Notes on Sobolev Spaces

Peter Lindqvist Norwegian University of Science and Technology

1 Lp-SPACES

1.1 Inequalities For any measurable function u : A [ , ], A Rn, we deﬁne → −∞ ∞ ∈ 1 p p u p = u p, A = u(x) dx k k k k  | |  ZA    and, if this quantity is ﬁnite, we say that u Lp(A). In most cases of interest p 1.   For p = we set ∈ ≥ ∞ u = u , A = ess sup u(x) . k k∞ k k∞ x A | | ∈ The essential supremum is the smallest number M such that u(x) M for a.e. x A. | | ≤ ∈ For example1, if u : [a, b] Rn, is continuous, then it is easy to see that → 1 b p u(x) p dx max u(x) .  | |  p−→ a x b| | Za  →∞ ≤ ≤   The fundamental inequalities    1 1 uv u v , + = 1, HOLDER¨ k k1 ≤k kpk kq p q u + v u + v , MINKOWSKI k kp ≤k kp k kp 1 An interesting application of this fact in connection with the heat equation ut = uxx is given in [BB, Ch.24, pp 145-150].

1 where p 1, can be derived in many ways. For example, taking the “elementary inequality”≥ ap bq ab + (a, b 0, p + q = pq) YOUNG ≤ p q ≥ for granted, we obtain the H¨older inequality (choose a = u(x)/ u p, b = v(x)/ u q and integrate the resulting inequality). The H¨older inequality impliesk k the Minkowskik k inequality. Remarks: 1) The special case

uv dx u2 dx v2 dx CAUCHY | | ≤ ZA tvZA tvZA is called the Cauchy inequality. 2) If 1 > p > 0, then the Minkowski inequality is reversed for positive functions! That is why one usually has p 1. ≥ 3) If mes (A) < , then the H¨older inequality shows that Lp1 (A) Lp2 (A), if ∞ ⊂ p1 p2. This is not true in general, if mes (A) = . (Find a simple ex.!) However,≥ if 1 p < p < p , then the H¨older inequality∞ implies that ≤ 1 2 ≤∞ λ 1 λ u p,A u u − k k ≤k kp1,Ak kp2,A where 1 λ 1 λ = + − . p p1 p2 That is, if u Lp1 Lp2 , then u Lp for all intermediate p. ∈ ∩ ∈ 4) Suppose that 0 < mes (A) < . The function ∞ 1 p 1 φ(p) = u(x) p dx , 0 < p < , mes (A) | |  | | ∞  ZA     1  φ(0) = exp log u(x) dx , mes (A) | |  ZA   =  + =  φ( ) ess inf u(x) , φ( ) ess sup u(x) −∞ A | | ∞ A | |

2 is increasing2 as ≦ p ≦ + . −∞ ∞ Let us ﬁnally mention that

1 1 Λ u(x) dx Λ(u(x)) dx JENSEN mes (A)  ≤ mes (A) ZA ZA     whenever Λ: R R is a convex function (0 < mes (A) < , u : A [ , ] is measurable). → ∞ → −∞ ∞ Example: 1 1 u(x) dx u(x) e 0 e dx R ≤ Z0 p 1 1 u(x) dx u(x) p dx mes (A) | |  ≤ mes (A) | |  ZA  ZA     Remark: Discrete versions of the above inequalities are

| {z }p p p p ak + bk |p a{zk + p } ak . k | | ≤ k | | k | | sX sX sX p p = Especially, lim k ak sup ak . p →∞ | | k | | Example: Draw thepP curves p x p + y p = 1 (0 < p ≦ ) in the xy-plane. | | | | ∞ 2J. Moser’s celebrated methodp to relate the maximum and the minimum of the solution to a partial diﬀerential equations is based on this.

3 For p = 2 we have the parallelogram3 law

u + v 2 + u v 2 = 2 u 2 + 2 v 2. k k2 k − k2 k k2 k k2

Remark: So-called reverse H¨older inequalities like

1 p 1 C u u mes (Q) | | ≤ mes (Q) | |  ZQ  ZQ     valid for every cube with some ﬁxed p > 1 (!) play a central role for solutions u to certain partial diﬀerential equations. (Such an inequality cannot hold for arbitrary functions.) The method is due to F. W. GEHRING.

1.2 (Strong) convergence in Lp

n p Let A R and ﬁx p 1. Then p deﬁnes a semi-norm in the vector space L (A), i.e., ∈ ≥ k k

(i)’ 0 u < , u Lp(A) ≤k kp ∞ ∈ (ii) λu = λ u , u Lp(A), < λ < k kp | |k kp ∈ −∞ ∞ (iii) u + v u + v , u, v Lp(A). k kp ≤k kp k kp ∈

However, this is, strictly speaking, not a norm, the reason being that u p,A = 0 u(x) = 0 for a.e. x A. We agree to say that u = v in Lp(A), if u(xk) k= v(x) for⇐⇒ a.e. x A. With this convention∈ is a norm, i.e. (i)’ can be replaced by ∈ k kp (i) 0 u < , and u = 0 u = 0 in Lp. ≤k kp ∞ k kp ⇐⇒ (Strictly speaking, this Lp-space consists of equivalence classes of functions, but here there is no point in maintaining this distinction.)

Theorem 1 (RIESZ-FISCHER) The Lp-spaces, p 1, are Banach spaces. That p ≥ p is, if u1, u2,... is a Cauchy sequence in L (A), then there is a function u L (A) such that u u 0, as k . ∈ k k − kp,A → →∞ 3The counterpart for general exponents p is more involved. A pair of inequalities (CLARK- SON’s inequalities) will replace the parallelogram law [A, p.37].

