Normed and Banach Spaces

Normed and Banach spaces In this chapter we introduce the basic setting of functional analysis, in the form of normed spaces and bounded linear operators. We are particularly interested in complete, i.e. Banach, spaces and the process of completion of a normed space to a Banach space. In lectures I proceed to the next chapter, on Lebesgue integration after Section 7 and then return to the later sections of this chapter at appropriate points in the course. There are many good references for this material and it is always a good idea to get at least a couple of different views. The treatment here, whilst quite brief, does cover what is needed later. 1. Vector spaces You should have some familiarity with linear, or I will usually say ‘vector’, spaces. Should I break out the axioms? Not here I think, but they are included in Section 14 at the end of the chapter. In short it is a space V in which we can add elements and multiply by scalars with rules quite familiar to you from the the basic examples of Rn or Cn . Whilst these special cases are (very) important below, this is not what we are interested in studying here. What we want to come to grips with are spaces of functions hence the name of the course. Note that for us the ‘scalars’ are either the real numbers or the complex numbers – usually the latter. To be neutral we denote by K either R or C, but of course consistently. Then our set V – the set of vectors with which we will deal, comes with two ‘laws’. These are maps (1.1) + : V × V −→ V, · : K × V −→ V. which we denote not by +(v, w) and ·(s, v) but by v + w and sv. Then we impose the axioms of a vector space – see Section 14 below! These are commutative group axioms for +, axioms for the action of K and the distributive law linking the two. The basic examples: • The field K which is either R or C is a vector space over itself. • The vector spaces Kn consisting of ordered n-tuples of elements of K. Addition is by components and the action of K is by multiplication on all components. You should be reasonably familiar with these spaces and other finite dimensional vector spaces. • Seriously non-trivial examples such as C([0, 1]) the space of continuous functions on [0, 1] (say with complex values). In these and many other examples we will encounter below, the ‘component addition’ corresponds to the addition of functions. Lemma 1.1. If X is a set then the spaces of all functions (1.2) F(X; R) = {u : X −→ R}, F(X; C) = {u : X −→ C} are vector spaces over R and C respectively. Non-Proof. Since I have not written out the axioms of a vector space it is hard to check this – and I leave it to you as the first of many important exercises. In fact, better do it more generally as in Problem 1.2 – then you can sound sophisticated by saying ‘if V is a linear space then F(X; V ) inherits a linear structure’. The main point to make sure you understand is precisely this; because we do know how to add and multiply in either R and C, we can add functions and multiply them by constants (we can multiply functions by each other but that is not part of the definition of a vector space so we ignore it for the moment since many of the spaces of functions we consider below are not multiplicative in this sense):(1.3) (c1 f1 + c2 f2 )(x) = c1 f1 (x) + c2 f2 (x) defines the function c1 f1 + c2 f2 if c1 , c2 ∈ K and f1 , f2 ∈ F(X; K).  You should also be familiar with the notions of linear subspace and quotient space. These are discussed a little below and most of the linear spaces we will meet are either subspaces of these function-type spaces, or quotients of such subspaces – see Problems 1.3 and 1.5. Although you are probably most comfortable with finite-dimensional vector spaces it is the infinite-dimensional case that is most important here. The notion of dimension is based on the concept of the linear independence of a subset of a vector space. Thus a subset E ⊂ V is said to be linearly independent if for any finite collection of distinct elements vi ∈ E, i = 1, . . . , N, and any collection of ‘constants’ ai ∈ K, i = 1, . . . , N we have the following implication (1.4) N X ai vi = 0 =⇒ ai = 0 ∀ i. i=1 That is, it is a set in which there are ‘no non-trivial finite linear dependence relations between the elements’. A vector space is finite-dimensional if every linearly independent subset is finite. It follows in this case that there is a finite and maximal linearly independent subset – a basis – where maximal means that if any new element is added to the set E then it is no longer linearly independent. A basic result is that any two such ‘bases’ in a finite dimensional vector space have the same number of elements – an outline of the finite-dimensional theory can be found in Problem 1.1. Still it is time to leave this secure domain behind, since we are most interested in the other case, namely infinite-dimensional vector spaces. As usual with such mysterious-sounding terms as ‘infinite-dimensional’ it is defined by negation. Definition 1.1. A vector space is infinite-dimensional if it is not finite dimensional, i.e. for any N ∈ N there exist N elements with no, non-trivial, linear dependence relation between them. Thus the infinite-dimensional vector spaces, which you may be quite keen to understand, appear just as the non-existence of something. That is, it is the ‘residual’ case, where there is no finite basis. This means that it is ‘big’. So, finite-dimensional vector spaces have finite bases, infinite-dimensional vector spaces do not. Make sure that you see the gap between these two cases, i.e. either a vector space has a finite-dimensional basis or else it has an infinite linearly independent set. In particular if there is a linearly independent set with N elements for any N then there is an infinite one, there is a point here – if an independent finite set has the property that there is no element of the space which can be added to it so that it remains independent then it already is a basis and any other independent set has the same or fewer elements. The notion of a basis in an infinite-dimensional vector spaces needs to be modified to be useful analytically. Convince yourself that the vector space in Lemma 1.1 is infinite dimensional if and only if X is infinite. 1 2. Normed spaces We need to deal effectively with infinite-dimensional vector spaces. To do so we need the control given by a metric (or even more generally a non-metric topology, but we will only get to that much later in this course; first things first). A norm on a vector space leads to a metric which is ‘compatible’ with the linear structure. Definition 1.2. A norm on a vector space is a function, traditionally denoted k · k : V −→ [0, ∞), (1.5) with the following properties (Definiteness) (1.6) v ∈ V, kvk = 0 =⇒ v = 0. (Absolute homogeneity) For any λ ∈ K and v ∈ V, (1.7) kλvk = |λ|kvk. (Triangle Inequality) For any two elements v, w ∈ V (1.8) kv + wk ≤ kvk + kwk. Note that (1.7) implies that k0k = 0. Thus (1.6) means that kvk = 0 is equivalent to v = 0. This definition is based on the same properties holding for the standard norm(s), |z|, on R and C. You should make sure you understand that ( x if x ≥ 0 R 3 x −→ |x| = ∈ [0, ∞) is a norm as is −x if x ≤ 0 (1.9) 1 C 3 z = x + iy −→ |z| = (x2 + y 2 ) 2 . Situations do arise in which we do not have (1.6):Definition 1.3. A function (1.5) which satisfes (1.7) and (1.8) but possibly not (1.6) is called a seminorm. 1Hint: For each point y ∈ X consider the function f : X −→ C which takes the value 1 at y and 0 at every other point. Show that if X is finite then any function X −→ C is a finite linear combination of these, and if X is infinite then this is an infinite set with no finite linear relations between the elements. A metric, or distance function, on a set is a map (1.10) d : X × X −→ [0, ∞) satisfying three standard conditions (1.11) d(x, y) = 0 ⇐⇒ x = y, (1.12) d(x, y) = d(y, x) ∀ x, y ∈ X and (1.13) d(x, y) ≤ d(x, z) + d(z, y) ∀ x, y, z ∈ X. As you are no doubt aware, a set equipped with such a metric function is called a metric space. If you do not know about metric spaces, then you are in trouble. I suggest that you take the appropriate course now and come back next year. You could read the first few chapters of Rudin’s book [4] before trying to proceed much further but it will be a struggle to say the least. The point is Proposition 1.1. If k · k is a norm on V then (1.14) d(v, w) = kv − wk is a distance on V turning it into a metric space. Proof. Clearly (1.11) corresponds to (1.6), (1.12) arises from the special case λ = −1 of (1.7) and (1.13) arises from (1.8).  