L P and Sobolev Spaces

Home , Mollifier, Norm (mathematics), Sobolev inequality, Trace operator

NOTES ON Lp AND SOBOLEV SPACES

STEVE SHKOLLER

1. Lp spaces 1.1. Definitions and basic properties. Definition 1.1. Let 0 < p < ∞ and let (X, M, µ) denote a measure space. If f : X → R is a measurable function, then we define 1 Z p p kfkLp(X) := |f| dx and kfkL∞(X) := ess supx∈X |f(x)| . X

Note that kfkLp(X) may take the value ∞. Definition 1.2. The space Lp(X) is the set p L (X) = {f : X → R | kfkLp(X) < ∞} . The space Lp(X) satisfies the following vector space properties: (1) For each α ∈ R, if f ∈ Lp(X) then αf ∈ Lp(X); (2) If f, g ∈ Lp(X), then |f + g|p ≤ 2p−1(|f|p + |g|p) , so that f + g ∈ Lp(X). (3) The triangle inequality is valid if p ≥ 1. The most interesting cases are p = 1, 2, ∞, while all of the Lp arise often in nonlinear estimates. Definition 1.3. The space lp, called “little Lp”, will be useful when we introduce Sobolev spaces on the torus and the Fourier series. For 1 ≤ p < ∞, we set ( ∞ ) p ∞ X p l = {xn}n=1 | |xn| < ∞ . n=1 1.2. Basic inequalities. Lemma 1.4. For λ ∈ (0, 1), xλ ≤ (1 − λ) + λx. Proof. Set f(x) = (1 − λ) + λx − xλ; hence, f 0(x) = λ − λxλ−1 = 0 if and only if λ(1 − xλ−1) = 0 so that x = 1 is the critical point of f. In particular, the minimum occurs at x = 1 with value f(1) = 0 ≤ (1 − λ) + λx − xλ . Lemma 1.5. For a, b ≥ 0 and λ ∈ (0, 1), aλb1−λ ≤ λa + (1 − λ)b with equality if a = b.

Date: March 19, 2009. 1 2 STEVE SHKOLLER

Proof. If either a = 0 or b = 0, then this is trivially true, so assume that a, b > 0. Set x = a/b, and apply Lemma 1 to obtain the desired inequality. Theorem 1.6 (H¨older’sinequality). Suppose that 1 ≤ p ≤ ∞ and 1 < q < ∞ with 1 1 p q 1 p + q = 1. If f ∈ L and g ∈ L , then fg ∈ L . Moreover,

kfgkL1 ≤ kfkLp kgkLq .

Note that if p = q = 2, then this is the Cauchy-Schwarz inequality since kfgkL1 = |(f, g)L2 |. Proof. We use Lemma 1.5. Let λ = 1/p and set |f|p |g|q a = p , and b = q kfkLp kgkLp for all x ∈ X. Then aλb1−λ = a1/pb1−1/p = a1/pb1/q so that |f| · |g| 1 |f|p 1 |g|q ≤ p + q . kfkLp kgkLq p kfkLp q kgkLq Integrating this inequality yields Z |f| · |g| Z 1 |f|p 1 |g|q 1 1 dx ≤ p + q dx = + = 1 . X kfkLp kgkLq X p kfkLp q kgkLq p q p 1 1 Deﬁnition 1.7. q = p−1 or q = 1 − p is called the conjugate exponent of p. Theorem 1.8 (Minkowski’s inequality). If 1 ≤ p ≤ ∞ and f, g ∈ Lp then

kf + gkLp ≤ kfkLp + kgkLp . Proof. If f + g = 0 a.e., then the statement is trivial. Assume that f + g 6= 0 a.e. Consider the equality |f + g|p = |f + g| · |f + g|p−1 ≤ (|f| + |g|)|f + g|p−1 , and integrate over X to ﬁnd that Z Z |f + g|pdx ≤ (|f| + |g|)|f + g|p−1 dx X X H¨older’s p−1 p p ≤ (kfkL + kgkL ) |f + g| Lq . p Since q = p−1 , 1 Z q p−1 p |f + g| Lq = |f + g| dx , X from which it follows that 1− 1 Z q p |f + g| dx ≤ kfkLp + kgkLq , X 1 1 which completes the proof, since p = 1 − q . Corollary 1.9. For 1 ≤ p ≤ ∞, Lp(X) is a normed linear space. NOTES ON Lp AND SOBOLEV SPACES 3

Example 1.10. Let Ω denote a (Lebesgue) measure-1 subset of Rn. If f ∈ L1(Ω) satisfies f(x) ≥ M > 0 for almost all x ∈ Ω, then log(f) ∈ L1(Ω) and satisfies Z Z log fdx ≤ log( fdx) . Ω Ω To see this, consider the function g(t) = t − 1 − log t for t > 0. Compute g0(t) = 1 00 1 − t = 0 so t = 1 is a minimum (since g (1) > 0). Thus, log t ≤ t − 1 and letting 1 t 7→ t we see that 1 1 − ≤ log t ≤ t − 1 . (1.1) t Since log x is continuous and f is measurable, then log f is measurable for f > 0. Let t = f(x) in (1.1) to find that kfkL1

kfkL1 f(x) 1 − ≤ log f(x) − log kfkL1 ≤ − 1 . (1.2) f(x) kfkL1 Since g(x) ≤ log f(x) ≤ h(x) for two integrable functions g and h, it follows that log f(x) is integrable. Next, integrate (1.2) to ﬁnish the proof, as R f(x) − 1 dx = X kfkL1 0.

p 1.3. The space (L (X), k·kLp (X) is complete. Recall the a normed linear space is a Banach space if every Cauchy sequence has a limit in that space; furthermore, recall that a sequence xn → x in X if limn→∞ kxn − xkX = 0. The proof of completeness makes use of the following two lemmas which are restatements of the Monotone Convergence Theorem and the Dominated Conver- gence Theorem, respectively.

1 Lemma 1.11 (MCT). If fn ∈ L (X), 0 ≤ f1(x) ≤ f2(x) ≤ · · ·, and kfnkL1(X) ≤ 1 C < ∞, then limn→∞ fn(x) = f(x) with f ∈ L (X) and kfn − fkL1 → 0 as n → 0. 1 Lemma 1.12 (DCT). If fn ∈ L (X), limn→∞ fn(x) = f(x) a.e., and if ∃ g ∈ 1 1 L (X) such that |fn(x)| ≤ |g(x)| a.e. for all n, then f ∈ L (X) and kfn−fkL1 → 0.

Proof. Apply the Dominated Convergene Theorem to the sequence hn = |fn −f| → 0 a.e., and note that |hn| ≤ 2g. Theorem 1.13. If 1≤ p < ∞ then Lp (X) is a Banach space.

∞ Proof. Step 1. The Cauchy sequence. Let {fn}n=1 denote a Cauchy sequence in Lp, and assume without loss of generality (by extracting a subsequence if neces- −n sary) that kfn+1 − fnkLp ≤ 2 . Step 2. Conversion to a convergent monotone sequence. Deﬁne the se- ∞ quence {gn}n=1 as

g1 = 0, gn = |f1| + |f2 − f1| + ··· + |fn − fn−1| for n ≥ 2 . It follows that 0 ≤ g1 ≤ g2 ≤ · · · ≤ gn ≤ · · · so that gn is a monotonically increasing sequence. Furthermore, {gn} is uniformly bounded in Lp as Z ∞ !p p p X p gndx = kgnkLp ≤ kf1kLp + kfi − fi−1kLp ≤ (kfkLp + 1) ; X i=2 4 STEVE SHKOLLER

p p p thus, by the Monotone Convergence Theorem, gn % g a.e., g ∈ L , and gn ≤ g a.e.

Step 3. Pointwise convergence of {fn}. For all k ≥ 1,

Therefore, fn → f a.e. Since n X |fn| ≤ |f1| + |fi − fi−1| ≤ gn ≤ g for all n ∈ N , i=2 p p p p p p it follows that |f| ≤ g a.e. Hence, |fn| ≤ g , |f| ≤ g , and |f − fn| ≤ 2g , and by the Dominated Convergence Theorem, Z Z p p lim |f − fn| dx = lim |f − fn| dx = 0 . n→∞ X X n→∞ p p 1.4. Convergence criteria for L functions. If {fn} is a sequence in L (X) p which converges to f in L (X), then there exists a subsequence {fnk } such that fnk (x) → f(x) for almost every x ∈ X (denoted by a.e.), but it is in general not true that the entire sequence itself will converge pointwise a.e. to the limit f, without some further conditions holding. Example 1.14. Let X = [0, 1], and consider the subintervals 1 1 1 1 2 2 1 1 2 2 3 3 1 0, , , 1 , 0, , , , , 1 , 0, , , , , , , 1 , 0, , ··· 2 2 3 3 3 3 4 4 4 4 4 4 5 th Let fn denote the indicator function of the n interval of the above sequence. Then kfnkLp → 0, but fn(x) does not converge for any x ∈ [0, 1].

Example 1.15. Set X = R, and for n ∈ N, set fn = 1[n,n+1]. Then fn(x) → 0 as p n → ∞, but kfnkLp = 1 for p ∈ [1, ∞); thus, fn → 0 pointwise, but not in L .

Example 1.16. Set X = [0, 1], and for n ∈ , set fn = n1 1 . Then fn(x) → 0 N [0, n ] p a.e. as n → ∞, but kfnkL1 = 1; thus, fn → 0 pointwise, but not in L . p Theorem 1.17. For 1 ≤ p < ∞, suppose that {fn} ⊂ L (X) and that fn(x) → p f(x) a.e. If limn→∞ kfnkLp(X) = kfkLp(X), then fn → f in L (X). a+b p 1 p p p Proof. Given a, b ≥ 0, convexity implies that 2 ≤ 2 (a +b ) so that (a+b) ≤ p−1 p p p p−1 p p 2 (a +b ), and hence |a−b| ≤ 2 (|a| +|b| ). Set a = fn and b = f to obtain the inequality p−1 p p p 0 ≤ 2 (|fn| + |f| ) − |fn − f| .

Since fn(x) → f(x) a.e., Z Z p p p−1 p p p 2 |f| dx = lim 2 (|fn| + |f| ) − |fn − f| dx . X X n→∞ Thus, Fatou’s lemma asserts that Z Z p p p−1 p p p 2 |f| dx ≤ lim inf 2 (|fn| + |f| ) − |fn − f| dx X n→∞ X NOTES ON Lp AND SOBOLEV SPACES 5

Since kfnkLp(X) → kfkLp(X), we see lim supn→∞ kfn − fkLp(X) = 0.

1.5. The space L∞(X).

Deﬁnition 1.18. With kfkL∞(X) = inf{M ≥ 0 | |f(x)| ≤ M a.e.}, we set

∞ L (X) = {f : X → R | kfkL∞(X) < ∞} .

∞ Theorem 1.19. (L (X), k · kL∞(X)) is a Banach space.

∞ Proof. Let fn be a Cauchy sequence in L (X). It follows that |fn − fm| ≤ kfn − fmkL∞(X) a.e. and hence fn(x) → f(x) a.e., where f is measurable and essentially bounded. Choose > 0 and N() such that kfn − fmkL∞(X) < for all n, m ≥ N(). Since |f(x) − fn(x)| = limm→∞ |fm(x) − fn(x)| ≤ holds a.e. x ∈ X, it follows that kf − fnkL∞(X) ≤ for n ≥ N(), so that kfn − fkL∞(X) → 0.

Remark 1.20. In general, there is no relation of the type Lp ⊂ Lq. For example, − 1 1 2 suppose that X = (0, 1) and set f(x) = x 2 . Then f ∈ L (0, 1), but f 6∈ L (0, 1). On the other hand, if X = (1, ∞) and f(x) = x−1, then f ∈ L2(1, ∞), but f 6∈ L1(1, ∞).

Lemma 1.21 (Lp comparisons). If 1 ≤ p < q < r ≤ ∞, then (a) Lp ∩ Lr ⊂ Lq, and (b) Lq ⊂ Lp + Lr.

Proof. We begin with (b). Suppose that f ∈ Lq, deﬁne the set E = {x ∈ X : |f(x)| ≥ 1}, and write f as

f = f1E + f1Ec = g + h .

p r p p q Our goal is to show that g ∈ L and h ∈ L . Since |g| = |f| 1E ≤ |f| 1E and r r q |h| = |f| 1Ec ≤ |f| 1Ec , assertion (b) is proven. For (a), let λ ∈ [0, 1] and for f ∈ Lq,

1 1 Z q Z q q λq (1−λ)q kfkLq = |f| dx = |f| |f| dµ X X 1 λq (1−λ)q q λ (1−λ) ≤ kfkLp kfkLr = kfkLp kfkLr .

Theorem 1.22. If µ(X) ≤ ∞ and q > p, then Lq ⊂ Lp.

Proof. Consider the case that q = 2 and p = 1. Then by the Cauchy-Schwarz inequality, Z Z p |f|dx = |f| · 1 dx ≤ kfkL2(X) µ(X) . X X 6 STEVE SHKOLLER

1.6. Approximation of Lp(X) by simple functions. Pn Lemma 1.23. If p ∈ [1, ∞), then the set of simple functions f = i=1 ai1Ei , where p each Ei is an element of the σ-algebra A and µ(Ei) < ∞, is dense in L (X, A, µ). p ∞ Proof. If f ∈ L , then f is measurable; thus, there exists a sequence {φn}n=1 of simple functions, such that φn → f a.e. with

0 ≤ |φ1| ≤ |φ2| ≤ · · · ≤ |f|, i.e., φn approximates f from below. p p p p 1 Recall that |φn − f| → 0 a.e. and |φn − f| ≤ 2 |f| ∈ L , so by the Dominated Convergence Theorem, kφn − fkLp → 0. Now, suppose that the set Ei are disjoint; then b deﬁnition of the Lebesgue integral, Z n p X p φndx = |ai| µ(Ei) < ∞ . X i=1 If ai 6= 0, then µ(Ei) < ∞. 1.7. Approximation of Lp(Ω) by continuous functions.

Lemma 1.24. Suppose that Ω ⊂ Rn is bounded. Then C0(Ω) is dense in Lp(Ω) for p ∈ [1, ∞). Proof. Let K be any compact subset of Ω. The functions 1 F , (x) = ∈ C0(Ω) satisfy F , ≤ 1 , K n 1 + n dist(x, K) K n and decrease monotonically to the characteristic function 1K . The Monotone Con- vergence Theorem gives p fK ,n → 1K in L (Ω), 1 ≤ p < ∞ . Next, let A ⊂ Ω be any measurable set, and let λ denote the Lebesgue measure. Then λ(A) = sup{µ(K): K ⊂ A, K compact} .

It follows that there exists an increasing sequence of Kj of compact subsets of A such that λ(A\ ∪j Kj) = 0. By the Monotone Convergence Theorem, 1Kj → 1A in Lp(Ω) for p ∈ [1, ∞). According to Lemma 1.23, each function in Lp(Ω) is a norm limit of simple functions, so the lemma is proved. 1.8. Approximation of Lp(Ω) by smooth functions. For Ω ⊂ Rn open, for > 0 taken suﬃciently small, deﬁne the open subset of Ω by

Ω := {x ∈ Ω | dist(x, ∂Ω) > } . Definition 1.25 (Mollifiers). Define η ∈ C∞(Rn) by 2 −1 Ce(|x| −1) if |x| < 1 η(x) := , 0 if |x| ≥ 1 R with constant C > 0 chosen such that n η(x)dx = 1. R For > 0, the standard sequence of mollifiers on Rn is defined by −n η(x) = η(x/) , R and satisfy n η(x)dx = 1 and spt(η) ⊂ B(0, ). R NOTES ON Lp AND SOBOLEV SPACES 7

Deﬁnition 1.26. For Ω ⊂ Rn open, set p p ˜ ˜ Lloc(Ω) = {u :Ω → R | u ∈ L (Ω) ∀ Ω ⊂⊂ Ω} , where Ω˜ ⊂⊂ Ω means that there exists K compact such that Ω˜ ⊂ K ⊂ Ω. We say that Ω˜ is compactly contained in Ω.

