q − 1)   2kq μ(E 2k ) k ∈Z q+1 ≤2 (4.5) 2kq μ(Kk )   q 2kq Capp (Kk ; Gn ) p k ∈Z q+1 ≤ C1 2   q 2kq Capp (E 2k ; Gn ) p k ∈Z q+1 ≤ C1 2 k ∈Z We shall use the following inequality: for any nonnegative sequence {a j }j ∈Z ,    κ  a κj ∀ κ ∈ (0, 1]. aj ≤ j ∈Z j ∈Z This estimation, along with the fact p ≤ q, further yields  (4.6) kq 2 k ∈Z  n Capp (E 2k ; G )  qp ≤   2kp Capp (E 2k ; Gn ) = (1 − 2−p )−1 k ∈Z ∫ −p −1 ≤ (1 − 2 ) = (1 − 2−p )−1 ∫ k ∈Z ∞  qp 2k 2k −1 ∫ k ∈Z (4.7) Capp (E 2k ; G ) 2 k ∈Z Again, using Proposition 3.1.4(ii) we obtain  n kp Capp (E 2k ; Gn ) dt p 2k 2k −1 Capp (Et ; Gn ) dt p Capp (Et ; Gn ) dt p . 0 Combining (4.5), (4.6), (4.7), and Theorem 3.2.6 yields ∫ Rn q q+1 | f | dμ ≤ C 1 2 q −p − p (1 − 2 ) ∫ 0 ∞ n Capp (Et ; G ) dt p q p , 4.3 Trace Inequality for W 1,p (Gn ) Under 0 < q < p < ∞ 59 which, together with Lemma 4.1.1, yields (4.4). Thus, (ii) holds and q q C 2 ≤ 2q+1 (1 − 2−p )− p C Lemma 4.1.1C 1 , where C Lemma 4.1.1 is the positive constant determined in Lemma 4.1.1. Now we show that (ii) implies (i). By Theorem 3.2.6, it suffices to prove that (i) holds for Cap0,p (·; Gn ). Let K be a compact subset of Rn . For any f ∈ A(K) we have f ∈ Cc1 (Rn ) & f |K ≥ 1. Clearly, (4.4) holds for such f . From this and the fact that μ is nonnegative, it follows that  μ(K)  q1 ∫ ≤ 1 q | f | dμ Rn q ≤ C 2  f W 1,p (Gn ) . Taking infimum over all f ∈ A(K) yields  μ(K) q Thus (i) holds with C 1 ≤ C 2 .  q1 1  ≤ C 2 Cap0,p (E; Gn ) p .  4.3 Trace Inequality for W 1,p (Gn ) Under 0<q <p <∞ The second restriction/trace result for W 1,p (Gn ) is presented below. Theorem 4.3.1. Let p ∈ [1, ∞), 0 < q < p < ∞, and μ be a nonnegative Radon measure. Then the following two conditions are equivalent. (i) The function   (0, ∞) ∋ t → h μ,p (t) := inf Capp (K; Gn ) : Rn ⊇ K is compact with μ(K) ≥ t satisfies (4.8) p ∫ ∞ ds p−q   h μ,p  :=  q    p−q 0 h μ,p (s)   p−q p < ∞. (ii) There exists a positive constant C such that (4.9) ∫ Rn 1 q | f | dμ q ≤ C  f W 1,p (Gn ) ∀ f ∈ C 0 (Rn ) ∩ W 1,p (Gn ). 60 4 Restriction of Gaussian Sobolev p-Space Moreover, h μ,p  ≃ C q whose implicit constants depend only on p and q. Proof. (i) ⇒ (ii) Fix f ∈ C 0 (Rn ) ∩ W 1,p (Gn ).   