Real Analysis 1 Fall Homework 3. a n.

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1 eal Analysis Fall 06 Homework 3. Let and consider the measure sace N, P, µ, where µ is counting measure. That is, if N, then µ equals the number of elements in if is finite; µ = otherwise. One usually denotes the sace L N, PN, µ by l. More generally, if is any set and µ is counting measure on, one writes l for L µ. The most common set used is N, followed by Z. I will use the usual notation for functions defined on N, namely the sequence notation: If a : N S S some set, I ll write a n for an. If you have never done this before you may want to verify that a function a : N C is integrable iff and only if a n < and then N a dµ = a n. The sace c 0 is defined as the set of all sequences a of comlex or real if one wants to stay real numbers such that n a n = 0. Show that c 0 is a closed subsace of l, thus a Banachsace in its own right. Prove: c 0 = l.. Let, M be a measurable sace and let ν : M C be a comlex measure. ecall that the total variation measure determined by ν is defined by ν = su{ ν i :,,... M, n m = if n m, n }. i= It is always a ositive, finite measure on M a similar definition if ν is a signed measure that can take on the value or will roduce a not necessarily finite measure. Prove: There exists h : C measurable such that hx = for all x and ν = h d ν for all M. The hint, of course, is adon-nikodym. More hints are given at the end of this exercise set. 3. Let, M, µ be as usual a measure sace and let f L µ. Let λ : M C be defined by λ = f dµ. Then λ = f dµ for all M. Hint: xercise 4. Let = {a, b} be a set consisting of two oints, a b. Let M = {, {a}, {b}, } = P and define the measure µ on M by µ = 0, µ{a} =, µ{b} = µ =. Convince yourself that, M, µ is a measure sace. Prove: L, M, µ L, M, µ.

2 Why does this not violate what F calls the Kantorovitch eresentation Theorem? 5. Let f L µ, where, M, µ is a measure sace and < <. In class we roved that L µ = L µ where + = if, M, µ is σ-finite. xtend the roof to the non-σ-finite case. 6. This exercise is about the Banach-Saks theorem, but we begin taking another swie at F The last theorem of Section 9.4 of F is the Banach-Saks Theorem. The text states that the roof is identical to the roof for the case of Lebesgue measure, and refers to Chater 4 for the roof. In Chater 4 the first statement of the roof is to assume = and a reference to the original aer by Stehan Banach and Stanislaw Saks, which I ersonally find extremely annoying. In the sace occuied by stating the same theorem twice and giving a articular roof, it could have been roved in generality once and for all. So I thought it would be a good idea to learn the roof of the Banach-Saks Theorem from the original aer by the two masters, Sur la convergence forte forte dans les chams L, ublished in the Polish ournal Studia Mathematica Studia Math, 930, Banach and Saks work with Lebesgue measure but, as F so wisely observes, the roof is the same in any measure sace, so I will state it in general: Theorem Banach-Saks Let, M, µ be a measure sace and let,. Assume {f n } is a bounded sequence in L µ. Then there is a subsequence {f nk } of {f n } such that the sequence { k k = f n } converges strongly in the L norm. This is a rather imortant and useful theorem. There are two differences with the statement in F. oyden assumes the sequence converges weakly, Banach and Saks assume that the sequence is bounded. However, the first ste in their roof is to say that a bounded sequence in L has a weakly convergent subsequence. This is due to the fact that L is reflexive if < <. Second, F assumes that the measure sace is σ-finite. That is not necessary, esecially not if the assumtion is that the sequence converges weakly. The only reason for assuming that the measure is σ- finite is because the roof that the dual of L is L if < as done in F and in class requires a σ-finite measure. So if they actually roved the theorem assuming only that the sequence is bounded, they would need a σ-finite measure to extract the weakly convergent subsequence. But, having already the subsequence, and not roving it excet for = it isn t clear why F adds an additional hyothesis. Moreover, while the assumtion that the measure is σ-finite cannot be lifted in the roof that L = L see xercise 4, it can be lifted see xercise 5 if < <. Well, let s get on with the roof. We have this sequence {f n } in L µ, where, ; by reflexivity we may assume {f n } converges weakly to some f L µ; subtracting f from each f n we may assume the weak convergence is to 0. So our task now is to rove: Let < < and let {f n } be a sequence in L µ such that f n g dµ = 0 n