We will not use any special notation for the metric, nor usually mention it explicitly – we just subsume all of metric space theory from now on. So kv − wk is the distance between two points in a normed space. Now, we need to talk about a few examples; there are more in Section 7. The most basic ones are the usual finite-dimensional spaces Rn and Cn with their Euclidean norms ! 12 X (1.15) |x| = |xi |2 i where it is at first confusing that we just use single bars for the norm, just as for R and C, but you just need to get used to that. There are other norms on Cn (I will mostly talk about the complex case, but the real case is essentially the same). The two most obvious ones are (1.16) |x|∞ = max |xi |, x = (x1 , . . . , xn ) ∈ Cn , X |x|1 = |xi | i but as you will see (if you do the problems) there are also the norms X 1 (1.17) |x|p = ( |xi |p ) p , 1 ≤ p < ∞. i In fact, for p = 1, (1.17) reduces to the second norm in (1.16) and in a certain sense the case p = ∞ is consistent with the first norm there. In lectures I usually do not discuss the notion of equivalence of norms straight away. However, two norms on the one vector space – which we can denote k · k(1) and k · k(2) are equivalent if there exist constants C1 and C2 such that (1.18) kvk(1) ≤ C1 kvk(2) , kvk(2) ≤ C2 kvk(1) ∀ v ∈ V. The equivalence of the norms implies that the metrics define the same open sets – the topologies induced are the same. You might like to check that the reverse is also true, if two norms induced the same topologies (just meaning the same collection of open sets) through their associated metrics, then they are equivalent in the sense of (1.18) (there are more efficient ways of doing this if you wait a little). Look at Problem 1.6 to see why we are not so interested in norms in the finitedimensional case – namely any two norms on a finite-dimensional vector space are equivalent and so in that case a choice of norm does not tell us much, although it certainly has its uses. One important class of normed spaces consists of the spaces of bounded continuous functions on a metric space X : (1.19) C∞ (X) = C∞ (X; C) = {u : X −→ C, continuous and bounded} . That this is a linear space follows from the (pretty obvious) result that a linear combination of bounded functions is bounded and the (less obvious) result that a linear combination of continuous functions is continuous; this we are supposed to know. The norm is the best bound (1.20) kuk∞ = sup |u(x)|. x∈X That this is a norm is straightforward to check. Absolute homogeneity is clear, kλuk∞ = |λ|kuk∞ and kuk∞ = 0 means that u(x) = 0 for all x ∈ X which is exactly what it means for a function to vanish. The triangle inequality ‘is inherited from C’ since for any two functions and any point, (1.21) |(u + v)(x)| ≤ |u(x)| + |v(x)| ≤ kuk∞ + kvk∞ by the definition of the norms, and taking the supremum of the left gives ku + vk∞ ≤ kuk∞ + kvk∞ . Of course the norm (1.20) is defined even for bounded, not necessarily continuous functions on X. Note that convergence of a sequence un ∈ C∞ (X) (remember this means with respect to the distance induced by the norm) is precisely uniform convergence (1.22) kun − vk∞ → 0 ⇐⇒ un (x) → v(x) uniformly on X. Other examples of infinite-dimensional normed spaces are the spaces lp , 1 ≤ p ≤ ∞ discussed in the problems below. Of these l2 is the most important for us. It is in fact one form of Hilbert space, with which we are primarily concerned:X (1.23) l2 = {a : N −→ C; |a(j)|2 < ∞}. j It is not immediately obvious that this is a linear space, nor that   21 X (1.24) kak2 =  |a(j)|2  j is a norm. It is. From now on we will generally use sequential notation and think of a map from N to C as a sequence, so setting a(j) = aj . Thus the ‘Hilbert space’ l2 consists of the square summable sequences. 3. Banach spaces You are supposed to remember from metric space theory that there are three crucial properties, completeness, compactness and connectedness. It turns out that normed spaces are always connected, so that is not very interesting, and they are never compact (unless you consider the trivial case V = {0}) so that is not very interesting either – in fact we will ultimately be very interested in compact subsets. So that leaves completeness. This is so important that we give it a special name in honour of Stefan Banach who first emphasized this property. Definition 1.4. A normed space which is complete with respect to the induced metric is a Banach space. Lemma 1.2. The space C∞ (X), defined in (1.19) for any metric space X, is a Banach space. Proof. This is a standard result from metric space theory – basically that the uniform limit of a sequence of (bounded) continuous functions on a metric space is continuous. However, it is worth recalling how one proves completeness at least in outline. Suppose un is a Cauchy sequence in C∞ (X). This means that given δ > 0 there exists N such that (1.25) n, m > N =⇒ kun − um k∞ = sup |un (x) − um (x)| < δ. x∈X Fixing x ∈ X this implies that the sequence un (x) is Cauchy in C. We know that this space is complete, so each sequence un (x) must converge (we say the sequence of functions converges pointwise). Since the limit of un (x) can only depend on x, we may define u(x) = limn un (x) in C for each x ∈ X and so define a function u : X −→ C. Now, we need to show that this is bounded and continuous and is the limit of un with respect to the norm. Any Cauchy sequence is bounded in norm – take δ = 1 in (1.25) and it follows from the triangle inequality that (1.26) kum k∞ ≤ kuN +1 k∞ + 1, m > N and the finite set kun k∞ for n ≤ N is certainly bounded. Thus kun k∞ ≤ C, but this means |un (x)| ≤ C for all x ∈ X and hence |u(x)| ≤ C by properties of convergence in C and thus kuk∞ ≤ C, so the limit is bounded. The uniform convergence of un to u now follows from (1.25) since we may pass to the limit in the inequality to find n > N =⇒ |un (x) − u(x)| = lim |un (x) − um (x)| ≤ δ m→∞ (1.27) =⇒ kun − uk∞ ≤ δ. The continuity of u at x ∈ X follows from the triangle inequality in the form |u(y) − u(x)| ≤ |u(y) − un (y)| + |un (y) − un (x)| + |un (x) − u(x)| ≤ 2ku − un k∞ + |un (x) − un (y)|. Given δ > 0 the first term on the far right can be make less than δ/2 by choosing n large using (1.27) and then, having chosen n, the second term can be made less than δ/2 by choosing d(x, y) small enough, using the continuity of un .  I have written out this proof (succinctly) because this general structure arises often below – first find a candidate for the limit and then show it has the properties that are required. There is a space of sequences which is really an example of this Lemma. Consider the space c0 consisting of all the sequences {aj } (valued in C) such that limj→∞ aj = 0. As remarked above, sequences are just functions N −→ C. If we make {aj } into a function α : D = {1, 1/2, 1/3, . . . } −→ C by setting α(1/j) = aj then we get a function on the metric space D. Add 0 to D to get D = D ∪ {0} ⊂ [0, 1] ⊂ R; clearly 0 is a limit point of D and D is, as the notation dangerously indicates, the closure of D in R. Now, you will easily check (it is really the definition) that α : D −→ C corresponding to a sequence, extends to a continuous function on D vanishing at 0 if and only if limj→∞ aj = 0, which is to say, {aj } ∈ c0 . Thus it follows, with a little thought which you should give it, that c0 is a Banach space with the norm kak∞ = sup kaj k. (1.28) j What is an example of a non-complete normed space, a normed space which is not a Banach space? These are legion of course. The simplest way to get one is to ‘put the wrong norm’ on a space, one which does not correspond to the definition. Consider for instance the linear space T of sequences N −→ C which ‘terminate’, i.e. each element {aj } ∈ T has aj = 0 for j > J, where of course the J may depend on the particular sequence. Then T ⊂ c0 , the norm on c0 defines a norm on T but it cannot be complete, since the closure of T is easily seen to be all of c0 – so there are Cauchy sequences in T without limit in T . Make sure you are not lost here – you need to get used to the fact that we often need to discuss the ‘convergence of sequences of convergent sequences’ as here. One result we will exploit below, and I give it now just as preparation, concerns absolutely summable series. Recall that a series is just a sequence where we ‘think’ about adding the terms. Thus if vn is a sequence in some vector space V then there N P is the corresponding sequence of partial sums wN = vi . I will say that {vn } is a i=1 series if I am thinking about summing it. Definition 1.5. A series {vn } with partial sums {wN } is said to be absolutely summable if X X (1.29) kvn kV < ∞, i.e. kwN − wN −1 kV < ∞. n N >1 Proposition 1.2. The sequence of partial sums of any absolutely summable series in a normed