Sobolev Spaces. Chapter 10

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1 Chapter 1 Sobolev Spaces We now define spaces H 1,p (R n ), known as Sobolev spaces. For u to belong to H 1,p (R n ), we require that u L p (R n ) and that u have weak derivatives of first order in L p (R n ): (1.1) j u = f j L p (R n ), where (1.1) means (1.2) ϕ u dx = ϕf j dx, ϕ C (Rn ). If u C (Rn ), we see that j u = u/, by integrating by parts, using (7.67). We define a norm on H 1,p (R n ) by (1.3) u H 1,p = u L p + j j u L p. We claim that H 1,p (R n ) is complete, hence a Banach space. Indeed, let (u ν ) be a Cauchy sequence in H 1,p (R n ). Then (u ν ) is Cauchy in L p (R n ); hence it has an L p -norm limit u L p (R n ). Also, for each j, j u ν = f jν is Cauchy in L p (R n ), so there is a limit f j L p (R n ), and it is easily verified from (1.2) that j u = f j. We can consider convolutions and products of elements of H 1,p (R n ) with elements of C (Rn ) and readily obtain identities (1.4) j (ϕ u) = ( j ϕ) u = ϕ ( j u), j (ψu) = ψ u + ψ( j u), 129

2 13 1. Sobolev Spaces and estimates (1.5) ϕ u H 1,p ϕ L 1 u H 1,p, ψu H 1,p ψ L u H 1,p + j ψ L u L p, for ϕ, ψ C (Rn ), u H 1,p (R n ). For example, the first identity in (1.4) is equivalent to ψ (x) ϕ(x y)u(y) dy dx = ψ(x) ϕ (x y) u(y) dy dx, ψ C (R n ), an identity that can be established by using Fubini s Theorem (to first do the x-integral) and integration by parts, via (7.67). If p <, u H 1,p (R n ), and (ϕ j ) is an approximate identity of the form (7.64), with ϕ j C (Rn ), then we can show that (1.6) ϕ j u u in H 1,p -norm, using (1.4) and (7.65). Given ε >, we can take j such that ϕ j u u H 1,p < ε. Then we can pick ψ C (Rn ) such that ψ(ϕ j u) ϕ j u H 1,p < ε. Of course, ϕ j u is smooth, so ψ(ϕ j u) C (Rn ). We have established Proposition 1.1. For p [1, ), the space C (Rn ) is dense in H 1,p (R n ). Sobolev spaces are very useful in analysis, particularly in the study of partial differential equations. We will establish just a few results here, some of which will be useful in Chapter 11. More material can be found in [EG], [Fol], [T1], and [Yo]. The following result is known as a Sobolev Imbedding Theorem. Proposition 1.2. If p > n or if p = n = 1, then (1.7) H 1,p (R n ) C(R n ) L (R n ). For now we concentrate on the case p (n, ). Since C (Rn ) is then dense in H 1,p (R n ), it suffices to establish the estimate (1.8) u L C u H 1,p, for u C (Rn ).

3 1. Sobolev Spaces 131 In turn, it suffices to establish (1.9) u() C u H 1,p, for u C (Rn ). To get this, it suffices to show that, for a given ϕ C ( B 1 ) with ϕ() = 1, (1.1) u() C (ϕu) L p, where v = ( 1 v,..., n v) or, equivalently, that (1.11) u() C u L p, u C ( B 1 ). In turn, this will follow from an estimate of the form (1.12) u() u(ω) C u L p (B 1 ), u C (R n ), given ω R n, ω = 1. Thus we turn to a proof of (1.12). Without loss of generality, we can take ω = e n = (,...,, 1). We will work with the set Σ = {z R n 1 : z 3/2}. For z Σ, let γ z be the path from to e n consisting of a line segment from to (z, 1/2), followed by a line segment from (z, 1/2) to e n, as illustrated in Figure 1.1. Then (with A = Area Σ) ( ) dz (1.13) u(e n ) u() = du A = u(x) ψ(x) dx, Σ γ z where the last identity applies the change of variable formula. The behavior of the Jacobian determinant of the map (t, z) γ z (t) yields (1.14) ψ(x) C x (n 1) + C x e n (n 1). B 1 Thus (1.15) B 1/2 ψ(x) q dx C 1/2 r nq+q r n 1 dr. It follows that (1.16) ψ L q (B 1 ), q < n n 1. Thus (1.17) u(e n ) u() u L p (B 1 ) ψ L p (B 1 ) C u L p (B 1 ),

