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L P and Sobolev Spaces

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L P and Sobolev Spaces NOTES ON Lp AND SOBOLEV SPACES STEVE SHKOLLER 1. Lp spaces 1.1. Definitions and basic properties. Definition 1.1. Let 0 < p < 1 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) := jfj dx and kfkL1(X) := ess supx2X jf(x)j : X Note that kfkLp(X) may take the value 1. Definition 1.2. The space Lp(X) is the set p L (X) = ff : X ! R j kfkLp(X) < 1g : The space Lp(X) satisfies the following vector space properties: (1) For each α 2 R, if f 2 Lp(X) then αf 2 Lp(X); (2) If f; g 2 Lp(X), then jf + gjp ≤ 2p−1(jfjp + jgjp) ; so that f + g 2 Lp(X). (3) The triangle inequality is valid if p ≥ 1. The most interesting cases are p = 1; 2; 1, 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 < 1, we set ( 1 ) p 1 X p l = fxngn=1 j jxnj < 1 : n=1 1.2. Basic inequalities. Lemma 1.4. For λ 2 (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 2 (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 ≤ 1 and 1 < q < 1 with 1 1 p q 1 p + q = 1. If f 2 L and g 2 L , then fg 2 L . Moreover, kfgkL1 ≤ kfkLp kgkLq : Note that if p = q = 2, then this is the Cauchy-Schwarz inequality since kfgkL1 = j(f; g)L2 j. Proof. We use Lemma 1.5. Let λ = 1=p and set jfjp jgjq a = p ; and b = q kfkLp kgkLp for all x 2 X. Then aλb1−λ = a1=pb1−1=p = a1=pb1=q so that jfj · jgj 1 jfjp 1 jgjq ≤ p + q : kfkLp kgkLq p kfkLp q kgkLq Integrating this inequality yields Z jfj · jgj Z 1 jfjp 1 jgjq 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 ≤ 1 and f; g 2 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 jf + gjp = jf + gj · jf + gjp−1 ≤ (jfj + jgj)jf + gjp−1 ; and integrate over X to find that Z Z jf + gjpdx ≤ (jfj + jgj)jf + gjp−1 dx X X H¨older's p−1 p p ≤ (kfkL + kgkL ) jf + gj Lq : p Since q = p−1 , 1 Z q p−1 p jf + gj Lq = jf + gj dx ; X from which it follows that 1− 1 Z q p jf + gj dx ≤ kfkLp + kgkLq ; X 1 1 which completes the proof, since p = 1 − q . Corollary 1.9. For 1 ≤ p ≤ 1, 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 2 L1(Ω) satisfies f(x) ≥ M > 0 for almost all x 2 Ω, then log(f) 2 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!1 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 2 L (X), 0 ≤ f1(x) ≤ f2(x) ≤ · · ·, and kfnkL1(X) ≤ 1 C < 1, then limn!1 fn(x) = f(x) with f 2 L (X) and kfn − fkL1 ! 0 as n ! 0. 1 Lemma 1.12 (DCT). If fn 2 L (X), limn!1 fn(x) = f(x) a.e., and if 9 g 2 1 1 L (X) such that jfn(x)j ≤ jg(x)j a.e. for all n, then f 2 L (X) and kfn−fkL1 ! 0. Proof. Apply the Dominated Convergene Theorem to the sequence hn = jfn −fj ! 0 a.e., and note that jhnj ≤ 2g. Theorem 1.13. If 1≤ p < 1 then Lp (X) is a Banach space. 1 Proof. Step 1. The Cauchy sequence. Let ffngn=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- 1 quence fgngn=1 as g1 = 0; gn = jf1j + jf2 − f1j + ··· + jfn − fn−1j for n ≥ 2 : It follows that 0 ≤ g1 ≤ g2 ≤ · · · ≤ gn ≤ · · · so that gn is a monotonically increasing sequence. Furthermore, fgng is uniformly bounded in Lp as Z 1 !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 2 L , and gn ≤ g a.e. Step 3. Pointwise convergence of ffng. For all k ≥ 1, jfn+k − fnj = jfn+k − fn+k−1 + fn+k−1 + · · · − fn+1 + fn+1 − fnj k+1 X ≤ jfi − fi−1j = gn+k − gn −! 0 a.e. i=n+1 Therefore, fn ! f a.e. Since n X jfnj ≤ jf1j + jfi − fi−1j ≤ gn ≤ g for all n 2 N ; i=2 p p p p p p it follows that jfj ≤ g a.e. Hence, jfnj ≤ g , jfj ≤ g , and jf − fnj ≤ 2g , and by the Dominated Convergence Theorem, Z Z p p lim jf − fnj dx = lim jf − fnj dx = 0 : n!1 X X n!1 p p 1.4. Convergence criteria for L functions. If ffng is a sequence in L (X) p which converges to f in L (X), then there exists a subsequence ffnk g such that fnk (x) ! f(x) for almost every x 2 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 2 [0; 1]. Example 1.15. Set X = R, and for n 2 N, set fn = 1[n;n+1]. Then fn(x) ! 0 as p n ! 1, but kfnkLp = 1 for p 2 [1; 1); thus, fn ! 0 pointwise, but not in L . Example 1.16. Set X = [0; 1], and for n 2 , set fn = n1 1 . Then fn(x) ! 0 N [0; n ] p a.e. as n ! 1, but kfnkL1 = 1; thus, fn ! 0 pointwise, but not in L . p Theorem 1.17. For 1 ≤ p < 1, suppose that ffng ⊂ L (X) and that fn(x) ! p f(x) a.e. If limn!1 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 ja−bj ≤ 2 (jaj +jbj ). Set a = fn and b = f to obtain the inequality p−1 p p p 0 ≤ 2 (jfnj + jfj ) − jfn − fj : Since fn(x) ! f(x) a.e., Z Z p p p−1 p p p 2 jfj dx = lim 2 (jfnj + jfj ) − jfn − fj dx : X X n!1 Thus, Fatou's lemma asserts that Z Z p p p−1 p p p 2 jfj dx ≤ lim inf 2 (jfnj + jfj ) − jfn − fj dx X n!1 X NOTES ON Lp AND SOBOLEV SPACES 5 Since kfnkLp(X) ! kfkLp(X), we see lim supn!1 kfn − fkLp(X) = 0.
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