space is Cauchy and a normed space is complete if and only if every absolutely summable series in it converges, meaning that the sequence of partial sums converges. Proof. The sequence of partial sums is (1.30) wn = n X vj . j=1 Thus, if m > n then (1.31) wm − wn = m X j=n+1 vj . It follows from the triangle inequality that m X kwn − wm kV ≤ (1.32) kvj kV . j=n+1 So if the series is absolutely summable then ∞ X kvj kV < ∞ and lim n→∞ j=1 ∞ X kvj kV = 0. j=n+1 Thus {wn } is Cauchy if {vj } is absolutely summable. Hence if V is complete then every absolutely summable series is summable, i.e. the sequence of partial sums converges. Conversely, suppose that every absolutely summable series converges in this sense. Then we need to show that every Cauchy sequence in V converges. Let un be a Cauchy sequence. It suffices to show that this has a subsequence which converges, since a Cauchy sequence with a convergent subsequence is convergent. To do so we just proceed inductively. Using the Cauchy condition we can for every k find an integer Nk such that (1.33) n, m > Nk =⇒ kun − um k < 2−k . Now choose an increasing sequence nk where nk > Nk and nk > nk−1 to make it increasing. It follows that (1.34) kunk − unk−1 k ≤ 2−k+1 . Denoting this subsequence as u0k = unk it follows from (1.34) and the triangle inequality that ∞ X (1.35) ku0n − u0n−1 k ≤ 4 n=1 so the sequence v1 = u01 , vk = u0k − u0k−1 , k > 1, is absolutely summable. Its sequence of partial sums is wj = u0j so the assumption is that this converges, hence the original Cauchy sequence converges and V is complete.  Notice the idea here, of ‘speeding up the convergence’ of the Cauchy sequence by dropping a lot of terms. We will use this idea of absolutely summable series heavily in the discussion of Lebesgue integration. 4. Operators and functionals The vector spaces we are most interested in are, as already remarked, spaces of functions (or something a little more general). The elements of these are the objects of primary interest but we are especially interested in the way they are related by linear maps. The sorts of maps we have in mind here are differential and integral operators. For example the indefinite Riemann integral of a continuous function f : [0, 1] −→ C is also a continuous function of the upper limit: Z x (1.36) I(f )(x) = f (s)ds. 0 So, I : C([0, 1]) −→ C([0, 1]) it is an ‘operator’ which turns one continuous function into another. You might want to bear such an example in mind as you go through this section. A map between two vector spaces (over the same field, for us either R or C) is linear if it takes linear combinations to linear combinations:(1.37) T : V −→ W, T (a1 v1 +a2 v2 ) = a1 T (v1 )+a2 T (v2 ), ∀ v1 , v2 ∈ V, a1 , a2 ∈ K. In the finite-dimensional case linearity is enough to allow maps to be studied. However in the case of infinite-dimensional normed spaces we will require continuity, which is automatic in finite dimensions. It makes perfectly good sense to say, demand or conclude, that a map as in (1.37) is continuous if V and W are normed spaces since they are then metric spaces. Recall that for metric spaces there are several different equivalent conditions that ensure a map, T : V −→ W, is continuous: (1.38) vn → v in V =⇒ T vn → T v in W (1.39) O ⊂ W open =⇒ T −1 (O) ⊂ V open (1.40) C ⊂ W closed =⇒ T −1 (C) ⊂ V closed. For a linear map between normed spaces there is a direct characterization of continuity in terms of the norm. Proposition 1.3. A linear map (1.37) between normed spaces is continuous if and only if it is bounded in the sense that there exists a constant C such that (1.41) kT vkW ≤ CkvkV ∀ v ∈ V. Of course bounded for a function on a metric space already has a meaning and this is not it! The usual sense would be kT vk ≤ C but this would imply kT (av)k = |a|kT vk ≤ C so T v = 0. Hence it is not so dangerous to use the term ‘bounded’ for (1.41) – it is really ‘relatively bounded’, i.e. takes bounded sets into bounded sets. From now on, bounded for a linear map means (1.41). Proof. If (1.41) holds then if vn → v in V it follows that kT v − T vn k = kT (v − vn )k ≤ Ckv − vn k → 0 as n → ∞ so T vn → T v and continuity follows. For the reverse implication we use the second characterization of continuity above. Denote the ball around v ∈ V of radius  > 0 by BV (v, ) = {w ∈ V ; kv − wk < }. Thus if T is continuous then the inverse image of the the unit ball around the origin, T −1 (BW (0, 1)) = {v ∈ V ; kT vkW < 1}, contains the origin in V and so, being open, must contain some BV (0, ). This means that (1.42) T (BV (0, )) ⊂ BW (0, 1) so kvkV <  =⇒ kT vkW ≤ 1. Now proceed by scaling. If 0 6= v ∈ V then kv 0 k <  where v 0 = v/2kvk. So (1.42) shows that kT v 0 k ≤ 1 but this implies (1.41) with C = 2/ – it is trivially true if v = 0.  Note that a bounded linear map is in fact uniformly continuous – given δ > 0 there exists  > 0 such that (1.43) kv − wkV = dV (v, w) <  =⇒ kT v − T wkW = dW (T v, T W ) < δ namely  = δ/C. One consequence of this is that a linear map T : U −→ W into a Banach space, defined and continuous on a linear subspace, U ⊂ V. (with respect to the restriction of the norm from V to U ) extends uniquely to a continuous map T : U −→ W on the closure of U. As a general rule we drop the distinguishing subscript for norms, since which norm we are using can be determined by what it is being applied to. So, if T : V −→ W is continous and linear between normed spaces, or from now on ‘bounded’, then kT k = sup kT vk < ∞. (1.44) kvk=1 Lemma 1.3. The bounded linear maps between normed spaces V and W form a linear space B(V, W ) on which kT k defined by (1.44) or equivalently (1.45) kT k = inf{C; (1.41) holds} is a norm. Proof. First check that (1.44) is equivalent to (1.45). Define kT k by (1.44). Then for any v ∈ V, v 6= 0, (1.46) kT k ≥ kT ( kT vk v )k = =⇒ kT vk ≤ kT kkvk kvk kvk since as always this is trivially true for v = 0. Thus C = kT k is a constant for which (1.41) holds. Conversely, from the definition of kT k, if  > 0 then there exists v ∈ V with kvk = 1 such that kT k −  < kT vk ≤ C for any C for which (1.41) holds. Since  > 0 is arbitrary, kT k ≤ C and hence kT k is given by (1.45). From the definition of kT k, kT k = 0 implies T v = 0 for all v ∈ V and for λ 6= 0, (1.47) kλT k = sup kλT vk = |λ|kT k kvk=1 and this is also obvious for λ = 0. This only leaves the triangle inequality to check and for any T, S ∈ B(V, W ), and v ∈ V with kvk = 1 (1.48) k(T + S)vkW = kT v + SvkW ≤ kT vkW + kSvkW ≤ kT k + kSk so taking the supremum, kT + Sk ≤ kT k + kSk.  Thus we see the very satisfying fact that the space of bounded linear maps between two normed spaces is itself a normed space, with the norm being the best constant in the estimate (1.41). Make sure you absorb this! Such bounded linear maps between normed spaces are often called ‘operators’ because we are thinking of the normed spaces as being like function spaces. You might like to check boundedness for the example, I, of a linear operator in (1.36), namely that in terms of the supremum norm on C([0, 1]), kT k ≤ 1. One particularly important case is when W = K is the field, for us usually C. Then a simpler notation is handy and one sets V 0 = B(V, C) – this is called the dual space of V (also sometimes denoted V ∗ ). Proposition 1.4. If W is a Banach space then B(V, W ), with the norm (1.44), is a Banach space. Proof. We simply need to show that if W is a Banach space then every Cauchy sequence in B(V, W ) is convergent. The first thing to do is to find the limit. To say that Tn ∈ B(V, W ) is Cauchy, is just to say that given  > 0 there exists N such that n, m > N implies kTn − Tm k < . By the definition of the norm, if v ∈ V then kTn v − Tm vkW ≤ kTn − Tm kkvkV so Tn v is Cauchy in W for each v ∈ V. By assumption, W is complete, so (1.49) Tn v −→ w in W. However, the limit can only depend on v so we can define a map T : V −→ W by T v = w = limn→∞ Tn v as in (1.49). This map defined from the limits is linear, since Tn (λv) = λTn v −→ λT v and Tn (v1 + v2 ) = Tn v1 + Tn v2 −→ T v2 + T v2 = T (v1 + v2 ). Moreover, |kTn k − kTm k| ≤ kTn − Tm k so kTn k is Cauchy in [0, ∞) and hence converges, with limit S, and (1.50) kT vk = lim kTn vk ≤ Skvk n→∞ so kT k ≤ S shows that T is bounded. Returning to the Cauchy condition above and passing to the limit in kTn v − Tm vk ≤ kvk as m → ∞ shows that kTn − T k ≤  if n > M and hence Tn → T in B(V, W ) which is therefore complete.  