4 p Proof: Suppose that u1, u2,... is a Cauchy sequence in L (A), i.e., given ε > 0 there is an index Nε such that uk u j p < ε, when k, j Nε. Let us consider the case p , k − k ≥ ∞ (the case p = is simpler, and its proof will be skipped). We can construct indices ∞ 1 n < n <... such that ≤ 1 2 1 = un un + p < (k 1, 2, 3,...). k k − k 1 k 2k Deﬁne N = = = ∞ gN (x) un un + , g(x) lim gN (x) un un + k k 1 N k k 1 = | − | = | − | Xk 1 →∞ Xk 1 Then 0 g (x) g (x) ... g(x) ≦ ≤ 1 ≤ 2 ≤ ≤ ∞ and p p p p g = lim gN = lim gN lim gN | | N N | | ≤ N | | ZA ZA →∞ ZA →∞ →∞ ZA

1 + 1 + + 1 p by Fatou’s lemma. By the construction gN p 2 4 ... 2N < 1 and so g 1. Thus k k ≤ A | | ≤ g(x) < + for a.e. x A. This means that the series un (x) un (x) + un (Rx) un (x) +... ∞ ∈ | 2 − 1 | | 3 − 2 | converges for a.e. x A. So does a fortiori, the series ∈ + ∞ un2 (x) (unk+1 unk ). = − Xk 1

The partial sums of this series are plainly un2 (x), un3 (x), un4 (x),... Hence

u(x) = lim un (x) k k →∞ exists and is ﬁnite for a.e. x A (If you insist on having an everywhere deﬁned function, ∈ set u(x)=0 in a subset of measure zero.) By Fatou’s lemma (again!)

p = p p p u(x) u j(x) dx lim unk (x) u j(x) dx lim unk (x) u j(x) dx ε | − | k | − | ≤ k | − | ≤ ZA ZA →∞ →∞ ZA whenever j > Nε. This shows that u u j p,A 0 as j . k − k → →∞ The proof yields information about pointwise behaviour.

Corollary 2 Suppose that u uinLp(A), i.e., u u 0. Then there is a k → k k − kp,A → subsequence that converges a.e. in A : u(x) = limun (x) for a.e. x A. k k →∞ ∈

5 1.3 Thedualof Lp

If X is any Banach space, its dual X∗ is the collection of all continuous linear functions (functionals) l : X R . The norm of l is defined as → = l(x) = l X∗ sup | | sup l(x) k k x X x X x X 1| | ∈ k k ! k k ≤ and the continuity is equivalent to l < . Note that k kX∗ ∞ l(x) l x . | |≤k kX∗ k k Let 1 p < and fix a function g Lq(A). Then ≤ ∞ ∈ l( f ) = f (x)g(x) dx ( f Lp(A)) ∈ ZA is well-defined and linear. By Hölder’s inequality l( f ) f p g q. Hence l | |≤kq 2 k k k k k∗ ≤ g q. Here equality is attained for the choice f = q − g. (The case q = is slightlyk k different.) Thus | | ∞ l = g q. k k∗ k k The essential fact here is that virtually all continuous functionals in Lp (p , ) are of this form! ∞ Theorem 3 (F. RIESZ’ representation thm) Let l : Lp(A) R be a continuous linear functional, 1 p < . Then there is a unique g Lq(→A) such that ≤ ∞ ∈ l( f ) = f (x)g(x) dx

ZA p for all f L (A). Moreover, l = g q,A. ∈ k k∗ k k Proof: The Radon-Nikodym theorem is used in most proofs of the representation theorem. A more direct proof in the one dimensional case is given in [R. p.121]. Remark:

1) There are continuous linear functionals in L∞(A) that do not have this simple form. 2) One says that Lq is the dual of Lp (1 p < ). ≤ ∞

6 1.4 Weak convergence in Lp Weak convergence in Lp is important for many applications, for example, it leads to the existence theorems for partial diﬀerential equations. The so-called direct methods in the Calculus of Variations are based on this concept. p Suppose that u1, u2, u3,... are functions in L . We say that uk u weakly in Lp, if → liml(uk) = l(u) k →∞ for every continuous linear functional l : Lp R . Using Riesz’ representation theorem we can state the deﬁnition in a more convenient→ form.

p Deﬁnition 4 Let 1 p < . Suppose that u1, u2,... and u are functions in L (A). We say that u u≤ weakly∞ in Lp(A), if k → q lim ukv dx = uvdx foreachv L (A). k →∞ ∈ ZA ZA

Remark:

1) Sometimes the notation uk ⇀ u is used to indicate weak convergence.

2) Strong convergence implies weak convergence: if uk u p 0, then p k − k → uk ⇀ u weakly in L . 3) The weak limit is unique, that is, unique in Lp.

2 Example: The functions un(x) = sin(nx) converge weakly in L ([0, 2π]) to zero. By the Riemann-Lebesgue lemma

2π lim v(x) sin(nx) dx = 0 n →∞ Z0 for every v in L2([0, 2π]). (This follows easily from Bessel’s inequality.) Example: (Warning!) Suppose that f : [a, b] R is a bounded measurable function, for example, assume 0 f (x) 2. Then→ there is a sequence of functions v : [a, b] 0, 2 converging weakly≤ in≤L2([a, b]) to f. Observe that v (x) = 0 or n →{ } n = 2 ( vn takes no other values!!) This is not diﬃcult to realise in the special case f (x) 1. ≡ 7 R = 3 = Example: Deﬁne ui :]0, 1[ by ui iχ]0,i− [ for i 1, 2, 3,... Then ui 0 p → ε /3 → strongly in L ([0, 1]), if 1 p < 3. We have ui 3+ε = i for every ε > 0. As we shall see, weakly convergent≤ sequences arek boundedk in Lp-norm. Therefore p u1, u2,... does not converge weakly in L (]0, 1[), if p > 3. —Show that ui ⇀ 0 weakly in L3(]0, 1[)! (This convergence is not strong.) p If ui ⇀ u weakly in L (A), then there is a constant M such that ui p,A M for all i = 1, 2, 3,... This follows from the following lemma. k k ≤ ≤ ∞ p Lemma 5 Suppose that u1, u2,... are functions in L (A), 1 < p < . If the sequence u is unbounded, there is a w Lq(A) such that ∞ k ikp,A ∈

lim ui (x)w(x) dx = + k k →∞ ∞ ZA for some subsequence. Proof: The function w is constructed and written down in [S, pp.25-28]. The most important fact about Lp-spaces seems to be the following weak compactness property. (Not valid for p = 1).

p Theorem 6 (Weak Compactness) Let u1, u2,... be functions in L (A), 1 < p < . If there is a constant M such that ui p,A M for each index i, then there exists ∞ p k k ≤ p a function u L (A) such that uik ⇀ u weakly in L (A) for some subsequence. Moreover ∈

ui p,A lim uik p,A. k k ≤ k k k →∞ Proof: Advanced books on Functional Analysis usually contain a proof. For example, the above mentioned book of Sobolev [S] gives a proof on pp. 29-30. See also [A]. Usually, the lower semicontinuity comes as a by-product of the proof, but the following simple argument also yields this property. Since x p is a convex function of x, we have | |

8 Remarks:

1) The theorem is not true for p = 1.

