18.156 Lecture 18 Notes

Lecture 18 Embedding Theorems for Sobolev spaces Let Ω a bounded domain in Rn , and 1 ≤ p < ∞. Sobolev Embedding Theorem.  np  L n−p (Ω), p < n     W01,p (Ω) ⊆ C 0,α (Ω), α = 1 − np , p > n,      i.e in particular ⊆ C 0 (Ω). Furthermore, those embeddings are continuous in the following sense: there exists C(n, p, Ω) such that for u ∈ W01,p (Ω) ||u|| np L n−p (Ω) ≤ C · ||∇u||Lp (Ω) , 1 ∀p < n 1 sup |u| ≤ C ′ · Vol(Ω) n − p · ||Du||Lp (Ω) , Ω ∀p > n. We start with a function whose derivative and itself belong to Lp . The above theorem gives us n more regularity for the function – it belongs to Lp· n−p – based on its regular derivative. Proof. C01 (Ω) is dense in W01,p (Ω). We prove first for u ∈ C01 (Ω) and will later justify why the proof actually extends to the larger space. Case p = 1. fix an index i ∈ {1, . . . , n} and observe u(x) = Z xi Di u(x1 , . . . , t, . . . , xn )dt. −∞ From which 1 xi |u(x)| ≤ Z ∞ ≤ Z −∞ −∞ |Di u|(x1 , . . . , t, . . . , xn )dt |Di u|(x1 , . . . , t, . . . , xn )dt. (1) Write this down for each i, take a product of the terms and take the n − 1th root of the result to yield altogether |u(x)| n n−1 n Z Y ≤ ∞ −∞ i=1 |Di u|dxi 1  n−1 . Quick Reminder. Hölder’s inequality (HI) tells us 1 1 + =1 p q ⇒ Z Z u·v ≤  p1  Z  1q u · vp , p or more generally 1 1 + ... + =1 p1 pk Z ⇒ u1 · · · u k ≤ Z up11  p1 1 ··· Z upkk  p1 k . Coming back to our inequality, we integrate over the x1 axis and subsequently apply the Hölder inequality with k = n − 1, pi = n − 1 – Z ∞ −∞ |u(x)| n n−1 dx1 ≤ = Z ∞ n Z Y −∞ i=1 Z ≤ Hölder’s Ineq. = −∞ |Di u|dxi ∞ −∞ Z ∞ ∞ −∞ |D1 u|dx1 Z Z 1  n−1 · ∞ −∞ 1  n−1 |D1 u|dx1 ∞ dx1 . n Z Y −∞ i=2 ∞ −∞ n Z 1  n−1 Y · i=2 1 Z  n−1 · |D1 u|dx1 ∞ −∞ Z ∞ −∞ 1  n−1 −∞ −∞ |Di u|dxi |D2 u|dx2 dx1 n hZ Y i=3 dx1 . ∞ hZ ∞ · 2 |Di u|dxi ∞ −∞ 1  n−1 Z n−1 i n−1 dx1 1  n−1 . · ∞ −∞ 1 i n−1  . |Di u|dxi dx1 Now courageously continuing with this confusing calculation, we integrate over the x2 axis. This is the reason we singled out the second terms from the n−2 others ones; if we would have integrated now over the xj axis we would have choosen a term involving integration over that axis. And indeed now the middle term is a constant wrt this operation, that is only the other two terms appear in this integral, hence – Z ∞ −∞ ≤ Z ∞ −∞ Z n |u(x)| n−1 dx1 dx2 ∞ −∞ Z ∞ −∞ |D2 u|dx2 dx1 1  n−1 Z · ∞ −∞ n Z ∞ −∞ n hZ 1  n−1 Y |D1 u|dx1 · i=3 ∞ −∞ i |Di u|dxi dx2 o 1 ! n−1 . and using the Hölder Inequality the second term transforms, and we have = Z ∞ −∞ Z ∞ −∞ 1 Z  n−1 · |D2 u|dx2 dx1 ∞ hZ −∞ · n Z Y i=3 ∞ −∞ ∞ −∞ Z |D1 u|dx1 ∞ −∞ Z n−1 i n−1 dx2 1  n−1 ∞ −∞ |Di u|dxi dx1 dx2 · 1  n−1 . In the same vein, we now isolate among the n terms the only term involving integration over the x3 axis, integrate over that axis and then once again apply the Hölder Inequality for the remaining n − 1 terms (at each stage we always have n − 1 terms except from the isolated one; the Hölder