1 1 Definition 1.27 (Mollification of L ). If f ∈ Lloc(Ω), define its mollification f = η ∗ f in Ω , so that Z Z f (x) = η(x − y)f(y)dy = η(y)f(x − y)dy ∀x ∈ Ω . Ω B(0,) Theorem 1.28 (Mollification of Lp(Ω)). ∞ (A) f ∈ C (Ω). (B) f → f a.e. as → 0. (C) If f ∈ C0(Ω), then f → f uniformly on compact subsets of Ω. p p (D) If p ∈ [1, ∞) and f ∈ Lloc(Ω), then f → f in Lloc(Ω). Proof. Part (A). We rely on the difference quotient approximation of the partial derivative. Fix x ∈ Ω, and choose h sufficiently small so that x + hei ∈ Ω for i = 1, ..., n, and compute the difference quotient of f : f (x + he ) − f(x) Z 1 x + he − y x − y i = −n η i − η f(y)dy Ω h h Z 1 x + he − y x − y = −n η i − η f(y)dy Ω˜ h for some open set Ω˜ ⊂⊂ Ω. On Ω,˜ 1 x + he − y x − y 1 ∂η x − y ∂η x − y lim η i − η = = n , h→0 h ∂xi ∂xi so by the Dominated Convergence Theorem, ∂f Z ∂η (x) = (x − y)f(y)dy . ∂xi Ω ∂xi A similar argument for higher-order partial derivatives proves (A). Step 2. Part (B). By the Lebesgue differentiation theorem, 1 Z lim |f(y) − f(x)|dy for a.e. x ∈ Ω . →0 |B(x, )| B(x,) Choose x ∈ Ω for which this limit holds. Then Z |f(x) − f(x)| ≤ η(x − y)|f(y) − f(x)|dy B(x,) 1 Z = n η((x − y)/)|f(y) − f(x)|dy B(x,) C Z ≤ |f(x) − f(y)|dy −→ 0 as → 0 . |B(x, )| B(x,) 8 STEVE SHKOLLER

Step 3. Part (C). For Ω˜ ⊂ Ω, the above inequality shows that if f ∈ C0(Ω) and hence uniformly continuous on Ω,˜ then f (x) → f(x) uniformly on Ω.˜ p Step 4. Part (D). For f ∈ Lloc(Ω), p ∈ [1, ∞), choose open sets U ⊂⊂ D ⊂⊂ Ω; then, for > 0 small enough,

kf kLp(U) ≤ kfkLp(D) . To see this, note that Z |f (x)| ≤ η(x − y)|f(y)|dy B(x,) Z (p−1)/p 1/p = η(x − y) η(x − y) |f(y)|dy B(x,) Z !(p−1)/p Z !1/p p ≤ η(x − y)dy η(x − y)|f(y)| dy , B(x,) B(x,) so that for > 0 suﬃciently small Z Z Z p p |f (x)| dx ≤ η(x − y)|f(y)| dydx U U B(x,) Z Z ! Z p p ≤ |f(y)| η(x − y)dx dy ≤ |f(y)| dy . D B(y,) D

0 p 0 Since C (D) is dense in L (D), choose g ∈ C (D) such that kf − gkLp(D) < δ; thus

kf − fkLp(U) ≤ kf − g kLp(U) + kg − gkLp(U) + kg − fkLp(U) ≤ 2kf − gkLp(D) + kg − gkLp(U) ≤ 2δ + kg − gkLp(U) .

1.9. Continuous linear functionals on Lp(X). Let Lp(X)0 denote the dual p p 0 space of L (X). For φ ∈ L (X) , the operator norm of φ is deﬁned by kφkop = supLp(X)=1 |φ(f)|.

p q p Theorem 1.29. Let p ∈ (1, ∞], q = p−1 . For g ∈ L (X), deﬁne Fg : L (X) → R as Z Fg(f) = fgdx . X p Then Fg is a continuous linear functional on L (X) with operator norm kFgkop = kgkLp(X).

Proof. The linearity of Fg again follows from the linearity of the Lebesgue integral. Since Z Z

|Fg (f)| = fgdx ≤ |fg| dx ≤ kfkLp kgkLq , X X with the last inequality following from H¨older’s inequality, we have that sup |Fg (f)| ≤ kfkLp =1 kgkLq . NOTES ON Lp AND SOBOLEV SPACES 9

For the reverse inequality let f = |g|q−1 sgn g. f is measurable and in Lp since q |f|p = |f| q−1 = |g|q and since fg = |g|q,

1 + 1 Z Z Z p q q q Fg(f) = fgdx = |g| dx = |g| dx X X X 1 1 Z p Z q p q = |f| dx |g | dx = kfkLp kgkLq X X

Fg (f) so that kgkLq = ≤ kFgkop. kfkLp Remark 1.30. Theorem 1.29 shows that for 1 < p ≤ ∞, there exists a linear q p 0 p isometry g 7→ Fg from L (X) into L (X) , the dual space of L (X). When p = ∞, 1 ∞ 0 ∞ 0 1 g 7→ Fg : L (X) → L (X) is rarely onto (L (X) is strictly larger than L (X)); on the other hand, if the measure space X is σ-finite, then L∞(X) = L1(X)0. 1.10. A theorem of F. Riesz. Theorem 1.31 (Representation theorem). Suppose that 1 < p < ∞ and φ ∈ p 0 q p L (X) . Then there exists g ∈ L (X), q = p−1 such that Z φ (f) = fgdx ∀f ∈ Lp(X) , X and kφkop = kgkLq . Corollary 1.32. For p ∈ (1, ∞) the space Lp (X, µ) is reflexive, i.e., Lp(X)00 = Lp(X). The proof Theorem 1.31 crucially relies on the Radon-Nikodym theorem, whose statement requires the following definition. Definition 1.33. If µ and ν are measure on (X,A) then ν µ if ν(E) = 0 for every set E for which µ(E) = 0. In this case, we say that ν is absolutely continuous with respect to µ. Theorem 1.34 (Radon-Nikodym). If µ and ν are two finite measures on X, i.e., µ(X) < ∞, ν(X) < ∞, and ν µ, then Z Z F (x) dν(x) = F (x)h(x)dµ(x) (1.3) X X holds for some nonnegative function h ∈ L1(X, µ) and every positive measurable function F . Proof. Define measures α = µ + 2ν and ω = 2µ + ν, and let H = L2 (X, α) (a Z Hilbert space) and suppose φ : L2 (X, α) → R is defined by φ (f) = fdω. We X show that φ is a bounded linear functional since Z Z Z

|φ(f)| = f d(2µ + ν) ≤ |f| d(2µ + 4ν) = 2 |f| dα X X X p ≤ kfkL2(x,α) α(X) . 10 STEVE SHKOLLER

Thus, by the Riesz representation theorem, there exists g ∈ L2(X, α) such that Z Z φ(f) = f dω = fg dα , X X which implies that Z Z f(2g − 1)dν = f(2 − g)dµ . (1.4) X X F 2−g Given 0 ≤ F a measurable function on X, if we set f = 2g−1 and h = 2g−1 then R R X F dν = X F h dx which is the desired result, if we can prove that 1/2 ≤ g(x) ≤ 2. Deﬁne the sets 1 1 1 E1 = x ∈ X | g(x) < − and E2 = x ∈ X | g(x) > 2 + . n 2 n n n

By substituting f = 1 j , j = 1, 2 in (1.4), we see that En j j µ(En) = ν(En) = 0 for j = 1, 2 , from which the bounds 1/2 ≤ g(x) ≤ 2 hold. Also µ({x ∈ X | g(x) = 1/2}) = 0 1 and ν({x ∈ X | g(x) = 2}) = 0. Notice that if F = 1, then h ∈ L (X). Remark 1.35. The more general version of the Radon-Nikodym theorem. Suppose that µ(X) < ∞, ν is a ﬁnite signed measure (by the Hahn decomposition, ν = ν− + + 1 R R ν ) such that ν µ; then, there exists h ∈ L (X, µ) such that X F dν = X F h dµ. Lemma 1.36 (Converse to H¨older’sinequality). Let µ(X) < ∞. Suppose that g is measurable and fg ∈ L1(X) for all simple functions f. If Z

M(g) = sup fg dµ : f is a simple function < ∞ , (1.5) kfkLp =1 X q then g ∈ L (X), and kgkLq (X) = M(g).

Proof. Let φn be a sequence of simple functions such that φn → g a.e. and |φn| ≤ |g|. Set q−1 |φn| sgn (φn) fn = q−1 kφnkLq so that kfnkLp = 1 for p = q/(q − 1). By Fatou’s lemma, Z kgkLq (X) ≤ lim inf kφnkLq (X) = lim inf |fnφn|dµ . n→∞ n→∞ X

Since φn → g a.e., then Z Z kgkLq (X) ≤ lim inf |fnφn|dµ ≤ lim inf |fng|dµ ≤ M(g) . n→∞ X n→∞ X The reverse inequality is implied by H¨older’s inequality. Proof of the Lp(X)0 representation theorem. We have already proven that there exists a natural inclusion ι : Lq(X) → Lp(X)0 which is an isometry. It remains to show that ι is surjective. Let φ ∈ Lp(X)0 and deﬁne a set function ν on measurable subsets E ⊂ X by Z ν(E) = 1Edν =: φ(1E) . X NOTES ON Lp AND SOBOLEV SPACES 11

Thus, if µ(E) = 0, then ν(E) = 0. Then Z f dν =: φ(f) X for all simple functions f, and by Lemma 1.23, this holds for all f ∈ Lp(X). By the Radon-Nikodym theorem, there exists 0 ≤ g ∈ L1(X) such that Z Z f dν = fg dµ ∀ f ∈ Lp(X) . X X But Z Z φ(f) = f dν = fg dµ (1.6) X X and since φ ∈ Lp(X)0, then M(g) given by (1.5) is finite, and by the converse to q Hölder’sinequality, g ∈ L (X), and kφkop = M(g) = kgkLq (X). 1.11. Weak convergence. The importance of the Representation Theorem 1.31 is in the use of the weak-* topology on the dual space Lp(X)0. Recall that for a 0 ∗ 0 Banach space B and for any sequence φj in the dual space B , φj * φ in B weak-*, if hφj, fi → hφ, fi for each f ∈ B, where h·, ·i denotes the duality pairing between B0 and B. Theorem 1.37 (Alaoglu’s Lemma). If B is a Banach space, then the closed unit ball in B0 is compact in the weak -* topology. p Definition 1.38. For 1 ≤ p < ∞, a sequence {fn} ⊂ L (X) is said to weakly converge to f ∈ Lp(X) if Z Z q p fn(x)φ(x)dx → f(x)φ(x)dx ∀φ ∈ L (X), q = . X X p − 1 p We denote this convergence by saying that fn * f in L (X) weakly. Given that Lp(X) is reflexive for p ∈ (1, ∞), a simple corollary of Alaoglu’s Lemma is the following

p Theorem 1.39 (Weak compactness for L , 1 < p < ∞). If 1 < p < ∞ and {fn} p is a bounded seequence in L (X), then there exists a subsequence {fnk} such that p fnk * f in L (X) weakly. ∞ Deﬁnition 1.40. A sequence {fn} ⊂ L (X) is said to converge weak-* to f ∈ L∞(X) if Z Z 1 fn(x)φ(x)dx → f(x)φ(x)dx ∀φ ∈ L (X) . X X ∗ ∞ We denote this convergence by saying that fn * f in L (X) weak-*. ∞ Theorem 1.41 (Weak-* compactness for L ). If {fn} is a bounded sequence in ∞ ∗ ∞ L (X), then there exists a subsequence {fnk} such that fnk * f in L (X) weak-*. p p Lemma 1.42. If fn → f in L (X), then fn * f in L (X). Proof. By H¨older’sinequality, Z

g(fn − f)dx ≤ kfn − fkLp kgkLq . X 12 STEVE SHKOLLER

Note that if fn is weakly convergent, in general, this does not imply that fn is strongly convergent.

2 Example 1.43. If p = 2, let fn denote any orthonormal sequence in L (X). From Bessel’s inequality ∞ Z X 2 fngdx ≤ kgk 2 , L (X) n=1 X 2 we see that fn * 0 in L (X).

This example shows that the map f 7→ kfkLp is continuous, but not weakly continuous. It is, however, weakly lower-semicontinuous.

p Theorem 1.44. If fn * f weakly in L (X), then kfkLp ≤ lim infn→∞ kfnkLp . Proof. As a consequence of Theorem 1.31, Z Z

kfkLp(X) = sup fgdx = sup lim fngdx n→∞ kgkLq (X)=1 X kgkLq (X)=1 X

≤ sup lim inf kfnkLp kgkLq . n→∞ kgkLq (X)=1

p p Theorem 1.45. If fn * f in L (X), then fn is bounded in L (X).

Theorem 1.46. Suppose that Ω ⊂ Rn is a bounded. Suppose that

sup kfnkLp(Ω) ≤ M < ∞ and fn → f a.e. n p If 1 < p < ∞, then fn * f in L (Ω). Proof. Egoroff’s theorem states that for all > 0, there exists E ⊂ Ω such that c p µ(E) < and fn → f uniformly on E . By definition, fn * f in L (Ω) for R q p p ∈ (1, ∞) if Ω(fn − f)gdx → 0 for all g ∈ L (Ω), q = p−1 . We have the inequality Z Z Z (fn − f)gdx ≤ |fn − f| |g| dx + |fn − f| |g| dx . Ω E Ec c Choose n ∈ N sufficiently large, so that |fn(x) − f(x)| ≤ δ for all x ∈ E . By Hölder’sinequality, Z c |fn − f| |g| dx ≤ kfn − fkLp(Ec)kgkLq (Ec) ≤ δµ(E )kgkLq (Ω) ≤ Cδ Ec for a constant C < ∞. By the Dominated Convergence Theorem, kfn − fkLp(Ω) ≤ 2M so by Hölder’s inequality, the integral over E is bounded by 2MkgkLq (E). Next, we use the fact that the integral is continuous with respect to the measure of the set over which the integral is taken. In particular, if 0 ≤ h is integrable, then for all δ > 0, there R exists > 0 such that if the set E has measure µ(E) < , then hdx ≤ δ. To see E this, either approximate h by simple functions, or use the Dominated Convergence R theorem for the integral Ω 1E (x)h(x)dx. NOTES ON Lp AND SOBOLEV SPACES 13

Remark 1.47. The proof of Theorem 1.46 does not work in the case that p = 1, as H¨older’sinequality gives Z |fn − f| |g| dx ≤ kfn − fkL1(Ω)kgkL∞(E) , E so we lose the smallness of the right-hand side.

Remark 1.48. Suppose that E ⊂ X is bounded and measurable, and let g = 1E. p If fn * f in L (X), then Z Z fn(x)dx → f(x)dx; E E hence, if fn * f, then the average of fn converges to the average of f pointwise. 1.12. Integral operators. If u : Rn → R satisfies certain integrability conditions, then we can define the operator K acting on the function u as follows: Z Ku(x) = k(x, y)u(y)dy , n R where k(x, y) is called the integral kernel.The mollification procedure, introduced in Definition 1.27, is one example of the use of integral operators; the Fourier transform is another. Definition 1.49. Let L(Lp(Rn),Lp(Rn)) denote the space of bounded linear operators from Lp(Rn) to itself. Using the Representation Theorem 1.31, the natural norm on L(Lp(Rn),Lp(Rn)) is given by Z

p n p n kKkL(L (R ),L (R )) = sup sup Kf(x)g(x)dx . n kfkLp =1 kgkLq =1 R R Theorem 1.50. Let 1 ≤ p < ∞, Ku(x) = n k(x, y)u(y)dy, and suppose that R Z Z n n |k(x, y)|dx ≤ C1 ∀y ∈ R and |k(x, y)|dy ≤ C2 ∀x ∈ R , n n R R p n p n where 0 < C1,C2 < ∞. Then K : L (R ) → L (R ) is bounded and 1 p−1 p p p n p n kKkL(L (R ),L (R )) ≤ C1 C2 . In order to prove Theorem 1.50, we will need another well-known inequality. 1 1 Lemma 1.51 (Cauchy-Young Inequality). If p + q = 1, then for all a, b ≥ 0, ap bq ab ≤ + . p q Proof. Suppose that a, b > 0, otherwise the inequality trivially holds. ab = exp(log(ab)) = exp(log a + log b) (since a, b > 0) 1 1 = exp log ap + log bq p q 1 1 ≤ exp(log ap) + exp(log bq) (using the convexity of exp) p q ap bq = + p q 1 1 where we have used the condition p + q = 1. 14 STEVE SHKOLLER

1 1 Lemma 1.52 (Cauchy-Young Inequality with δ). If p + q = 1, then for all a, b ≥ 0, p q ab ≤ δ a + Cδb , δ > 0 , −q/p −1 with Cδ = (δp) q . Proof. This is a trivial consequence of Lemma 1.51 by setting b ab = a · (δp)1/p . (δp)1/p |f(y)|p |g(x)|q Proof of Theorem 1.50. According to Lemma 1.51, |f(y)g(x)| ≤ p + q so that Z Z

k(x, y)f(y)g(x)dydx n n R R Z Z |k(x, y)| Z Z |k(x, y)| ≤ dx|f(y)|pdy + dy|g(x)|qdx n n p n n q R R R R C1 p C2 q ≤ kfk p + kgk q . p L q L To improve this bound, notice that Z Z

k(x, y)f(y)g(x)dydx n n R R Z Z |k(x, y)| Z Z |k(x, y)| ≤ dx|tf(y)|pdy + dy|t−1g(x)|qdx n n p n n q R R R R p −q C1t p C2t q ≤ kfk p + kgk q =: F (t) . p L q L Find the value of t for which F (t) has a minimum to establish the desired bounded.