Et := x ∈ Rn : | f (x)| > t . For t ∈ (0, ∞), let Then ∫ | f |q dμ ≤ (2q − 1) Rn  2kq μ(E 2k ) k ∈Z = (2q − 1) (4.10)   i ∈Z iq  2 ∞  2kq μ(E 2i \ E 2i +1 )  μ(E 2i ) − μ(E 2i +1 ) . k ∈Z i=k Using q < p and the Hölder inequality, we get    2iq μ(E 2i ) − μ(E 2i +1 ) i ∈Z (4.11) =  2ip Capp (E 2i ; Gn ) i ∈Z ≤    qp μ(E i ) − μ(E i +1 ) 2 2  q Capp (E 2i ; Gn ) p 2ip Capp (E 2i ; Gn ) i ∈Z  qp   μ(E 2i ) − μ(E 2i +1 )     q   i ∈Z  Capp (E 2i ; Gn ) p   It follows from Proposition 3.1.4(ii) and Lemma 4.1.1 that  ip −p −1 n 2 Capp (E 2i ; G ) = (1 − 2 ) ∫ ≤ (1 − 2−p )−1 = (1 − 2−p )−1 2i −1 ∫ i ∈Z i ∈Z 2i ∫ i ∈Z ∞ 2i 2i −1 Capp (Et ; Gn ) dt p Capp (Et ; Gn ) dt p   f W 1,p (Gn ) , so that (4.12)   i ∈Z ip n 2 Capp (E 2i ; G )  qp     Capp (E 2i ; Gn ) dt p 0 p p ( q )′ q   f W 1,p (Gn ) . p−q p . 4.3 Trace Inequality for W 1,p (Gn ) Under 0 < q < p < ∞ Now we consider   μ(E 2i ) − μ(E 2i +1 )     q   i ∈Z  Capp (E 2i ; Gn ) p   p ( q )′ in (4.11). Since f ∈ C 0 (Rn ),     61 p−q p it follows that every E 2i is open. As μ is a Radon measure, there exists a compact set Ki ⊆ E 2i such that 2−1 μ(E 2i ) ≤ μ(Ki ) ≤ μ(E 2i ). According to the definition of the function h μ,p , we see that   h μ,p 2−1 μ(E 2i ) ≤ Capp (Ki ; Gn ) ≤ Capp (E 2i ; Gn ), where the second inequality holds because of (3.17). From this, p ′ p > 1, = q p −q the monotone increasing property of h μ,p , and (4.8), it follows that   μ(E 2i ) − μ(E 2i +1 )   q  i ∈Z  Capp (E 2i ; Gn ) p  (4.13) p ( q )′ p  μ(E 2i ) − μ(E 2i +1 ) p−q ≤    q i ∈Z h μ,p 2−1 μ(E 2i ) p−q p p  μ(E 2i ) p−q − μ(E 2i +1 ) p−q ≤    q h μ,p 2−1 μ(E 2i ) p−q i ∈Z p ∫ 2−1 μ(E i ) p  2 ds p−q p−q =2    q −1 i ∈Z 2 μ(E 2i +1 ) h μ,p 2−1 μ(E 2i ) p−q p ∫ ∞ p ds p−q p−q ≤2   q 0 h μ,p (s) p−q p p = 2 p−q h μ,p  p−q . Inserting the estimates of (4.12) and (4.13) into (4.11) yields    q 2iq μ(E 2i ) − μ(E 2i +1 )  h μ,p   f W 1,p (Gn ) , i ∈Z which combined with (4.10) further gives us that ∫ q | f |q dμ  h μ,p   f W 1,p (Gn ) . Rn 62 4 Restriction of Gaussian Sobolev p-Space Thus (ii) holds with 1 C  h μ,p  q . (ii) ⇒ (i) According to Theorem 3.2.6, we may replace Capp in the definition of h μ,p by Cap0,p . Observe that if μ(Rn ) ≤ t then we take it for granted that h μ;p (t) = ∞. Thus p−q p p ∫ μ(Rn ) p−q ds   h μ,p  =  .  q   p−q 0 h (s) μ,p   n For μ(R ) < ∞, there exists a unique J0 ∈ Z such that 2 J0 < μ(Rn ) ≤ 2 J0 +1 . For μ(Rn ) = ∞, let J0 = ∞. Since h μ,p is an increasing function, it follows that J0 ∫ 2 j +1 ds p−q   h μ,p  =  q    p−q j 2 h μ,p (s) j=−∞  p (4.14) ≤ (2 p p−q − 1) p−q p p−q p 0 2j p−q    q    j p−q j=−∞ h μ,p (2 )  J p p−q p . For each j ∈ Z such that j < J0 + 1, by the definition of h μ,p and Theorem 3.2.6, for any ǫ ∈ (0, ∞), there exists a compact set K j ⊆ Rn such that μ(K j ) ≥ 2j and Cap0,p (K j ; Gn ) − ǫh μ,p (2j ) ≤ h μ,p (2j ) ≤ Cap0,p (K j ; Gn ). Further, according to the definition of Cap0,p (K j ; Gn ), there exists a function f j ∈ A(K j ) such that (4.15) p p  f j W 1,p (Gn ) − 2−j ǫ ≤ Cap0,p (K j ; Gn ) ≤  f j W 1,p (Gn ) . We define the function Fi,m := max γj f j i ≤j ≤m 2j with γj := h μ,p (2j ) Notice that f j ∈ C 0 (Rn ) ⇒ Fi,m ∈ C 0 (Rn ). 