Inequality allows us to lift the 1 n−1 exponent and let another new dxi come in to the integral of those n − 1 terms). Finally, therefore, we will arrive at Z ∞ −∞ ··· Z ∞ −∞ |u(x)| n n−1 dx1 · · · dxn ≤ n Z Y j=1 ∞ −∞ ··· Z ∞ −∞ |Dj u|dx1 · · · dxn In other words if we restrict to Ω n Z 1 n  n−1  n−1  Y n |D u|dx ≤ . ||u||L n−1 j (Ω) j=1 3 Ω 1  n−1 . or still ||u|| n L n−1 (Ω) ≤ Z Z Z n n  n1 1 X 1 X |Du|dx |Dj u|dx ≤ · |Du|dx = . . |Dj u|dx ≤ · n j=1 Ω n j=1 Ω Ω Ω n Z Y j=1 = ||∇u||L1 (Ω) . This concludes the p = 1 < n case. Let us remark that of course we neglected at the last steps to seek the best possible Sobolev constant and contented ourselves with the constant 1: ||u|| n·1 L n−1 (Ω) ≤ 1 · ||∇u||L1 (Ω) . In fact the best possible Sobolev constant c is achieved for Ω = B(0, r), u = IB(0,r) (IA is the characteristic function on the set A, evaluating to 1 on A and 0 otherwise); believing that, we compute Vol(B(0, r)) n n−1 =c· Z B(0,r) |DIB(0,r) |dx = c · Z B(0,r) |δ∂B(0,r) |dx = c · Area(S(r)), i.e n (ωn r n ) n−1 = c · nωn r n−1 ⇒ c= 1 √ . n n ωn Case 1 < p < ∞. A little trick will make our previous work apply to this case as well. Let γ > 1 be a constant to be specified. We have by our previous case γ n ≤ || |u| ||L n−1 (Ω) Let q be such that γ Ω D|u| dx ≤ γ Z Ω |u|γ−1 · |Du|dx. 1 1 + = 1. One has using the Hölder Inequality p q Z Ω We have q = Z p p−1 . n γ· n−1 |u| n  n−1  q1  Z  p1 Z (γ−1)q dx |u| dx · |Du|p dx . ≤ Choose γ = Ω n−1 n−p Ω · p in order to have (γ − 1)q = 4 n n−1 · γ. Hence Z n−1 n ( n−p ·p)· n−1 Ω |u| n  p1 Z Z  n−1  p−1 p n−1 q ( n−p ·p)−1)·( p−1 ) |u| |Du|p dx , ≤ · dx dx Ω Ω or succintly ||u|| np L n−p (Ω) = nZ np Ω |u| n−p p−1 o n−1 n − p ≤ n−1 · p||∇u||Lp (Ω) . n−p This deals with the case p < n indeed. We remark that characteristic functions no longer give the best Sobolev constants in the case 1 < p < n. Remark. The above proof holds and is valid for u ∈ C01 (Ω)! We did not prove for distributional coefficient . If u is only in W01,p (Ω), take a sequence {um } ⊆ C01 (Ω) such that um → u in the W01,p (Ω)-norm. This means that also ||ui − uj || np L n−p (Ω) np ≤ c · ||Dui − duj ||Lp (Ω) → 0. np {um } is thus a Cauchy sequence in L n−p (Ω). L n−p (Ω) is a Banach space d’aprês Riesz-Fischer, i.e np u′ := lim{um } ∈ L n−p (Ω); ⇔ u′ = u is in that space too. Now ||um || np L n−p (Ω) ≤ c · ||Dum ||Lp (Ω) → 0. ↓ ↓ ||u|| np L n−p (Ω) ≤ c · ||Du||Lp (Ω) → 0. So we our Theorem applies equally well to functions in the larger space. In this last line we needed also mention that Dum → Du, but this is true since {Dum } is a Cauchy sequence from same computation as above. Its limit lies in Lp again as this is a Banach space and so indeed Dum → Du in Lp and hence also ||Dum ||Lp (Ω) → ||Du||Lp (Ω) . Case p > n. We postpone the proof of this case to state a Corollary. 