2) The existence of solutions to partial diﬀerential equations is often a direct consequence of the weak compactness.

3) Let ui p M for i = 1, 2, 3,... According to the BANACH-SAKS theorem k k ≤ p there are indices i1 < i2 < i3 <...and a function u in L such that u + u + ... + u i1 i2 iν u 0 ν − p −→

+ as ν , that is, the arithmetic means converge strongly. (The Banach- Saks→ theorem∞ is valid also for p = 1.)

1.5 Approximation in Lp and some other things The essential fact is that the functions in Lp can be approximated in the Lp-norm by smooth functions, if 1 p < . These are constructed as convolutions. Let us ﬁrst state some auxiliary results.≤ ∞

Lemma 7 (“Continuity in the Lp-norm”) Let 1 p < . If f Lp(A), then ≤ ∞ ∈ lim f (x + h) f (x) p dx = 0, h 0 | − | → ZA where f is regarded as 0 outside A.

The proof is not quite simple. Usually one uses the theorem of Lusin, valid for measurable functions.

Theorem 8 (Lusin) Let f : A [ , ] be a measurable function that is ﬁnite a.e. → −∞ ∞ in A. Suppose that (A) < . Given ε > 0, there is a compact set Kε A such that the ∞ ⊂ restriction f K is continuous and mes(A Kε) <ε. | ε \ Proof: See, [EG, p.15]. If A is open, then Lusin’s theorem can be combined with the extension theorem of Urysohn-Tietze [R, p.148].

9 Theorem 9 (Urysohn-Tietze) Suppose that A is open and mes (A) < . Let K A be a ∞ ⊂ compact set. If f : K R is a continuous function, then there is a function ϕ C (A) → ∈ 0 such that ϕ(x) = f (x), when x K. Moreover, max ϕ = max f ∈ A | | A | | p Lemma 10 C0(Ω) is dense in L (Ω), 1 p < . Here Ω is open and C0(Ω) denotes all continuous functions with compact≤ support∞ on Ω. In other words, if u Lp(Ω), then there are functions ϕ C (Ω) such that ∈ k ∈ 0 lim u ϕk p,Ω = 0. k →∞k − k Remarks: 1) Of course, this implies that C(Ω) is dense in Lp(Ω).

2) Functions in L∞(Ω) cannot, in general, be uniformly approximated by continuous functions. 3) If ϕ : Rn R is continuous, then the closure in Rn of the set where ϕ(x) , 0 is called the→ support of ϕ. Thus supp(ϕ) = x Rn ϕ(x) , 0 . { ∈ | } If supp(ϕ) is compact (it is closed by deﬁnition) and if supp(ϕ) Ω, then ⊂ we say that ϕ C0(Ω). In this case the distance between supp(ϕ) and the boundary ∂Ω is∈ positive.

4) A deep result for Lp-functions is related to the Lebesgue points. Suppose that f Lp (Rn), that is, f Lp(B) for every ball B in Rn . Then ∈ loc ∈ 1 lim f (x) f (y) p dy = 0 r o+ mes (B(x, r)) | − | → B(Zx,r) for a.e. x.

Deﬁne 1 Ce− 1 x 2 , x < 1 ρ(x) = −| | | |  0, x 1 | | ≥  and choose the constant C > 0 such that ρ(x) dx = 1. “Friedrichs’ molliﬁer” is  Rn R C ε2 − ε2 x 2 1 x n e −| | , when x < ε, ρε(x) = ρ = ε | | εn ε   0, when x ε.  | | ≥   10 The constant C depends only on the dimension n. Now we have

ρε(x) dx = 1, ε > 0. (1) RZn Rn Observe that ρε C0∞( ). The support of ρε is the closed ball x ε. The convolution∈ | | ≤

u (x) = (ρ u)(x) = ρ (x y)u(y) dy ε ε ∗ ε − RZn

1 Rn Ω is well-defined for u Lloc( ). If u is defined only in the domain , then we regard u as extended to∈ zero outside Ω : u(x) = 0, when x Rn Ω. Hence we can 1 Ω ∈ \ calulate the convolution uε for any u in Lloc( ). Rn 1 Rn Observe that uε is always a smooth function: uε C∞( ), if u Lloc( ). For differentiation we have the rule ∈ ∈

Dαu = (Dαρ ) u. (2) ε ε ∗

Here α = (α1, α2,...,αn) and

α1+α2+...+αn α = ∂ D α1 α2 αn . ∂x1 ∂x2 ...∂xn If u was deﬁned in the domain Ω, then the formula (2) holds for the original (unextended) u at all points x with dist(x, ∂Ω) > ε. Analogously, if supp u Ω, ⊂ then u C∞(Ω), when ε < dist(supp u, ∂Ω). ε ∈ 0 p p Lemma 11 If u L (Ω), 1 p < , then uε L (Ω) and ρε u p u p. Moreover ∈ ≤ ∞ ∈ k ∗ k ≤ k k lim uε u p = 0. ε 0+ → k − k Lemma 12 If u C(Ω) and if K Ω is compact then ∈ ⊂⊂ + max uε(x) u(x) 0 as ε 0 . x K ∈ | − | → → In other words, the convergence u u is uniform on compact subsets. ε →

11 Proof: The locally uniform convergence for a continuous u follows form

u(x) u (x) = u(x) ρ (x y)u(y) dy − ε − ε − Z = (u(x) u(y))ρ (x y) dy − ε − x Zy <ε | − | u(x) uε(x) max u(x) u(y) 1 (ε < dist(x, ∂Ω)) | − | ≤ y | − |· x y ≦ε | − | in accordance with Weierstrass theorem ( a function that is continuous on a compact set us uniformly continuous). If u Lp(Ω), then ∈ 1 1 p q u ρ (x y) u(y) dy ρ (x y) u(y) p dy ρ (x y) dy | ε| ≤ ε − | | ≤ ε − | | ε − Z (Z ) (Z ) =1 and so | {z }

u (x) p dx ρ (x y) u(y) p dy dx | ε | ≤ ε − | | Z Z Z ! = u(y) p ρ (x y) dx dy = u(y) p dy. | | ε − | | Z Z ! Z =1