Theorem 1.53 (Simple version of Young’s inequality). Suppose that k ∈ L1(Rn) and f ∈ Lp(Rn). Then kk ∗ fkLp ≤ kkkL1 kfkLp . Proof. Deﬁne Z Kk(f) = k ∗ f := k(x − y)f(y)dy . n R p n 1 n Let C1 = C2 = kkkL (R ). Then according to Theorem 1.50, Kk : L (R ) → p n p n p n L (R ) and kKkkL(L (R ),L (R )) ≤ C1. Theorem 1.50 can easily be generalized to the setting of integral operators K : q n r n p n 1 1 1 L (R ) → L (R ) built with kernels k ∈ L (R ) such that 1 + r = p + q . Such a generalization leads to

Theorem 1.54 (Young’s inequality). Suppose that k ∈ Lp(Rn) and f ∈ Lq(Rn). Then 1 1 1 kk ∗ fk r ≤ kkk p kfk q for 1 + = + . L L L r p q NOTES ON Lp AND SOBOLEV SPACES 15

1.13. Appendix 1: The Fubini and Tonelli Theorems. Let (X, A, µ) and (Y, B, ν) denote two ﬁxed measure spaces. The product σ-algebra A × B of subsets of X × Y is deﬁned by

A × B = {A × B : A ∈ A,B ∈ B}. The set function µ × ν : A × B → [0, ∞] defined by (µ × ν)(A × B) = µ(A) · ν(B) for each A × B ∈ A × B is a measure. Theorem 1.55 (Fubini). Let f : X × Y → R be a µ × ν-integrable function. Then both iterated integrals exist and Z Z Z Z Z f d(µ × ν) = f dµdν = f dνdµ . X×Y Y X X Y The existence of the iterated integrals is by no means enough to ensure that the function is integrable over the product space. As an example, let X = Y = [0, 1] and µ = ν = λ with λ the Lebesgue measure. Set ( 2 2 x −y , (x, y) 6= (0, 0) f(x, y) = (x2+y2)2 . 0, (x, y) = (0, 0) Then compute that Z 1 π Z 1 π f(x, y)dxdy = − , f(x, y)dydx = . 0 4 0 4 Fubini’s theorem shows, of course, that f is not integrable over [0, 1]2 There is a converse to Fubini’s theorem, however, according to which the existence of one of the iterated integrals is sufficient for the integrability of the function over the product space. The theorem is known as Tonelli’s theorem, and this result is often used. Theorem 1.56 (Tonelli). Let (X, A, µ) and (Y, B, ν) denote two σ-finite measure spaces, and let f : X × Y → R be a µ × ν-measurable function. If one of the R R R R iterated integrals X Y |f|dνdµ or Y X |f|dµdν exists, then the function f is µ×ν- integrable and hence, the other iterated integral exists and Z Z Z Z Z f d(µ × ν) = f dµdν = f dνdµ . X×Y Y X X Y 2. Sobolev Spaces 2.1. Weak derivatives. Definition 2.1 (Test functions). For Ω ⊂ Rn, set ∞ ∞ C0 (Ω) = {u ∈ C (Ω) | spt(u) ⊂ V ⊂⊂ Ω}, the smooth functions with compact support. Traditionally D(Ω) is often used to ∞ denote C0 (Ω), and D(Ω) is often referred to as the space of test functions. 1 du For u ∈ C (R), we can define dx by the integration-by-parts formula; namely, Z du Z dφ (x)φ(x)dx = − u(x) (x)dx ∀φ ∈ C∞( ) . dx dx 0 R R R 1 Notice, however, that the right-hand side is well-defined, whenever u ∈ Lloc(R) 16 STEVE SHKOLLER

Definition 2.2. An element α ∈ Zn is called a multi-index. For such an α = α ∂α1 ∂αn (α1, ..., αn), we write D = ··· and |α| = α1 + ···αn. ∂xα1 ∂xαn Example 2.3. Let n = 2. If |α| = 0, then α = (0, 0); if |α| = 1, then α = (1, 0) or α = (0, 1). If |α| = 2, then α = (1, 1). 1 α 1 Definition 2.4 (Weak derivative). Suppose that u ∈ Lloc(Ω). Then v ∈ Lloc(Ω) is called the αth weak derivative of u, written vα = Dαu, if Z Z α |α| α ∞ u(x) D φ(x)dx = (−1) v (x)φ(x)dx ∀φ ∈ C0 (Ω) . Ω Ω Example 2.5. Let n = 1 and set Ω = (0, 2). Define the function x, 0 ≤ x < 1 u(x) = . 1, 1 ≤ x ≤ 2 Then the function 1, 0 ≤ x < 1 v(x) = 0, 1 ≤ x ≤ 2 ∞ is the weak derivative of u. To see this, note that for φ ∈ C0 (0, 2), Z 2 dφ Z 1 dφ Z 2 dφ u(x) (x)dx = x (x)dx + (x)dx 0 dx 0 dx 1 dx Z 1 Z 1 1 2 = − φ(x)dx + xφ|0 + φ|1 = − φ(x)dx 0 0 Z 2 = − v(x)φ(x)dx . 0 Example 2.6. Let n = 1 and set Ω = (0, 2). Define the function x, 0 ≤ x < 1 u(x) = . 2, 1 ≤ x ≤ 2 Then the weak derivative does not exist! 1 To prove this, assume for the sake of contradiction that there exists v ∈ Lloc(Ω) ∞ such that for all φ ∈ C0 (0, 2), Z 2 Z 2 dφ v(x)φ(x)dx = − u(x) (x)dx . 0 0 dx Then Z 2 Z 1 dφ Z 2 dφ v(x)φ(x)dx = − x (x)dx − 2 (x)dx 0 0 dx 1 dx Z 1 = φ(x)dx − φ(1) + 2φ(1) 0 Z 1 = φ(x)dx + φ(1) . 0 ∞ Suppose that φj is a sequence in C0 (0, 2) such that φj(1) = 1 and φj(x) → 0 for x 6= 1. Then Z 1 Z 2 Z 1 1 = φj(1) = φj(x)dx = v(x)φj(x)dx − φj(x)dx → 0 , 0 0 0 which provides the contradiction. NOTES ON Lp AND SOBOLEV SPACES 17

Definition 2.7. For p ∈ [1, ∞], define W 1,p(Ω) = {u ∈ Lp(Ω) | weak derivative exists , Du ∈ Lp(Ω)}, where Du is the weak derivative of u. Example 2.8. Let n = 1 and set Ω = (0, 1). Define the function f(x) = sin(1/x). 1 du 2 1 1,p Then u ∈ L (0, 1) and dx = − cos(1/x)/x ∈ Lloc(0, 1), but u 6∈ W (Ω) for any p. In the case p = 2, we set H1(Ω) = W 1,p(Ω).

Example 2.9. Let Ω = B(0, 1) ⊂ R2 and set u(x) = |x|−α. We want to determine the values of α for which u ∈ H1(Ω). −α P3 −α/2 −α −α−2 Since |x| = j=1(xjxj) , then ∂xi |x| = −α|x| xi is well-deﬁned away from x = 0. 1 R −α R 2π R 1 −α Step 1. We show that u ∈ Lloc(Ω). To see this, note that Ω |x| dx = 0 0 r rdrdθ < ∞ whenever α < 2. −α−2 Step 2. Set v(x) = −α|x| xi. We show that Z Z ∞ u(x)Dφ(x)dx = − v(x)φ(x)dx ∀φ ∈ C0 (B(0, 1)) . B(0,1) B(0,1)

To see this, let Ωδ = B(0, 1)−B(0, δ), let n denote the inward-pointing unit normal to ∂Ωδ. Integration by parts yields Z Z 2π Z |x|−αDφ(x)dx = δ−αφ(x)n(x)δdθ + α |x|−α−2x φ(x)dx . Ωδ 0 Ωδ 1−α R 2π Since limδ→0 δ 0 φ(x)n(x)dθ = 0 if α < 1, we see that Z Z lim |x|−αDφ(x)dx = lim α |x|−α−2x φ(x)dx δ→0 δ→0 Ωδ Ωδ R 2π R 1 −α−1 Since 0 0 r rdrdθ < ∞ if α < 1, the Dominated Convergence Theorem shows that v is the weak derivative of u. 2 R 2π R 1 −2α−2 Step 3. v ∈ L (Ω), whenever 0 0 r rdrdθ < ∞ which holds if α < 0. Remark 2.10. Note that if the weak derivative exists, it is unique. To see this, R suppose that both v1 and v2 are the weak derivative of u on Ω. Then Ω(v1 − ∞ v2)φdx = 0 for all φ ∈ C0 (Ω), so that v1 = v2 a.e. 2.2. Definition of Sobolev Spaces. Definition 2.11. For integers k ≥ 0 and 1 ≤ p ≤ ∞, k,p 1 α p W (Ω) = {u ∈ Lloc(Ω) |D u exists and is in L (Ω) for |α| ≤ k}. Definition 2.12. For u ∈ W k,p(Ω) define 1   p X α p kukW k,p(Ω) =  kD ukLp(Ω) for 1 ≤ p < ∞ , |α|≤k and X α kukW k,∞(Ω) = kD ukL∞(Ω) . |α|≤k p The function k · kW k,p(Ω) is clearly a norm since it is a finite sum of L norms. k,p Definition 2.13. A sequence uj → u in W (Ω) if limj→∞ kuj − ukW k,p(Ω) = 0. 18 STEVE SHKOLLER

Theorem 2.14. W k,p(Ω) is a Banach space. k,p Proof. Let uj denote a Cauchy sequence in W (Ω). It follows that for all |α| ≤ k, α p p D uj is a Cauchy sequence in L (Ω). Since L (Ω) is a Banach space (see Theorem 1.19), for each α there exists uα ∈ Lp(Ω) such that α α p D uj → u in L (Ω) . (0,...,0 p When α = (0, ..., 0) we set u := u ) so that uj → u in L (Ω). We must show that uα = Dαu. ∞ For each φ ∈ C0 (Ω), Z Z α α uD φdx = lim ujD φdx Ω j→∞ Ω Z |α| α = (−1) lim D ujφdx j→∞ Ω Z = (−1)|α| uαφdx ; Ω α α α α p thus, u = D u and hence D uj → D u in L (Ω) for each |α| ≤ k, which shows k,p that uj → u in W (Ω). Definition 2.15. For integers k ≥ 0 and p = 2, we define Hk(Ω) = W k,2(Ω) . k P α α H (Ω) is a Hilbert space with inner-product (u, v)Hk(Ω) = |α|≤k(D u, D v)L2(Ω). 2.3. A simple version of the Sobolev embedding theorem. For two Banach spaces B1 and B2, we say that B1 is embedded in B2 if kukB2 ≤ CkukB1 for some k,p constant C and for u ∈ B1. We wish to determine which Sobolev spaces W (Ω) can be embedded in the space of continuous functions. To motivate the type of analysis that is to be employed, we study a special case. Theorem 2.16 (Sobolev embedding in 2-D). For kp ≥ 2, ∞ max |u(x)| ≤ Ckuk k,p 2 ∀u ∈ C (Ω) . (2.1) 2 W (R ) 0 x∈R ∞ Proof. Given u ∈ C0 (Ω), we prove that for all x ∈ spt(u), α |u(x)| ≤ CkD u(x)kL2(Ω) ∀|α| ≤ k . By choosing a coordinate system centered about x, we can assume that x = 0; thus, it suffices to prove that α |u(0)| ≤ CkD u(x)kL2(Ω) ∀|α| ≤ k . ∞ 1 Let 0 ≤ g ∈ C ([0, ∞)) such that g(x) = 1 for x ∈ [0, 2 ] and g(x) = 0 for 3 x ∈ [ 4 , ∞). By the fundamental theorem of calculus, Z 1 Z 1 u(0) = − ∂r[g(r)u(r, θ)]dr = − ∂r(r) ∂r[g(r)u(r, θ)]dr 0 0 Z 1 2 = r ∂r [g(r)u(r, θ)]dr 0 k Z 1 k Z 1 (−1) k−1 k (−1) k−2 k = r ∂r [g(r)u(r, θ)]dr = r ∂r [g(r)u(r, θ)]rdr (k − 1)! 0 (k − 1)! 0 NOTES ON Lp AND SOBOLEV SPACES 19

Integrating both sides from 0 to 2π, we see that k Z 2π Z 1 (−1) k−2 k u(0) = r ∂r [g(r)u(r, θ)]rdrdθ . 2π(k − 1)! 0 0 The change of variables from Cartesian to polar coordinates is given by x(r, θ) = r cos θ , y(r, θ) = r sin θ . By the chain-rule,

∂ru(x(r, θ), y(r, θ)) = ∂xu cos θ + ∂yu sin θ , 2 2 2 2 2 2 ∂r u(x(r, θ), y(r, θ)) = ∂xu cos θ + 2∂xyu cos θ sin θ + ∂y u sin θ . . k P α It follows that ∂r = |α|≤k aα(θ)D so that (−1)k Z X u(0) = rk−2 a (θ)Dα[g(r)u(x)]dx 2π(k − 1)! α B(0,1) |α|≤k k−2 X α ≤ kr kLq (B(0,1)) kD (gu)kLp(B(0,1)) |α|≤k p−1 1 p Z p(k−2) ≤ C r p−1 rdr kuk k,p 2 . W (R ) 0 p(k−2) Hence, we require p−1 + 1 > −1 or kp > 2. k,p 2.4. Approximation of W (Ω) by smooth functions. Recall that Ω = {x ∈ Ω | dist(x, ∂Ω) > }. Theorem 2.17. For integers k ≥ 0 and 1 ≤ p < ∞, let u = η ∗ u in Ω , where η is the standard mollifier defined in Definition 1.25. Then ∞ (A) u ∈ C (Ω) for each > 0, and k,p (B) u → u in Wloc (Ω) as → 0. k,p k,p ˜ Definition 2.18. A sequence uj → u in Wloc (Ω) if uj → u in W (Ω) for each Ω˜ ⊂⊂ Ω. Proof of Theorem 2.17. Theorem 1.28 proves part (A). Next, let vα denote the the α α αth weak partial derivative of u. To prove part (B), we show that D u = η ∗ v in Ω. For x ∈ Ω, Z α α D u (x) = D η(x − y)u(y)dy Ω Z α = Dx η(x − y)u(y)dy Ω Z |α| α = (−1) Dy η(x − y)u(y)dy Ω Z α α = η(x − y)v (y)dy = (η ∗ v )(x) . Ω α α p By part (D) of Theorem 1.28, D u → v in Lloc(Ω). 20 STEVE SHKOLLER

It is possible to reﬁne the above interior approximation result all the way to the boundary of Ω. We record the following theorem without proof.

Theorem 2.19. Suppose that Ω ⊂ Rn is a smooth, open, bounded subset, and that u ∈ W k,p(Ω) for some 1 ≤ p < ∞ and integers k ≥ 0. Then there exists a sequence ∞ uj ∈ C (Ω) such that k,p uj → u in W (Ω) .

It follows that the inequality (2.1) holds for all u ∈ W k,p(R2).

2.5. H¨olderSpaces. Recall that for Ω ⊂ Rn open and smooth, the class of Lips- chitz functions u :Ω → R satisﬁes the estimate |u(x) − u(y)| ≤ C|x − y| ∀x, y ∈ Ω for some constant C.

Deﬁnition 2.20 (Classical derivative). A function u :Ω → R is diﬀerentiable at x ∈ Ω if there exists f :Ω → L(Rn; Rn) such that |u(x) − u(y) − f(x) · (x − y)| → 0 . |x − y| We call f(x) the gradient of u(x), and denote it by Du(x).

Deﬁnition 2.21. If u :Ω → R is bounded and continuous, then

kuk 0 = max |u(x)| . C (Ω) x∈Ω If in addition u has a continuous and bounded derivative, then

kukC1(Ω) = kukC0(Ω) + kDukC0(Ω) .

The H¨olderspaces interpolate between C0(Ω) and C1(Ω).

