MIT OpenCourseWarehttp://ocw.mit.edu 18.102 Introduction to Functional Analysis Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.� � � � � � � 8 LECTURE NOTES FOR 18.102, SPRING 2009 Lecture 3. Tuesday, 10 Feb. Recalled the proof from last time that the bounded operators from a normed space into a Banach space form a Banach space – mainly to suggest that it is not so hard to remember how such a proof goes. Then proved that a normed space is Banach iff every ‘absolutely summable’ series is convergent. Absolute summability means that the sum of the norms is finite. Then did most of the proof that every normed space can be completed to a Banach space using this notion of absolutely summable sequences. The last part – and a guide to how to attempt the part of the proof that is the first question on the next homework. The proof of the result about completeness from the early part of the leture is in (1) Wilde:- Proposition 1.6 (2) Chen:- I didn’t find it. (3) Ward:- Lemma 2.1 (easy way only) Here is a slightly abbreviated version of what I did in lecture today on the completion of a normed space. The very last part I asked you to finish as the first part of the second problem set, not due until February 24 due to the vagaries of the MIT calendar (but up later today). This problem may seem rather heavy sledding but if you can work through it all you will understand, before we get to it, the main sorts of arguments needed to prove most of the integrability results we will encounter later. Let V be a normed space with norm �·�V . A completion of V is a Banach space B with the following properties:-(1) There is an injective (1-1)linear map I : V −→ B (2) The norms satisfy (3.1) �I(v)�B = �v�V ∀ v ∈ V. (3) The range I(V ) ⊂ B is dense in B. Notice that if V is itself a Banach space then we can take B = V with I the identity map. So, the main result is: Theorem 1. Each normed space has a completion. ‘Proof’ (the last bit is left to you). First we introduce the rather large space ∞(3.2) V�= {uk}∞k=1; uk ∈ V and �uk� < ∞ k=1 the elements of which I called the absolutely summable series in V. Now, I showed in the earlier result that each element of V�is a Cauchy sequence N– meaning the corresponding sequence of partial sums vN = uk is Cauchy if k=1 {uk} is absolutely summable. Now V�is a linear space, where we add sequences, and multiply by constants, by doing the operations on each component:-(3.3) t1{uk} + t2{u�k} = {t1uk + t2u�k}. This always gives an absolutely summable series by the triangle inequality: (3.4) �t1uk + t2u�k� ≤ |t1| �uk� + |t2| �u�k�. k k k�� � � � � � � LECTURE NOTES FOR 18.102, SPRING 2009 9 Within �V (3.5) consider the linear subspace � S = {uk}; � k �uk� < ∞, � � k uk = 0 of those which converge to 0. As always for a linear subspace of a linear space we can form the quotient (3.6) B = V /S the elements of which are the ‘cosets’ of the form {uk} + S ⊂ V�V . where {uk} ∈ �We proceed to check the following properties of this B. (1) A norm on B is defined by n(3.7) �b�B = lim uk�, {uk} ∈ b. n→∞ � k=1 (2) The original space V is imbedded in B by (3.8) V � v �−→ I(v) = {uk} + S, u1 = v, uk = 0 ∀ k > 1 and the norm satisfies (3.1). (3) I(V ) ⊂ B is dense. (4) B is a Banach space with the norm (3.7). So, first that (3.7) is a norm. The limit on the right does exist since the limit of the norm of a Cauchy sequence always exists – namely the sequence of norms is itself Cauchy but now in R. Moreover, adding an element of S to {uk} does not change the norm of the sequence of partial sums, since the addtional term tends to zero in norm. Thus �b�B is well-defined for each element b ∈ B and �b�B = 0 means exactly that the sequence {uk} used to define it tends to 0 in norm, hence is in S hence b = 0. The other two properties of norm are reasonably clear, since if b, b� ∈ B are represented by {uk}, {u�k} in V�then tb and b + b� are reprented by {tuk} and {uk + u�k} and n nlim tuk� = t lim n→∞ � k=1 | | n→∞ � k=1 uk�, nlim (uk + u�)� = A = ⇒kn→∞ � k=1 n(3.9) for � > 0 ∃ N s.t. ∀ n ≥ N, A − � ≤ � (uk + u�k)� = ⇒ k=1 n nA − � ≤ � uk� + � uk�)� ∀ n ≥ N = ⇒ k=1 k=1 A − � ≤ �b�B + �b��B ∀ � > 0 =⇒ �b + b��B ≤ �b�B + �b��B . Now the norm of the element I(v) = v, 0, 0, ··· , is the limit of the norms of the sequence of partial sums and hence is �v�V so �I(v)�B = �v�V and I(v) = 0 therefore implies v = 0 and hence I is also injective. So, we need to check that B is complete, and also that I(V ) is dense. Here is an extended discussion of the difficulty – of course maybe you can see it directly� � � � � � � � � � � � � 10 LECTURE NOTES FOR 18.102, SPRING 2009 yourself (or have a better scheme). Note that I want you to write out your own version of it carefully for the next problem set. Okay, what does it mean for B to be a Banach space, well as we saw in class today it means that every absolutely summable series in B is convergent. Such a series {bn} is given by bn = {uk (n)} + S where {uk (n)} ∈ V�and the summability condition is that N(3.10) = lim u(n)∞ > n �bn�B nN→∞ � k=1 k �V . So, we want to show that bn = b converges, and to do so we need to find the n limit b. It is supposed to be given by an absolutely summable series. The ‘problem’ is that this series should look like �� u(kn) in some sense – because it is supposed nk to represent the sum of the bn’s. Now, it would be very nice if we had the estimate (3.11) �u(kn)�V < ∞ nk since this should allow us to break up the double sum in some nice way so as to get an absolutely summable series out of the whole thing. The trouble is that (3.11) need not hold. We know that each of the sums over k – for given n – …