Note that this proof is structurally the same as that of Lemma 1.2. One simple consequence of this is:Corollary 1.1. The dual space of a normed space is always a Banach space. However you should be a little suspicious here since we have not shown that the dual space V 0 is non-trivial, meaning we have not eliminated the possibility that V 0 = {0} even when V 6= {0}. The Hahn-Banach Theorem, discussed below, takes care of this. One game you can play is ‘what is the dual of that space’. Of course the dual is the dual, but you may well be able to identify the dual space of V with some other Banach space by finding a linear bijection between V 0 and the other space, W, which identifies the norms as well. We will play this game a bit later. 5. Subspaces and quotients The notion of a linear subspace of a vector space is natural enough, and you are likely quite familiar with it. Namely W ⊂ V where V is a vector space is a (linear) subspace if any linear combinations λ1 w1 + λ2 w2 ∈ W if λ1 , λ2 ∈ K and w1 , w2 ∈ W. Thus W ‘inherits’ its linear structure from V. Since we also have a topology from the metric we will be especially interested in closed subspaces. Check that you understand the (elementary) proof of Lemma 1.4. A subspace of a Banach space is a Banach space in terms of the restriction of the norm if and only if it is closed. There is a second very important way to construct new linear spaces from old. Namely we want to make a linear space out of ‘the rest’ of V, given that W is a linear subspace. In finite dimensions one way to do this is to give V an inner product and then take the subspace orthogonal to W. One problem with this is that the result depends, although not in an essential way, on the inner product. Instead we adopt the usual ‘myopia’ approach and take an equivalence relation on V which identifies points which differ by an element of W. The equivalence classes are then ‘planes parallel to W ’. I am going through this construction quickly here under the assumption that it is familiar to most of you, if not you should think about it carefully since we need to do it several times later. So, if W ⊂ V is a linear subspace of V we define a relation on V – remember this is just a subset of V × V with certain properties – by (1.51) v ∼W v 0 ⇐⇒ v − v 0 ∈ W ⇐⇒ ∃ w ∈ W s.t. v = v 0 + w. This satisfies the three conditions for an equivalence relation: (1) v ∼W v (2) v ∼W v 0 ⇐⇒ v 0 ∼W v (3) v ∼W v 0 , v 0 ∼W v 00 =⇒ v ∼W v 00 which means that we can regard it as a ‘coarser notion of equality.’ Then V /W is the set of equivalence classes with respect to ∼W . You can think of the elements of V /W as being of the form v + W – a particular element of V plus an arbitrary element of W. Then of course v 0 ∈ v + W if and only if v 0 − v ∈ W meaning v ∼W v 0 . The crucial point here is that (1.52) V /W is a vector space. You should check the details – see Problem 1.5. Note that the ‘is’ in (1.52) should really be expanded to ‘is in a natural way’ since as usual the linear structure is inherited from V : (1.53) λ(v + W ) = λv + W, (v1 + W ) + (v2 + W ) = (v1 + v2 ) + W. The subspace W appears as the origin in V /W. Now, two cases of this are of special interest to us. Proposition 1.5. If k · k is a seminorm on V then E = {v ∈ V ; kvk = 0} ⊂ V (1.54) is a linear subspace and kv + EkV /E = kvk (1.55) defines a norm on V /E. Proof. That E is linear follows from the properties of a seminorm, since kλvk = |λ|kvk shows that λv ∈ E if v ∈ E and λ ∈ K. Similarly the triangle inequality shows that v1 + v2 ∈ E if v1 , v2 ∈ E. To check that (1.55) defines a norm, first we need to check that it makes sense as a function k · kV /E −→ [0, ∞). This amounts to the statement that kv 0 k is the same for all elements v 0 = v + e ∈ v + E for a fixed v. This however follows from the triangle inequality applied twice: (1.56) kv 0 k ≤ kvk + kek = kvk ≤ kv 0 k + k − ek = kv 0 k. Now, I leave you the exercise of checking that k·kV /E is a norm, see Problem ??.  The second application is more serious, but in fact we will not use it for some time so I usually do not do this in lectures at this stage. Proposition 1.6. If W ⊂ V is a closed subspace of a normed space then (1.57) kv + W kV /W = inf kv + wkV w∈W defines a norm on V /W ; if V is a Banach space then so is V /W. For the proof see Problems ?? and ??. 6. Completion A normed space not being complete, not being a Banach space, is considered to be a defect which we might, indeed will, wish to rectify. Let V be a normed space with norm k · kV . A completion of V is a Banach space B with the following properties:(1) There is an injective (i.e. 1-1) linear map I : V −→ B (2) The norms satisfy kI(v)kB = kvkV ∀ v ∈ V. (1.58) (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 1.1. Each normed space has a completion. There are several ways to prove this, we will come across a more sophisticated one (using the Hahn-Banach Theorem) later. In the meantime I will describe two proofs. In the first the fact that any metric space has a completion in a similar sense is recalled and then it is shown that the linear structure extends to the completion. A second, ‘hands-on’, proof is also outlined with the idea of motivating the construction of the Lebesgue integral – which is in our near future. Proof 1. One of the neater proofs that any metric space has a completion is to use Lemma 1.2. Pick a point in the metric space of interest, p ∈ M, and then define a map (1.59) M 3 q 7−→ fq ∈ C∞ (M ), fq (x) = d(x, q) − d(x, p) ∀ x ∈ M. That fq ∈ C∞ (M ) is straightforward to check. It is bounded (because of the second term) by the reverse triangle inequality |fq (x)| = |d(x, q) − d(x, p)| ≤ d(p, q) and is continuous, as the difference of two continuous functions. Moreover the distance between two functions in the image is (1.60) sup |fq (x) − fq0 (x)| = sup |d(x, q) − d(x, q 0 )| = d(q, q 0 ) x∈M x∈M using the reverse triangle inequality (and evaluating at x = q). Thus the map (1.59) is well-defined, injective and even distance-preserving. Since C∞ (M ) is complete, the closure of the image of (1.59) is a complete metric space, X, in which M can be identified as a dense subset. Now, in case that M = V is a normed space this all goes through. The disconcerting thing is that the map q −→ fq is not linear. Nevertheless, we can give X a linear structure so that it becomes a Banach space in which V is a dense linear subspace. Namely for any two elements fi ∈ X, i = 1, 2, define (1.61) λ1 f1 + λ2 f2 = lim fλ1 pn +λ2 qn n→∞ where pn and qn are sequences in V such that fpn → f1 and fqn → f2 . Such sequences exist by the construction of X and the result does not depend on the choice of sequence – since if p0n is another choice in place of pn then fp0n − fpn → 0 in X (and similarly for qn ). So the element of the left in (1.61) is well-defined. All of the properties of a linear space and normed space now follow by continuity from V ⊂ X and it also follows that X is a Banach space (since a closed subset of a complete space is complete). Unfortunately there are quite a few annoying details to check!  ‘Proof 2’ (the last bit is left to you). Let V be a normed space. First we introduce the rather large space ( ) ∞ X ∞ (1.62) Ve = {uk }k=1 ; uk ∈ V and kuk k < ∞ k=1 the elements of which, if you recall, are said to be absolutely summable. Notice that the elements of Ve are sequences, valued in V so two sequences are equal, are the same, only when each entry in one is equal to the corresponding entry in the other – no shifting around or anything is permitted as far as equality is concerned. We think of these as series (remember this means nothing except changing the name, a series is a sequence and a sequence is a series), the only difference is that we ‘think’ of taking the limit of a sequence but we ‘think’ of summing the elements of a series, whether we can do so or not being a different matter. Now, each element of Ve is a Cauchy series – meaning the corresponding seN P quence of partial sums vN = uk is Cauchy if {uk } is absolutely summable. As k=1 noted earlier, this is simply because if M ≥ N then (1.63) kvM − vN k = k M X M X uj k ≤ j=N +1 j=N +1 gets small with N by the assumption that P X kuj k ≤ kuj k j≥N +1 kuj k < ∞. j Moreover, Ve is a linear space, where we add sequences, and multiply by constants, by doing the operations on each component:t1 {uk } + t2 {u0k } = {t1 uk + t2 u0k }. (1.64) This always gives an absolutely summable series by the triangle inequality: X X X (1.65) kt1 uk + t2 u0k k ≤ |t1 | kuk k + |t2 | ku0k k. k k k Within Ve consider the linear subspace ( ) X X (1.66) S = {uk }; kuk k < ∞, uk = 0 k k of those which sum to 0. As discussed in Section 5 above, we can form the quotient (1.67) B = Ve /S the elements of which are the ‘cosets’ of the form {uk } + S ⊂ Ve where {uk } ∈ Ve . This is our completion, we proceed to check the following properties of this B. (1) A norm on B (via a seminorm on Ṽ ) is defined by (1.68) kbkB = lim k n→∞ n X k=1 uk k, b = {uk } + S ∈ B. (2) The original space V is imbedded in B by (1.69) V 3 v 7−→ I(v) = {uk } + S, u1 = v, uk = 0 ∀ k > 1 and the norm satisfies (1.58). (3) I(V ) ⊂ B is dense. (4) B is a Banach space with the norm (1.68). So, first that (1.68) 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 additional term tends to zero in norm. Thus kbkB is well-defined for each element b ∈ B and kbkB = 0 means exactly that the sequence {uk } used to define it tends to 0 in norm, hence is in S hence b = 0 in B. The other two properties of norm are reasonably clear, since if b, b0 ∈ B are represented by {uk }, {u0k } in Ve then tb and b + b0 are represented by {tuk } and {uk + u0k } and (1.70) n n X X uk k, =⇒ ktbk = |t|kbk tuk k = |t| lim k lim k n→∞ n→∞ k=1 lim k n→∞ n X k=1 (uk + u0k )k = A =⇒ k=1 for  > 0 ∃ N s.t. ∀ n ≥ N, A −  ≤ k n X (uk + u0k )k =⇒ k=1 A−≤k n X k=1 uk k + k n X u0k )k ∀ n ≥ N =⇒ A −  ≤ kbkB + kb0 kB ∀  > 0 =⇒ k=1 kb + b0 kB ≤ kbkB + kb0 kB . 