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
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 ).
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 ),
132 1. 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).
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.
134 1. 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.
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.
136 1. 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.
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.