This proves the contraction |uε p {zu p. } p k k ≤k k For the L -convergence, we ﬁrst note that if ϕ C0(Ω), then it is easily seen that ∈

p p ϕ(x) ϕε(x) dx sup ϕ(x) ϕ(y) (ε < dist(supp ϕ, ∂Ω)) | − | ≤ x y ≦ε| − | Z | − | p and so ϕ ϕε p 0. (Doing some more work, we could replace C0 by C L .) Now k − k → ∩ =(ϕ u)ε − u u u ϕ + ϕ ϕ + ϕ u k − εkp ≤k − kp k − εkp k ε − ε kp

ϕ u p z≤k}|− {k and so | {z } lim u uε p 2 u ϕ p. ε 0+ → k − k ≤ k − k 12 Ω p Ω Ω Since C0∞( ) is dense in L ( ), we can choose ϕ C0( ) so that u ϕ p is as ∈ p k − k small as we please. This concludes theproof for the L -convergence u uε p 0. k − k → As an application we mention the Variational Lemma. 1 Ω Ω Rn Lemma 13 (Variational Lemma) Let u Lloc( ), denoting an open set in . If ∈ u(x)ϕ(x) dx = 0 ZΩ whenever ϕ C∞(Ω), then u = 0 a.e. in Ω. ∈ 0 Proof: Takex ¯ Ω and choose ε so small that 0 < ε < dist(x ¯, ∂Ω). Then ρ (x ¯ x) will do as∈ a test function so that ε − u (x ¯) = u(x)ρ (x ¯ x) dx = 0. ε ε − Z Let B be any closed ball in Ω, i.e. B Ω. Then ⊂⊂ u = u u 0 k k1,B k − εk1,B → as ε 0+. Hence u = 0 a.e. in B and hence a.e. also in Ω. → Remark: The lemma is fundamental in the Calculus of Variations. If u is continuous, the proof is more elementary and u 0 in this case. We shall need the lemma to establish that the Sobolev derivatives≡ of a function are unique up to sets of measure zero. Example: (Hermann WEYL) Let u L1 (Rn) and suppose that ∈ loc u(x) ∆ϕ(x) dx = 0 Z Rn whenever ϕ C0∞( ). Then u has a continuous representative which is a harmonic ∈ 2 ∆ = ∂2u + + ∂2u = function, i.e. the continuous u belongs to C and u 2 ... 2 0. (In fact, ∂x ∂xn n 1 u C∞(R ).) ∈ Take ϕ(x) = ρ (x z) for some ﬁxed z. Then ∆u (z) = 0. Indeed, ε − ε u (z) = (ρ u)(z) = ρ (z x) u(x) dx ε ε ∗ ε − Z ∆u (z) = (∆ρ u)(z) = ∆ ρ (z x) u(x) dx ε ε ∗ z ε − Z = [∆ ρ (z x)]u(x) dx = 0 x ε − Z 13 by assumption. Note that

2 2 ∂ u = ∂ u 2 ρε(z x) 2 ρε(z x) ∂xk − ∂zk − by direct calculation (ρ (z x) is a function of z x 2). Since z was arbitrary, ε − | − | ∆uε 0. Thus the function uε is harmonic. One≡ does not seem to get any further without using some deeper property of harmonic functions. By the mean value property 1 u (x) = u (y) dy (Actually u = u!) ε mes (B(x, r)) ε ε B(Zx,r)

Since u u in L1(B(x, r)), we have ε → 1 lim uε(x) = u(y) dy. ε 0+ mes (B(x, r)) → B(Zx,r)

For a.e. x, u(x) = lim uε(x), at least when ε approaches zero through a subsequence ε1,ε2,ε3,... (Corollary 2). Redefining u in a set of measure zero, we get a function that satifies the mean value property. Hence (the redefined) u is harmonic in Rn .

14 2 SOBOLEV SPACES

The situation with the derivatives belonging to some Lp-space was studied by Tonelli, B. Levi, Sobolev, Kondrachev et consortes. The corresponding spaces are named after Sobolev.

2.1 Wm,p and Hm,p Throughout this chapter Ω denotes a domain or an open subset of Rn . Suppose u C1(Ω), where Ω. Then integration by parts yields ∈ ∂ϕ ∂u u(x) dx = ϕ(x) dx ∂xk − ∂xk ZΩ ZΩ Ω when ϕ C0∞( ). This formula is the starting point for the deﬁnition of weak (distributional,∈ generalized) partial derivatives.

Deﬁnition 14 Assume u L1 (Ω) . We say that v L1 (Ω) is the weak partial ∈ loc k ∈ loc derivative of u with respect to xk in Ω if

∂ϕ u(x) dx = vk(x)ϕ(x) dx ∂xk − ZΩ ZΩ

1 ∂u ∂u ∂u for all ϕ C (Ω). We write vk = Dku = and u = ,..., , provided the 0 ∂xk ∂x1 ∂xn weak derivatives∈ exists. ∇ The weak partial derivative is uniquely deﬁned a.e. in Ω (by the Variational lemma). Notice the requirement that the weak derivative is a function (to which Lebesgue’s theory applies), not merely a distribution. It is suﬃcient to consider Ω all ϕ C0∞( ) in the integration-by-parts formula. We∈ say that the function u belongs to the SOBOLEV SPACE W1,p(Ω), if u Lp(Ω) and the weak partial derivatives ∈

∂u ∂u ∂u , ,..., ∂x1 ∂x2 ∂xn

p Ω 1,p Ω exist and belong to L ( ) . Here 1 p . We say that u Wloc ( ) if 1,p ≤ ≤ ∞ ∈ u W (U) for each open U Ω . (In this case we may have u Ω = ∈ ⊂⊂ k kp, ∞ 15 1,p or u Ω = .) For u W (Ω) we deﬁne the norm k∇ kp, ∞ ∈ n ∂u u W1,p(Ω) = u 1,p,Ω = u p,Ω + ∂x Ω k k k k k k k= k p, X1