3 for all g L µ. Then there is a subsequence {f nk } of {f n } such that k k f n k = 0. = This will be done essentially by inequalities. Ste. Let ϕ : \{0} be defined by [ + x + x + ] = x ϕx = x Prove: There exists a constant A deending only on such that ϕx A for all x 0. This result requires only some Calculus notions. Continuous functions on, 0 0, that are bounded near 0 and near have an extremely high robability of being bounded everywhere. Moreover, if the it for x 0 is 0, they can even be extended to be also continuous at 0. Here is the integer art of ; the sum from = to is interreted as 0 if < <. It may be useful to remember that one defines for all, = 0,,,...,, if = 0, =, if =, +!, if N,, and then + x = + x = =0 if x < ; moreover, for every N N {0}, + x N =0 x 0 x x x N = 0. Ste Deduce from the inequality in Ste that for all a, b one has a + b a + sgn a a b + = a b + A b. sgn a = sign of a. Ste 3. The next inequality to be roved is as follows. n Let M be such that f n M for all n For n N set s n = f k. Then there is another constant B deending only on such that for all n N, s n A M + B n + s n sgn s n f n dµ + s n. If g is measurable, then g q sgn g is a function equal to g q at all oints where g is non negative, equal to g q where g is negative. It is an easy exercise to see it is measurable The roof of one simly set a = s n x, b = f n x in and one integrates over. The terms in = can be estimated as follows. First 3 k=

4 of all, alying Hölder s inequality with r = / which is >, seeing that then r = /, one gets s n f n dµ s n f n n f k f n k= n M M = n M. Next = n n = n n. So B = M works. Ste 3. It is time to construct the subsequence. One goes by induction, selecting n =. Assume n < n <... < n k found. 7. Here is a rather easy one. Let, M, µ be a measure sace and let,..., n [, ] be such that n i= i =. Let f i L i µ for i =,..., n. Prove f f n L µ and n f f n dµ f i i. i= 8. Let µ be Lebesgue measure on. Let, q, r [, ] be such that r = + q. For examle,,, is such a trile. So is,, for any [, ]. Another examle is,,,. Let f L, g L q. Show: a The ma x, y fx ygy : n n C is measurable. b The ma y fx ygy : n C is integrable for almost every x. c By the revious oint it makes sense to define a ma usually denoted by f g from to C by f gx = fx ygy dy for x. Prove f g L r µ and 3 f g r f g q. Hints rovided 4

5 Aendix: Hints For xercise. Clearly ν ν so that by adon-nikodymdν = h d ν for some h L ν. One has to show one can have hx = for all x. To show that h a.e[ nu ] is fairly easy since all one needs to see is that h d ν ν for all M. Suose we let now be the set of all x where hx > and assume ν > 0. Then also ν ɛ > 0 for some ɛ > 0, where ɛ is the set of all x where hx > + ɛ. Prove it isn t hard that there is a sequence {z n } of comlex numbers z n + ɛ such that the sequence of discs {D n = {z C : z z n < ɛ/} satisfies {z C : z > } D n. Suose ν {x : hx D n } > 0. Then h d ν = D n h z n d ν + z n ν D n z n ɛ ν D n D n + ɛ ν D n. This should contradict something. If h = a.e. [ ν ] one can modify it so a.e. becomes e. For xercise 8 The roof of 3 known as Young s inequality is fairly easy if, q, r =, q, q, esecially if q = ; it is then a simle alication of Fubini lus Hölder. It is also easy if, q, r =,, ; then it is ust Hölder. Other cases seem harder. Because you can relace the functions by their absolute values, one may assume all functions are non-negative at first; this minimizes roblems with the integration. In articular, f g is well defined, non-negative, ossibly equal to a lot of the time. Here is where it gets tricky. We have f gx r dx = f gx r f gx dx = f gx r fx ygy dy dx. All integrals here make sense, the worst that can haen is some are ; the equality is due to Tonelli s Theorem. The integrand of the last integral can be rewritten as follows f gx r fx ygy = f gx r/q fx y /q fx y /r gy q/r gy q/ f gx r/. This works because r q + r = r q + = r = r, r q + = r q + = r =, q r + q = q r + = q q =. Notice next that q + r + =, so we may aly the extended Hölder inequality of xercise 7 and after a bit of elementary algebra get 3 for non-negative functions, thus also the stronger version with f, g relaced by f, g. After that it is a simle ste or two, or three to comlete the exercise. One figures this out by toying with equalities for a while. Not having done this in years, it took me quite awhile to come u with the right exonents. 5