LECTURE 4, 18.155, 20 SEPTEMBER 2011So now for some ‘hard analysis’ rather than softer measure theory.(1) I will define the spaces Lp(Rn) by completion and then showthat they have concrete realizations as spaces of (equivalenceclasses of) functions. We can start with S(Rn), or C∞c(Rn) orC0c(Rn) – it doesn’t make much difference. Let me work withC∞c(Rn). On this space we have norms, one for each 1 ≤ p < ∞(and there is a similar norm for p = ∞ but the present approachdoes not work directly in that case – can you see why?)kφkLp=ZRn|φ|p)1/p.Here the integral really is a Riemann, or iterated Riemann ifyou only know the 1-D case, integral since there exists R > 0such that φ(x) = 0 if |xj| ≥ R for any j.(2) The first two properties of a norm – that kφkLp≥ 0 with equal-ity iff φ ≡ 0 follow from basic properties of the Riemann inte-gral, as does kcφkLp= |c|kφkLp. The triangle inequality is calledMinkowkski’s inequality:-kφ + ψkLp≤ kφkLp+ kψkLp∀ φ, ψ ∈ C∞c(Rn)needs proving! If you don’t know it you should look it up. Theusual chain of reasoning is to start with Young’s inequality (fornon-negative real numbers)ab ≤app+bqq,1p+1q= 1(which follows from calculus) then integrate this to getZ|af(x)||bg(x)|dx ≤1papZ|f(x)|p+1qbqZ|g(x)|q.In non-trivial cases take apR|f(x)|p= 1, bqR|g(x)|q= 1 inwhich case we see that|Zf(x)g(x)dx| ≤Z|f(x)g(x)|dx ≤ kfkLpkgkLqwhich is H¨older’s inequality. Then Minkowski’s inequality fol-lows by observing that|φ(x) + ψ(x)|p≤ |φ(x)||φ(x) + ψ(x)|p−1+ |ψ(x)||φ(x) + ψ(x)|p−112 LECTURE 4, 18.155, 20 SEPTEMBER 2011and applying H¨older’s inequality to the two products on theright.(3) It is easy to see that these norms are not complete on C∞c(Rn)(or even on continuous functions with compact support). Wecan always ‘complete’ a normed space to a Banach space ab-stractly. The standard methods are to embed it into its bidual(the dual of its dual) which uses the Hahn-Banach theorem orto embed it in a complete space – such as the space of contin-uous bounded functions on the space itself or, as I will quicklyoutline, to take the linear space of all Cauchy sequences, give ita seminorm and pass to the quotient Banach space.Let V be a normed space. The space of maps N −→ N has alinear structure given by addition in N. The Cauchy sequencesin N form a subspace, which we can temporarily denote c(N).Since the norm is continuous on N it follows that if {aj} ∈ c(N)then kajkNis Cauchy as a sequence in R and hence converges,with limit necessarily non-negative. Set k{aj}kN= limjkajkN.This is a seminorm on c(N) but not a norm. However, for anyseminorm on a linear space the null elements, those on whichthe seminorm vanishes, form a linear subspace S and the quo-tient, in this case c(N)/S is a normed space – the propertiesof the norm are inherited from the seminorm. For c(N), S isprecisely the subspace of c(N) consisting of the sequences whichconverge to 0 and B(N) = c(N)/S is a Banach space. This isthe usual sort of diagonalization argument. If {bj} is a Cauchysequence in B(N) then, by definition, each bj= {aj,k}kis rep-resented by a Cauchy sequence in N, i.e. an element of c(N).If k(j) is a sufficiently rapidly increasing sequence of integersthen cj= aj,k(j)is Cauchy in N, i.e. is an element of c(N) andbj−→ [c] converges in B(N) which is therefore complete, i.e. aBanach space. Now, one should embed N −→ B(N) isometri-cally by mapping each a ∈ N to the constant sequence cj= ato see that N can be realized as a dense subspace. You shouldalso know/check that any two Banach spaces in which a givennormed space N embed as isometric dense subspaces are natu-rally isometrically isomorphic – are ‘the same’ from this pointof view.(4) So, in this way we can define spaces Lp(Rn) by completion ofC∞c(Rn) with respect to k · kLp. Of course the result is only anabstract Banach space since we are essentially just ‘declaring’that Cauchy sequences converge. The Lebesgue integral realizesthe new ‘ideal’ elements in terms of actual functions on Rn.LECTURE 4, 18.155, 20 SEPTEMBER 2011 3However, this is just a little tricky since the assumption thatφjis Cauchy with respect to k · kLpdoes not imply that φj(x)converges (in C) for any one x ∈ Rn! This pointwise sequencemay not converge at all.(5) This statement notwithstanding, we can get some convergenceby ‘speeding up the Cauchy condition’. In a normed space (itworks even in a metric space) we can demand that a sequenceφjbe absolutely summable, meaning just thatXj≥1kφj+1− φjk < ∞.Note that this condition does imply that the sequence is Cauchy,since by the triangle inequality, if k > l thenkφk− φlk ≤k−lXp=1kφl+p− φl+p−1k ≤Xj≥lkφj+1− φjkand the convergence above implies that this last sum tends tozero as l → ∞.Now, if one replaces ‘Cauchy’ by ‘absolutely summable’ inthe discussion above, nothing significant changes and one getsthe ‘same’ Banach space.(6) So, the point is that for absolutely summable sequences withrespect to the Lpnorms we do get some pointwise convergence.In the homework you are supposed to think a bit about rectan-gles, meaning sets of the form{x ∈ Rn; aj≤ xj< bj}, aj≤ bj,and sets of measure zero. The volume (Lebesgue measure) of arectangle is vol(R) =Qj(bj− aj).A set E is said to be of measure zero if for each  > 0 it hasa countable coverging by rectanglesE ⊂XkRks.t.Xkvol(Rk) < .(7) Then we get a glimmer of the concrete realization by observingthatTheorem 1. If φj∈ C∞c(Rn) is absolutely summable with re-spect to k · kLpfor some 1 ≤ p < ∞, thenE = {x ∈ Rn;Xj≥1|φj+1(x) − φj(x)| = ∞} has measure zero;4 LECTURE 4, 18.155, 20 SEPTEMBER 2011conversely, for any set of measure zero there exists an abso-lutely summable sequence φj∈ C∞c(Rn) such thatPj≥1|φj+1(x) −φj(x)| = ∞ at each point of the set.(8) Proof. For the given sequence, set ψ1= |φ1| and ψj= |φj−φj−1|for j ≥ 2. Thus, our assumption is thatXjkψjkLp= C < ∞.Let T > 0 be chosen – eventually we will choose it large to getthe  in the definition of measure zero. Now, consider x ∈ E.There must exist M such thatPj≤Mψj(x) ≥ T + 1. Since thissum is finite, it is a continuous function so there is a rectanglecontaining x, R(x), (hence non-empty) on whichPj≤Mψj≥ T.We can choose this rectangle to have rational ‘end-points’ ajand bj. The choices