MIT OpenCourseWarehttp://ocw.mit.edu 18.102 Introduction to Functional Analysis Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.� � � � � � � � � � 41 LECTURE NOTES FOR 18.102, SPRING 2009 Lecture 7. Thursday, Feb 26 So, what was it with my little melt-down? I went too cheap on the monotonicity theorem and so was under-powered for Fatou’s Lemma. In my defense, I was trying to modify things on-the-fly to conform to how we are doing things here. I should also point out that at least one person in the audience made a comment which amounted to pointing out my error. So, here is something closer to what I should have said – it is not far from what I did say of course. Proposition 12. [Montonicity again] If fj ∈ L1(R) is a monotone sequence, either fj (x) ≥ fj+1(x) for all x ∈ R and all j or fj (x) ≤ fj+1(x) for all x ∈ R and all j, and fj is bounded then (7.1) {x ∈ R; lim fj (x) is finite} = R \ E j→∞ where E has measure zero and f = lim fj (x) a.e. is an element of L1(R) (7.2) j→∞ � with lim f − fj = 0. j→∞ | | Moral of the story – drop the assumption of positivity and replace it with the bound on the integral. In the approach through measure theory this is not necessary because one has the concept of a measureable, non-negative, function for which the integral ‘exists but is infinite’ – we do not have this. Proof. Since we can change the sign of the fi (now) it suffices to assume that the fi are monotonically increasing. The sequence of integrals is therefore also montonic increasing and, being bounded, converges. Thus we can pass to a subsequence gi = fni with the property that (7.3) |gj − gj−1| = gj − gj−1 < 2−j ∀ j > 1. This means that the series h1 = g1, hj = gj − gj−1, j > 1, is absolutely summable. So we know for the result last time that it converges a.e., that the limit, f, is integrable and that � � j� � (7.4) f = lim hk = lim gj = lim fj . j→∞ k=1 j→∞ n→∞ In fact, everywhere that the series hj (x), which is to say the sequence gk(x), j converges so does fn(x), since the former is a subsequence of the latter which is monotonic. So we have (7.1) and the first part of (7.2). The second part, corresponding to convergence for the equivalence classes in L1(R) follows from monotonicity, since (7.5) |f − fj | = f − fj → 0 as j → ∞.� � � � � � � � � � 42 LECTURE NOTES FOR 18.102, SPRING 2009 Now, to Fatou’s Lemma. This really just takes the monotonicity result and applies it to a general sequence of integrable functions with bounded integral. You should recall – as I did – that the max and min of two integrable functions is integrable and that (7.6) min(f, g) ≤ min( f, g). Lemma 3. [Fatou]. Let fj L1(R) be a sequence of non-negative (so real-valued integrable) functions such that fj is bounded above in R, then f(x) = lim inf fn(x) exists a.e., f ∈ L1(R) and n→∞(7.7) � � lim inf fn ≤ lim inf fn. Proof. You should remind yourself of the properties of liminf as necessary! Fix k and consider (7.8) Fk,n = min fp1(R) k≤p≤k+n (x) ∈ Las discussed briefly above. Moreover, this is a decreasing sequence, as n increases, because the minimum is over an increasing set an all elements are non-negative. Thus the integrals are bounded below by 0 so the monotonicity result above applies and shows that (7.9) gk(x) = inf fp(x) ∈ L1(R), gk ≤ fn ∀ n ≥ k. p≥k Note that for a decreasing sequence of non-negative numbers the limit exists every-where and is indeed the infimum. Thus in fact, (7.10) gk ≤ lim inf fn. Now, let k vary. Then, the infimum in (7.9) is over a set which decreases as k increases. Thus the gk(x) are increasing. The integral is always bounded by one of the fn� and hence is bounded above independent of k since we assumed a bound on the fn’s. So, now we can apply the monotonicity result again to see that f(x) = lim gk(x) exists a.e and f ∈ L1(R) has (7.11) k→∞ � � f ≤ lim inf fn. Since f(x) = lim inf fn(x), by definition of the latter, we have proved the Lemma. Now, we apply Fatou’s Lemma to prove what we are really after:-Theorem 2. [Lebesgue’s dominated convergence]. Suppose fj ∈ L1(R) is a se-quence of integrable functions such that (7.12) ∃ h ∈ L1(R) with |fj (x)| ≤ h(x) a.e. and f(x) = lim fj (x) exists a.e. � n→∞� Then f ∈ L1(R) and f = limn→∞ fn (including the assertion that this limit exists).� � � � � � � � � � � � � � � � � � 43 LECTURE NOTES FOR 18.102, SPRING 2009 Proof. First, we can assume that the fj are real since the hypotheses hold for its real and imaginary parts and together give the desired result. Moroever, we can change all the fj ’s to make them zero on the set on which the estimate in (7.12) does not hold. Then this bound on the fj ’s becomes (7.13) −h(x) ≤ fj (x) ≤ h(x) ∀ x ∈ R. In particular this means that gj = g − fj is a non-negative sequence of integrable functions and the sequence of integrals is also bounded, since (7.12) also implies that |fj | ≤ h, so gj ≤ 2 h. Thus Fatou’s Lemma applies to the gj . Since we have assumed that the sequence gj (x) converges a.e. to f we know that h − f(x) = lim inf gj (x) a.e. and (7.14) � � � � � h − f ≤ lim inf (h − fj ) = h − lim sup fj . Notice the change on the right from liminf to limsup because of the sign. Now we can apply the same argument to gj� (x) = h(x) + fj (x) since this is also non-negative and has integrals bounded above. This converges a.e. to h(x) + f(x) so this time we conclude that (7.15) h + f ≤ lim inf (h + fj ) = h + lim inf fj . In both inequalities (7.14) and (7.15) we can cancel and h and combining them we find (7.16) lim sup fj ≤ f ≤ lim inf fj . In particular the limsup on the left is smaller than, or equal to, the liminf on the right, for the same real sequence. This however implies that the are equal and that the sequence fj converges (look up properties of liminf and limsup if necessary ...). Thus indeed (7.17) f = lim fn. n→∞