Partial Differential Equations Mollification (With one colorful figure) 1 Lp-spaces, local integrability I will assume you know what it means for a function to be measurable, integrable, etc. In this course, measurability always means Lebesgue measurability, integrability is Lebesgue integrability. One very nice feature of Lebesgue integrability is that if E is a measurable subset of Rn of positive measure1 and f : E ! R is measurable and non-negative (f(x) ≥ 0 for all x 2 E), then Z f(x) dx E is always defined, possibly equal to 1. A function f : E ! R is integrable over E, in symbols f 2 L1(E), if and only if Z jf(x)j dx < 1: E More generally, one defines Lp(E) for 1 ≤ p; 1 by: f : E ! R is in Lp(E), if and only if Z jf(x)jp dx < 1: E 1 One also defines f 2 L (E) iff ess sup x2Ejf(x)j < 1. The essential supremum or ess sup of a function f over a set E can be defined as the infimum (strangely enough) of the sup of all functions g that are a.e. to f: ess sup x2Ef = inffsup g(x): g(x) = f(x) for a:e:x 2 Eg: x2E This is a notation I'll be using on occasions: Assume p 2 [1; 1]. The conjugate index of p will be denoted by p0; it is defined to be the number in [1; 1] so that the equation 1 1 + = 1 p p0 holds. If p = 1, one interprets 1=p0 = 0 as meaning p0 = 1; if p = 1, one interprets 1=p = 0, so p0 = 1. It would have been, perhaps, easier to simply say: p p0 = ; p − 1 but the equation 1=p + 1=p0 = 1 is usually preferred; it makes it clear at once that p; p0 play symmetric roles and that (p0)0 = p. A few things to notice are: 20 = 2; if 1 ≤ p < 2, then 2 < p0 ≤ 1, and vice-versa: 2 < p ≤ 1 implies 1 ≤ p0 < 2. One defines a norm in the spaces Lp(E) by ( 1=p R jf(x)jp dx ; if p < 1, kfkp = kfkLp(E) = E ess sup x2Ejf(x)j if p = 1. To see it is a norm is not totally trivial. That is, it is trivially a norm (if we interpret, as one does, f = 0 to mean f = 0 a.e.) for p = 1 and for p = 1. Otherwise it is not so obvious. To prove it is a norm also for 1 < p < 1, one 1Lebesgue integration over null sets is a rather stupid endeavor, done only by people who love the number zero so much they can't keep their hands off it. 1 LP -SPACES, LOCAL INTEGRABILITY 2 first proves a fundamental inequality: H¨older'sinequality. It states: Let E be a measurable subset of n and let 0 R f 2 Lp(E), g 2 Lp (E), then fg 2 L1(E) and Z jf(x)g(x)j dx ≤ kfkLp(E)kgkLp0 (E): (1) n R The proof of (1) is based on the following result, which could be a nice exercise in a Calculus course: Assume a; b are two non-negative real numbers. Then, if 1 < p < 1 and p0 = p=(p − 1), 1 1 0 ab ≤ ap + bp p p0 The hint one can give for this exercise is that it suffices to prove it for a; b > 0, to prove first that '(x) = 0 0 xp=p + x−p =p0 ≥ 1 for all x > 0, and to apply it with x = a1=p b−1=p. Once one has this inequality, given now p p0 f 2 L (E), g 2 L (E), one applies it with a = jf(x)j=kfkp, b = jg(x)j=kgkp0 to get 0 1 jf(x)jp jg(x)jp jf(x)g(x)j ≤ p + p0 kfk kgk 0 kfk p p p kgkp0 for a.e. x 2 E. Integrating over E gives 1 Z 1 1 jf(x)g(x)j dx ≤ + 0 = 1: kfkpkgkp0 E p p This proves (1) if 1 < p < 1; the proof when p = 1 (or equivalently when p = 1) is trivial. If p = 2, (1) is known as the Cauchy-Schwarz inequality in the Western part of the world. The Russians prefer to call it the Bunyakovsky inequality, since Viktor Bunyakovsky (1804-1889) had it some 25 years before Cauchy or Schwarz. With (1) it is now easy to prove also for 1 < p < 1 that kf + gkLp(E) ≤ kfkLp(E) + kgkLp(E) (2) p for all f; g 2 L (E); in other words, the main step in seeing k · kLp(E) is a