Any equivalent norm, as

1 p u = [ u p + u p] dx , k k Ω | | |∇ | (Z ) 1,p will do. We say that uk u (strongly) in W (Ω), if uk u W1,p(Ω) 0. Provided with this norm (or any equivalent→ norm) W1,p(Ω) is ak Banach− k space.→ Example: Let u(x) = x for < x < . It is easily veriﬁed that | | −∞ ∞ 1, x > 0 v(x) = 1, x < 0 ( − is the weak derivative of u. Show that ϕv dx = uϕ′ dx! (We can set v(0) = 0 or v(0) = A or even v(0) = . Sets of measure zero− do not count.) ∞ R R Example: Let v(x) = 1, when x 0 and v(x) = 1, when x < 0. Then the weak derivative of v does not exist in≥ ( 2, 2), for example.− The origin is the crucial point. (Dirac’s delta is not a function.− )

Higher weak derivatives are deﬁned in a similar way. If α = (α1, α2,...,αn), β = (β1,β2,...,βn) are multi-indices, we write

α = α1 + α2 + ... + αn, α! = α1!α2! ...αn! | | n α α! αk = = xα = xα1 xα2 ... xαn β (α β)!β! β 1 2 n ! k=1 k ! − Y + + + ∂α1 α2 ... αn α = α1 αn = D D1 ... Dn α1 α2 αn ∂x1 ∂x2 ...∂xn In this notation

α α α β β D (ϕψ) = D − ϕ D ψ, LEIBNIZ’ RULE β · 0≦ β ≦ α X| | | | ! hα ϕ(x + h) = ϕ(x) + Dαϕ(x) . TAYLOR’S FORMULA · α! 0< α X| | 16 If u L1 (Ω) and v L1 (Ω) are related by ∈ loc α ∈ loc

α α u(x) D ϕ(x) dx = ( 1)| | v (x)ϕ(x) dx − α ZΩ ZΩ Ω = α for all ϕ C0∞( ), we write vα D u. This is a weak derivative of order α . We say∈ that u Wm,p(Ω), if u Lp(Ω) and if Dαu exists and belongs to| L| p(Ω) for each multi-index∈ α with α k∈. Provided with the norm | | ≤ 1 p α p u Wm,p(Ω) = u m,p,Ω = D u dx , k k k k  | |  α m ZΩ  |X|≤   m,p   the Sobolev space W (Ω) isa Banach space. The termwithindex α = (0, 0,..., 0) 1 p is interpreted as u p dx . The integer m counts the order of the highest weak | | Ω  derivative. R    Theorem 15 Assume that u Wm,p(Ω) for some 1 p < . Then there exists a ∈ ≤ m,p∞ sequence of functions ϕ C∞(Ω) such that ϕ u in W (Ω), i.e., k ∈ k → k u ϕ m,p Ω →∞ 0. k − kkW ( ) → About the proof. The case Ω = Rn is relatively simple. The general case was is in [A, pp.52-53]. Let us just mention that the proof4 uses the partition of unity.

Partition of unity The partition of unity is frequently used in the theory of distributions. Suppose that Ω ⊂ ∞ n U j where each U j is open. Then there are functions ϕ C∞(R ) such that: j∪=1 ∈ 0 1) 0 ϕ 1. ≤ ≤ 2) ϕ1(x) + ϕ2(x) + ... = 1 at each point x in Ω .

3) If K Ω is any compact set, then only ﬁnitely many of the functions ϕk are not ⊂ identically zero in K.

4This was proved by N. Meyers and J. Serrin in 1964. However, this was not, as it were, the ﬁrst proof.

17 4) Each ϕ j C (Uk) for some k = k( j). ∈ 0∞ m,p m We define H (Ω) as the completion of C∞(Ω) in the norm m,p,Ω. Often H means Hm,2, p = 2 being the most important special case. Tokk be more precise, p m,p u L (Ω) belongs to the space H (Ω), if there are functions ϕi C∞(Ω) such ∈ p α p ∈ that ϕi u inL (Ω) and D ϕi is a Cauchy sequence in L (Ω) for each multi-index α, α →m. |It|is ≤ not difficult to see that Dαu exists, α m, if u Hm,p(Ω) . Moreover, Dαϕ Dαϕ in Lp(Ω) . The central result is [A,| | ≤pp.52-53]:∈ i → Theorem 16 Hm,p(Ω) = Wm,p(Ω), 1 p < , m = 1, 2, 3,... ≤ ∞ There is also a characterization of the Sobolev space in terms of integrated difference quotients. To this end, let ei = (0,..., 1,..., 0) denote the unit vector in the ith direction. If u W1,p(Ω), then ∈ u(x + he ) u(x) p i − dx D u p dx h ≤ | i | ΩZ ZΩ ′ for any subdomain Ω′ Ω, when 0 < h < dist(Ω′, ∂ Ω). (For smooth functions ⊂⊂ϕ this follows from the identity

h ϕ(x + he ) ϕ(x) 1 i − = D ϕ(x ,..., x + t,..., x ) dt h h i 1 i n Z0 and the general case follows by approximation.) Remark: For smooth functions in convex domains the formula 1 1 d ϕ(y) ϕ(x) = ϕ(x + t(y x)) dt = (y x) ϕ(x + t(y x)) dt − dt − − · ∇ − Z0 " # Z0 is the source of many useful inequalities. Theorem 17 Let u Lp(Ω), 1 < p < . Suppose that for any subdomain ∈ ∞ Ω′ Ω we have ⊂⊂ u(x + he ) u(x) p i − dx K p < h ≤ ∞ ΩZ ′ whenever 0 < h < dist(Ω′ , ∂ Ω). Then the weak derivative Diu exists and Diu p,Ω K. k k ≤

18 Proof: See [GT, p.169]. In conclusion, there are three5 equivalent deﬁnitions for the Sobolev space: I The deﬁnition based on the integration-by-parts formula.

II The deﬁnition based on approximation by smooth functions with respect to the Sobolev norm.

III The characterization in terms of the integrability of difference quotients. Often, II is used to prove auxiliary inequalities and imbedding theorems, III is sometimes used to prove the existence of weak derivatives. But I is the Main Definition. When we say that u W1,p(Ω) is, for example, continuous, we mean that there exists a continuous function∈ ϕ such that u(x) = ϕ(x) for a.e. x Ω . Thus u can be made continuous after a redefinition in a set of measure zero. ∈(Remember that, in general, functions in Lp are defined only almost everywhere.)

