Math630-10

# Math630-10 - Real Analysis Math 630 Homework Set#10 Chapter...

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Unformatted text preview: Real Analysis - Math 630 Homework Set #10 - Chapter 6 by Bobby Rohde 11-17-00 Problem 7 a) For 1 p that v 1 v v v p , we denote l p the space of all sequences v v 1 such . Prove the Minkowski ineqality for sequences p v p v p. Here we have 1 v p p p v1 , v p and v sup p Proof Case 1 v p p v1 v p 1 p v v p = v1 v v p 1 p . If either v or v has norm 0, the proof is trivial so we may assume that each has norm greater than v v 1, 1. 1 and then define real numbers , > 0 such that p p v1 v1 v v p p 1 p v v1 v v p 1 p v p v1 1 p = p v1 v v v p v 1 p = 1 p p p , since p 1 and the quantities inside the distribution are each less than 1. , with 1. However expnading this expression and summing we see that it is in turn = 1 pp So v 1 v += v v p+ v p . QED. MATH630-10.nb 2 Case p v v v sup v v p sup v v sup 1 p 1 q v + sup v = . QED. v v b) Show that if v1 v v l p and p v v l q with q 1, then Proof Case p v1 v 1, q v v1 v sup v = v1 v v = v 1 v . Case 1 p q Define v p . We may chosoe to assume that only on their absolute value. By Lemma 3 p t Summing v v and v v v are positive since the norm is dependant p p1 p =p t vt v v v v1 p v both t v pp sides. v p p v1 pt p t v p v p p p v p = v t p v v v p Differentiating with respect to t we have that v p v q . QED. v1 p p1 p = Problem 10 Let f n be a sequence of functions in L . Prove that f n converges to f in L if and only if there is a set E of measure 0 such that fn converges uniformly to f on E. Proof fn f in L imples f n f except on a set E of measure 0, fn 0. Thus 0 N such that n N , f n f f . Thus fn is uniformly converges to f on E. . If fn uniformly converges to f except on a set E of measure 0 then and n N , fn x f x . except on E, f x < fn x bounded a.e. and hence f is in L . > 0, N such that x E ess sup fn x . Thus f is MATH630-10.nb 3 Problem 11 Prove that L is complete. Proof In order to show this we need that every Cauchy sequence of functions some f in L . Given > 0, N such that m, n N , f n fm norm we know that fn fm is bounded a.e. by . Define the pointwise limit f x It suffices to show that ess sup f limn fn in L converges to , by definition of Cauchy. Since this is L fn x , if the limit exists . , otherwise , since if fn converges then it must converge to f a.e. Since fn fm is bounded a.e. by , except on a set M of measure 0. However if we consider the collection of all such M m > n, we have a countable collection of sets of measure zero so there if limn fn x exists. Since fn is in L it union is also a set of measure 0. Thus fn x f x is bounded by some ess sup . Then where the limit exists, f is bounded by + . Thus we only need to show that the limit must exist upto a set of measure 0. MATH630-10.nb 4 Problem 16 Let f n be a sequence of functions in L p, 1 p , which converge a.e. p . Show that to a function f in L f n converges to f in L p if and only if fn f. Proof If fn fn fn p 1 p fn f p p f 1 p fn p f p 1 p fn 0, since p f 0 fn p 1. However f p f , for p 1 fn f 0 fn f p 1 p fn converges to f in L p f p. 0. However, fn f p ?????????????????????????????????? Demonstrate Proposition 8 fails at p = . Show that f , g functions , f L , such that step functions 0 and g and continuous 0. Counter-example on step functions. Define f x sin 1 , x 0 x . 0, x 0 Clearly f x is bounded by 1 and thus in L . Consider a step function on partition 0 1 . Let denote the size of the smallest interval in the partition. 0 1 2 ... N Consider intervals of the form An = n1 , n 1 , for n . Clearly An 0, 1 , and infinitely 1 . Clearly by many An of length < . So choose n, m such that An m , m 1 and length An 1 construction f An 0, 1 , so regardless of the value of over m, m 1 , f 2 over An . 1 Hence f 2 . Thus showing the counter example to Proposition 8. MATH630-10.nb 5 Counter-example on continuous functions. Define g x 0, x 1, x 1 2 1 2 . approximating g. By continuity at x . 1 2 1 2 Clearly g is in L . Consider a continuous function 1 >0 > 0 such that x 1 ,x 2 2 Suppose that 1 0, then choose = 2 1 over this interval so g 2. Suppose that So g 8 at p = 1 2 1 2 , and the interval , 1 2 where x 1 2 , yet g = 1 . QED. 2 1 2 0, then choose . Thus g = 2 1 2 and the interval 1 2 , 1 2 where x 2 1 2 . > 0. Thus showing the counter example for Proposition ...
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## This note was uploaded on 03/23/2010 for the course MATH 515 taught by Professor Staff during the Spring '08 term at Iowa State.

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