SIXTH ASSIGNMENT, SOLUTIONS18.155 FALL 2001RICHARD MELROSEProblem 1. Hilbert space and the Riesz representation theorem. If you need helpwith this, it can be found in lots of places – for instance [1] has a nice treatment.i) A pre-Hilbert space is a vector space V (over C) with a ‘positive definitesesquilinear inner pro duct’ i.e. a functionV × V 3 (v, w) 7→ hv, wi ∈ Csatisfying• hw, vi = hv, wi• ha1v1+ a2v2, wi = a1hv1, wi + a2hv2, wi• hv, vi ≥ 0• hv, vi = 0 ⇒ v = 0.Prove Schwarz’ inequality, that|hu, vi| ≤ hu, ui12hv, vi12∀ u, v ∈ V.Hint: Reduce to the case hv, vi = 1 and then expandhu − hu, viv , u − hu, vivi ≥ 0.Solution. If v = 0 then hu, vi = 0 and Schwarz’ inequality certainly holds.If v 6= 0 then hv, vi > 0 so we can divide through by hv, vi12and replace v byˆv = v/hv , vi12which has hˆv, ˆvi = 1. Thus we may as well assume hv, vi = 1.Now using the linearity and anti-linearity which follow from the condi-tions above,hu − hu, viv , u − hu, vivi= hu, ui − hu, vihu, vi − hu, vihv, ui + |hu, vi|2= hu, ui − |hu, vi|2≥ 0.This proves Schwarz’ inequality. ii) Show that kvk = hv, vi1/2is a norm and that it satisfies the parallelogramlaw:(1) kv1+ v2k2+ kv1− v2k2= 2kv1k2+ 2kv2k2∀ v1, v2∈ V.Solution. As with the special case of L2that I did in class, the triangleinequality follows from Schwarz’ inequality:ku + vk2= hu + v, u + vi ≤ kuk2+ 2|hu, vi|2+ kvk2≤ (kuk + kvk)2.12 RICHARD MELROSEThe other properties of a norm follow directly. For the parallelogram lawexpand out the left side:kv1+ v2k2+ kv1− v2k2=kv1k2+ hv1, v2i + hv2, v1i + kv2k2+ kv1k2− hv1, v2i − hv2, v1i + kv2k2= 2kv1k2+ 2kv2k2. iii) Conversely, suppose that V is a linear space over C with a norm whichsatisfies (1). Show that4hv, wi = kv + wk2− kv − wk2+ ikv + iwk2− ikv − iwk2defines a pre-Hilbert inner product which gives the original norm.Solution. With hu, vi defined this way4hv, vi = 4kvk2+ i|1 + i|kvk2− i|1 − i|kvk2= 4kvk2and it follows directly that h−u, vi = −hu, vi, hiu, vi = ihu, vi, hc2u, c2vi =c4hu, vi, c > 0. Furthermore4h2u, vi = k2u + vk2− k2u − vk2+ ik2u + ivk2− ik2u − ivk2= 2ku + vk2+ 2kvk2− kuk2− 2ku − vk2− 2kvk2+ kuk2+ 2iku + ivk2+ 2ikvk2− ikuk2− 2i|u − ivk2− 2ikvk2+ ikuk2= 8hu, vi.Iterating this it follows that hku, lvi = klhu, vi for integers k, l and hencethat hau, vi = ahu, vi for all rational a. The triangle inequality gives conti-nuity so that hzu, vi = zhu, vi for all z ∈ C. Addivity in the first variablethen also follows from the parallelogram law:4hu1+u2, vi = ku1+u2+vk2−ku1+u2−vk2+iku1+u2+ivk2−iku1+u2−ivk2= 2ku1+12vk2+2ku2+12vk2−ku1−u2k2−2ku1−12vk2−2ku2−12vk2+ku1−u2k2+2iku1+i12vk2+2iku2+i12vk2−iku1−u2k2−2i|u1−i12vk2−2iku2−12vk2+iku1−u2k2= 8hu1,v2i + 8hu2,12vi = 4hu1, vi + 4hu2, vi.Thus h, i is pre-Hilbert inner product giving the original norm. iv) Let V be a Hilbert space, so as in (i) but complete as well. Let C ⊂ V be aclosed non-empty convex subset, meaning v, w ∈ C ⇒ (v + w)/2 ∈ C. Showthat there exists a unique v ∈ C minimizing the norm, i.e. such thatkvk = infw∈Ckwk.Hint: Use the parallelogram law to show that a norm minimizing se-quence is Cauchy.3Solution. By