5 Corollary. By iterating, ∀k ≥ 2 holds W0k,p (Ω) ⊆  np  L n−k·p (Ω),  Proof. For instance, if k = 2, u ∈ W 2,p np np u, Du ∈ L n−p . That means Du ∈ W 1, n−p ′ p = np n · ( n−p ) np n − ( n−p ) Cm, kp < n 0 ≤ m ≤ k − np . u, Du ∈ W 1,p . By the k = 1 case above we have ⇒ ′ ⇒ (by k = 1 case once again) u ∈ W 1,p where = np n2 p = . 2 n − np − np n − 2p This proof repeated carries over ∀k ∈ N. Now for the second inclusion, the promised postponed. We will need the following lemma en passant. Lemma. Let Ω be a bounded domain, B :=Ball ⊆ Ω, u ∈ W 1,1 . Then for all x ∈ Ω 1 u(x) − Vol(B) Z B Z Z udx ≡ u(x)− 6 udx ≤ c · B B |Du(y)| dy. |x − y|n−1 Proof. By our density theorem C01 (Ω) is dense in W01,p (Ω) and thus work with u in the former. Take x, y ∈ Ω. Let ω := y−x , |y−x| u(x) − u(y) = Z |x−y| Dr u(x + rω)dr. 0 Integrating over some ball B Vol(B) · u(x) − Z u(y) = B Z B Put 6 Z |x−y| 0
Dr u(x + rω)dr dy. v(x) =  Dr u(x), 0, x∈Ω x 6∈ Ω. Take now a particular ball B(x, R) ⊆ Ω to get Z u(x)− 6 u(y)dy ≤ B 1 Vol(B) Z |x−y|<2R Z 0 ∞
|v(x + rω)|dr dy. Switch order of integration, and change coordinates to spherical ones 1 = Vol(B) Z ∞ 0 Z 2R 0 Z S n−1 (1)
 |v(x + rω)|ρn−1 dωS n−1 (1) dρ dr after rescaling, where (ρ, ω) are the spherical coordinates, i.e ω are coordinates on the unit sphere. Now (2R)n = nVol(B) Z (2R)n = nVol(B) Z ∞ 0 ∞ 0 Z Z
|v(x + rω)|dωS n−1 (1) dr S n−1 (1)
|v(x + rω)| n−1 r dωS n−1 (1) dr. n−1 r S n−1 (1) Set z := x + rω, → r = |rω| = |x − z|, r n−1 dωS n−1 (1) dr = dz, and as B(x, R) ⊆ Ω (2R)n = nVol(B) Z |v(z)| dz, |x − z|n−1 (2R)n nVol(B) Z |Dr u(z)| dz. |x − z|n−1 B ⇒ ≤ Ω Claim. Z BR n |x − y|1−n |Du(y)|dy ≤ CR1− p ||Du||Lp (BR ) , for BR := B(x0 , R) ⊆ Rn . 7 ∀p > n. Proof. By the Hölder inequality, ∀q such that Z BR 1−n |x − y| |Du(y)|dy ≤ 1 q nZ x∈Ω =1 |x − y|(1−n)q dy BR ≤ sup 1 p + nZ BR =c· nZ R =c· nZ R =c· nZ · ||Du||Lp (BR ) |x0 − y|(1−n)q dy BR r (1−n)q r n−1 dr 0 n−1 r 1−p dr 0 = C(n, p)R o q1 · ||Du||Lp (BR ) o q1 · ||Du||Lp (BR ) |x − y|(1−n)q dy o 1q
n−1 =c· +1 R 1−p as o q1 p−n p o q1 · ||Du||Lp (BR ) · ||Du||Lp (BR ) n−1 +1 1−p
 q · ||Du||Lp (BR ) · ||Du||Lp (BR )
1
n−1 p n−1 p−1 p−n n−1 +1 · = +1 · =− + = . 1−p q 1−p 1−p p p p Now we can finally, combining those last two results conclude the second inclusion in our Theorem as well as the estimate therein. First, using the triangle inequality together with the first lemma we have Z Z Z |u(x) − u(y)| ≤ u(x)− 6 udx + 6 udy − u(y) ≤ 2c · B B B |Du(y)| dy |x − y|n−1 which in turn is n ≤ c(n, p)|x − y|1− p ||Du||Lp (B) once we choose a ball B = B(x, |x − y|) and apply the Claim. Since this is for any x, y ∈ Ω, and n u ∈ W 1,p (Ω) then u ∈ C 1− p (Ω), if p > n. Second and finally, we have as well 8 Z Z |u(x)| ≤ u(x)− 6 udx ≤ 2c · B Ω n |Du(y)| dy ≤ c(n, p) · diam(Ω)1− p ||Du||Lp (Ω) n−1 |x − y| 1 1 = c′ (n, p) · Vol(Ω) n − p ||Du||Lp (Ω) which gives the desired sup norm. Indeed for k ≥ 2 the Corollary follows by iterating: we get first Hölder regularity of u, then we have Du is W01,p (Ω) so we apply the first Theorem to it and get Du is Hölder and so on. 9