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 kvkV so kI(v)kB = kvkV and I(v) = 0 therefore implies v = 0 and hence I is also injective. 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 yourself (or have a better scheme). Note that I suggest that you to write out your own version of it carefully in Problem ??. Okay, what does it mean for B to be a Banach space, as discussed above it means that every absolutely summable series in B is convergent. Such a series {bn } (n) (n) is given by bn = {uk } + S where {uk } ∈ Ve and the summability condition is that ∞> X So, we want to show that P (1.71) kbn kB = n X n lim k N →∞ N X (n) uk k V . k=1 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’ P P (n) is that this series should look like uk in some sense – because it is supposed n k to represent the sum of the bn ’s. Now, it would be very nice if we had the estimate X X (n) (1.72) kuk kV < ∞ n k 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 (1.72) need not hold. We know that each of the sums over k – for given n – converges, but not the sum of the sums. All we know here is that the sum of the ‘limits of the norms’ in (1.71) converges. So, that is the problem! One way to see the solution is to note that we do not (n) have to choose the original {uk } to ‘represent’ bn – we can add to it any element (n) of S. One idea is to rearrange the uk – I am thinking here of fixed n – so that it ‘converges even faster.’ I will not go through this in full detail but rather do it later when we need the argument for the completeness of the space of Lebesgue integrable functions. Given  > 0 we can choose p1 so that for all p ≥ p1 , X (n) X (n) (1.73) |k uk kV − kbn kB | ≤ , kuk kV ≤ . k≤p k≥p Then in fact we can choose successive pj > pj−1 (remember that little n is fixed here) so that X (n) X (n) (1.74) |k uk kV − kbn kB | ≤ 2−j , kuk kV ≤ 2−j  ∀ j. k≤pj k≥pj (n) Now, ‘resum the series’ defining instead v1 = p1 P k=1 (n) (n) uk , vj = pj P k=pj−1 +1 (n) uk and do this setting  = 2−n for the nth series. Check that now X X (n) kvk kV < ∞. (1.75) n k (n) Of course, you should also check that bn = {vk } + S so that these new summable series work just as well as the old ones. After this fiddling you can now try to find a limit for the sequence as X (p) (1.76) b = {wk } + S, wk = vl ∈ V. l+p=k+1 So, you need to check that this {wk } is absolutely summable in V and that bn → b as n → ∞. Finally then there is the question of showing that I(V ) is dense in B. You can do this using the same idea as above – in fact it might be better to do it first. Given an element b ∈ B we need to find elements in V, vk such that kI(vk ) − bkB → 0 as Nj P k → ∞. Take an absolutely summable series uk representing b and take vj = uk k=1 where the pj ’s are constructed as above and check that I(vj ) → b by computing X X uk kV ≤ kuk kV . (1.77) kI(vj ) − bkB = lim k →∞ k>pj k>pj  7. More examples Let me collect some examples of normed and Banach spaces. Those mentioned above and in the problems include: • c0 the space of convergent sequences in C with supremum norm, a Banach space. • lp one space for each real number 1 ≤ p < ∞; the space of p-summable series with corresponding norm; all Banach spaces. The most important of these for us is the case p = 2, which is (a) Hilbert space. • l∞ the space of bounded sequences with supremum norm, a Banach space with c0 ⊂ l∞ as a closed subspace with the same norm. • C([a, b]) or more generally C(M ) for any compact metric space M – the Banach space of continuous functions with supremum norm. • C∞ (R), or more generally C∞ (M ) for any metric space M – the Banach space of bounded continuous functions with supremum norm. • C0 (R), or more generally C0 (M ) for any metric space M – the Banach space of continuous functions which ‘vanish at infinity’ (see Problem ??) with supremum norm. A closed subspace, with the same norm, in C∞ (M ). • C k ([a, b]) the space of k times continuously differentiable (so k ∈ N) functions on [a, b] with norm the sum of the supremum norms on the function and its derivatives. Each is a Banach space – see Problem ??. • The space C([0, 1]) with norm Z 1 (1.78) kukL1 = |u|dx 0 given by the Riemann integral of the absolute value. A normed space, but not a Banach space. We will construct the concrete completion, L1 ([0, 1]) of Lebesgue integrable ‘functions’. • The space R([a, b]) of Riemann integrable functions on [a, b] with kuk defined by (1.78). This is only a seminorm, since there are Riemann integrable functions (note that u Riemann integrable does imply that |u| is Riemann integrable) with |u| having vanishing Riemann integral but which are not identically zero. This cannot happen for continuous functions. So the quotient is a normed space, but it is not complete. • The same spaces – either of continuous or of Riemann integrable functions but with the (semi- in the second case) norm Z (1.79) kukLp = ! p1 b p |u| . a Not complete in either case even after passing to the quotient to get a norm for Riemann integrable functions. We can, and indeed will, define Lp (a, b) as the completion of C([a, b]) with respect to the Lp norm. However we will get a concrete realization of it soon. • Suppose 0 < α < 1 and consider the subspace of C([a, b]) consisting of the ‘Hölder continuous functions’ with exponent α, that is those u : [a, b] −→ C which satisfy (1.80) |u(x) − u(y)| ≤ C|x − y|α for some C ≥ 0. (1.81) Note that this already implies the continuity of u. As norm one can take the sum of the supremum norm and the ‘best constant’ which is the same as |u(x) − u(y)| kukC α = sup |u(x)| + sup ; |x − y|α x∈[a,b]| x6=y∈[a,b] it is a Banach space usually denoted C α ([a, b]). • Note the previous example works for α = 1 as well, then it is not denoted C 1 ([a, b]), since that is the space of once continuously differentiable functions; this is the space of Lipschitz functions Λ([a, b]) – again it is a Banach space. • We will also talk about Sobolev spaces later. These are functions with ‘Lebesgue integrable derivatives’. It is perhaps not easy to see how to define these, but if one takes the norm on C 1 ([a, b]) 1  du 2 2 2 (1.82) kukH 1 = kukL2 + k kL2 dx and completes it, one gets the Sobolev space H 1 ([a, b]) – it is a Banach space (and a Hilbert space). In fact it is a subspace of C([a, b]) = C([a, b]). Here is an example to see that the space of continuous functions on [0, 1] with norm (1.78) is not complete; things are even worse than this example indicates! It is a bit harder to show that the quotient of the Riemann integrable functions is not complete, feel free to give it a try. Take a simple non-negative continuous function on R for instance ( 1 − |x| if |x| ≤ 1 (1.83) f (x) = 0 if |x| > 1. R1 Then −1 f (x) = 1. Now scale it up and in by setting fN (x) = N f (N 3 x) = 0 if |x| > N −3 . R1 So it vanishes outside [−N −3 , N −3 ] and has −1 fN (x)dx = N −2 . It follows that the sequence {fN } is absolutely summable with respect to the integral norm in (1.78) P on [−1, 1]. The pointwise series fN (x) converges everywhere except at x = 0 – (1.84) N since at each point x 6= 0, fN (x) = 0 if N 3 |x| > 1. The resulting function, even if we ignore the problem at x = 0, is not Riemann integrable because it is not bounded. You might respond that the sum of the series is ‘improperly Riemann integrable’. This is true but does not help much. It is at this point that I start doing Lebesgue integration in the lectures. The following material is from later in the course but fits here quite reasonably. 