Now, the above series has all non-negative terms, and hence is an upper bound for any fixed term in the series. That is to say, for fixed n 0 N,

Transcription

1 l p IS COMPLETE Let 1 p, and recall the definition of the metric space l p : { } For 1 p <, l p = sequences a = (a n ) in R such that a n p < ; whereas l consists of all those sequences a = (a n ) such that sup n a n <. We defined the p-norm as the function p : l p [0, ), given by a p = a n p, for 1 p <, and a = sup n a n. In class, we showed that the function d p : l p l p [0, ) given by d p (a, b) = a b p is actually a metric. We now proceed to show that (l p, d p ) is a complete metric space for 1 p. For convenience, we will wor with the case p <, as the case p = requires slightly different language (although the same ideas apply). Suppose that a 1, a 2, a 3,... is a Cauchy sequence in l p. ote, each term a in the sequence is a point in l p, and so is itself a sequence: a = (a 1, a 2, a 3,...). ow, to say that (a ) =1 is a Cauchy sequence in l p is precisely to say that ɛ > 0 K s.t., m K, a a m p < ɛ. That is, for given ɛ > 0 and sufficiently large, m, we have a m p m p < ɛ p n a n = a a p. ow, the above series has all non-negative terms, and hence is an upper bound for any fixed term in the series. That is to say, for fixed n 0, m m a n0 a n0 a n a n p < ɛ p, and so we see that the sequence (a n ) 0 =1 is a Cauchy sequence in R. But we now that R is a complete metric space, and thus there is a limit a n0 R to this sequence. This holds for each n 0. The following diagram illustrates what s going on. a 1 = a 1 1 a 1 2 a 1 3 a a 1 a 2 a 3 a a 2 = a 2 1 a 2 2 a 2 3 a 2 a 3 = a 3 1 a 3 2 a 3 3 a 3 a 4 = a 4 1 a 4 2 a 4 3 a 4 So, we have shown that, in this l p -Cauchy sequence of horizontal sequences, each vertical sequence actually converges. Hence, there is a sequence a = (a 1, a 2, a 3, a 4,...) to which a converges in a vague sense. The sense is the point-wise convergence along vertical 1

2 2 lines in the above diagram. To be more precise, recall that a sequence a is a function a : R, where we customarily write a(n) = a n. What we have shown is that, if (a 1, a 2, a 3,...) is a Cauchy sequence of such l p functions, then there is a function a: R such that a converges to a point-wise; i.e. a (n) a(n) for each n. ow, our goal is to find a point b l p such that a b as in the sense of l p ; that is, such that a b p 0 as. The putative choice for this b is the sequence a given above. In order to show that one wors, we need to show first that it is actually an l p sequence, and second that a converges to a in l p sense, not just point-wise. To do this, it is convenient to first pass to a family of subsequences of the (a n ), as follows. Since (a 1 ) =1 converges to a 1, we can choose 1 so that for 1, a 1 1 a 1 < 1. 2 Having done so, and nowing that a a 2, we can choose a larger 2 so that for 2, 1 1 we have a < and a 2 1 a 1 < 4 2 a 2. Continuing this way iteratively, we can find an 4 increasing sequence of integers 1 < 2 < 3 < such that for each j, a n a n < 2 j for n = 1, 2,..., j and j. (1) In particular, we have a n j a n < 2 j for j n. That gives us the following. Lemma 1. The sequence a = (a n ) of point-wise limits of (a ) =1 is in lp. Proof. Fix, and recall that the finite-dimensional versions of the l p -norms, (a 1,..., a ) p = a n p also satisfy the triangle inequality (i.e. d p (x, y) = x y p is a metric on R ). Hence, we can estimate the initial-segment of terms of a as follows: and so ( ) 1/p ( ( a n p a n a n p + a n p. (2) ow, the last term in Equation 2 is bounded by the actual l p -norm of the whole sequence a ; that is, we can tac on the infinitely many more terms, ( ) 1/p a n p a n p = a p. a n = (a n a ) + a n n, Recall that (a ) =1 is a Cauchy sequence in the metric space lp. We have proved that any Cauchy sequence in a metric space is bounded. Thus, there is a constant R independent of such that a p R. Combining this with Equation 1, we can therefore estimate the right-hand-side of Equation 2 by ( ( a n p (2 ) p + R = 2 p + R.

