Problem Set #10 Math 471 Real Analysis Assignment: Chapter 8 #2, 3, 6, 8
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1 Poblem Set #0 Math 47 Real Analysis Assignment: Chate 8 #2, 3, 6, 8 Clayton J. Lungstum Decembe, 202
2 xecise 8.2 Pove the convese of Hölde s inequality fo = and =. Show also that fo eal-valued f / L ), thee exists a function g L ), + =, such that fg / L ). Conside the case whee = so that =. Then, by Hölde s inequality, we cetainly have the left-hand side majoizes the ight-hand side, and we have equality if we let g = sgnf). Fo then we have fg = f = f. By definition, g, hence it is clea that fo =, thee exists a g L ) such that f = fg. su g Altenatively, conside the case whee =. Then we have thee cases. If f = 0, we have f = 0 a.e. which imlies fg = 0 a.e., i.e. fg = 0 fo all g. Now, if f is ositive and finite, we may assume that f = without loss of geneality. Then define the set n = {x : fx) > and x < n} and note that 0 < n n < fo each intege n. Define { g n x) = n : x n. 0 : x / n Clealy n g n =. Now obseve that f g n = f g n ) g n = n n n. Thus, taking the suemum ove all such g with g, we have the desied equality. When f =, aly the same agument just used on the set F n = {x : fx) > n}. This gives us a lowe bound of n, and taking the suemum ove all ositive integes yields the equality. To show that fo f / L ), thee exists a g L ) such that fg / L ), conside the following. Without loss of geneality, we can assume all functions ae nonnegative. Now, suose thee is a sequence {g k } k= L ) with g k = and fg k > 3 k. Set g = 2 k g k and obseve that, by Minkowski s inequality, g. Note that ) k 3 fg = f 2 k g k > =. 2 k= k= k=
3 Thus, g L ) but fg / L ). Hence, we have educed the oblem to showing that such a sequence exists. Fist note that if f = on any set A of ositive measue, then we can simly take : x B g = B, 0 : x / B whee B A has ositive, finite measue. Thus g has the desied oeties. This imlies we can assume f is finite a.e., and in aticula, fo any ositive eal numbe c, we can find a set F with finite measue such that f = c. With this in mind, we can find a nested F sequence of sets { k } k=, each with finite measue, such that k= k = and k f > 3 k. Now, take : x k g k = k. 0 : x / k By constuction, g k =, and it is easy to check that fg k as k. Hence, we have the existence of the sequence, and we ve demonstated the existence of such functions g. Q..D. 2
4 xecise 8.3 Pove Theoems 8.2) and 8.3). Show that Minkowski s inequality fo seies fails when <. Let us ecall Theoem 8.2 and Theoem 8.3. Theoem 8.2, i.e., Hölde s inequality fo seies, states the following: Suose that, + =, a = {a k } k=, b = {b k} k=, and ab = {a kb k } k=. Then ) a k b k a k k= k= k= ) ) a k b k su a k b k k= k= k= k= b k ) when < < when =,. The esult is clea as the oof fo the integal vesion goes though. Fo agument s sake, we esent it hee. The second inequality is tivial since ) ) a k b k su a k ) b k = su a k b k. By symmety of agument, this oves the inequality fo the cases =,. Now suose < <. Then by Young s inequality, we have ak a k b k + b ) k k= k= = a k + b k k= = a + b k= k= = + =, whee we have assumed a = b =. Obseve this is enough as we can set A = a and B = b b and check that A = = B. Thus, Hölde s inequality fo seies holds as stated fo. Theoem 8.3, i.e., Minkowski s inequality fo seies, states the following: Suose that, a = {a k } k=, b = {b k} k=, and a + b = {a k + b k } k=. Then k= a k + b k ) su k= a k + b k su ) ) a k + b k a k + su k= b k when =. 3 when < a
5 Again, the oof fo the integal vesion goes though, but fo agument s sake, we ll esent it hee. Obseve that case = is just the standad tiangle inequality fo eal numbes, so that case is finished. When =, note that by the tiangle inequality, we always have a k + b k a k + b k. By the definition of suemum, we must have a k su a k and b k su b k. Thus a k + b k su a k + su b k. Since the ight-hand side is an ue bound, it must be at least as geat as the least ue bound, in aticula, su a k + b k su a k + su b k, as desied. Now, suose < <. Then, a + b = a k + b k a k + b k k= a k + b k a k + k= k= + ) a k + b k k= = a + b a k + b k b k k= k= ) a k + b k a k ) k= a + a + b b. b k ) Now, to show the inequality cannot be imoved uon, simly conside the sequences given by { { : k = : k = 2 a k = and b k = 0 : k > 0 : k 2. It is not had to see Minkowski s inequality is not tue fo any 0 < < fo these sequences, fo obseve ) a k + b k = 2. Since 0 < <, we know < k= <, hence 2 > 2 = + = k= ) ) a k + b k. k= Q..D. 4
6 xecise 8.6 Pove the following genealization of Hölde s inequality. If k i= = with i,, then f f k f f k k. We ll oceed by induction by fist showing the inequality to be tue when k = 2. To this end, conside fg = fg whee o equivalently, f ) = f g 2, + 2 =, + 2 =. g 2 ) 2 Thus, the genealization holds when k = 2. Assume the inequality is tue fo some n N. Then, alying the inequality above with f = f f n and g = f n+, we have f f n+ f f n f n+ n+. n ), Now, by the induction hyothesis, taking = i= i we have the inequality fo the nom of f f n, and the genealization follows as desied. Q..D. 5
7 xecise 8.8 Pove the following integal vesion of Minkowski s inequality fo < : [ fx, y)dx ] [ dy ] fx, y) dy dx. Suose fist that =. Then, fx, y)dx dy fx, y) dx dy = fx, y) dy dx, whee the last equality is achieved using Fubini s theoem fo nonnegative measuable functions. Now, let < <. If the left-hand side is zeo, we have nothing moe to ove. Now, define F y) = fz, y) dz and obseve fx, y)dx dy F y)) fx, y) dx dy = F y)) fx, y) dy dx ) fx, y) dy ) F y)) dy dx. Notice that F y)) dy is, in fact, a constant so any owes of it doesn t change the fact that it is a constant), so we can ull it out of the integal by the integal s lineaity, divide by it since we obseved above that it is ositive, and we get as was to be shown. [ fx, y)dx ] [ dy ] fx, y) dy dx, Q..D. 6
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