8. Baire’s theorem At least once I wrote a version of the following material on the blackboard during the first mid-term test, in an an attempt to distract people. It did not work very well – its seems that MIT students have already been toughened up by this stage. Baire’s theorem will be used later (it is also known as ‘Baire category theory’ although it has nothing to do with categories in the modern sense). This is a theorem about complete metric spaces – it could be included in the earlier course ‘Real Analysis’ but the main applications are in Functional Analysis. Theorem 1.2 (Baire). If M is a non-empty complete metric space and Cn ⊂ M, n ∈ N, are closed subsets such that [ (1.85) M= Cn n then at least one of the Cn ’s has an interior point, i.e. contains a non-empty ball in M. Proof. We will assume that each of the Cn ’s has empty interior, hoping to arrive at a contradiction to (1.85) using the other properties. Thus if p ∈ M and  > 0 the open ball B(p, ) is not contained in any one of the Cn . We start by choosing p1 ∈ M \ C1 which must exist since M is not empty and otherwise C1 = M. Now, there must exist 1 > 0 such that B(p1 , 1 ) ∩ C1 = ∅, since C1 is closed. No open ball around p1 can be contained in C2 so there exists p2 ∈ B(p1 , 1 /3) which is not in C2 . Again since C2 is closed there exists 2 > 0, 2 < 1 /3 such that B(p2 , 2 ) ∩ C2 = ∅. Proceeding inductively we suppose there is are k points pi , i = 1, . . . , k and positive numbers (1.86) 0 < k < k−1 /3 < k−2 /32 < · · · < 1 /3k−1 such that (1.87) pj ∈ B(pj−1 , j−1 /3), B(pj , j ) ∩ Cj = ∅. Then we can add another pk+1 by using the properties of Ck+1 – it has empty interior so there is some point in B(pk , k /3) which is not in Ck+1 and then B(pk+1 , k+1 ) ∩ Ck+1 = ∅ where k+1 > 0 but k+1 < k /3. Thus, we have a sequence {pk } in M satisfying (1.86) and (1.87) for all k. Since d(pk+1 , pk ) < k /3 this is a Cauchy sequence, in fact (1.88) d(pk , pk+l ) < k /3 + · · · + k+l−1 /3 < 2k . Since M is assumed to be complete this sequence converges to a limit, q ∈ M. Notice however that pl ∈ B(pk , 2k /3) for all k > l so d(pk , q) ≤ 2k /3 which implies that q∈ / Ck for any k. This is the desired contradiction to (1.85). Thus, at least one of the Cn must have non-empty interior.  In applications one might get a complete metric space written as a countable union of subsets [ (1.89) M= En , En ⊂ M n where the En are not necessarily closed. We can still apply Baire’s theorem however, just take Cn = En to be the closures – then of course (1.85) holds since En ⊂ Cn . The conclusion from (1.89) for a complete M is (1.90) For at least one n the closure of En has non-empty interior. 9. Uniform boundedness One application of Baire’s theorem is often called the uniform boundedness principle or Banach-Steinhaus Theorem. Theorem 1.3 (Uniform boundedness). Let B be a Banach space and suppose that Tn is a sequence of bounded (i.e. continuous) linear operators Tn : B −→ V where V is a normed space. Suppose that for each b ∈ B the set {Tn (b)} ⊂ V is bounded (in norm of course) then supn kTn k < ∞. Proof. This follows from a pretty direct application of Baire’s theorem to B. Consider the sets (1.91) Sp = {b ∈ B, kbk ≤ 1, kTn bkV ≤ p ∀ n}, p ∈ N. Each Sp is closed because Tn is continuous, so if bk → b is a convergent sequence in Sp then kbk ≤ 1 and kTn (b)k ≤ p. The union of the Sp is the whole of the closed ball of radius one around the origin in B : [ (1.92) {b ∈ B; d(b, 0) ≤ 1} = Sp p because of the assumption of ‘pointwise boundedness’ – each b with kbk ≤ 1 must be in one of the Sp ’s. So, by Baire’s theorem one of the sets Sp has non-empty interior, it therefore contains a closed ball of positive radius around some point. Thus for some p, some v ∈ Sp , and some δ > 0, (1.93) w ∈ B, kwkB ≤ δ =⇒ kTn (v + w)kV ≤ p ∀ n. Since v ∈ Sp is fixed it follows that kTn wk ≤ kTn (v + w)k + kTn vk ≤ 2p for all w with kwk ≤ δ. This however implies that the norms are uniformly bounded: (1.94) kTn k ≤ 2p/δ as claimed.  10. Open mapping theorem The second major application of Baire’s theorem is to Theorem 1.4 (Open Mapping). If T : B1 −→ B2 is a bounded and surjective linear map between two Banach spaces then T is open: (1.95) T (O) ⊂ B2 is open if O ⊂ B1 is open. This is ‘wrong way continuity’ and as such can be used to prove the continuity of inverse maps as we shall see. The proof uses Baire’s theorem pretty directly, but then another similar sort of argument is needed to complete the proof. Note however that the proof is considerably simplified if we assume that B1 is a Hilbert space. There are more direct but more computational proofs, see Problem ??. I prefer this one because I have a reasonable chance of remembering the steps. Proof. What we will try to show is that the image under T of the unit open ball around the origin, B(0, 1) ⊂ B1 contains an open ball around the origin in B2 . The first part, of the proof, using Baire’s theorem shows that the closure of the image, so in B2 , has 0 as an interior point – i.e. it contains an open ball around the origin in B2 : (1.96) T (B(0, 1) ⊃ B(0, δ), δ > 0. To see this we apply Baire’s theorem to the sets (1.97) Cp = clB2 T (B(0, p)) the closure of the image of the ball in B1 of radius p. We know that [ (1.98) B2 = T (B(0, p)) p since that is what surjectivity means – every point is the image of something. Thus one of the closed sets Cp has an interior point, v. Since T is surjective, v = T u for some u ∈ B1 . The sets Cp increase with p so we can take a larger p and v is still an interior point, from which it follows that 0 = v − T u is an interior point as well. Thus indeed (1.99) Cp ⊃ B(0, δ) for some δ > 0. Rescaling by p, using the linearity of T, it follows that with δ replaced by δ/p, we get (1.96). If we assume that B1 is a Hilbert space (and you are reading this after we have studied Hilbert spaces) then (1.96) shows that if v ∈ B2 , kvk < δ there is a sequence un with kun k ≤ 1 and T un → v. As a bounded sequence un has a weakly convergent subsequence, unj * u, where we know this implies kuk ≤ 1 and Aunj * Au = v since Aun → v. This strengthens (1.96) to T (B(0, 1) ⊃ B(0, δ/2) and proves that T is an open map. If B1 is a Banach space but not a Hilbert space (or you don’t yet know about Hilbert spaces) we need to work a little harder. Having applied Baire’s thereom, consider now what (1.96) means. It follows that each v ∈ B2 , with kvk = δ, is the limit of a sequence T un where kun k ≤ 1. What we want to find is such a sequence, un , which converges. To do so we need to choose the sequence more carefully. Certainly we can stop somewhere along the way and see that δ 1 (1.100) v ∈ B2 , kvk = δ =⇒ ∃ u ∈ B1 , kuk ≤ 1, kv − T uk ≤ = kvk 2 2 where of course we could replace 2δ by any positive constant but the point is the last inequality is now relative to the norm of v. Scaling again, if we take any v 6= 0 in B2 and apply (1.100) to v/kvk we conclude that (for C = p/δ a fixed constant) 1 (1.101) v ∈ B2 =⇒ ∃ u ∈ B1 , kuk ≤ Ckvk, kv − T uk ≤ kvk 2 where the size of u only depends on the size of v; of course this is also true for v = 0 by taking u = 0. Using this we construct the desired better approximating sequence. Given w ∈ B1 , choose u1 = u according to (1.101) for v = w = w1 . Thus ku1 k ≤ C, and w2 = w1 − T u1 satisfies kw2 k ≤ 12 kwk. Now proceed by induction, supposing that we have constructed a sequence uj , j < n, in B1 with kuj k ≤ C2−j+1 kwk and kwj k ≤ 2−j+1 kwk for j ≤ n, where wj = wj−1 − T uj−1 – which we have for n = 1. Then we can choose un , using (1.101), so kun k ≤ Ckwn k ≤ C2−n+1 kwk and such that wn+1 = wn − T un has kwn+1 k ≤ 12 kwn k ≤ 2−n kwk to extend the induction. Thus we get a sequence un which is absolutely summable in B1 , since P kun k ≤ 2Ckwk, and hence converges by the assumed completeness of B1 this n time. Moreover (1.102) w − T( n X j=1 uj ) = w1 − n X (wj − wj+1 ) = wn+1 j=1 so T u = w and kuk ≤ 2Ckwk. Thus finally we have shown that each w ∈ B(0, 1) in B2 is the image of some u ∈ B1 with kuk ≤ 2C. Thus T (B(0, 3C)) ⊃ B(0, 1). By scaling it follows that the image of any open ball around the origin contains an open ball around the origin. Now, the linearity of T shows that the image T (O) of any open set is open, since if w ∈ T (O) then w = T u for some u ∈ O and hence u + B(0, ) ⊂ O for  > 0 and then w + B(0, δ) ⊂ T (O) for δ > 0 sufficiently small.  One important corollary of this is something that seems like it should be obvious, but definitely needs completeness to be true. Corollary 1.2. If T : B1 −→ B2 is a bounded linear map between Banach spaces which is 1-1 and onto, i.e. is a bijection, then it is a homeomorphism – meaning its inverse, which is necessarily linear, is also bounded. Proof. The only confusing thing is the notation. Note that T −1 is generally used both for the inverse, when it exists, and also to denote the inverse map on sets even when there is no true inverse. The inverse of T, let’s call it S : B2 −→ B1 , is certainly linear. If O ⊂ B1 is open then S −1 (O) = T (O), since to say v ∈ S −1 (O) means S(v) ∈ O which is just v ∈ T (O), is open by the Open Mapping theorem, so S is continuous.  