Deﬁnition 2.22. For 0 < γ ≤ 1, the space C0,γ (Ω) consists of those functions for which

kukC0,γ (Ω) := kukC0(Ω) + [u]C0,γ (Ω) < ∞ , where the γth H¨oldersemi-norm [u]C0,γ (Ω) is deﬁned as |u(x) − u(y)| [u]C0,γ (Ω) = max . x,y∈Ω |x − y| x6=y

The space C0,γ (Ω) is a Banach space.

2.6. Morrey’s inequality. We can now oﬀer a reﬁnement and extension of the simple version of the Sobolev Embedding Theorem 2.16.

Theorem 2.23 (Morrey’s inequality). For n < p ≤ ∞, let B(x, r) ⊂ Rn and let y ∈ B(x, r). Then

1− n 1 n |u(x) − u(y)| ≤ Cr p kDukLp(B(x,2r))∀u ∈ C (R ) . NOTES ON Lp AND SOBOLEV SPACES 21

Notation 2.24 (Averaging). Let B(0, 1) ⊂ n. The volume of B(0, 1) is given by n R π 2 n−1 αn = n and the surface area is |S | = nαn. We deﬁne Γ( 2 +1)

Z 1 Z − f(y)dy = n f(y)dy B(x,r) αnr B(x,r) Z 1 Z − f(y)dS = n−1 f(y)dS . ∂B(x,r) nαnr ∂B(x,r)

Lemma 2.25. For B(x, r) ⊂ Rn, y ∈ B(x, r) and u ∈ C1(B(x, r)),

Z Z |Du(y)| − |u(y) − u(x)|dy ≤ C n−1 dy . B(x,r) B(x,r) |x − y|

Proof. For some 0 < s < r, let y = x + sω where ω ∈ Sn−1 = ∂B(0, 1). By the fundamental theorem of calculus, for 0 < s < r,

Z s d u(x + sω) − u(x) = u(x + tω)dt 0 dt Z s = Du(x + tω) ωdt . 0

Since |ω| = 1, it follows that

Z s |u(x + sω) − u(x)| ≤ |Du(x + tω)|dt . 0

Thus, integrating over Sn−1 yields

p−1 1 Z ! p Z ! p p(1−n) n−1 p |u(x) − u(y)| ≤ C s p−1 s dsdω |Du(z)| dz B(0,2r) B(x,2r) Morrey’s inequality implies the following embedding theorem. Theorem 2.26 (Sobolev embedding theorem for k = 1). There exists a constant C = C(p, n) such that 1,p n n 1,p n kuk 0,1− n ≤ CkukW ( ) ∀u ∈ W (R ) . C p (R ) R Proof. First assume that u ∈ C1(Rn). Given Morrey’s inequality, it suﬃces to show n 1,p n that max |u| ≤ CkukW (R ). Using Lemma 2.25, for all x ∈ R , Z Z |u(x)| ≤ − |u(x) − u(y)|dy + − |u(y)|dy B(x,1) B(x,1) Z |Du(y)| p n ≤ C n−1 dy + CkukL (R ) B(x,1) |x − y|

1,p n ≤ CkukW (R ) , the last inequality following whenever p > n. Thus, 1 n n 1,p n kuk 0,1− n ≤ CkukW ( ) ∀u ∈ C (R ) . (2.2) C p (R ) R ∞ n 1,p n ∞ n By the density of C0 (R ) in W (R ), there is a sequence uj ∈ C0 (R ) such that 1,p n uj → u ∈ W (R ) . By (2.2), for j, k ∈ N, 0,1− n n p 1,p n kuj − ukkC (R ) ≤ Ckuj − ukkW (R ) . 0,1− n 0,1− n Since C p (Rn) is a Banach space, there exists a U ∈ C p (Rn) such that 0,1− n n uj → U in C p (R ) . NOTES ON Lp AND SOBOLEV SPACES 23

It follows that U = u a.e. in Ω. By the continuity of norms with respect to strong convergence, we see that

n 1,p n kUk 0,1− n ≤ CkukW ( ) C p (R ) R which completes the proof. Remark 2.27. By approximation, Morrey’s inequality holds for all u ∈ W 1,p(B(x, 2r)) for n < p < ∞. You are asked to prove this. As a consequence of Morrey’s inequality, we extract information about the classical diﬀerentiability properties of weak derivatives.

n 1,p Theorem 2.28 (Differentiability a.e.). If Ω ⊂ R , n < p ≤ ∞ and u ∈ Wloc (Ω), then u is differentiable a.e. in Ω, and its gradient equals its weak gradient almost everywhere. Proof. We first restrict n < p < ∞. By Lebesgue’s differentiation theorem, for almost every x ∈ Ω, Z lim − |Du(x) − Du(z)|pdz = 0 . (2.3) r→0 B(x,r) Fix x ∈ Ω for which (2.3) holds, and define the function

wx(y) = u(y) − u(x) − Du(x) · (y − x) .

Notice that wx(x) = 0 and that

Dywx(y) = Du(y) − Du(x) . Set r = |x − y|. An application of Morrey’s inequality then yields the estimate

1 Z ! p ≤ Cr − |Du(z) − Du(x)|pdz B(x,r) = o(r) as r → 0 .

1,∞ 1,p The case that p = ∞ follows from the inclusion Wloc (Ω) ⊂ Wloc (Ω) for all 1 ≤ p < ∞. 2.7. The Gagliardo-Nirenberg-Sobolev inequality. In the previous section, we considered the embedding for the case that p > n.

∗ np Theorem 2.29 (Gagliardo-Nirenberg inequality). For 1 ≤ p < n, set p = n−p . Then 1,p n kuk p∗ n ≤ C kDuk p n ∀u ∈ W ( ) . L (R ) p,n L (R ) R 24 STEVE SHKOLLER

Proof for the case n = 2. Suppose first that p = 1 in which case p∗ = 2, and we must prove that 1 2 2 2 1 2 kukL (R ) ≤ CkDukL (R ) ∀u ∈ C0 (R ) . (2.4) Since u has compact support, by the fundamental theorem of calculus, Z x1 Z x2 u(x1, x2) = ∂1u(y1, x2)dy1 = ∂2u(x1, y2)dy2 −∞ −∞ so that Z ∞ Z ∞ |u(x1, x2)| ≤ |∂1u(y1, x2)|dy1 ≤ |Du(y1, x2)|dy1 −∞ −∞ and Z ∞ Z ∞ |u(x1, x2)| ≤ |∂2u(x1, y2)|dy2 ≤ |Du(x1, y2)|dy2 . −∞ −∞ Hence, it follows that Z ∞ Z ∞ 2 |u(x1, x2)| ≤ |Du(y1, x2)|dy1 |Du(x1, y2)|dy2 −∞ −∞ Integrating over R2, we find that Z ∞ Z ∞ 2 |u(x1, x2)| dx1dx2 −∞ −∞ Z ∞ Z ∞ Z ∞ Z ∞ ≤ |Du(y1, x2)|dy1 |Du(x1, y2)|dy2 dx1dx2 −∞ −∞ −∞ −∞ Z ∞ Z ∞ 2 ≤ |Du(x1, x2)|dx1dx2 −∞ −∞ which is (2.4). Next, if 1 ≤ p < 2, substitute |u|γ for u in (2.4) to find that

1 Z 2 Z |u|2γ dx ≤ Cγ |u|γ−1|Du|dx 2 2 R R p−1 Z p(γ−1) p p 2 p−1 ≤ CγkDukL (R ) |u| dx 2 R p(γ−1) p Choose γ so that 2γ = p−1 ; hence, γ = 2−p , and

2−p Z 2p 2p 2−p p 2 |u| dx ≤ CγkDukL (R ) , 2 R so that kuk 2p ≤ Cp,nkDukLp( n) (2.5) n R L 2−p (R ) 1 2 for all u ∈ C0 (R ). ∞ 2 1,p 2 ∞ 2 Since C0 (R ) is dense in W (R ), there exists a sequence uj ∈ C0 (R ) such that 1,p 2 uj → u in W (R ) . Hence, by (2.5), for all j, k ∈ N,

kuj − ukk 2p ≤ Cp,nkDuj − DukkLp( n) n R L 2−p (R ) NOTES ON Lp AND SOBOLEV SPACES 25

2p so there exists U ∈ L 2−p (Rn) such that 2p n uj → U in L 2−p (R ) . Hence U = u a.e. in R2, and by continuity of the norms, (2.5) holds for all 1,p 2 u ∈ W (R ). It is common to employ the Gagliardo-Nirenberg inequality for the case that p = 2; as stated, the inequality is not well-deﬁned in dimension two, but in fact, we have the following theorem.

Theorem 2.30. Suppose that u ∈ H1(R2). Then for all 1 ≤ q < ∞, √ q 2 1 2 kukL (R ) ≤ C qkukH (R ) . Proof. Let x and y be points in R2, and write r = |x − y|. Let θ ∈ S1. Introduce spherical coordinates (r, θ) with origin at x, and let g be the same cut-oﬀ function that was used in the proof of Theorem 2.16. Deﬁne U := g(r)u(r, θ). Then Z 1 ∂U Z 1 ∂U u(x) = − (r, θ)dr − |x − y|−1 (r, θ)rdr 0 ∂r 0 ∂r and Z 1 |u(x)| ≤ |x − y|−1|DU(r, θ)|rdr . 0 Integrating over S1, we obtain: Z 1 −1 |u(x)| ≤ 1B(x,1)|x − y| |DU(y)|dy := K ∗ |DU| , 2π 2 R 1 −1 where the integral kernel K(x) = 2π 1B(0,1)|x| . Using Young’s inequality from Theorem 1.54, we obtain the estimate 1 1 1 kK ∗ fk q 2 ≤ kKk k 2 kfk 2 2 for = − + 1 . (2.6) L (R ) L (R ) L (R ) k q 2 Using the inequality (2.6) with f = |DU|, we see that

1 "Z # k −k q 2 2 2 kukL (R ) ≤ CkDUkL (R ) |y| dy B(0,1)

1 Z 1 k 1−k 2 2 ≤ CkDUkL (R ) r dr 0 1 q + 2 k = Ckuk 1 2 . H (R ) 4 1 1 When q → ∞, k → 2 , so 1 q 2 2 1 2 kukL (R ) ≤ Cq kukH (R ) .

∞ n 1,n n Evidently, it is not possible to obtain the estimate kukL (R ) ≤ CkukW (R ) with a constant C < ∞. The following provides an example of a function in this borderline situation. 26 STEVE SHKOLLER

Example 2.31. Let Ω ⊂ R2 denote the open unit ball in R2. The unbounded 1 1 function u = log log 1 + |x| belongs to H (B(0, 1)). First, note that Z Z 2π Z 1 h 1i2 |u(x)|2dx = log log 1 + rdrdθ . Ω 0 0 r The only potential singularity of the integrand occurs at r = 0, but according to L’Hospital’s rule, h 1i2 lim r log log 1 + = 0, (2.7) r→0 r so the integrand is continuous and hence u ∈ L2(Ω). In order to compute the partial derivatives of u, note that ∂ x d f(x) df |x| = j , and |f(z)| = dz , ∂xj |x| dz |f(z)| where f : R → R is diﬀerentiable. It follows that for x away from the origin, −x Du(x) = , (x 6= 0) . 1 2 log(1 + |x| )(|x| + 1)|x| ∞ Let φ ∈ C0 (Ω) and ﬁx > 0. Then Z ∂φ Z ∂u Z u(x) (x)dx = − (x)φ(x)dx + uφNidS , Ω−B(0) ∂xi Ω−B(0,) ∂xi ∂B(0,) where N = (N1, ..., Nn) denotes the inward-pointing unit normal on the curve ∂B(0, ), so that N dS = (cos θ, sin θ)dθ. It follows that Z Z u(x)Dφ(x)dx = − Du(x)φ(x)dx Ω−B(0) Ω−B(0) Z 2π 1 − (cos θ, sin θ) log log 1 + φ(, θ)dθ . (2.8) 0 We claim that Du ∈ L2(Ω) (and hence also in L1(Ω)), for Z Z 2π Z 1 1 |Du(x)|2dx = drdθ h i2 Ω 0 0 2 1 r(r + 1) log 1 + r Z 1/2 1 Z 1 1 ≤ π dr + π dr 2 h i2 0 r(log r) 1/2 2 1 r(r + 1) log 1 + r

1 1 where we use the inequality log(1 + r ) ≥ log r = − log r ≥ 0 for 0 ≤ r ≤ 1. The second integral on the right-hand side is clearly bounded, while Z 1/2 Z − log 2 Z − log 2 1 1 t 1 2 dr = 2 t e dt = 2 dx < ∞ , 0 r(log r) −∞ t e −∞ x so that Du ∈ L2(Ω). Letting → 0 in (2.8) and using (2.7) for the boundary integral, by the Dominated Convergence Theorem, we conclude that Z Z ∞ u(x)Dφ(x)dx = − Du(x)φ(x)dx ∀φ ∈ C0 (Ω) . Ω Ω NOTES ON Lp AND SOBOLEV SPACES 27

2.8. Local coordinates near ∂Ω. Let Ω ⊂ Rn denote an open, bounded subset 1 K with C boundary, and let {Ul}l=1 denote an open covering of ∂Ω, such that for each l ∈ {1, 2, ..., K}, with

Vl = B(0, rl), denoting the open ball of radius rl centered at the origin and, + Vl = Vl ∩ {xn > 0} , − Vl = Vl ∩ {xn < 0} , 1 there exist C -class charts θl which satisfy 1 θl : Vl → Ul is a C diﬀeomorphism , (2.9) + θl(Vl ) = Ul ∩ Ω ,

θl(Vl ∩ {xn = 0}) = Ul ∩ ∂Ω . 2.9. Sobolev extensions and traces. Let Ω ⊂ Rn denote an open, bounded domain with C1 boundary. Theorem 2.32. Suppose that Ω˜ ⊂ Rn is a bounded and open domain such that Ω ⊂⊂ Ω˜. Then for 1 ≤ p ≤ ∞, there exists a bounded linear operator 1,p 1,p n E : W (Ω) → W (R ) such that for all u ∈ W 1,p(Ω), (1) Eu = u a.e. in Ω; (2) spt(u) ⊂ Ω˜;

1,p n 1,p ˜ (3) kEukW (R ) ≤ CkukW (Ω) for a constant C = C(p, Ω, Ω). Theorem 2.33. For 1 ≤ p < ∞, there exists a bounded linear operator T : W 1,p(Ω) → Lp(Ω) such that for all u ∈ W 1,p(Ω) 1,p 0 (1) T u = u|∂Ω for all u ∈ W (Ω) ∪ C (Ω); (2) kT ukLp(∂Ω) ≤ CkukW 1,p(Ω) for a constant C = C(p, Ω). Proof. Suppose that u ∈ C1(Ω), z ∈ ∂Ω, and that ∂Ω is locally flat near z. In particular, for r > 0 sufficiently small, B(z, r) ∪ ∂Ω ⊂ {xn = 0}. Let 0 ≤ ξ ∈ ∞ + C0 (B(z, r) such that ξ = 1 on B(z, r/2). Set Γ = ∂Ω ∪ B(z, r/2), B (z, r) = B(z, r) ∪ Ω, and let dxh = dx1 ··· dxn−1. Then Z Z p p |u| dxh ≤ ξ|u| dxh Γ {xn=0} Z ∂ = − (ξ|u|p)dx B+(z,r) ∂xn Z ∂ξ Z ∂u ≤ − |u|pdx − p ξ|u|p−2u dx B+(z,r) ∂xn B+(z,2δ) ∂xn Z p p−1 ∂u ≤ C |u| dx + Ck|u| k p L p−1 (B+(z,r)) B+(z,r) ∂xn Lp(B+(z,r)) Z ≤ C (|u|p + |Du|p)dx . (2.10) B+(z,r) On the other hand, if the boundary is not locally flat near z ∈ ∂Ω, then we use a C1 diffeomorphism to locally straighten the boundary. More specifically, suppose 28 STEVE SHKOLLER