3 ( Finally, the term 2 ) p 1/p = 1/p 2 converges to 0 as (remember your calculus!), and hence this sequence is also bounded by some constant S. In total, then, we have a n p R + S for all. p In other words, a n (R + S) p. The constant on the right does not depend on ; it is an upper bound for the increasing sequence of partial sums of the series a n p = a p p. Thus, we have a p R + S, and so a l p. So, we have shown that the putative limit a (the point-wise limit of the sequence (a ) =1 of points in l p ) is actually an element of the metric space l p. But we have yet to show that it is the limit of the sequence (a ) in l p. That somewhat involved proof now follows. Proposition 2. Let (a ) =1 be a Cauchy sequence in lp, and let a be its point-wise limit (which is in l p, by Lemma 1). Then a a p 0 as. Proof. Let ɛ > 0. Lemma 1 shows that a l p, which means that a n p <. Hence, by the Cauchy criterion, there is an 1 so that a n p < ɛ p. n= 1 In addition, we now that (a ) is l p -Cauchy, so there is 2 so that, whenever, m 2, =1 a a m p < ɛ. Letting = max{ 1, 2 }, we therefore have a n p < ɛ p and a a p < ɛ. (3) n= ow, the sequence a is in l p, and so we can apply the Cauchy criterion again: select large enough so that a n p < ɛ p. (4) n= ote, we can always increase and still maintain this estimate, so we are free to chose. We now use the constant we defined above in the bounds we will need later. Since a n a n for each fixed n, we can choose K 1 so that a < ɛ p / 1 a 1 for K 1. Liewise, we can choose K 2 so that a 2 a 2 < ɛ p / for K 2. Continuing this way for steps, we can tae K = max{k 1, K 2,..., K } and then we have ɛ p a n a n <, for K and n. (5) For good measure, we will also (increasing it if necessary) mae sure that K. ow, for any K, brea up b = a a as follows: (b n ) = (b n ) 1 + (b n ) n=. 3

4 4 (To be a little more pedantic, we are expressing b n = x n + y n where x n = b n when n < and = 0 when n, and y n = 0 when n < and = b n when n.) The triangle inequality for the p-norm then gives ( 1 ( a a p a n a n p + a n a n p. (6) n= Equation 5 shows that, for K, the first term here is ( ) 1/p ( 1 ) 1/p 1 p ɛ p 1 a n a n = ɛ < ɛ. For the second term in Equation 6, we use the triangle inequality for the l p -norm restricted to the range n to get ( ) 1/p a n a n p a n p + a n p. n= n= n= Since, Equation 3 shows that the second term here is < ɛ. So, summing up the last two estimates, we have a a p 2ɛ + a n p, (7) n= whenever K. So we need only show this final term is small. Here we mae one more decomposition: a n = a n a n + a n, and so once again applying the triangle inequality, p p p a n a n a n + a. n= n= n= The first of these terms is a sum of non-negative terms over n, and so it is bounded above by the sum over n 1 which is equal to a a p, which is < ɛ by Equation 3 (since K ). And the second term is also < ɛ, by Equation 4. Whence, the last term in Equation 7 is also < 2ɛ, and so we have shown that ɛ > 0, K such that K a a p < 4ɛ. Of course, we should have been more clever and chosen all our constants in terms of ɛ/4 to get a clean ɛ in the end, but such tidying is not really necessary; 4ɛ is also arbitrarily small, and so we have shown that (a ) =1 does converge to a in lp. This concludes the proof that l p is complete. Whew! Let us conclude by remaring that a very similar (though somewhat simpler) proof wors for p = ; the details are left to the reader. n

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