norm. The cases p = 1; 1 being, as mentioned, trivial (I hope that if I say a sufficient number of times that it is trivial you will believe it), one assumes 1 < p < 1 and proceeds as follows (notice that (p − 1)p0 = p): Z Z Z Z p p p−1 p−1 p−1 kf + gkp = jf + gj dx = jf + gjjf + gj dx ≤ jfj jf + gj dx + jgj jf + gj dx E E E E 1=p 1=p0 1=p 1=p0 Z Z p0 Z Z p0 ≤ jfjp dx jf + gjp−1 dx + jgjp dx jf + gjp−1 dx E E E E p=p0 p=p0 = kfkp kf + gkp + kgkp kf + gkp : 0 p=p 0 Dividing both sides by kf + gkp , (2) follows, since p − (p=p ) = p − (p − 1) = 1. p FACT: The spaces L (E) are complete in the metric defined by the k · kLp(E) norm, thus Banach spaces. Definition 1 Let U be open in Rn and let f : U ! R be measurable. We say f is locally integrable in U and write f 2 L1 (U) iff loc Z jf(x)j dx < 1 K for all compact sets K ⊂ U. Notice that if the restriction of f to a compact set K satisfies kfkLp(K) < 1; then Z jf(x)j dx < 1 K This is clear if p = 1, since then both statements say the same thing. Otherwise, with χK denoting the characteristic function of K, jKj the Lebesgue measure of K, Z Z 1=p0 jf(x)j dx = χK (x)jf(x)j dx ≤ kχK kLp0 (K)kfkLp(K) = jKj kfkLp(K) < 1: K K 2 CONVOLUTIONS 3 2 Convolutions Let f; g : Rn ! R be measurable. Then it is not too hard to show that for almost all x 2 Rn, the function y 7! f(x − y)g(y) is measurable. If (and only if) it is also integrable for almost all x 2 Rn, one defines a function f ∗ g : n ! by R R Z f ∗ g(x) = f(x − y)g(y) dy n R for almost all x 2 Rn. I do not know (and I don't really care to know) the EXACT conditions on f; g so that this product is defined. But it is useful to know some basic conditions under which the convolution product is defined. p n q n r n 1 1 1 • If f 2 L (R ); g 2 L (R ) and 1=p + 1=q ≥ 1, then f ∗ g is defined, f ∗ g 2 L (R ) where r = p + q − 1. Moreover, r n p n q n kf ∗ gkL (R ) ≤ kfkL (R ) kgkL (R ): (3) Important subcases of this fact are: The case p = q = 1. In this case r = 1 and one sees that L1(Rn) is an algebra under the convolution product. It is also frequently used with p = 1. Then r = q and it becomes q n 1 n q n kf ∗ gkL (R ) ≤ kfkL (R ) kgkL (R ): The case p = q = 2. In this case r = 1 so f ∗ g is bounded. One can prove a bit more: In this case, f ∗ g is continuous. More generally, if q = p0, then r = 1 and one can prove also that f ∗ g is continuous. 1 n • If f; g 2 Lloc(R ), and at least one of them vanishes outside of a compact set, then f ∗ g is defined. If both vanish outside of compact sets, then f ∗g is locally integrable and vanishes outside of a compact set. Generally speaking, if f(x) = 0 for x2 = A and g(x) = 0 for x2 = B, then f ∗ g(x) = 0 for x2 = A + B. Specifically, supp f ∗ g ⊂ supp f + supp g: Proof of (3). Inequality (3) is known as the Hausdorff, sometimes Hausdorff-Young, inequality. Its proof consists in a moderately clever application of H¨older'sinequality, more precisely of a simple generalization of H¨older's inequality: Let p1; : : : ; pn 2 [1; 1] be such that 1 1 + ··· + = 1: p1 pn Then Z jf1f2 ··· fnj dx ≤ kf1kLp1 (E)kf2kLp2 (E) · · · kfnkLpn (E): (4) E This is easily proved by induction on n; the case n = 2 being just regular H¨older.We need the case n = 3. 1 1 1 1 1 Assume thus that p; q 2 [1; 1], that p + q ≥ 1, and set r = p + q − 1 (notice, incidentally, that 1=p + 1=q ≤ 2 so that 0 ≤ 1=r ≤ 1 and r 2 [1; 1]). Let f 2 Lp(Rn); g 2 Lq(Rn); we may assume that f; g only assume non-negative values. Now 1 1 1 1 1 1 + + = 2 − − + = 1 q0 p0 r q p r and we can use them as powers for an application of H¨older.