1 that z ∈ ∂Ω ∪ Ul for some l ∈ {1, ..., K} and consider the C chart θl deﬁned in + (2.9). Deﬁne the function U = u◦θl; then U : Vl → R. Setting Γ = Vl ∪{xn = 0k, we see from the inequality (2.10), that Z Z p p p |U| dxh ≤ Cl (|U| + |DU| )dx . + Γ Vl + Using the fact that Dθl is bounded and continuous on Vl , the change of variables formula shows that Z Z p p p |u| dS ≤ Cl (|u| + |Du| )dx . + Ul∪∂Ω Ul Summing over all l ∈ {1, ..., K} shows that Z Z |u|pdS ≤ C (|u|p + |Du|p)dx . (2.11) ∂Ω Ω The inequality (2.11) holds for all u ∈ C1(Ω). According to Theorem 2.19, for 1,p ∞ 1,p u ∈ W (Ω) there exists a sequence uj ∈ C (Ω) such that uj → u in W (Ω). By inequality (2.11),

kT uk − T ujkLp(∂Ω) ≤ Ckuk − ujkW 1,p(Ω) , p p so that T uj is Cauchy in L (∂Ω), and hence a limit exists in L (∂Ω) We deﬁne the trace operator T as this limit:

lim kT u − T ujkLp(∂Ω) = 0 . j→0 0 Since the sequence uj converges uniformly to u if u ∈ C (Ω), we see that T u = 1,p 0 u|∂Ω for all u ∈ W (Ω) ∪ C (Ω). Sketch of the proof of Theorem 2.32. Just as in the proof of the trace theorem, first suppose that u ∈ C1(ω) and that near z ∈ ∂Ω, ∂Ω is locally flat, so that for + some r > 0, ∂Ω ∪ B(z, r) ⊂ {xn = 0}. Letting B = B(z, r) ∪ {xn ≥ 0} and − B = B(z, r) ∪ {xn ≤ 0} , we define the extension of u by u(x) if x ∈ B+ u¯(x) = xn − −3u(x1, ..., xn−1, −xn) + 4u(x1, ..., xn−1, − 2 ) if x ∈ B . + − Define u =u ¯|B+ and u =u ¯|B− . + − It is clear that u = u on {xn = 0}, and by the chain-rule, it follows that − − − ∂u ∂u ∂u xn (x) = 3 (x1, ..., −xn) − 2 (x1, ..., − ) , ∂xn ∂xn ∂xn 2 ∂u+ ∂u− 1 so that = on {xn = 0}. This shows thatu ¯ ∈ C (B(z, r). using the charts ∂xn ∂xn ∞ 1,p θl to locally straighten the boundary, and the density of the C (Ω) in W (Ω), the theorem is proved. 1,p 2.10. The subspace W0 (Ω). 1,p ∞ 1,p Definition 2.34. We let W0 (Ω) denote the closure of C0 (Ω) in W (Ω). Theorem 2.35. Suppose that Ω ⊂ Rn is bounded with C1 boundary, and that u ∈ W 1,p(Ω). Then 1,p u ∈ W0 (Ω) iff T u = 0 on ∂Ω . NOTES ON Lp AND SOBOLEV SPACES 29

We can now state the Sobolev embedding theorems for bounded domains Ω. Theorem 2.36 (Gagliardo-Nirenberg inequality for W 1,p(Ω)). Suppose that Ω ⊂ Rn is open and bounded with C1 boundary, 1 ≤ p < n, and u ∈ W 1,p(Ω). Then

kuk np ≤ CkukW 1,p(Ω) for a constant C = C(p, n, Ω) . L n−p (Ω)

Proof. Choose Ω˜ ⊂ Rn bounded such that Ω ⊂⊂ Ω,˜ and let Eu denote the Sobolev n ˜ 1,p n extension of u to R such that Eu = u a.e., spt(Eu) ⊂ Ω, and kEukW (R ) ≤ CkukW 1,p(Ω). Then by the Gagliardo-Nirenberg inequality, kuk np ≤ kEuk np ≤ CkD(Eu)k p n ≤ CkEuk 1,p n ≤ Ckuk 1,p . n−p n−p n L (R ) W (R ) W (Ω) L (Ω) L (R ) 1,p Theorem 2.37 (Gagliardo-Nirenberg inequality for W0 (Ω)). Suppose that Ω ⊂ n 1 1,p R is open and bounded with C boundary, 1 ≤ p < n, and u ∈ W0 (Ω). Then for np all 1 ≤ q ≤ n−p ,

kukLq (Ω) ≤ CkDukLp(Ω) for a constant C = C(p, n, Ω) . (2.12) ∞ Proof. By definition there exists a sequence uj ∈ C0 (Ω) such that uj → u in 1,p c W (Ω). Extend each uj by 0 on Ω . Applying Theorem 2.29 to this extension, and using the continuity of the norms, we obtain kuk pn ≤ CkDukLp(Ω). Since L n−p (Ω) Ω is bounded, the assertion follows by Hölder’sinequality. Theorem 2.38. Suppose that Ω ⊂ R2 is open and bounded with C1 boundary, and 1 u ∈ H0 (Ω). Then for all 1 ≤ q < ∞, √ kukLq (Ω) ≤ C qkDukL2(Ω) for a constant C = C(Ω) . (2.13) Proof. The proof follows that of Theorem 2.30. Instead of introducing the cut-off function g, we employ a partition of unity subordinate to the finite covering of the bounded domain Ω, in which case it suffices that assume that spt(u) ⊂ spt(U) with U also defined in the proof Theorem 2.30. Remark 2.39. Inequalities (2.12) and (2.13) are commonly referred to as Poincaré inequalities. They are invaluable in the study of the Dirichlet problem for Poisson’s equation, since the right-hand side provides an H1(Ω)-equivalent norm for all u ∈ 1 H0 (Ω). In particular, there exists constants C1,C2 such that

C1kDukL2(Ω) ≤ kukH1(Ω) ≤ C2kDukL2(Ω) . 2.11. Weak solutions to Dirichlet’s problem. Suppose that Ω ⊂ Rn is an open, bounded domain with C1 boundary. A classical problem in the linear theory of partial diﬀerential equations consists of ﬁnding solutions to the Dirichlet problem:

−∆u = f in Ω , (2.14a) u = 0 on ∂Ω , (2.14b) Pn ∂2 where ∆ = i=1 2 denotes the Laplace operator or Laplacian. As written, ∂xi (2.14) is the so-called strong form of the Dirichlet problem, as it requires that u to possess certain weak second-order partial derivatives. A major turning-point in the modern theory of linear partial differential equations was the realization that weak 30 STEVE SHKOLLER solutions of (2.14) could be defined, which only require weak first-order derivatives of u to exist. (We will see more of this idea later when we discuss the theory of distributions.)

1 −1 Deﬁnition 2.40. The dual space of H0 (Ω) is denoted by H (Ω). For f ∈ H−1(Ω),

kfkH−1(Ω) = sup hf, ψi , kψk 1 =1 H0 (Ω) −1 1 where hf, ψi denotes the duality pairing between H (Ω) and H0 (Ω). 1 Deﬁnition 2.41. A function u ∈ H0 (Ω) is a weak solution of (2.14) if Z 1 Du · Dv dx = hf, vi ∀v ∈ H0 (Ω) . Ω Remark 2.42. Note that f can be taken in H−1(Ω). According to the Sobolev embedding theorem, this implies that when n = 1, the forcing function f can be taken to be the Dirac Delta distribution.

∞ Remark 2.43. The motivation for Deﬁnition 2.41 is as follows. Since C0 (Ω) is dense in H1(Ω), multiply equation (2.14a) by φ ∈ C∞(Ω), integrate over Ω, and 0 R 0 R employ the integration-by-parts formula to obtain Ω Du · Dφ dx = Ω fφ dx; the boundary terms vanish because φ is compactly supported. Theorem 2.44 (Existence and uniqueness of weak solutions). For any f ∈ H−1(Ω), there exists a unique weak solution to (2.14).

1 Proof. Using the Poincaréinequality, kDukL2(Ω) is an H -equivalent norm for all 1 1 u ∈ H0 (Ω), and (Du, Dv)L2(Ω) defines the inner-product on H0 (Ω). As such, 1 according to the definition of weak solutions to (2.14), we are seeking u ∈ H0 (Ω) such that 1 (u, v) 1 = hf, vi ∀v ∈ H (Ω) . (2.15) H0 (Ω) 0 1 The existence of a unique u ∈ H0 (Ω) satisfying (2.15) is provided by the Riesz representation theorem for Hilbert spaces. Remark 2.45. Note that the Riesz representation theorem shows that there exists a distribution, denote −∆u ∈ H−1(Ω) such that 1 h−∆u, vi = hf, vi ∀v ∈ H0 (Ω) . 1 −1 The operator −∆ : H0 (Ω) → H (Ω) is thus an isomorphism. A fundamental question in the theory of linear partial differential equations is commonly referred to as elliptic regularity, and can be explained as follows: in order to develop an existence and uniqueness theorem for the Dirichlet problem, we have significantly generalized the notion of solution to the class of weak solutions, which permitted very weak forcing functions in H−1(Ω). Now suppose that the forcing function is smooth; is the weak solution smooth as well? Furthermore, does the weak solution agree with the classical solution? The answer is yes, and we will develop this regularity theory in Chapter 6, where it will be shown that for k 1 k−2 integers k ≥ 2, −∆ : H (Ω) ∩ H0 (Ω) → H (Ω) is also an isomorphism. An −1 k−2 k 1 important consequence of this result is that (−∆) : H (Ω) → H (Ω) ∩ H0 (Ω) is a compact linear operator, and as such has a countable set of eigenvalues, a fact NOTES ON Lp AND SOBOLEV SPACES 31 that is eminently useful in the construction of solutions for heat- and wave-type equations. For this reason, as well as the consideration of weak limits of nonlinear combi- nations of sequences, we must develop a compactness theorem, which generalizes the well-known Arzela-Ascoli theorem to Sobolev spaces.

2.12. Strong compactness. In Section 1.11, we deﬁned the notion of weak con- verence and weak compactness for Lp-spaces. Recall that for 1 ≤ p < ∞, a se- p p p quence uj ∈ L (Ω) converges weakly to u ∈ L (Ω), denoted uj * u in L (Ω), if R R q p Ω ujvdx → Ω uvdx for all v ∈ L (Ω), with q = p−1 . We can extend this deﬁnition to Sobolev spaces.

1,p Deﬁnition 2.46. For 1 ≤ p < ∞, uj * u in W (Ω) provided that uj * u in p p L (Ω) and Duj * Du in L (Ω). Alaoglu’s Lemma (Theorem 1.37) then implies the following theorem.

Theorem 2.47 (Weak compactness in W 1,p(Ω)). For Ω ⊂ Rn, suppose that

sup kujkW 1,p(Ω) ≤ M < ∞ for a constant M 6= M(j) . 1,p Then there exists a subsequence ujk * u in W (Ω). It turns out that weak compactness often does not suﬃce for limit processes involving nonlinearities, and that the Gagliardo-Nirenberg inequality can be used to obtain the following strong compactness theorem.

Theorem 2.48 (Rellich’s theorem). Suppose that Ω ⊂ Rn is an open, bounded domain with C1 boundary, and that 1 ≤ p < n. Then W 1,p(Ω) is compactly embedded q np in L (Ω) for all 1 ≤ q < n−p , i.e. if

sup kujkW 1,p(Ω) ≤ M < ∞ for a constant M 6= M(j) , q then there exists a subsequence ujk → u in L (Ω). In the case that n = 2 and p = 2, H1(Ω) is compactly embedded in Lq(Ω) for 1 ≤ q < ∞. In order to prove Rellich’s theorem, we need two lemmas.

0 Lemma 2.49 (Arzela-Ascoli Theorem). Suppose that uj ∈ C (Ω), kujkC0(Ω) ≤

M < ∞, and uj is equicontinuous. Then there exists a subsequence ujk → u uniformly on Ω. Lemma 2.50. Let 1 ≤ r ≤ s ≤ t ≤ ∞, and suppose that u ∈ Lr(Ω) ∩ Lt(Ω). Then 1 a 1−a for s = r + t a 1−a kukLs(Ω) ≤ kukLr (Ω)kukLt(Ω) . Proof. By H¨older’sinequality, Z Z |u|sdx = |u|as|u|(1−a)sdx Ω Ω as (1−a)s Z r Z t r (1−a)s t (1−a)s as as (1−a)s as ≤ |u| dx |u| dx = kukLr (Ω)kukLt(Ω) . Ω Ω 32 STEVE SHKOLLER

Proof of Rellich’s theorem. Let Ω˜ ⊂ Rn denote an open, bounded domain such that Ω ⊂⊂ Ω.˜ By the Sobolev extension theorem, the sequence uj satisﬁes spt(uj) ⊂ Ω,˜ and

1,p n sup kEujkW (R ) ≤ CM. Denote the sequence Euj byu ¯j. By the Gagliardo-Nirenberg inequality, if 1 ≤ q < np n−p ,

q q n 1,p n sup kukL (Ω) ≤ sup ku¯kL (R ) ≤ C sup ku¯jkW (R ) ≤ CM. For > 0, let η denote the standard molliﬁers and setu ¯j = η ∗ Euj. By ∞ ˜ choosing > 0 suﬃciently small,u ¯j ∈ C0 (Ω). Since Z Z 1 y u¯j = n η( )¯uj(x − y)dy = η(z)¯uj(x − z)dz , B(0,) B(0,1) and ifu ¯j is smooth, Z 1 d Z 1 u¯j(x − z) − u¯j(x) = u¯j(x − tz)dt = − Du¯j(x − tz) · z dt . 0 dt 0 Hence, Z Z 1 |u¯j(x) − u¯j(x)| = η(z) |Du¯j(x − tz)| dzdt , B(0,1) 0 so that Z Z Z 1 Z |u¯j(x) − u¯j(x)|dx = η(z) |Du¯j(x − tz)| dxdzdt Ω˜ B(0,1) 0 Ω˜

≤ kDu¯jkL1(Ω)˜ ≤ kDu¯jkLp(Ω)˜ < CM . Using the Lp-interpolation Lemma 2.50, a 1−a ku¯j − u¯jkLq (Ω)˜ ≤ ku¯j − u¯jk 1 ˜ ku¯j − u¯jk np L (Ω) L n−p (Ω)˜ ≤ CM kDu¯ − Du¯ k1−a j j Lp(Ω)˜ ≤ CMM 1−a (2.16)

q The inequality (2.16) shows thatu ¯j is arbitrarily close tou ¯j in L (Ω) uniformly in j ∈ N; as such, we attempt to use the smooth sequenceu ¯j to construct a convergent subsequenceu ¯ . Our goal is to employ the Arzela-Ascoli Theorem, so we show that jk for > 0 ﬁxed, ˜ ku¯jkC0(Ω)˜ ≤ M < ∞ andu ¯j is equicontinous. For x ∈ Rn, Z sup ku¯jkC0(Ω) ≤ sup η(x − y)|u¯j(y)|dy j j B(x,) −n ∞ n ≤ kηkL (R ) sup ku¯jkL1(Ω)˜ ≤ C < ∞ , j and similarly −n−1 ¯ ∞ n sup kDujkC0(Ω) ≤ kDηkL (R ) sup ku¯jkL1(Ω)˜ ≤ C < ∞ . j j NOTES ON Lp AND SOBOLEV SPACES 33

The latter inequality proves equicontinuity of the sequenceu ¯j, and hence there ˜ exists a subsequence ujk which converges uniformly on Ω, so that lim sup ku¯ − u¯ k = 0 . jk jl Lq (Ω)˜ k,l→∞ It follows from (2.16) and the triangle inequality that

lim sup ku¯jk − u¯jl kLq (Ω)˜ ≤ C . k,l→∞

1 1 Letting C = 1, 2 , 3 , etc., and using the diagonal argument to extract further subsequences, we can arrange to ﬁnd a subsequence again denoted by {u¯jk } of {u¯j} such that

lim sup ku¯jk − u¯jl kLq (Ω)˜ = 0 , k,l→∞ and hence

q lim sup kujk − ujl kL (Ω) = 0 , k,l→∞ The case that n = p = 2 follows from Theorem 2.30.

3. The Fourier Transform 3.1. Fourier transform on L1(Rn) and the space S(Rn).

Definition 3.1. For all f ∈ L1(Rn) the Fourier transform F is defined by Z − n −ix·ξ Ff(ξ) = fˆ(ξ) = (2π) 2 f(x)e dx . n R By Hölder’sinequality, F : L1(Rn) → L∞(Rn). Definition 3.2. The space of Schwartz functions of rapid decay is denoted by

n ∞ n β α ∞ n n S(R ) = {u ∈ C (R ) | x D u ∈ L (R ) ∀α, β ∈ Z+}. It is not diﬃcult to show that

n n F : S(R ) → S(R ) , and that α β ˆ |α| |β| α β ξ Dξ f = (−i) (−1) F(Dx x f) .