1 p−q . 4.3 Trace Inequality for W 1,p (Gn ) Under 0 < q < p < ∞ 63 So, applying f j ∈ W 1,p (Gn ) and Lemma 3.1.2 yields that ∇Fi,m exists a. e. on Rn and that a. e. on Rn . |∇Fi,m | ≤ max γj |∇f j | i ≤j ≤m It is easy to verify Fi,m W 1,p (Gn ) ≤ m  γj  f j W 1,p (Gn ) < ∞. j=i Hence Fi,m ∈ W 1,p (Gn ). Accordingly, (ii) yields (4.16) ∫ q Rn |Fi,m |q dμ ≤ C q Fi,m W 1,p (Gn ) . For the left side of (4.16), using the nonincreasing rearrangement of Fi,m implies ∫ ∞ ∫ q   inf {s > 0 : μ {x ∈ Rn : |Fi,m (x)| > s} ≤ t } dt |Fi,m |q dμ = = ∫ ≥ 2 Rn 0 (4.17) 2j j −1 j ∈Z 2 m   j−1 j=i  q   inf {s > 0 : μ {x ∈ Rn : |Fi,m (x)| > s} ≤ t } dt q inf {s > 0 : μ({x ∈ Rn : |Fi,m (x)| > s}) ≤ 2j } . For every x ∈ K j with i ≤ j ≤ m, we have Fi,m (x) ≥ γj f j (x) ≥ γj , thereby getting via (4.3) that for any small number η > 0,   μ {x ∈ Rn : |Fi,m (x)| > γj − η} ≥ μ(K j ) ≥ 2j , so that     inf s > 0 : μ {x ∈ Rn : |Fi,m (x)| > s} ≤ 2j ≥ γj − η. Letting η → 0 gives   inf s > 0 : μ({x ∈ Rn : |Fi,m (x)| > s}) ≤ 2j ≥ γj ∀ i ≤ j ≤ m. This, plus (4.17), implies (4.18) ∫ Rn q |Fi,m | dμ ≥ m  j=i q 2j−1 γj −1 =2 m  j=i p 2j p−q   q . h μ,p (2j ) p−q 64 4 Restriction of Gaussian Sobolev p-Space Now we compute the right side of (4.16). Clearly, q Fi,m W 1,p (Gn ) ≤ ≤ =    p Fi,m Lp (Gn ) m   p +  |∇Fi,m |  p L (Gn ) p p γj  f j Lp (Gn ) + j=i m  m  j=i p p γj  f j W 1,p (Gn ) j=i  qp  qp p p γj  |∇f j | Lp (Gn )  qp . By using (4.3) and (4.15), together with the definition of γj , we see m  p p γj  f j W 1,p (Gn ) ≤ j=i m  p γj h μ,p (2j ) ≤ (1 + 2ǫ) j=i m  j=i whence  q Fi,m W 1,p (Gn ) ≤ (1 + 2ǫ)  (4.19) m  j=i p 2j p−q   q , h μ,p (2j ) p−q q p 2  . q    h μ,p (2j ) p−q  p j p−q Inserting the estimates of (4.18) and (4.19) into (4.16), we obtain 2−1 m  j=i  2j p−q   ≤ C q (1 + 2ǫ) q    j=i h μ,p (2 j ) p−q   p p m 2j p−q  q  h μ,p (2j ) p−q Since q < p, it follows that m 2j p−q    q    p−q j  j=i h μ,p (2 )  p q p p−q p q ≤ 2C q (1 + 2ǫ) p . Taking supremum over all i, m ∈ Z such that −∞ < i ≤ m < J0 + 1, we use (4.14) to deduce J0 2j p−q   h μ,p  ≤ Cp,q  q    j p−q j=−∞ h μ,p (2 )  p p−q p q ≤ 2Cp,q C q (1 + 2ǫ) p . Letting ǫ → 0, we obtain (i), thereby completing the proof of the theorem.