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7 Pages

### duality

Course: MATH 6327, Fall 2008
School: Georgia Tech
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Word Count: 1540

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on Notes Norms and Duality by Eric Carlen In the previous section we saw how a certain class of convex functions , the Orlicz functions, induced norms on ceretain subspaces V of the space of measurable functions on some measure space (, S, ). We saw that for each such , V is complete in its norm, and hence is a Bancah space. So far in our analysis though we did not use all of the properties required of an Orlicz...

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on Notes Norms and Duality by Eric Carlen In the previous section we saw how a certain class of convex functions , the Orlicz functions, induced norms on ceretain subspaces V of the space of measurable functions on some measure space (, S, ). We saw that for each such , V is complete in its norm, and hence is a Bancah space. So far in our analysis though we did not use all of the properties required of an Orlicz function . In particular, we have not made any use of the requirement that (1) = 1. Nor have we made full use of the requirement that lims (s)/s = : lims (s) = would have sufficed in our proofs so far. We now explain the reasons for these requirements. Let (t) = sup{ts - (s)} s0 denote the convex function conjuagte to . Then we know that for any s, t 0 s (s) = (s) + ( (s)) provided is differentiable at s. In particular, under our assumptions, 1 = (1) + (1) . Since for all t, taking t = 1 we see ( ) (1) = (( ) (1)) + (1) . Now s = (s) + (1) t( ) (t) = (( ) (t)) + (t) (1) has at most one solution if is strictly convex. Since s = 1 is a solution, it follows in this case that ( ) (1) = 1 . also, it follws from (1) and (0) = (0) = 0 that (0) = 0. Since we also have t (t) = (t) + (( ) (t)) , (2) for all t, taking t = 0 we have that ( ) (0) = 0. We conclude that is an Orlicz function whenever is a strictly convex Orlicz function. We now apply Young's inequality for the pair , in order to deduce a generalization of Hlder's inequality. To do this, let a and b be any positive numbers, and let o a,b (s) = b(as) . 1 Then (t) = sup{ts - b(as)} a,b s0 = sup b s0 t as - b(as) ab . = b Young's inequality then gives us t ab st b(as) + b t ab for all s and t. Also, we know that there is equality only if t = ab (as) . Then for any f and g, |f ||g|d b (|af |)d + b |g| ab d . Now, by definition, |f | f d = (1) and |g| g d = (1) . Therefore, if we choose a= This gives us 1 f g and b= f g , |f ||g|d f ((1) + (1)) = f g . There is equality if and only if |g(x)| = ab (a|f (x)|) = g for almost every x. This prove the following Theorem 1 (Generalized Hlder's Inequality) Let be any strictly convex Orlicz o function and its conjugate function. Then for all measurabel functions f and g, |f ||g|d f 2 |f (x)| f g . (3) There is equlaity in (3) if and only if |g(x)| = g for almost very x. Now let f V be given. Is there a function g in V with |f ||g|d = f |f (x)| f (4) g (5) or is there not? If there is, we have just seen that we must have In case (s) = sp /p, (s) = sp-1 . Therefore, for any f in Lp , |f (x)| f = |f (x)| f p p-1 and the function on the right is clearly a unit vector Lp where p = p/(p - 1). Indeed, |f (x)| f p p-1 p/(p-1) d = |f (x)|p d = 1 . f p p Hence, for any f Lp , there is a unit vector g in Lp so that eqaulity holds in (5). This has an important consequence. Define p-1 |f (x)| , (6) h(x) = sgn(f (x)) f p where for any complex number , sgn() = /|| if = 0 , 0 if = 0 p and denotes the complex conjugate. Then h f hd = = 1, and we have p 1 f p |f |p = f . Now since f gd |f ||g|d with equality only in case sgn(g(x)) = sgn)f (x)) for all x such that |g(x)||f (x)| = 0. Then from the contiitions for eqaulity in Hlder's inequality, we see that if g is a unit vector in o p L and f hd = f p , then g = h where h is given by (6). 3 This shows that for any f in Lp , 1 < p < , there is one and only one unit vector h in Lp , p = p/(p - 1), so that f p = f hd . (7) It is also easy to generalize this result to p = 1 with p = in that case. We will save the discussion of L1 and L for the end of the section; the results in these cases are realtively straightforward. For now we restrict our attention to the case 1 < p < . There is a related result, which combined with these observations, is very useful. It is a sort of converse to Hlder's inequality due to Landau. We first state and prove it for o the classical Lp spaces. Theorem (Landau's 2: Thoerem) Let f be a measurable function on a sigma finite measure space (, S, ), and suppose that for some p with 1 < p < , |f h|d < (8) for all unit vectors h in Lp where p = p/(p - 1). Then f Lp . Proof: Let An be an increasing sequence of sets of finite measure so that = n An . Let f be any measurable complex valued function, and define Bn = |{ x : |f (x) n } An . Then = n An . Therefore, if we define fn (x) = f (x)1Bn (x) , we have that limn |fn (x)| = |f (x)| for all x. Clearly fn p n(An )1/p and so fn Lp for all n. By the Monotone Convergence Theorem, |f |p d = lim n |fn |p d Now since each fn Lp , we have that fn hn d = fn p, where hn is a unit vector in Lp . (In fact, hn is given in terms of fn by (6)). By the definition of fn , f hn d = fn hn d = fn 4 p . Now if sup |f h|d : sup fn h p 1 <, (9) then n0 p < and so |f |p d < , which is what we want to show in case (8) holds. It therefore suffices to show that (8) implies (9). Therefore, suppose that (8) holds, but (9) does not. Then we can choose a sequence {gn } of unit vectors in Lp so that f gn 0 and f gn d 2n . Now define h= 2-n gn . j=1 Then by convexity the pth power, |h(x)|p sum 2-n |gn (x)|p , j=1 and integration yields |h(x)|p d sum 2-n = 1 . j=1 But f hd = 2-n f gn d j=1 = j=1 2-n 2n = . Since h p 1, this contradicts (8). Thus (8) implies (9), and the result is proved. To apply this result, we first extend the norm function to all measurable functions f on (, S, ) by defining f p = if f inLp . With this definition, we have the following result: Theorem 3: (Dual Characterization of Lp Norms, 1 < p < ) Let f be a measurable function on a sigma finite measure space (, S, ). Then there is a unit vector h in Lp , p = p/(p - 1) such that f p = f hd . 5 (10) If f p is finite, this unit vector h is unique, and is given in terms of f by (6). Proof: If f Lp , so that f p < , this follows from the remarks above the statement of Landau's Theorem. If f p = , so that f Lp , it follows from Landau's Theorem that / p there is a unit vector h in L so that f hd = . Here is a simple application of this: Take any two vectors f and g in Lp . Then there is a unit vector h so that f +g But by Hlder's inequality, o f hd f p p = (f + g)hd = f hd + ghd . h p = f p and likewise ghd g f +g p h p ...

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