Deﬁnition 3.3. For all f ∈ L1(Rn), we deﬁne operator F ∗ by Z ∗ − n ix·ξ F f(x) = (2π) 2 f(ξ)e dξ . n R Lemma 3.4. For all u, v ∈ S(Rn), ∗ 2 n 2 n (Fu, v)L (R ) = (u, F v)L (R ) . Recall that the L2(Rn) inner-product for complex-valued functions is given by R (u, v)L2( n) = n u(x)v(x)dx. R R 34 STEVE SHKOLLER

Proof. Since u, v ∈ S(Rn), by Fubini’s Theorem, Z Z − n −ix·ξ 2 n 2 (Fu, v)L (R ) = (2π) u(x)e dx v(ξ) dξ n n ZR ZR − n = (2π) 2 u(x)eix·ξv(ξ) dξ dx n n ZR R Z − n ∗ 2 ix·ξ 2 n = (2π) u(x) e v(ξ) dξ dx = (u, F v)L (R ) , n n R R Theorem 3.5. F ∗ ◦ F = Id = F ◦ F ∗ on S(Rn). Proof. We ﬁrst prove that for all f ∈ S(Rn), F ∗Ff(x) = f(x).

Z Z F ∗Ff(x) = (2π)−n eiξ·x e−iy·ξf(y)dy dξ n n R R Z Z = (2π)−n ei(x−y)·ξf(y) dy dξ . n n R R By the dominated convergence theorem,

Z Z 2 F ∗Ff(x) = lim(2π)−n e−|ξ| ei(x−y)·ξf(y) dy dξ . →0 n n R R 2 For all > 0, the convergence factor e−|ξ| allows us to interchange the order of integration, so that by Fubini’s theorem,

Z Z 2 F ∗Ff(x) = lim(2π)−n f(y) e−|ξ| ei(y−x)·ξ dξ dy . →0 n n R R Deﬁne the integral kernel Z −n −|ξ|2+ix·ξ p(x) = (2π) e dξ n R Then Z ∗ F Ff(x) = lim p ∗ f := p(x − y)f(y)dy . →0 n R −n R −|ξ|2+ix·ξ Let p(x) = p1(x) = (2π) n e dξ. Then R √ Z 2 √ p(x/ ) = (2π)−n e−|ξ| +ix·ξ/ dξ n ZR −n −|ξ|2+ix·ξ n n = (2π) e 2 dξ = 2 p(x) . n R We claim that 1 |x|2 Z − 4 p(x) = n e and that p(x)dx = 1 . (3.1) 2 n (4π) R

n Given (3.1), then for all f ∈ S(R ), p ∗ f → f uniformly as → 0, which shows that F ∗F = Id, and similar argument shows that FF ∗ = Id. (Note that this follows from the proof of Theorem 1.28, since the standard molliﬁers η can be replaced by the sequence p and all assertions of the theorem continue to hold, for if (3.1) is NOTES ON Lp AND SOBOLEV SPACES 35

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0 -4 -2 0 2 4

Figure 1. As → 0, the sequence of functions p becomes more localized about the origin.

R true, then even though p does not have compact support, B(0,δ)c p(x)dx → 0 as → 0 for all δ > 0.) 1 Thus, it remains to prove (3.1). It suﬃces to consider the case = 2 ; then by deﬁnition Z 2 −n ix·ξ − |ξ| p 1 (x) = (2π) e e 2 dξ 2 n R 2 −n/2 − |ξ| = F (2π) e 2 .

2 −n/2 − |x| In order to prove that p 1 (x) = (2π) e 2 , we must show that with the Gauss- 2 2 −n/2 − |x| ian function G(x) = (2π) e 2 , G(x) = F(G(ξ)) . By the multiplicative property of the exponential,

−|ξ|2/2 −ξ2/2 −ξ2 /2 e = e 1 ··· e n , it suffices to consider the case that n = 1. Then the Gaussian satisfies the differential equation d G(x) + xG(x) = 0 . dx Computing the Fourier transform, we see that d −i Gˆ(x) − iξGˆ(x) = 0 . dξ Thus, 2 − ξ Gˆ(ξ) = Ce 2 . To compute the constant C,

Z 2 −1 x − 1 C = Gˆ(0) = (2π) e 2 dx = (2π) 2 R which follows from the fact that

Z x2 1 e 2 dx = (2π) 2 . (3.2) R 36 STEVE SHKOLLER

To prove (3.2), one can again rely on the multiplication property of the exponential to observe that 2 2 2 2 Z x1 Z x2 Z x1+x2 e 2 dx e 2 dx = e 2 dx 2 R R R 2π ∞ Z Z 2 = e−2r rdrdθ = 2π . 0 0 It follows from Lemma 3.4 that for all u, v ∈ S(Rn), ∗ 2 n 2 n 2 n (Fu, Fv)L (R ) = (u, F Fv)L (R ) = (u, v)L (R ) . Thus, we have established the Plancheral theorem on S(Rn). Theorem 3.6 (Plancheral’s theorem). F : S(Rn) → S(Rn) is an isomorphism with inverse F ∗ preserving the L2(Rn) inner-product. 3.2. The topology on S(Rn) and tempered distributions. An alternative to Definition 3.2 can be stated as follows: p Definition 3.7 (The space S(Rn)). Setting hxi = 1 + |x|2, n ∞ n k α S(R ) = {u ∈ C (R ) | hxi |D u| ≤ Ck,α ∀k ∈ Z+} . The space S(Rn) has a Fréchet topology determined by seminorms. n Definition 3.8 (Topology on S(R )). For k ∈ Z+, define the semi-norm k α pk(u) = sup hxi |D u(x)| , n x∈R ,|α|≤k and the metric on S(Rn) ∞ X pk(u − v) d(u, v) = 2−k . 1 + pk(u − v) k=0 The space (S(Rn), d) is a Fréchetspace. n n Definition 3.9 (Convergence in S(R )). A sequence uj → u in S(R ) if pk(uj − u) → 0 as j → ∞ for all k ∈ Z+. Definition 3.10 (Tempered Distributions). A linear map T : S(Rn) → C is continuous if there exists some k ∈ Z+ and constant C such that n |hT, ui| ≤ Cpk(u) ∀u ∈ S(R ) . The space of continuous linear functionals on S(Rn) is denoted by S0(Rn). Elements of S0(Rn) are called tempered distributions. 0 n 0 n Definition 3.11 (Convergence in S (R )). A sequence Tj *T in S (R ) if n hTj, ui → hT, ui for all u ∈ S(R ). For 1 ≤ p ≤ ∞, there is a natural injection of Lp(Rn) into S0(Rn) given by Z n hf, ui = f(x)u(x)dx ∀u ∈ S(R ) . n R Any finite measure on Rn provides an element of S0(Rn). The basic example of such a finite measure is the Dirac delta ‘function’ defined as follows: n hδ, ui = u(0) or, more generally, hδx, ui = u(x) ∀u ∈ S(R ) . NOTES ON Lp AND SOBOLEV SPACES 37

Definition 3.12. The distributional derivative D : S0(Rn) → S(Rn) is defined by the relation n hDT, ui = −hT, Dui ∀u ∈ S(R ) . More generally, the αth distributional derivative exists in S0(Rn) and is defined by α |α| α n hD T, ui = (−1) hT,D ui ∀u ∈ S(R ) . Multiplication by f ∈ S(Rn) preserves S0(Rn); in particular, if T ∈ S0(Rn), then fT ∈ S(Rn) and is defined by n hfT, ui = hT, fui ∀u ∈ S(R ) .

Example 3.13. Let H := 1[0,∞) denote the Heavyside function. Then dH = δ in S0( n) . dx R This follows since for all u ∈ S(Rn), dH du Z ∞ du h , ui = −hH, i = − dx = u(0) = hδ, ui . dx dx 0 dx Example 3.14 (Distributional derivative of Dirac measure). dδ du h , ui = − (0) ∀u ∈ S( n) . dx dx R 3.3. Fourier transform on S0(Rn). Definition 3.15. Define F : S0(Rn) → S0(Rn) by n hFT, ui = hT, Fui ∀u ∈ S(R ) , with the analogous definition for F ∗ : S0(Rn) → S0(Rn). Theorem 3.16. FF ∗ = Id = F ∗F on S0(Rn) . Proof. By Definition 3.15, for all u ∈ S(Rn) hFF ∗T, ui = hF ∗w, Fui = hT, F ∗Fui = hT, ui , the last equality following from Theorem 3.5.

− n Example 3.17 (Fourier transform of δ). We claim that Fδ = (2π) 2 . According to Deﬁnition 3.15, for all u ∈ S(Rn), Z − n hFδ, ui = hδ, Fui = Fu(0) = (2π) 2 u(x)dx , n R − n so that Fδ = (2π) 2 .

∗ − n ∗ n Example 3.18. The same argument shows that F δ = (2π) 2 so that F [(2π) 2 ] = − n 1. Using Theorem 3.16, we see that F(1) = (2π) 2 δ. This demonstrates nicely the identity |ξαuˆ(ξ)| = |Dαu(x)|. In other the words, the smoother the function x 7→ u(x) is, the faster ξ 7→ uˆ(ξ) must decay. 38 STEVE SHKOLLER

2 n ∞ n 3.4. The Fourier transform on L (R ). In Theorem 1.28, we proved that C0 (R ) p n ∞ n n n is dense in L (R ) for 1 ≤ p < ∞. Since C0 (R ) ⊂ S(R ), it follows that S(R ) is dense in Lp(Rn) as well. Thus, for every u ∈ L2(Rn), there exists a sequence n 2 n uj ∈ S(R ) such that uj → u in L (R ), so that by Plancheral’s Theorem 3.6,

2 n 2 n kuˆj − uˆkkL (R ) = kuj − ukkL (R ) < . 2 n It follows from the completeness of L (R ) that the sequenceu ˆj converges in L2(Rn). 2 n 2 n Deﬁnition 3.19 (Fourier transform on L (R )). For u ∈ L (R ) let uj denote an approximating sequence in S(Rn). Deﬁne the Fourier transform as follows:

Fu =u ˆ = lim uˆj . j→∞ Note well that F on L2(Rn) is well-deﬁned, as the limit is independent of the approximating sequence. In particular,

kuˆkL2( n) = lim kuˆjkL2( n) = lim kujkL2( n) = kukL2( n) . R j→∞ R j→∞ R R By the polarization identity

1 2 2 2 2 (u, v) 2 n = ku + vk 2 n − iku + ivk 2 n − (1 − i)kuk 2 n − (1 − i)kvk 2 n L (R ) 2 L (R ) L (R ) L (R ) L (R ) 1 we have proved the Plancheral theorem on L2(Rn): 2 n 2 n 2 n Theorem 3.20. (u, v)L (R ) = (Fu, Fv)L (R ) ∀u, v ∈ L (R ) . 3.5. Bounds for the Fourier transform on Lp(Rn). We have shown that for 1 n − n 2 n ∞ n 2 1 n 2 n u ∈ L (R ), kuˆkL (R ) ≤ (2π) kukL (R ), and that for u ∈ L (R ), kuˆkL (R ) = 2 n kukL (R ). Interpolating p between 1 and 2 yields the following result. Theorem 3.21 (Hausdorﬀ-Young inequality). If u ∈ Lp(Rn) for 1 ≤ p ≤ 2, the p−1 for q = p , there exists a constant C such that

q n p n kuˆkL (R ) ≤ CkukL (R ) . Returning to the case that u ∈ L1(Rn), not only is Fu ∈ L∞(Rn), but the transformed function decays at infinity. Theorem 3.22 (Riemann-Lebesgue “lemma”). For u ∈ L1(Rn), Fu is continuous and Fu(ξ) → 0 as |ξ| → ∞. n 1 n Proof. Let BM = B(0,M) ⊂ R . Since f ∈ L (R ), for each > 0, we can choose M sufficiently large such that fˆ(ξ) ≤ + R e−ix·ξ|f(x)|dx. Using Lemma 1.23, BM choose a sequence of simple functions φj(x) → f(x) a.e. on BM . For jnN chosen sufficiently large, Z ˆ −ix·ξ f(ξ) ≤ 2 + φj(x)e dx . BM PN Write φj(x) = l=1 Cl1El (x) so that N Z ˆ X −ix·ξ f(ξ) ≤ 2 + Cl φj(x)e dx . l=1 El

1The unitarity of the Fourier transform is often called Parseval’s theorem in science and engi- neering ﬁelds, based on an earlier (but less general) result that was used to prove the unitarity of the Fourier series. NOTES ON Lp AND SOBOLEV SPACES 39

By the regularity of the Lebesgue measure µ, for all > 0 and each l ∈ {1, ..., N}, there exists a compact set Kl and an open set Ol such that

µ(Ol) − /2 < µ(El) < µ(Kl) + /2 .

l α n Then Ol = {∪α∈Al Vα | Vl ⊂ R is open rectangle ,Al arbitrary set }, and Kl ⊂ Nl l ∪j=1Vj ⊂ Ol where {1, ..., Nl} ⊂ Al such that

Nl l |µ(El) − µ(∪j=1Vj )| < . It follows that

Z Z e−ix·ξdx − e−ix·ξdx < . Nl l El ∪j=1Vj

l R −ix·ξ On the other hand, for each rectangle Vj , l e dx| ≤ C/(ξ1 ··· ξn), so that Vj

1 fˆ(ξ) ≤ C + . ξ1 ··· ξn

Since > 0 is arbitrary, we see that fˆ(ξ) → 0 as |ξ| → ∞. Continuity of Fu follows easily from the dominated convergence theorem.

3.6. The Fourier transform and convolution.

Theorem 3.23. If u, v ∈ L1(Rn), then u ∗ v ∈ L1(Rn) and

n F(u ∗ v) = (2π) 2 Fu Fv .

Proof. Young’s inequality (Theorem 1.53) shows that u ∗ v ∈ L1(Rn) so that the Fourier transform is well-deﬁned. The assertion then follows from a direct compu- tation: Z − n −ix·ξ F(u ∗ v) = (2π) 2 e (u ∗ v)(x)dx n ZR Z − n −ix·ξ = (2π) 2 u(x − y)v(y)dy e dx n n ZR ZR − n −i(x−y)·ξ −iy·ξ = (2π) 2 u(x − y)e dx v(x) e dy n n R R n = (2π) 2 uˆvˆ (by Fubini’s theorem) .

By using Young’s inequality (Theorem 1.54) together with the Hausdorﬀ-Young inequality, we can generalize the convolution result to the following

Theorem 3.24. Suppose that u ∈ Lp( n) and v ∈ Lq( n), and let r satisfy R r R 1 1 1 r−1 n r = p + q − 1 for 1 ≤ p, q, r ≤ 2. Then F(u ∗ v) ∈ L (R ) and

n F(u ∗ v) = (2π) 2 Fu Fv . 40 STEVE SHKOLLER

4. The Sobolev Spaces Hs(Rn), s ∈ R The Fourier transform allows us to generalize the Hilbert spaces Hk(Rn) for s n k ∈ Z+ to H (R ) for all s ∈ R, and hence study functions which possess fractional derivatives (and anti-derivatives) which are square integrable. p Definition 4.1. For any s ∈ Rn, let hξi = 1 + |ξ|2, and set s n 0 n s 2 n H (R ) = {u ∈ S (R ) | hξi uˆ ∈ L (R )} 0 n s 2 n = {u ∈ S (R ) | Λ u ∈ L (R )} , where Λsu = F ∗(hξisuˆ). The operator Λs can be thought of as a “differential operator” of order s, and according to Rellich’s theorem, Λ−s is a compact operator, yielding the isomorphism s n −s 2 n H (R ) = Λ L (R ) . Definition 4.2. The inner-product on Hs(Rn) is given by s s s n s n 2 n (u, v)H (R ) = (Λ u, Λ v)L (R ) ∀u, v ∈ H (R ) . and the norm on Hs(Rn) is s s n kuk s n = (u, u) s n ∀u ∈ H ( ) . H (R ) H (R ) R s n s n The completeness of H (R ) with respect to the k · kH (R )) is induced by the completeness of L2(Rn). s n s n Theorem 4.3. For s ∈ R, (H (R ), k · kH (R )) is a Hilbert space. Example 4.4 (H1(Rn)). The H1(Rn) in Fourier representation is exactly the same as the that given by Definition 2.12: Z 2 2 2 kukH1( n) = hξi kuˆ(ξ)k dξ R n ZR = (1 + |ξ|2)kuˆ(ξ)k2dξ n ZR = (|u(x)|2 + |Du(x)|2)dx , n R the last equality following from the Plancheral theorem.