11. Closed graph theorem For the next application you should check, it is one of the problems, that the product of two Banach spaces, B1 × B2 , – which is just the linear space of all pairs (u, v), u ∈ B1 and v ∈ B2 , – is a Banach space with respect to the sum of the norms (1.103) k(u, v)k = kuk1 + kvk2 . Theorem 1.5 (Closed Graph). If T : B1 −→ B2 is a linear map between Banach spaces then it is bounded if and only if its graph (1.104) Gr(T ) = {(u, v) ∈ B1 × B2 ; v = T u} is a closed subset of the Banach space B1 × B2 . Proof. Suppose first that T is bounded, i.e. continuous. A sequence (un , vn ) ∈ B1 × B2 is in Gr(T ) if and only if vn = T un . So, if it converges, then un → u and vn = T un → T v by the continuity of T, so the limit is in Gr(T ) which is therefore closed. Conversely, suppose the graph is closed. This means that viewed as a normed space in its own right it is complete. Given the graph we can reconstruct the map it comes from (whether linear or not) in a little diagram. From B1 × B2 consider the two projections, π1 (u, v) = u and π2 (u, v) = v. Both of them are continuous since the norm of either u or v is less than the norm in (1.103). Restricting them to Gr(T ) ⊂ B1 × B2 gives Gr(T ) < (1.105) S B1 | π1 π2 T # / B2 . This little diagram commutes. Indeed there are two ways to map a point (u, v) ∈ Gr(T ) to B2 , either directly, sending it to v or first sending it to u ∈ B1 and then to T u. Since v = T u these are the same. Now, as already noted, Gr(T ) ⊂ B1 × B2 is a closed subspace, so it too is a Banach space and π1 and π2 remain continuous when restricted to it. The map π1 is 1-1 and onto, because each u occurs as the first element of precisely one pair, namely (u, T u) ∈ Gr(T ). Thus the Corollary above applies to π1 to show that its inverse, S is continuous. But then T = π2 ◦ S, from the commutativity, is also continuous proving the theorem.  You might wish to entertain yourself by showing that conversely the Open Mapping Theorem is a consequence of the Closed Graph Theorem. The characterization of continuous linear maps through the fact that their graphs are closed has led to significant extensions. For instance consider a linear map but only defined on a subspace (often required to be dense in which case it is said to be ‘densely defined’) D ⊂ B, where B is a Banach space, (1.106) A : D −→ B linear. Such a map is said to be closed if its graph Gr(A) = {(u, Au); u ∈ D} ⊂ B × B is closed. Check for example that if H 1 (R) ⊂ L2 (R) (I’m assuming that you are reading this near the end of the course . . . ) is the space defined in Chapter 4, as consisting of the elements with a strong derivative in L2 (R) then d : D = H 1 (R) −→ L2 (R) is closed. dx This follows for instance from the ‘weak implies strong’ result for differentiation. If un ∈ H 1 (R) is a sequence such that un → u in L2 (R) and dun /dx −→ v in L2 (which is convergence in L2 (R) × L2 (R)) then u ∈ H 1 (R) and v = du/dx in the same strong sense. Such a closed operator, A, can be turned into a bounded operator by changing the norm on the domain D to the ‘graph norm’ (1.107) (1.108) kukGr = kuk + kAuk. 12. Hahn-Banach theorem Now, there is always a little pressure to state and prove the Hahn-Banach Theorem. This is about extension of functionals. Stately starkly, the basic question is: Does a normed space have any non-trivial continuous linear functionals on it? That is, is the dual space always non-trivial (of course there is always the zero linear functional but that is not very amusing). We do not really encounter this problem since for a Hilbert space, or even a pre-Hilbert space, there is always the space itself, giving continuous linear functionals through the pairing – Riesz’ Theorem says that in the case of a Hilbert space that is all there is. If you are following the course then at this point you should also see that the only continuous linear functionals on a pre-Hilbert space correspond to points in the completion. I could have used the Hahn-Banach Theorem to show that any normed space has a completion, but I gave a more direct argument for this, which was in any case much more relevant for the cases of L1 (R) and L2 (R) for which we wanted concrete completions. Theorem 1.6 (Hahn-Banach). If M ⊂ V is a linear subspace of a normed space and u : M −→ C is a linear map such that |u(t)| ≤ CktkV ∀ t ∈ M (1.109) then there exists a bounded linear functional U : V −→ C with kU k ≤ C and U M = u. First, by computation, we show that we can extend any continuous linear functional ‘a little bit’ without increasing the norm. Lemma 1.5. Suppose M ⊂ V is a subspace of a normed linear space, x ∈ / M and u : M −→ C is a bounded linear functional as in (1.109) then there exists u0 : M 0 −→ C, where M 0 = {t0 ∈ V ; t0 = t + ax, a ∈ C}, such that (1.110) u0 M = u, |u0 (t + ax)| ≤ Ckt + axkV , ∀ t ∈ M, a ∈ C. Proof. Note that the decomposition t0 = t + ax of a point in M 0 is unique, since t + ax = t̃ + ãx implies (a − ã)x ∈ M so a = ã, since x ∈ / M and hence t = t̃ as well. Thus (1.111) u0 (t + ax) = u0 (t) + au(x) = u(t) + λa, λ = u0 (x) and all we have at our disposal is the choice of λ. Any choice will give a linear functional extending u, the problem of course is to arrange the continuity estimate without increasing the constant C. In fact if C = 0 then u = 0 and we can take the zero extension. So we might as well assume that C = 1 since dividing u by C arranges this and if u0 extends u/C then Cu0 extends u and the norm estimate in (1.110) follows. So we now assume that (1.112) |u(t)| ≤ ktkV ∀ t ∈ M. We want to choose λ so that (1.113) |u(t) + aλ| ≤ kt + axkV ∀ t ∈ M, a ∈ C. Certainly when a = 0 this represents no restriction on λ. For a 6= 0 we can divide through by −a and (1.113) becomes t t (1.114) |a||u(− ) − λ| = |u(t) + aλ| ≤ kt + axkV = |a|k − − xkV a a and since −t/a ∈ M we only need to arrange that (1.115) |u(t) − λ| ≤ kt − xkV ∀ t ∈ M and the general case will follow by reversing the scaling. A complex linear functional such as u can be recovered from its real part, as we see below, so set (1.116) w(t) = Re(u(t)), |w(t)| ≤ ktkV ∀ t ∈ M. We proceed to show the real version of the Lemma, that w can be extended to a linear functional w0 : M + Rx −→ R if x ∈ / M without increasing the norm. The same argument as above shows that the only freedom is the choice of λ = w0 (x) and we need to choose λ ∈ R so that |w(t) − λ| ≤ kt − xkV ∀ t ∈ M. (1.117) The norm estimate on w shows that (1.118) |w(t1 ) − w(t2 )| ≤ |u(t1 ) − u(t2 )| ≤ kt1 − t2 k ≤ kt1 − xkV + kt2 − xkV . Writing this out using the reality we find (1.119) w(t1 ) − w(t2 ) ≤ kt1 − xkV + kt2 − xkV =⇒ w(t1 ) − kt1 − xk ≤ w(t2 ) + kt2 − xkV ∀ t1 , t2 ∈ M. We can then take the supremum on the left and the infimum on the right and choose λ in between – namely we have shown that there exists λ ∈ R with (1.120) w(t) − kt − xkV ≤ sup (w(t1 ) − kt1 − xk) ≤ λ t2 ∈M ≤ inf (w(t1 ) + kt1 − xk) ≤ w(t) + kt − xkV ∀ t ∈ M. t2 ∈M This in turn implies that (1.121) −kt − xkV ≤ −w(t) + λ ≤ kt − xkV =⇒ |w(t) − λ| ≤ kt − xkV ∀ t ∈ M. So we have an extension of w to a real functional w0 : M + Rx −→ R with |w (t + ax)| ≤ kt + axkV for all a ∈ R. We can repeat this argument to obtain a further extension w00 : M + Cx = M + Rx + R(ix) −→ R without increasing the norm. Now we find the desired extension of u by setting 0 (1.122) u0 (t + cx) = w00 (t + ax + b(ix)) − iw00 (it − bx + a(ix)) : M + Cx −→ C where c = a + ib. This is certainly linear over the reals and linearity over complex coefficients follows since (1.123) u0 (it + icx) = w00 (it − bx + a(ix))) − iw00 (−t − ax − b(ix)) = i(w00 (t + ax + b(ix) − iw00 (it − bx + a(ix))) = iu0 (t + cx). The uniqueness of a complex linear functional with given real part also shows that u0 M = u. Finally, to estimate the norm of u0 notice that for each t ∈ M and c ∈ C there is a unique θ ∈ [0, 2π) such that (1.124) |u0 (t + cx)| = Re eiθ u0 (t + cx) = w00 (eiθ t + eiθ cx) ≤ keiθ t + eiθ cxkV = kt + cxkV . This completes the proof of the Lemma.  