1,p Ω 2.2 The Space W0 ( ) We wish to introduce functions with boundary values zero in Sobolev’s sense. 1,p Remember that functions in W (Ω) can be approximated by functions in C∞(Ω) with respect to the norm

n ∂u u 1,p,Ω = u p,Ω + . k k k k = ∂xk Xk 1

If the approximation can be done using merely functions with compact support Ω, 1,p Ω then the function itself is in a closed subspace denoted by W0 ( ).

Deﬁnition 18 Suppose that u W1,p(Ω). We say that u W1,p(Ω), if, given ε > 0, ∈ ∈ 0 there is a function ϕ C∞(Ω) such that u ϕ Ω <ε. ε ∈ 0 k − εk1,p, 1,p Ω Ω Hence W0 ( ) is the closure of C0∞( ) with respect to the corresponding Sobolev norm. Clearly,

C p(Ω) W1,p(Ω) W1,p(Ω). 0 ⊂ 0 ⊂ 5If Ω is the whole space Rn, a furtherdeﬁnition is possible. It is based on the Fourier transform. There is also a more advanced theory based on Bessel potentials.

19 1,p Ω Ω There are functions in W0 ( ), that do not have compact support in . (Ex- ample: n = 1, p = 2,⊂Ω =]0, π[, u(x) = sin x, supp u = [0, π] Ω .) If u C(Ω) W1,p(Ω) and if ⊃ ∈ ∩ lim u(x) = 0 x ξ → x Ω ∈ Ω 1,p Ω at each boundary point ξ ∂ , then u W0 ( ). However, there are continu- ∈ 1,p ∈Ω ous functions in the Sobolev space W0 ( ) that do not have “the right boundary values” zero in the classical sense: Example: Ω = x R3 0 < x < 1 . Here x = x2 + x2 + x2. Then { ∈ | | | } | | 1 2 3 q 1 u(x) = ln , x Ω, x ∈ | | 1,2 Ω is in W0 ( ). The origin is an isolated boundary point. We have 1 lim ln = + (, 0). x 0 x ∞ → | | (At all the other boundary points the function has the right boundary values in the classical sense.) Ω Rn 1,p Ω Remark: Let and suppose that p > n. Then every function in W0 ( ) is continuous and takes⊂ the boundary values zero in the classical sense. (In applications, one usually has p = 2 < 3 ≦ n, unfortunately.) 1,p Ω An extremely important property of the space W0 ( ) is that it is closed even 1,p Ω under weak convergence. That is, if u1, u2,... belong to W0 ( ) and if ui ⇀ u and p Ω 1,p Ω 1,p Ω ui ⇀ u weakly in L ( ) (by deﬁnition u W ( )), then u itself is in W0 ( ). ∇ Let∇f W1,p(Ω) and suppose that u W∈1,p(Ω). We say that u has the boundary ∈ ∈ 1,p Ω values f in Sobolev’s sense, if u f W0 ( ). (Sometimes this is written as + 1,p Ω − ∈ k,α Ω u f W0 ( ).) —There is a theory of so-called Trace Spaces like W (∂ ) for ∈ ∂u general boundary value problems. In particular,the normal derivative ∂n makes sense.

1,p Ω + 1,p Rn Lemma 19 If u W0 ( ), then the function uχΩ 0χRn Ω is in W0 ( ). ∈ \ If Ω Rn, then the cases 1 p < n, p = n (the borderline case), and p > n are very di⊂ﬀerent in some essential≤ estimates.

20 1,p Ω Ω Theorem 20 (the Sobolev inequality) Suppose that u W0 ( ), where is a domain in Rn . If 1 p < n, then there is a constant C depending∈ only on n and p such that ≤ n p 1 np− p np n p p u − dx C u dx . SOBOLEV  | |  ≤  |∇ |  ZΩ ZΩ      np   p∗ Ω  = =   Thus u L ( ); p∗ n p the Sobolev conjugate. ∈ − Ω 1,p Rn Proof: Extending u as zero outside we may assume that u W0 ( ). ﬃ ∈ Rn By approximation it is su cient to prove the inequality for ϕ C0∞( ). Let us, for instructive purposes, write down the proof in the two dimensional∈ case n = 2, x = (x1, x2). Multiply

x1 x2 ∂ϕ(t, x ) ∂ϕ(x , t) ϕ(x , x ) = 2 dt, ϕ(x , x ) = 1 dt, 1 2 ∂x 1 2 ∂x Z 1 Z 2 −∞ −∞ to get ∞ ∂ϕ(t , x ) ∞ ∂ϕ(x , t ) ϕ(x , x ) 2 1 2 dt 1 2 dt . | 1 2 | ≤ ∂x 1 · ∂x 2 Z 1 Z 2 −∞ −∞

Integrate with respect to x1 :

∞ ∞ ∂ϕ(t , x ) ∞ ∞ ϕ(x , x ) 2 dx 1 2 dt D ϕ dx dx . | 1 2 | 1 ≤ ∂x 1 | 2 | 1 2 Z Z 1 Z Z −∞ −∞ −∞ −∞

Integrate with respect to x2 :

2 ϕ 2 dx dx D ϕ dx dx D ϕ dx dx ϕ dx dx . " | | 1 2 ≤ " | 2 | 1 2 · " | 2 | 1 2 ≤ " |∇ | 1 2 !

Hence ϕ 2 ϕ 1. This proves the theorem in the case n = 2, p = 1. Fork thek general≤ k∇ kp, 1 < p < 2, set ψ = ϕ γ with γ > 1 as selected below. Then the above inequality applied on ψ yields | |

γ 1 γ+1 2 2 − HOLDER¨ 2γ 2γ γ 2γ 2 γ 1 2 2γ + ϕ γ ϕ − ϕ γ ϕ ϕ γ 1 " | | ≤ " | | |∇ | ≤ " | | " |∇ | ! ! !