1 n 1 n Example 4.5 (H 2 (R )). The H 2 (R ) can be viewed as interpolating between decay required for uˆ ∈ L2(Rn) and uˆ ∈ H1(Rn): Z 1 n 2 n p 2 H 2 (R ) = {u ∈ L (R ) | 1 + |ξ|2|uˆ(ξ)| dξ < ∞} . n R Example 4.6 (H−1(Rn)). The space H−1(Rn) can be heuristically described as those distributions whose anti-derivative is in L2(Rn); in terms of the Fourier representation, elements of H−1(Rn) possess a transforms that can grow linearly at inﬁnity: Z 2 −1 n 0 n |uˆ(ξ)| H (R ) = {u ∈ S (R ) | 2 dξ < ∞} . n 1 + |ξ| R NOTES ON Lp AND SOBOLEV SPACES 41

For T ∈ H−s(Rn) and u ∈ Hs(Rn), the duality pairing is given by −s s 2 n hT, ui = (Λ T, Λ u)L (R ) , from which the following result follows.

Proposition 4.7. For all s ∈ R, [Hs(Rn)]0 = H−s(Rn) . The ability to define fractional-order Sobolev spaces Hs(Rn) allows us to refine the estimates of the trace of a function which we previously stated in Theorem 2.33. That result, based on the Gauss-Green theorem, stated that the trace operator was 1 n 2 n−1 continuous from H (R+) into L (R ). In fact, the trace operator is continuous 1 n 1 n−1 from H (R+) into H 2 (R ). To demonstrate the idea, we take n = 2. Given a continuous function u : R2 → {x1 = 0}, we define the operator

T u = u(0, x2) . The trace theorem asserts that we can extend T to a continuous linear map from 1 1 H (R2) into H 2 (R) so that we only lose one-half of a derivative. 1 2 1 Theorem 4.8. T : H (R ) → H 2 (R), and there is a constant C such that

kT uk 1 ≤ CkukH1( 2) . H 2 (R) R Before we proceed with the proof, we state a very useful result.

2 Lemma 4.9. Suppose that u ∈ S(R ) and define f(x2) = u(0, x2). Then Z ˆ 1 f(ξ2) = √ uˆ(ξ1, ξ2)dξ1 . 2π Rξ1 ˆ 1 R 1 ∗ R Proof. f(ξ2) = √ uˆ(ξ1, ξ2)dξ1 if and only if f(ξ2) = √ F uˆ(ξ1, ξ2)dξ1, and 2π R 2π R 1 Z 1 Z Z √ F ∗ uˆ(ξ , ξ )dξ = uˆ(ξ , ξ )dξ eix2ξ2 dξ . 1 2 1 2π 1 2 1 2 2π R R R On the other hand, 1 Z Z u(x , x ) = F ∗[û(ξ , ξ )] = uˆ(ξ , ξ )eix1ξ1+ix2ξ2 dξ dξ , 1 2 1 2 2π 1 2 1 2 R R so that 1 Z Z u(0, x ) = F ∗[û(ξ , ξ )] = uˆ(ξ , ξ )eix2ξ2 dξ dξ . 2 1 2 2π 1 2 1 2 R R 2 Proof of Theorem 4.8. Suppose that u ∈ S(R ) and set f(x2) = u(0, x1). Accord- ing to Lemma 4.9, Z Z ˆ 1 1 −1 f(ξ2) = √ uˆ(ξ1, ξ2)dξ1 = √ uˆ(ξ1, ξ2)hξi hξi dξ1 2π 2π Rξ1 Rξ1 1 1 Z 2 Z 2 1 2 2 −2 ≤ √ |uˆ(ξ1, ξ2)| hξi dξ1 hξi dξ1 , 2π R R and hence Z Z 2 2 2 −2 |f(ξ2)| ≤ C |uˆ(ξ1, ξ2)| hξi dξ1 hξi dξ1 . R R 42 STEVE SHKOLLER

R −2 The key to this trace estimate is the explicit evaluation of the integral hξi dξ1: R +∞ tan−1 √ ξ1 Z 2 1 1+ξ2 1 2 − 2 dξ1 = ≤ π(1 + ξ2 ) . (4.1) 1 + ξ2 + ξ2 p1 + ξ2 R 1 2 2 −∞ R 2 − 1 ˆ 2 R 2 2 It follows that (1+ξ2 ) 2 |f(ξ2)| dξ2 ≤ C |uˆ(ξ1, ξ2)| hξi dξ1, so that integration R R 2 of this inequality over the set {ξ2 ∈ R} yields the result. Using the density of S(R ) 1 2 in H (R ) completes the proof. The proof of the trace theorem in higher dimensions and for general Hs(Rn) 1 1 n spaces, s > 2 , replacing H (R ) proceeds in a very similar fashion; the only dif- R −2 R −2s ference is that the integral hξi dξ1 is replaced by n−1 hξi dξ1 ··· dξn−1, and R R instead of obtaining an explicit anti-derivative of this integral, an upper bound is instead found. The result is the following general trace theorem. s n n Theorem 4.10 (The trace theorem for H (R )). For s > 2 , the trace operator s n s− 1 n T : H (R ) → H 2 (R ) is continuous. We can extend this result to open, bounded, C∞ domains Ω ⊂ Rn. ∞ K Definition 4.11. Let ∂Ω denote a closed C manifold, and let {ωl}l=1 denote an open covering of ∂Ω, such that for each l ∈ {1, 2, ..., K}, there exist C∞-class charts ϑl which satisfy n−1 ∞ ϑl : B(0, rl) ⊂ R → ωl is a C diffeomorphism . ∞ Next, for each 1 ≤ l ≤ K, let 0 ≤ ϕl ∈ C0 (Ul) denote a partition of unity so that PL l=1 ϕl(x) = 1 for all x ∈ ∂Ω. For all real s ≥ 0, we define s 2 H (∂Ω) = {u ∈ L (∂Ω) : kukHs(∂Ω) < ∞} , where for all u ∈ Hs(∂Ω), K 2 X 2 kuk s = k(ϕ u) ◦ ϑ k s n−1 . H (∂Ω) l l H (R ) l=1 s The space (H (∂Ω), k · kHs(∂Ω)) is a Hilbert space by virtue of the completeness of Hs(Rn−1); furthermore, any system of charts for ∂Ω with subordinate partition of unity will produce an equivalent norm. n Theorem 4.12 (The trace map on Ω). For s > 2 , the trace operator T :Ω → ∂Ω is continuous. K Proof. Let {Ul}l=1 denote an n-dimensional open cover of ∂Ω such that Ul ∩ ∂Ω = ∞ ωl. Define charts θl : Vl → Ul, as in (2.9) but with each chart being a C map, such that ϑl is equal to the restriction of θl to the (n − 1)-dimensional ball B(0, rl) ⊂ n−1 ∞ R ). Also, choose a partition of unity 0 ≤ ζl ∈ C0 (Ul) subordinate to the covering Ul such that ϕl = ζl|ωl . 1 Then by Theorem 4.10, for s > 2 , K K 2 X 2 X 2 2 kuk 1 = k(ϕ u) ◦ ϑ k 1 ≤ C k(ϕ u) ◦ ϑ k s n ≤ Ckuk s . s− l l s− n−1 l l H ( ) H (Ω) H 2 (∂Ω) H 2 (R ) R l=1 l=1 NOTES ON Lp AND SOBOLEV SPACES 43

n Remark 4.13. The restriction s > 2 arises from the requirement that Z −2s hξi dξ1 ··· dξn−1 < ∞ . n−1 R s− 1 n−1 One may then ask if the trace operator T is onto; namely, given f ∈ H 2 (R ) 1 s n for s > 2 , does there exist a u ∈ H (R ) such that f = T u? By essentially reversing the order of the proof of Theorem 4.8, it is possible to answer this question in the aﬃrmative. We ﬁrst consider the case that n = 2 and s = 1.

1 2 1 Theorem 4.14. T : H (R ) → H 2 (R) is a surjection.

Proof. With ξ = (ξ1, ξ2), we deﬁne (one of many possible choices) the function u on R2 via its Fourier representation: hξ i uˆ(ξ , ξ ) = Kfˆ(ξ ) 1 , 1 2 1 hξi2

1 1 for a constant K 6= 0 to be determined shortly. To verify that kukH (R ) ≤ kfk 1 , note that H 2 (R) Z ∞ Z ∞ Z ∞ Z ∞ 2 2 2 ˆ 2 hξ1i |uˆ(ξ1, ξ2)| hξi dξ1dξ2 = K |f(ξ1)| 2 dξ1dξ2 −∞ −∞ −∞ −∞ hξi Z ∞ Z ∞ ˆ 2 2 1 = K |f(ξ1)| (1 + ξ1 ) 2 2 dξ2 dξ1 −∞ −∞ 1 + ξ1 + ξ2 2 ≤ Ckfk 1 , H 2 (R) where we have used the estimate (4.1) for the inequality above. It remains to prove that u(x1, 0) = f(x1), but by Lemma 4.9, it suﬃces that Z ∞ √ ˆ uˆ(ξ1, ξ2)dξ2 = 2πf(ξ1) . −∞ Integratingu ˆ, we ﬁnd that Z ∞ q Z ∞ ˆ 2 1 ˆ uˆ(ξ1, ξ2)dξ2 = Kf(ξ1) 1 + ξ1 2 2 dξ2 ≤ Kπf(ξ1) −∞ −∞ 1 + ξ1 + ξ2 √ so setting K = 2π/π completes the proof. A similar construction yields the general result.

1 1 s n s− 2 n−1 Theorem 4.15. For s > 2 , T : H (R ) → H (R ) is a surjection. By using the system of charts employed for the proof of Theorem 4.12, we also have the surjectivity of the trace map on bounded domains.

1 1 s s− 2 Theorem 4.16. For s > 2 , T : H (Ω) → H (∂Ω) is a surjection. The Fourier representation provides a very easy proof of a simple version of the Sobolev embedding theorem.

Theorem 4.17. For s > n/2, if u ∈ Hs(Rn), then u is continuous and

s n max |u(x)| ≤ CkukH (R ). 44 STEVE SHKOLLER

Proof. By Theorem 3.6, u = F ∗uˆ; thus according to H¨older’sinequality and the Riemann-Lebesgue lemma (Theorem 3.22), it suﬃces to show that

1 n s n kuˆkL (R ) ≤ CkukH (R ) . But this follows from the Cauchy-Schwarz inequality since Z Z |uˆ(ξ)dξ = |uˆ(ξ)|hξishξi−sdξ n n R R 1 1 Z 2 Z 2 ≤ |uˆ(ξ)|2hξi2sdξ hξi−2sdξ n n R R s n ≤ CkukH (R ) , the latter inequality holding whenever s > n/2. Example 4.18 (Euler equation on T2). On some time interval [0,T ] suppose that u(x, t), x ∈ T2, t ∈ [0,T ], is a smooth solution of the Euler equations: 2 ∂tu + (u · D)u + Dp = 0 in T × (0,T ] , 2 div u = 0 in T × (0,T ] , 1 2 with smooth initial condition u|t=0 = u0. Written in components, u = (u , u ) i i j satisﬁes ut +u ,j j +p,i = 0 for i = 1, 2, where we are using the Einstein summation i i convention for summing repeated indices from 1 to 2 and where u ,j = ∂u /∂xj and p,i = ∂p/∂xi. Computing the L2(T2) inner-product of the Euler equations with u yields the equality Z Z Z 1 d 2 i j i i |u(x, t)| dx + u ,j u u dx + p,i u dx = 0 . 2 dt 2 2 2 T T T | {z } | {z } I1 I2 Notice that Z Z 1 2 j 1 2 I1 = (|u| ),j u dx = |u| div udx = 0 , 2 2 2 2 T T the second equality arising from integration by parts with respect to ∂/∂xj. In- tegration by parts in the integral I2 shows that I2 = 0 as well, from which the d 2 conservation law ku(·, t)k 2 2 follows. dt L (T ) To estimate the rate of change of higher-order Sobolev norms of u relies on the use of the Sobolev embedding theorem. In particular, we claim that on a short enough time interval [0,T ], we have the inequality

d 2 3 ku(·, t)k 3 2 ≤ Cku(·, t)k 3 2 (4.2) dt H (T ) H (T ) 2 from which it follows that ku(·, t)k 3 2 ≤ M for some constant M < ∞. H (T ) To prove (4.2), we compute the H3(T2) inner-product of the Euler equations with u: Z Z 1 d 2 X α i j α i X α α i ku(·, t)kH3( 2) + D u ,j u D u dx + D p,i D u dx = 0 . 2 dt T 2 2 |α|≤3 T |α|≤3 T The third integral vanishes by integration by parts and the fact that Dα div u = 0; thus, we focus on the nonlinearity, and in particular, on the highest-order derivatives |α| = 3, and use D3 to denote all third-order partial derivatives, as well as NOTES ON Lp AND SOBOLEV SPACES 45 the notation l.o.t. for lower-order terms. We see that Z Z Z Z 3 i j 3 i 3 i j 3 i i 3 j 3 i D (u ,j u )D u dx = D u ,j u D u dx + u ,j D u D u dx + l. o. t. dx . 2 2 2 2 T T T T | {z } | {z } K1 K2 R 3 By deﬁnition of being lower-order terms, 2 l. o. t. dx ≤ Ckuk 3 2 , so it remains T H (T ) to estimate the integrals K1 and K2. But the integral K1 vanishes by the same argument that proved I1 = 0. On the other hand, the integral K2 is estimated by H¨older’sinequality: i 3 j 3 i ∞ 2 3 2 3 2 |K2| ≤ ku ,j kL (T ) kD u kH (T ) kD u kH (T ) . Thanks to the Sobolev embedding theorem, for s = 2 (s needs only to be greater than 1), i i ∞ 2 2 2 3 2 ku ,j kL (T ) ≤ Cku ,j kH (T ) ≤ kukH (T ) , 3 from which it follows that K2 ≤ Ckuk 3 2 , and this proves the claim. H (T ) Note well, that it is the Sobolev embedding theorem that requires the use of the space H3(T2) for this analysis; for example, it would not have been possible to establish the inequality (4.2) with the H2(T2) norm replacing the H3(T2) norm.

5. The Sobolev Spaces Hs(Tn), s ∈ R 5.1. Fourier Series: Revisited.

Definition 5.1. For u ∈ L1(Tn), define Z −n −ik·x Fu(k) =u ˆk = (2π) e u(x)dx , Tn and ∗ X ik·x F uˆ(x) = uˆke . n k∈Z Note that F : L1(Tn) → l∞(Zn). If u is sufficiently smooth, then integration by parts yields α |α| α α α1 αn F(D u) = −(−i) k uˆk, k = k1 ··· kn . Example 5.2. Suppose that u ∈ C1(Tn). Then for j ∈ {1, ..., n}, ∂u Z ∂u F (k) = (2π)−n e−ik·xdx ∂x n ∂x j T j Z −n −ik·x = −(2π) u(x)(−ikj) e dx n T = ikjuˆk . Note that Tn is a closed manifold without boundary; alternatively, one may iden- tify Tn with the [0, 1]n with periodic boundary conditions, i.e., with opposite faces identified.

Deﬁnition 5.3. Let s = S(Zn) denote the space of rapidly decreasing functions uˆ on Zn such that for each N ∈ N, N pN (u) = sup hki |uˆk| < ∞ . n k∈Z 46 STEVE SHKOLLER

Then ∞ n ∗ ∞ n F : C (T ) → s , F : s → C (T ) , and F ∗F = Id on C∞(Tn) and FF ∗ = Id on s. These properties smoothly extend to the Hilbert space setting: F : L2(Tn) → l2 F ∗ : l2 → L2(Tn) F ∗F = Id on L2(Tn) FF ∗ = Id on l2 . Deﬁnition 5.4. The inner-products on L2(Tn) and l2 are Z − n 2 n 2 (u, v)L (T ) = (2π) u(x)v(x)dx n T and X (ˆu, vˆ)l2 = uˆkvˆk , n k∈Z respectively.

2 n 2 Parseval’s identity shows that kukL (T ) = kuˆkl . Definition 5.5. We set 0 n ∞ n 0 0 0 D (T ) = [C (T )] and s = [s] . The space D0(Tn) is termed the space of periodic distributions. In the same manner that we extended the Fourier transform from S(Rn) to S0(Rn) by duality, we may produce a similar extension to the periodic distributions: F : D0(Tn) → s0 F ∗ : s0 → D0(Tn) F ∗F = Id on D0(Tn) FF ∗ = Id on s0 . Definition 5.6 (Sobolev spaces Hs(Tn)). For all s ∈ R, the Hilbert spaces Hs(Tn) are defined as follows: s n 0 n s n H (T ) = {u ∈ D (T ) | kukH (T ) < ∞} , where the norm on Hs(Tn) is defined as 2 X 2 2s kuk s n = |uˆ | hki . H (T ) k n k∈Z s n s n The space (H (T ), k · kH (T )) is a Hilbert space, and we have that −s n s n 0 H (T ) = [H (T )] .