definition of the infimum, there is a sequence vj∈ C suchthat kvjk → I = infw∈Ckwk. By the parallelogram lawkvj− vkk2= 2kvjk2+ 2kvkk2− 4k12(vj+ vk)k2≤ 2kvjk2+ 2kvkk2− 4I → 0as j, k → ∞, where the convexity of C has been used. Thus the sequenceis Cauchy, so converges by the assumed completeness of the space. If v isthe limit, the continuity of the norm implies that |vk = I. If w ∈ C is anypoint with kwk = I then again by the parallelogram lawkv − wk2= 2kvk2+ 2kwk2− 4k12(v + w)k2≤ 0so v = w and the point is unique. v) Let u : H → C be a continuous linear functional on a Hilbert space, so|u(ϕ)| ≤ Ckϕk ∀ ϕ ∈ H. Show that N = {ϕ ∈ H; u(ϕ) = 0} is closed andthat if v0∈ H has u(v0) 6= 0 then each v ∈ H can be written uniquely inthe formv = cv0+ w, c ∈ C, w ∈ N.Solution. That N is c losed follows from the continuity of u, since N =u−1(0). If a = u(v0) 6= 0 and v ∈ H then w = v −u(v)av0∈ N so v = cv0+wwith c = u(v)/u(v0). vi) With u as in v), not the zero functional, show that there exists a uniquef ∈ H with u(f ) = 1 and hw, fi = 0 for all w ∈ N.Hint: Apply iv) to C = {g ∈ V ; u(g) = 1}.Solution. If u is not the zero functional there exists f ∈ H with u(f) =1. Thus C = {g ∈ H; u(g) = 1} is a closed set and it is convex sinceu(12(g1+ g2)) = 1 if g1, g2∈ C. Thus, by iv) there exists a unique f ∈ Cminimizing the norm. Now, by v) each g ∈ C can be written uniquelyg = f + w with u(w) = 0. Since f + sw ∈ C for all s ∈ R it follows thatddskf + swk2s=0= hf, wi + hw, f i = 0.The same is true of iw so hf, wi = 0 for all w ∈ N. Conversely if this holdsfor some point g ∈ C thenkg + wk2= kgk2+ kwk2is minimized on C at g; thus g = f by the uniquenss of the minimizer. vii) Prove the Riesz Representation theorem, that every continuous linear func-tional on a Hilbert space is of the formuf: H 3 ϕ 7→ hϕ, f i for a unique f ∈ H.Solution. If u is the ze ro functional this is clear, with f = 0. Thus we mayassume that u is not identically zero, so v) applies, as does vi). Now forevery v ∈ H, v = u(v)f + w with u(w) = 0. Thus hv, fi = u(v)kfk2andhenceu(v) = hv, gi, g = f/kfk2.This g is unique since f = g/kgk2satisfies hw, fi = 0 for all w ∈ N andu(f) = u(g)/kgk2= 1, the solution of which is unique by vi). 4 RICHARD MELROSEProblem 2. Density of C∞c(Rn) in Lp(Rn).i) Recall in a few words why simple integrable functions are dense in L1(Rn)with respect to the norm kfkL1=RRn|f(x)|dx.Solution. We may divide and f ∈ L1(Rn) into its real and imaginary partsand approximate them separately; this it suffices to consider real functions.We showed in class that any integrable function is the sum of a positiveintegrable function f+and a negative integrable function f−(namely thepositive and negative parts). For the positive function we showed that thereis an increasing sequence of positive simple functions gjconverging to f+almost everywhere, so kf+− gjkL1→ 0. Similarly for f−. Thus f is thelimit in L1of a sequence of simple integrable functions. ii) Show that simple functionsPNj=1cjχ(Uj) where the Ujare open and boundedare also dense in L1(Rn).Solution. We have seen that simple functions are