21 = 2p for γ > 1. Take 2γ 2 p . Then, after some arithmetic, − 2 p 2 −p p 2p p ϕ 2 p γ2 ϕ . " | | − ≤ " |∇ | ! ! This is the desired inequality for 1 < p < n = 2. See [EG, pp.138-140] or [GT, pp.155-156] for general n. A proof can also be based on the formula = 1 (x y) u(y) u(x) − · ∇n dy ωn Ω x y Z | − | valid for u W1,1(Ω). ∈ 0 Remark: The case n < p < . Let u W1,p(Ω), p > n. For each cube Q Ω we have ∞ ∈ ⊂ 2pn 1 n u(x) u(y) x y − p u | − | ≤ p n| − | k∇ kp,Q − for a.e. x, y Q. Hence u is locally Hölder continuous in Ω with Hölder exponent n ∈ 1/n 1/p α = 1 . If Ω is bounded, then u C(Ω¯ ) and u ,Ω C Ω − u p,Ω. − p ∈ k k∞ ≤ | | k∇ k The case p = n (borderline case6) is very special. 1,p 1,p For functions in W (but not in W0 ) the corresponding inequalities are more involved than the Sobolev inequality and they do require some additional regularity of the domain in question (balls, cubes, domains, smooth domains, Lipschitz domains etc). 1) 1 < p < n, u W1,p(Q), Q = a cube in Rn . Then ∈ n 1 1/n u p∗ − u + Q − u k kp∗,Q ≤ n k∇ kp,Q | | k kp,Q = np (The gain is that p∗ n p > p.) − 2) Poincaré’s inequality for u W1,p(Ω), 1 p < . ∈ ≤ ∞ p p/n p u(x) uΩ dx C(n, p) Ω u(x) dx. | − | ≤ | | |∇ | ZΩ ZΩ

6The Trudinger-Moser inequality

n/(n 1) u − exp | | dx C mes(Ω) Ω c u n ≤ Z k∇ k ! holds.

22 Ω = 1 Here is a convex domain and uΩ Ω u(x) dx is the average of u over | | Ω Ω . R 3) Poincar´e-Sobolev: 1 p < n, u W1,p(Ω), Ω = convex. ≤ ∈ n p np− 1/p np n p p u(x) uΩ − dx Cn,p u(x) dx  | − |  ≤  |∇ |  ZΩ  ZΩ          These inequalities constitute the main bulk in the Sobolev Imbedding Theorem. q 1,p For example, L (Ω) W (Ω), if q = p∗, and, for a domain of ﬁnite Lebesgue ⊃ 0 measure, this is valid if 1 q p∗. The Rellich-Kondrachev theorem is even stronger. ≤ ≤

Theorem 21 (Rellich7-Kondrachev) Let Ω be an arbitrary bounded domain in Rn 1,p Ω and consider the space W0 ( ). 1 p < n. Then W1,p(Ω) is compactly imbedded in Lq(Ω), where 1 q < ≤ 0 ≤ = np 1,p Ω p∗ n p (q need not be conjugate to p). In practical terms, if ui W0 ( ), and − ∈ u Ω M, i = 1, 2,..., k ik1,p, ≤ then there exists a function u W1,p(Ω) and a subsequence such that ∈ 0 u u Ω 0 as i k − ikq, → →∞ for each ﬁxed q, 1 q < p∗. p > n. Then W≤1,p(Ω) is compactly imbedded in Cα(Ω¯ ), α = 1 n . Now 0 − p u(x) u(y) K x y α. | − | ≤ | − | p = n. This is the borderline case. The same is true for W1,p(Ω), if the boundary of Ω is suﬃciently regular.

Remarks: The functions, not their derivatives, converge strongly in Lq(Ω). • The convergence u ⇀ u need not be strong in Lp∗ (Ω), but u ⇀ u weakly • iν iν also in Lp∗ (Ω).

7The case p = 2 is credited to Rellich.

23 2.3 About W1,p The ﬁrst order Sobolev space has some special properties not valid for higher derivatives. If u W1,p(Ω), so does u+ and u . As usually, u = u+ u , u = u+ + u . Hence, ∈ loc − − − | | − u W1,p(Ω). We have | |∈ loc u, when u > 0, u+ = ∇ ∇  0, when u ≦ 0,  with similar rules for u− and u . For example, for u we use | |  | | uν uν (u2 + ε2)1/2 ϕ dx = ϕ ∇ dx (ε , 0). ν ∇ − (u2 + ε2)1/2 Z Z ν The passage to the limit under the integral sign can be justiﬁed [R, Ch.4,Thm.6]. Now 1 2 1/2 ϕuν uν (u + ε )− ϕ uν . | ∇ | ν ≤ | ∇ | Finally, letting ε 0, we have by the Dominated Convergence Theorem → = + u ϕ dx ϕ 1χ u>0 1χ u<0 0χ u=0 u dx | |∇ − { } − { } { } ∇ Z Z i.e., u = u, when u > 0, = u, whenh u < 0, and = 0, wheniu = 0. ∇| | ∇ −∇ Lemma 22 Let u W1,p(Ω). Then u = 0 a.e. on any set where u is constant. ∈ ∇ Lemma 23 u, v W1,p(Ω) = max u, v , min u, v W1,p(Ω). ∈ ⇒ { } { }∈ The last lemma is not true with W1,p replaced by Wk,p, if k 2. Example!? ≥ 3 Theequation ∆u = 4αu3 + f (x), α 0 ≥ n Let Ω be a bounded domain in R . Suppose that f L∞(Ω) and that α 0is,for simplicity, a constant. Fix a function ϕ W1,2(Ω).∈Usually, ϕ is continuous≥ even in Ω¯ . It represents the boundary values. ∈ Problem: Minimize the variational integral 1 I(u) = u 2 + αu4 + f (x)u dx 2|∇ | ZΩ " # among all u W1,2(Ω) with boundary values ϕ (in Sobolev’s sense, i.e., u ϕ 1,2 Ω ∈ − ∈ W0 ( )). The problem is interesting even for the case ϕ 0. Before proving the existence of a unique solution, let us observe that ≡