5.2. The Poisson Integral Formula and the Laplace operator. For f : S1 → R, denote by PI(f)(r, θ) the harmonic function on the unit disk D = {x ∈ R2 : |x| < 1} with trace f: ∆ PI(f) = 0 in D 1 PI(f) = f on ∂D = S . PI(f) has an explicit representation via the Fourier series X ˆ |k| ikθ PI(f)(r, θ) = fkr e r < 1, 0 ≤ θ < 2π , (5.1) k∈Z NOTES ON Lp AND SOBOLEV SPACES 47 as well as the integral representation 1 − r2 Z f(φ) PI(f)(r, θ) = 2 dφ r < 1, 0 ≤ θ < 2π . (5.2) 2π 1 r − 2r cos(θ − φ) + 1 S The dominated convergence theorem shows that if f ∈ C0(S1), then PI(f) ∈ C∞(D) ∩ C0(D).

k− 1 1 k Theorem 5.7. PI extends to a continuous map from H 2 (S ) to H (D) for all k ∈ Z+. Proof. Define u = PI(f). − 1 Step 1. The case that k = 0. Assume that f ∈ H 2 (Γ) so that X ˆ 2 −1 |fk| hki ≤ M0 < ∞ . k∈Z Since the functions {r|k|eikθ : k ∈ Z} are orthogonal with respect to the L2(D) inner-product, 2 Z 2π Z 1 2 X ˆ |k| ikθ kukL2(D) = fkr e r dr dθ 0 0 k∈Z Z 1 X ˆ 2 2|k|+1 X ˆ 2 −1 2 ≤ 2π |f | r dr = π |f | (1 + |k|) ≤ πkfk 1 , k k 1 0 H 2 (S ) k∈Z k∈Z where we have used the monotone convergence theorem for the first inequality. 1 Step 2. The case that k = 1. Next, suppose that f ∈ H 2 (Γ) so that X ˆ 2 1 |fk| hki ≤ M1 < ∞ . k∈Z 2 Since we have shown that u ∈ L (D), we must now prove that uθ = ∂θu and 2 ur = ∂ru are both in L (D). Notice that by definition of the Fourier transform and (5.1), ∂ PI(f) = PI(f ) . (5.3) ∂θ θ 1 1 − 1 1 By definition, ∂θ : H 2 (S ) → H 2 (S ) continuously, so that for some constant C,

1 1 kfθk − 1 ≤ Ckfk 1 . H 2 (S ) H 2 (S ) It follows from the analysis of Step 1 and (5.3) that (with u = PI(f)),

2 1 kuθkL (D) ≤ Ckfk 1 . H 2 (S )

Next, using the identity (5.1) notice that |rur| = |uθ|. It follows that

2 1 krurkL (D) ≤ Ckfk 1 . (5.4) H 2 (S )

By the interior regularity of −∆ proven in Theorem 6.1, ur(r, θ) is smooth on {r < 1}; hence the bound (5.4) implies that, in fact,

2 1 kurkL (D) ≤ Ckfk 1 , H 2 (S ) and hence

1 1 kukH (D) ≤ Ckfk 1 , H 2 (S ) 48 STEVE SHKOLLER

k− 1 1 Step 3. The case that k ≥ 2. Since f ∈ H 2 (S ), it follows that k 1 1 k∂θ fk − 1 ≤ Ckfk k− 1 H 2 (S ) H 2 (S ) and by repeated application of (5.3), we ﬁnd that

k 1 kukH (D) ≤ Ckfk k− 1 . H 2 (S )

6. The Laplacian and its regularity We have studied the regularity properties of the Laplace operator on D = B(0, 1) ⊂ R2 using the Poisson integral formula. These properties continue to hold on more general open, bounded, C∞ subsets Ω of Rn. We revisit the Dirichlet problem ∆u = 0 in Ω , (6.1a) u = f on ∂Ω . (6.1b)

k− 1 Theorem 6.1. For k ∈ N, given f ∈ H 2 (∂Ω), there exists a unique solution u ∈ Hk(Ω) to (6.1) satisfying

kukHk(Ω) ≤ Ckfk k− 1 ,C = C(Ω) . H 2 (∂Ω) Proof. Step 1. k = 1. We begin by converting (6.1) to a problem with homo- geneous boundary conditions. Using the surjectivity of the trace operator provided by Theorem 4.16, there exists F ∈ H1(Ω) such that T (F ) = f on ∂Ω, and 1 kF kH1(Ω) ≤ Ckfk 1 . Let U = u − F ; then U ∈ H (Ω) and by linearity of the H 2 (∂Ω) 1 trace operator, T (U) = 0 on ∂Ω. It follows from Theorem 2.35 that U ∈ H0 (Ω) 1 1 and satisﬁes −∆U = ∆F in H0 (Ω); that is h−∆U, vi = h∆F, vi for all v ∈ H0 (Ω). 1 −1 According to Remark 2.45, −∆ : H0 (Ω) → H (Ω) is an isomorphism, so that ∆F ∈ H−1(Ω); therefore, by Theorem 2.44, there exists a unique weak solution 1 U ∈ H0 (Ω), satisfying Z 1 DU · Dv dx = h∆F, vi ∀v ∈ H0 (Ω) , Ω with kUkH1(Ω) ≤ Ck∆F kH−1(Ω) , (6.2) and hence 1 u = U + F ∈ H (Ω) and kukH1(Ω) ≤ kfk 1 . H 2 (∂Ω) Step 2. k = 2. Next, suppose that f ∈ H1.5(∂Ω). Again employing Theorem 4.16, 2 we obtain F ∈ H (Ω) such that T (F ) = f and kF kH2(Ω) ≤ CkfkH1.5(∂Ω); thus, we see that ∆F ∈ L2(Ω) and that, in fact, Z Z 1 DU · Dv dx = ∆F v dx ∀v ∈ H0 (Ω) . (6.3) Ω Ω

We ﬁrst establish interior regularity. Choose any (nonempty) open sets Ω1 ⊂⊂ ∞ Ω2 ⊂⊂ Ω and let ζ ∈ C0 (Ω2) with 0 ≤ ζ ≤ 1 and ζ = 1 on Ω1. Let 0 = min dist(spt(ζ), ∂Ω2)/2. For all 0 < < 0, deﬁne U (x) = η ∗ U(x) for all x ∈ Ω2, and set 2 v = −η ∗ (ζ U ,j ),j . NOTES ON Lp AND SOBOLEV SPACES 49

1 Then v ∈ H0 (Ω) and can be used as a test function in (6.3); thus, Z Z 2 2 − U,i η ∗ (ζ U ,j ),ji dx = − U,i η ∗ [ζ U ,ij +2ζζ,i U ,j ],j dx Ω Ω Z Z 2 = ζ U ,ij U ,ij dx − 2 η ∗ [ζζ,i U ,j ],j U,i dx , Ω2 Ω and Z Z Z 2 2 ∆F v dx = − ∆F η ∗ (ζ U ,j ),j dx = − ∆F η ∗ [ζ U ,jj +2ζζ,j U ,j ] dx . Ω Ω2 Ω2 By Young’s inequality (Theorem 1.53), 2 2 2 2 kη ∗ [ζ U ,jj +2ζζ,j U ,j ]kL (Ω2) ≤ kζ U ,jj +2ζζ,j U ,j kL (Ω2); hence, by the Cauchy-Young inequality with δ, Lemma 1.52, for δ > 0, Z 2 2 2 2 ∆F v dx ≤ δkζD U k 2 + C [kDU k 2 + k∆F k 2 ] . L (Ω2) δ L (Ω2) L (Ω) Ω Similarly, Z 2 2 2 2 2 η ∗ [ζζ, U , ], U, dx ≤ δkζD U k 2 + C [kDU k 2 + k∆F k 2 ] . i j j i L (Ω2) δ L (Ω2) L (Ω) Ω By choosing δ < 1 and readjusting the constant Cδ, we see that 2 2 2 2 2 2 kD U k 2 ≤ kζD U k 2 ≤ C [kDU k 2 + k∆F k 2 ] L (Ω1) L (Ω2) δ L (Ω2) L (Ω) 2 ≤ Cδk∆F kL2(Ω) , the last inequality following from (6.2), and Young’s inequality. Since the right-hand side does not depend on > 0, there exists a subsequence 2 0 2 D U * W in L (Ω1) . 1 2 By Theorem 2.17, U → U in H (Ω1), so that W = D U on Ω1. As weak conver- 2 2 2 gence is lower semi-continuous, kD UkL (Ω1) ≤ Ck∆F kL (Ω). As Ω1 and Ω2 are 2 arbitrary, we have established that U ∈ Hloc(Ω) and that

kUk 2 ≤ Ck∆F k 2 . Hloc(Ω) L (Ω) 1 2 For any w ∈ H0 (Ω), set v = ζw in (6.3). Since u ∈ Hloc(Ω), we may integrate by parts to ﬁnd that Z 1 (−∆U − ∆F ) ζw dx = 0 ∀w ∈ H0 (Ω) . Ω Since w is arbitrary, and the spt(ζ) can be chosen arbitrarily close to ∂Ω, it follows that for all x in the interior of Ω, we have that −∆U(x) = ∆F (x) for almost every x ∈ Ω . (6.4) We proceed to establish the regularity of U all the way to the boundary ∂Ω. K Let {Ul}l=1 denote an open cover of Ω which intersects the boundary ∂Ω, and let K {θl}l=1 denote a collection of charts such that ∞ θl : B(0, rl) → Ul is a C diﬀeomorphism ,

det Dθl = 1 ,

θl(B(0, rl) ∩ {xn = 0}) → Ul ∩ ∂Ω ,

θl(B(0, rl) ∩ {xn > 0}) → Ul ∩ Ω . 50 STEVE SHKOLLER

∞ Let 0 ≤ ζl ≤ 1 in C0 (Ul) denote a partition of unity subordinate to the open covering Ul, and define the horizontal convolution operator, smoothing functions defined on Rn in the first 1, ..., n − 1 directions, as follows: Z ρ ∗h F (xh, xn) = ρ(xh − yh)F (yh, xn)dyh , n−1 R −(n−1) n−1 where ρ(xh) = ρ(xh/), ρ the standard mollifier on R , and xh = (x1, ..., xn−1). Let α range from 1 to n − 1, and substitute the test function 2 −1 1 v = − ρ ∗h [(ζl ◦ θl) ρ ∗h (U ◦ θl),α ],α ◦ θl ∈ H0 (Ω) into (6.3), and use the change of variables formula to obtain the identity Z Z k j Ai (U ◦ θl),k Ai (v ◦ θl),j dx = (∆F ) ◦ θl v ◦ θl dx , (6.5) B+(0,rl) B+(0,rl) ∞ −1 where the C matrix A(x) = [Dθl(x)] and B+(0, rl) = B(0, rl) ∩ {xn > 0}. We define

l U = U ◦ θl , and denote the horizontal convolution operator by H = ρ ∗h .

Then, with ξl = ζl ◦ θl, we can rewrite the test function as 2 l v ◦ θl = −H[ξl HU ,α ],α . Since diﬀerentiation commutes with convolution, we have that

2 l l (v ◦ θl),j = −H(ξl HU ,jα ),α −2H(ξlξl,j HU ,α ),α , and we can express the left-hand side of (6.5) as Z k j Ai (U ◦ θl),k Ai (v ◦ θl),j dx = I1 + I2 , B+(0,rl) where Z j k l 2 l I1 = − Ai Ai U ,k H(ξl HU ,jα ),α dx , B+(0,rl) Z j k l l I2 = −2 Ai Ai U ,k H(ξlξl,j HU ,α ),α dx . B+(0,rl) Next, we see that Z j k l 2 l I1 = [H(Ai Ai U ,k )],α (ξl HU ,jα ) dx = I1a + I1b , B+(0,rl) where Z j k l 2 l I1a = (Ai Ai HU ,k ),α ξl HU ,jα dx , B+(0,rl) Z j k l 2 l I1b = ([H,Ai Ai ]U ,k ),α ξl HU ,jα dx , B+(0,rl) and where j k l j k l j k l [H,Ai Ai ]U ,k = H(Ai Ai U ,k ) − Ai Ai HU ,k (6.6) NOTES ON Lp AND SOBOLEV SPACES 51 denotes the commutator of the horizontal convolution operator and multiplication. The integral I1a produces the positive sign-deﬁnite term which will allow us to build the global regularity of U, as well as an error term: Z 2 j k l l j k l 2 l I1a = [ξl Ai Ai HU ,kα HU ,jα +(Ai Ai ),α HU ,k ξl HU ,jα ] dx ; B+(0,rl) thus, together with the right hand-side of (6.5), we see that

Z Z 2 j k l l j k l 2 l ξl Ai Ai HU ,kα HU ,jα dx ≤ (Ai Ai ),α HU ,k ξl HU ,jα ] dx B+(0,rl) B+(0,rl)

+ |I1b| + |I2| + (∆F ) ◦ θl v ◦ θl dx . B+(0,rl) ∞ T Since each θl is a C diﬀeomorphism, it follows that the matrix AA is positive deﬁnite: there exists λ > 0 such that 2 j k n λ|Y | ≤ Ai Ai YjYk ∀Y ∈ R . It follows that

Z Z 2 ¯ l 2 j k l 2 l λ ξl |∂DHU | dx ≤ (Ai Ai ),α HU ,k ξl HU ,jα ] dx B+(0,rl) B+(0,rl)

+ |I1b| + |I2| + (∆F ) ◦ θl v ◦ θl dx , B+(0,rl) where D = (∂x1 , ..., ∂xn ) andp ¯ = (∂x1 , ..., ∂xn−1 ). Application of the Cauchy-Young inequality with δ > 0 shows that

Z Z j k l 2 l (Ai Ai ),α HU ,k ξl HU ,jα ] dx + |I2| + (∆F ) ◦ θl v ◦ θl dx B+(0,rl) B+(0,rl) Z 2 ¯ l 2 2 ≤ δ ξl |∂DHU | dx + Cδk∆F kL2(Ω) . B+(0,rl)

It remains to establish such an upper bound for |I1b|. jk j k To do so, we ﬁrst establish a pointwise bound for (6.6): for A = Ai Ai , Z j k l jk jk l [H,Ai Ai ]U ,k (x) = ρ(xh − yh)[A (yh, xn) − A (xh, xn)]U ,k (yh, xn) dyh B(xh,) jk jk 1,∞ By Morrey’s inequality, |[A (yh, xn) − A (xh, xn)]| ≤ CkAkW (B+(0,rl)). Since 1 x − h − y ∂ ρ (x − y ) = ρ0 h , xα h h 2 we see that Z 1 x − h − y ∂ [H ,AjAk]U l, (x) ≤ C ρ0 h |U l, (y , x )| dy xα i i k k h n h B(xh,) and hence by Young’s inquality,

j k l 1 2 ∂xα [H,A A ]U ,k ≤ CkUkH (Ω) ≤ Ck∆F kL (Ω) . i i 2 L (B+(0,rl) 52 STEVE SHKOLLER

It follows from the Cauchy-Young inequality with δ > 0 that Z 2 ¯ l 2 2 |I1b| ≤ δ ξl |∂DHU | dx + Cδk∆F kL2(Ω) . B+(0,rl) By choosing 2δ < λ, we obtain the estimate Z 2 ¯ l 2 2 ξl |∂DHU | dx ≤ Cδk∆F kL2(Ω) . B+(0,rl) Since the right hand-side is independent of , we ﬁnd that Z 2 ¯ l 2 2 ξl |∂DU | dx ≤ Cδk∆F kL2(Ω) . (6.7) B+(0,rl)

From (6.4), we know that ∆U(x) = ∆F (x) for a.e. x ∈ Ul. By the chain-rule this means that almost everywhere in B+(0, rl), jk l jk l −A U ,kj = A ,j U ,k +∆F ◦ θl , or equivalently, nn l jα l βk l jk l −A Unn = A U ,αj +A U ,kβ +A ,j U ,k +∆F ◦ θl . Since Ann > 0, it follows from (6.7) that Z 2 2 l 2 2 ξl |D U | dx ≤ Cδk∆F kL2(Ω) . (6.8) B+(0,rl) Summing over l from 1 to K and combining with our interior estimates, we have that kukH2(Ω) ≤ Ck∆F kL2(Ω) . Step 3. k ≥ 3. At this stage, we have obtained a pointwise solution U ∈ H2(Ω) ∩ 1 k−1 H0 (Ω) to ∆U = ∆F in Ω, and ∆F ∈ H . We diﬀerentiate this equation r times r 2 until D ∆F ∈ L (Ω), and then repeat Step 2.

Department of Mathematics, University of California, Davis, CA 95616, USA E-mail address: [email protected]