4 Sobolev Spaces Γ z e n Figure 1.1 as long as p < n/(n 1), which is the same as p > n. This proves (1.12) and hence Proposition 1.2, for p (n, ). For n = 1, (1.13) simplifies to u(1) u() = 1 u (x) dx, which immediately gives the estimate (1.12) for p = n = 1. We can refine Proposition 1.2 to the following. Proposition 1.3. If p (n, ), then every u H 1,p (R n ) satisfies a Hölder condition: (1.18) H 1,p (R n ) C s (R n ), s = 1 n p. Proof. Applying (1.12) to v(x) = u(rx), we have, for ω = 1, (1.19) u(rω) u() p Cr p u(rx) p dx = Cr p n u(x) p dx. This implies B 1 (1.2) u(x) u(y) C x y 1 n/p( B r u(z) p dz) 1/p, r = x y, B r(x) which gives (1.18).

5 1. Sobolev Spaces 133 If u H 1, (R n ) and if ϕ C (Rn ), then ϕu H 1,p (R n ) for all p [1, ), so Proposition 1.3 applies. We next show that in fact H 1, (R n ) coincides with the space of Lipschitz functions: (1.21) Lip(R n ) = {u L (R n ) : u(x) u(y) K x y }. Proposition 1.4. We have the identity (1.22) H 1, (R n ) = Lip(R n ). Proof. First, suppose u Lip(R n ). Thus (1.23) h 1[ u(x + he j ) u(x) ] is bounded in L (R n ). Hence, by Proposition 9.4, there is a sequence h ν and f j L (R n ) such that (1.24) h 1 [ ν u(x + hν e j ) u(x) ] f j weak in L (R n ). In particular, for all ϕ C (Rn ), (1.25) h 1 ν ϕ(x) [ u(x + h ν e j ) u(x) ] dx ϕ(x)f j (x) dx. But the left side of (1.25) is equal to [ϕ(x (1.26) h 1 ν hν e j ) ϕ(x) ] u(x) dx ϕ u(x) dx. This shows that j u = f j. Hence Lip(R n ) H 1, (R n ). Next, suppose u H 1, (R n ). Let ϕ j (x) = j n ϕ(jx) be an approximate identity as in (1.6), with ϕ C (Rn ). We do not get ϕ j u u in H 1, - norm, but we do have u j = ϕ j u bounded in H 1, (R n ); in fact, each u j is C, and we have (1.27) u j L K 1, u j L K 2. Also u j u locally uniformly. The second estimate in (1.27) implies (1.28) u j (x) u j (y) K 2 x y, since u j (x) u j (y) = 1 (x y) u(tx + (1 t)y) dt. Thus in the limit j, we get also u(x) u(y) K 2 x y. This completes the proof. We next show that, when p [1, n), H 1,p (R n ) is contained in L q (R n ) for some q > p. One technical tool which is useful for our estimates is the following generalized Hölder inequality.

6 Sobolev Spaces Lemma 1.5. If p j [1, ], p 1 j = 1, then (1.29) u 1 u m dx u 1 L p 1 (M) u m L pm (M). M The proof follows by induction from the case m = 2, which is the usual Hölder inequality. Proposition 1.6. For p [1, n), (1.3) H 1,p (R n ) L np/(n p) (R n ). In fact, there is an estimate (1.31) u L np/(n p) C u L p for u H 1,p (R n ), with C = C(p, n). Proof. It suffices to establish (1.31) for u C (Rn ). Clearly (1.32) u(x) j u dy j, where the integrand, written more fully, is j u(x 1,..., y j,..., x n ). (Note that the right side of (1.32) is independent of x j.) Hence (1.33) u(x) n/(n 1) n ( ) 1/(n 1). j u dy j j=1 We can integrate (1.33) successively over each variable x j, j = 1,..., n, and apply the generalized Hölder inequality (1.29) with m = p 1 = = p m = n 1 after each integration. We get { n (1.34) u L n/(n 1) j=1 R n } 1/n j u dx C u L 1. This establishes (1.31) in the case p = 1. We can apply this to v = u γ, γ > 1, obtaining (1.35) u γ L n/(n 1) C u γ 1 u L 1 C u γ 1 L p u L p. For p < n, pick γ = (n 1)p/(n p). Then (1.35) gives (1.31) and the proposition is proved.