Proof of Hahn-Banach. This is an application of Zorn’s Lemma. I am not going to get into the derivation of Zorn’s Lemma from the Axiom of Choice, but if you believe the latter – and you are advised to do so, at least before lunchtime – you should believe the former. See also the discussion in Section 2.9 Zorn’s Lemma is a statement about partially ordered sets. A partial order on a set E is a subset of E × E, so a relation, where the condition that (e, f ) be in the relation is written e ≺ f and it must satisfy the three conditions e≺e∀e∈E (1.125) e ≺ f and f ≺ e =⇒ e = f e ≺ f and f ≺ g =⇒ e ≺ g. So, the missing ingredient between this and an order is that two elements need not be related at all, either way. A subset of a partially ordered set inherits the partial order and such a subset is said to be a chain if each pair of its elements is related one way or the other. An upper bound on a subset D ⊂ E is an element e ∈ E such that d ≺ e for all d ∈ D. A maximal element of E is one which is not majorized, that is e ≺ f, f ∈ E, implies e = f. Lemma 1.6 (Zorn). If every chain in a (non-empty) partially ordered set has an upper bound then the set contains at least one maximal element. So, we are just accepting this Lemma as axiomatic. However, make sure that you appreciate that it is true for countable sets. Namely if C is countable and has no maximal element then it must contain a chain which has no upper bound. To see this, write C as a sequence {ci }i∈N . Then x1 = c1 is not maximal so there exists some ck , k > 1, with c1 ≺ ck in terms of the order in C. From the properties of N it follows that there is a smallest k = k2 such that ck has this property, but k2 > 1. Let this be x2 = ck2 and proceed in the same way – x3 = ck3 where k3 > k2 is the smallest such integer for which ck2 ≺ ck3 . Assuming C is infinite in the first place this grinds out an infinite chain xi . Now you can check that this cannot have an upper bound because every element of C is either one of these, and so cannot be an upper bound, or else it is cj with kl < j < kl+1 for some l and then it is not greater than xl . The point of Zorn’s Lemma is precisely that it applies to uncountable sets. One consequence of Zorn’s Lemma is the existence of ‘Hamel bases’ for infinite dimensional vector spaces. This is pretty much irrelevant for us but I include it since you can use this to show the existence of non-continuous linear functionals on a Banach space; the proof is also an easier version of the proof of the Hahn-Banach theorem. Definition 1.6. A Hamel basis B ⊂ V of a vector space is a linearly independent subspace which spans, i.e. every element of V is a finite linear combination of elements of B. Hamel bases have a strong tendency to be big. Notice that for a finite dimensional space this is just the usual notion of a basis. Proof. Look at the collection X of all linearly independent subsets of V. This is non-empty – assuming V 6= {0} we can take {x} ∈ X for any x 6= 0 in V. Inclusion is a partial order on X . Suppose Y ⊂ X is a chain with respect to this partial order – so for any two elements one is contained in the other. Let L = ∪Y be the union of all the elements of Y. Each element of C is contained in L so L is an upper bounde for Y. Thus by Zorn’s lemma X just contain a maximal element, say B. Then B is a Hamel basis, since if some element w ∈ V were not a finite linear combination of elements of B then B ∪ {w} would also be linearly independent and hence an element of X which contains B which contradicts its maximality.  So, back to Hahn-Banach. We are given a functional u : M −→ C defined on some linear subspace M ⊂ V of a normed space where u is bounded with respect to the induced norm on M. We will apply Zorn’s Lemma to the set E consisting of all extensions (v, N ) of u with the same norm; it is generally non-countable. That is, V ⊃ N ⊃ M, v M = u and kvkN = kukM . This is certainly non-empty since it contains (u, M ) and has the natural partial order that (v1 , N1 ) ≺ (v2 , N2 ) if N1 ⊂ N2 and v2 N = v1 . You should check that 1 this is a partial order. Let C be a chain in this set of extensions. Thus for any two elements (vi , Ni ) ∈ C, i = 1, 2, either (v1 , N1 ) ≺ (v2 , N2 ) or the other way around. This means that [ (1.126) Ñ = {N ; (v, N ) ∈ C for some v} ⊂ V is a linear space. Note that this union need not be countable, or anything like that, but any two elements of Ñ are each in one of the N ’s and one of these must be contained in the other by the chain condition. Thus each pair of elements of Ñ is actually in a common N and hence so is their linear span. Similarly we can define an extension (1.127) ṽ : Ñ −→ C, ṽ(x) = v(x) if x ∈ N, (v, N ) ∈ C. There may be many pairs (v, N ) ∈ C satisfying x ∈ N for a given x but the chain condition implies that v(x) is the same for all of them. Thus ṽ is well defined, and is clearly also linear, extends u and satisfies the norm condition |ṽ(x)| ≤ kukM kxkV . Thus (ṽ, Ñ ) is an upper bound for the chain C. So, the set of all extensions, E, with the norm condition, satisfies the hypothesis of Zorn’s Lemma, so must – at least in the mornings – have a maximal element (ũ, M̃ ). If M̃ = V then we are done. However, in the contary case there exists x ∈ V \ M̃ . This means we can apply our little lemma and construct an extension (u0 , M̃ 0 ) of (ũ, M̃ ) which is therefore also an element of E and satisfies (ũ, M̃ ) ≺ (u0 , M̃ 0 ). This however contradicts the condition that (ũ, M̃ ) be maximal, so is forbidden by Zorn.  There are many applications of the Hahn-Banach Theorem. As remarked earlier, one significant one is that the dual space of a non-trivial normed space is itself non-trivial. Proposition 1.7. For any normed space V and element 0 6= v ∈ V there is a continuous linear functional f : V −→ C with f (v) = 1 and kf k = 1/kvkV . Proof. Start with the one-dimensional space, M, spanned by v and define u(zv) = z. This has norm 1/kvkV . Extend it using the Hahn-Banach Theorem and you will get a continuous functional f as desired.  13. Double dual Let me give another application of the Hahn-Banach theorem, although I have generally not covered this in lectures. If V is a normed space, we know its dual space, V 0 , to be a Banach space. Let V 00 = (V 0 )0 be the dual of the dual. Proposition 1.8. If v ∈ V then the linear map on V 0 : (1.128) Tv : V 0 −→ C, Tv (v 0 ) = v 0 (v) is continuous and this defines an isometric linear injection V ,→ V 00 , kTv k = kvk. Proof. The definition of Tv is ‘tautologous’, meaning it is almost the definition of V 0 . First check Tv in (1.128) is linear. Indeed, if v10 , v20 ∈ V 0 and λ1 , λ2 ∈ C then Tv (λ1 v10 + λ2 v20 ) = (λ1 v10 + λ2 v20 )(v) = λ1 v10 (v) + λ2 v20 (v) = λ1 Tv (v10 ) + λ2 Tv (v20 ). That Tv ∈ V 00 , i.e. is bounded, follows too since |Tv (v 0 )| = |v 0 (v)| ≤ kv 0 kV 0 kvkV ; this also shows that kTv kV 00 ≤ kvk. On the other hand, by Proposition 1.7 above, if kvk = 1 then there exists v 0 ∈ V 0 such that v 0 (v) = 1 and kv 0 kV 0 = 1. Then Tv (v 0 ) = v 0 (v) = 1 shows that kTv k = 1 so in general kTv k = kvk. It also needs to be checked that V 3 v 7−→ Tv ∈ V 00 is a linear map – this is clear from the definition. It is necessarily 1-1 since kTv k = kvk.  Now, it is definitely not the case in general that V 00 = V in the sense that this injection is also a surjection. Since V 00 is always a Banach space, one necessary condition is that V itself should be a Banach space. In fact the closure of the image of V in V 00 is a completion of V. If the map to V 00 is a bijection then V is said to be reflexive. It is pretty easy to find examples of non-reflexive Banach spaces, the most familiar is c0 – the space of infinite sequences converging to 0. Its dual can be identified with l1 , the space of summable sequences. Its dual in turn, the bidual of c0 , is the space l∞ of bounded sequences, into which the embedding is the obvious one, so c0 is not reflexive. In fact l1 is not reflexive either. There are useful characterizations of reflexive Banach spaces. You may be interested enough to look up James’ Theorem:- A Banach space is reflexive if and only if every continuous linear functional on it attains its supremum on the unit ball. 14. Axioms of a vector space In case you missed out on one of the basic linear algebra courses, or have a poor memory, here are the axioms of a vector space over a field K (either R or C for us). A vector space structure on a set V is a pair of maps (1.129) + : V × V −→ V, · : K × V −→ V satisfying the conditions listed below. These maps are written +(v1 , v2 ) = v1 + v2 and ·(λ, v) = λv, λ ∈ K, V, v1 , v2 ∈ V. additive commutativity v1 + v2 = v2 + v1 for all v1 , v2 ∈ V. additive associativity v1 + (v2 + v3 ) = (v1 + v2 ) + v3 for all v1 , v2 , v3 ∈ V. existence of zero There is an element 0 ∈ V such that v + 0 = v for all v ∈ V. additive invertibility For each v ∈ V there exists w ∈ V such that v + w = 0. distributivity of scalar additivity (λ1 + λ2 )v = λ1 v + λ2 v for all λ1 , λ2 ∈ K and v ∈ V. multiplicativity λ1 (λ2 v) = (λ1 λ2 )v for all λ1 , λ2 ∈ K and v ∈ V. action of multiplicative identity 1v = v for all v ∈ V. distributivity of space additivity λ(v1 + v2 ) = λv1 + λv2 for all λ ∈ K v1 , v2 ∈ V.