24 [I] The function u W1,2(Ω), u ϕ W1,2(Ω), is minimizing if and only if ∈ − ∈ 0

u η + 4αu3η + f (x)η dx = 0 EULER-LAGRANGE EQN in weak form ∇ · ∇ Z h i 1,2 Ω for all test-functions η W0 ( ). (Hence a partial integration and the Variational lemma leads to the Eqn∈ ∆u = 4αu3 + f (x), provided that u has a second derivatives.) To see this, note that for any real number ε 1 I(u + ε η) = I(u) + ε [ u η + 4αu3η + f (x)η] dx + ε2 η 2 dx ∇ · ∇ 2 |∇ | Z Z + ε2 α(6u2η2 + 4uη3 ε +η4 ε2) dx Z ( ) Suppose that u is minimizing. The u(x) + ε η(x) is admissible, and ⇒ I(u + ε η) I(u) ≥ by assumption. We must have ( u η + 4αu3η + f (x)η) = 0. (If I(u) + ε J + ε2(...) attains its minimum for∇ ε· ∇= 0, then J = 0!!!) R ( ) If Euler’s equation in weak form holds, then ⇐ 1 I(u + 1η) = I(u) + 0 + η 2 dx + α η2[u2 6 + 4uη + η2] dx I(u), 2 |∇ | · ≥ Z Z since η 2 0and6u2 +4uη+η2 2u2 0. This means that I(u+η) I(u), in other|∇ words| ≥ u is minimizing. (Any≥ admissible≥ function v can be written≥ as v = u + (v u), v u = η W1,2(Ω).) − − ∈ 0 [II] The minimizing function u is unique (if it exists). Suppose that there are two solutions, say u and u . Then u u and u u are in W1,2(Ω). Choose the test 1 2 1 − 2 2 − 1 0

25 function η = u u in the Euler equation for u and the test function u u in 2 − 1 2 1 − 2 the Euler equation for u1. Adding the two equations we get (DO IT)

2 1 2 2 (u1 u2) (u +u ) ≥ − 2 1 2 2 + 3 3 = u1 u2 dx 4 α (u1 u2)(u1 u2) dx 0. |∇ − ∇ | − − ZΩ ZΩ 0 z ≥}| {

Hence the ﬁrst integral is zero and so u|1 = u2 {za.e. in Ω .}Thus u1 u2 is constant a.e. in Ω (you may wish to prove this∇ in Sobolev’s∇ space). The boundary− values will force this constant to be zero. This shows that u1 = u2.

[III] There exists a minimizing function u. We need some estimates to begin with. Now

HOLDER¨ f (x)v dx = f (x)(v ϕ) dx + f (x)ϕ dx f v ϕ + f ϕ − ≤ k k2k − k2 k k1 Z Z Z SOBOLEV f CΩ (v ϕ) + f ϕ ≤ k k2 k∇ − k2 k k1 ≦ f CΩ v + f CΩ ϕ + f ϕ k k2 k∇ k2 k k2 k∇ k2 k k1 = A v + B k∇ k2 for all v. This means that 1 A I(v) v v B ≥ 2k∇ k2 k∇ k2 − 2 − and so I(v) A2 B. Hence we have ≥ − 32 − < Inf I(v) I(ϕ) < + . −∞ v ≤ ∞ Observe also that, for example, A v max 2 + , B + I(v) I(v). k∇ k2 ≤ { 2 } ≈

By the deﬁnition of the inﬁmum, there are admissible functions u1, u2, u3,... such that limI(ui) = I0 = inf I(v). i →∞ 26 This is called a minimizing sequence. We may assume that

I I(u ) < I + 1. 0 ≤ i 0 By the above bound

u max 2 + A , 1 + B + I = K k∇ ik2 ≤ 2 0 n o for all i = 1, 2, 3,... Now

u ϕ + u ϕ ϕ + CΩ u ϕ ϕ + CΩK + CΩ ϕ k ik2 ≤k k2 k i − k2 ≤k k2 k∇ i − ∇ k2 ≤k k2 k∇ k2 for all i = 1, 2, 3,... By the weak compactness of L2(Ω) there are functions u and w¯ in L2(Ω) such that u ⇀ u, u ⇀ w¯ iν ∇ iν 2 1,p weakly in L (Ω). We must have thatw ¯ = u and u W (Ω). Since ui ϕ 1,p Ω ∇ = ∈ − ∈ W0 ( ), so does u ϕ. Now we claim that I(u) I0, that is, u is the solution. It is suﬃcient to− establish that

I(u) limI(uν) ≤ v →∞ since I(u) I (u is admissible!). First, ≥ 0 u 2 dx u 2 dx + 2 u ( u u) dx |∇ iν | ≥ |∇ | ∇ · ∇ iν − ∇ ZΩ ZΩ ZΩ

0 by the→ weak convergence in L 2 ( Ω ) | {z } and so 2 2 lim uiν dx u dx. ν |∇ | ≥ |∇ | →∞ ZΩ ZΩ 2 By the Rellich-Kondrachev theorem ui ϕ u ϕ strongly in L (Ω) at least for some subsequence. Hence, passing again− → to some− subsequence, we have that u (x) u(x) at a.e. point x Ω . Thus i → ∈ αu4 dx lim αu4 dx ≤ i ZΩ ZΩ

27 by Fatous Lemma, at least for a subsequence. Finally,

f (x)u dx f (x)u dx iν → Z Z by weak convergence. Collecting results, we have the desired semicontinuity.

1) Is u continuous? What about u C2(Ω), that is, is u a classical solution? This is REGULARITY THEORY∈ (de Giorgi, Moser, Nash).

2) Are the boundary values attained in the classical sense; lim u(x) = ϕ(x), ξ x ξ → ∈ x Ω ∂ Ω? (This is true only in “regular domains.”) ∈

3) Stability? What do small changes of the data cause?

References

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[EG] L.Evans &R.Gariepy, Measure Theory and Fine Properties of Functions, CRC Press, Boca Raton 1992.

[G] E.Giusti, Direct Methods in the Calculus of Variations, World Scientiﬁc, Singapore 2003.

[GT] D.Gilbarg &N.Trudinger, Elliptic Partial Diﬀerential Equations of Sec- ond Order, 2nd Ed., Springer-Verlag, Berlin 1983.

[J] J.Jost, Partial Diﬀerential Equations, Springer, New York 2002. —Chapter 7.

[LL] E.Lieb &M.Loss, Analysis, American Mathematical Society, Providence 1977.

[R] H. Royden, Real Analysis, 2nd Ed., MacMillan, New York 1970.

[S] S. Sobolev, Applications of Functional Analysis in Mathematical Physics, Translations of Mathematical Monographs, American Mathematical Society, Providence 1963, pp.25-28.