7 1. Sobolev Spaces 135 There are also Sobolev spaces H k,p (R n ), for each k Z +. By definition u H k,p (R n ) provided (1.36) α u = f α L p (R n ), α k, where α = α 1 1 αn n, α = α 1 + +α n, and, as in (1.2), (1.36) means (1.37) ( 1) α α ϕ x α u dx = ϕf α dx, ϕ C (R n ). Given u H k,p (R n ), we can apply Proposition 1.6 to estimate the L np/(n p) -norm of k 1 u in terms of k u L p, where we use the notation (1.38) k u = { α u : α = k}, k u L p = α =k and proceed inductively, obtaining the following corollary. Proposition 1.7. For kp < n, (1.39) H k,p (R n ) L np/(n kp) (R n ). α u L p, The next result provides a generalization of Proposition 1.2. Proposition 1.8. We have (1.4) H k,p (R n ) C(R n ) L (R n ) for kp > n. Proof. If p > n, we can apply Proposition 1.2. If p = n and k 2, since it suffices to obtain an L bound for u H k,p (R n ) with support in the unit ball, just use u H 2,n ε (R n ) and proceed to the next step of the argument. If p [1, n), it follows from Proposition 1.6 that (1.41) H k,p (R n ) H k 1,p 1 (R n ), p 1 = np n p. Thus the hypothesis kp > n implies (k 1)p 1 > kp > n. Iterating this argument, we obtain H k,p (R n ) H l,q (R n ), for some l 1 and q > n, and again we can apply Proposition 1.2.

8 Sobolev Spaces Exercises 1. Write down the details for the proof of the identities in (1.4). 2. Verify the estimates in (1.14). Hint. Write the first integral in (1.13) as 1/A times Σ 1 v + (z) u(tz, 1 2 t) dt dz + Σ 1 v (z) u(tz, 1 1 2t) dt dz, where v ± (z) = (±z, 1/2). Then calculate an appropriate Jacobian determinant to obtain the second integral in (1.13). 3. Suppose 1 < p <. If τ y f(x) = f(x y), show that f belongs to H 1,p (R n ) if and only if τ y f is a Lipschitz function of y with values in L p (R n ), i.e., (1.42) τ y f τ z f L p C y z. Hint. Consider the proof of Proposition 1.4. What happens in the case p = 1? 4. Show that H n,1 (R n ) C(R n ) L (R n ). Hint. u(x) = 1 n u(x + y) dy 1 dy n. 5. If p j [1, ] and u j L p j, show that u 1 u 2 L r provided 1/r = 1/p 1 + 1/p 2 and (1.43) u 1 u 2 L r u 1 L p 1 u 2 L p 2. Show that this implies (1.29). 6. Given u L 2 (R n ), show that (1.44) u H k,2 (R n ) ( 1 + ξ ) kû L 2 (R n ). 7. Let f L 1 (R), and set g(x) = x f(y) dy. Continuity of g follows from the Dominated Convergence Theorem. Show that (1.45) 1 g = f.

9 1. Sobolev Spaces 137 Hint. Given ϕ C (R), start with (1.46) dϕ x g(x) dx = ϕ (x)f(y) dy dx, dx and use Fubini s Theorem. Then use y ϕ (x) dx = ϕ(y). Alternative. Write the left side of (1.46) as lim h 1 [ϕ(x ] + h) ϕ(x) g(x) dx = lim h h and use (4.64). x+h f(y)ϕ(x) dy dx, 8. If u H 1,p (R n ) for some p [1, ) and j u = on a connected open set U R n, for 1 j n, show that u is (equal a.e. to a) constant on U. Hint. Approximate u by (1.6), i.e., by u ν = ϕ ν u, where ϕ ν C (Rn ) has support in { x < 1/ν}, ϕν dx = 1. Show that j (ϕ ν u) = on U ν U, where U ν U as ν. More generally, if j u = f j C(U), 1 j n, show that u is equal a.e. to a function in C 1 (U). 9. In case n = 1, deduce from Exercises 7 and 8 that, if u L 1 loc (R), (1.47) 1 u = f L 1 (R) = u(x) = c + for some constant c. x x f(y) dy, a.e. x R, 1. Let g H 2,1 (R), I = [a, b], and f = g I. Show that the estimate (9.75) concerning the trapezoidal rule holds in this setting.