MATH 418 Assignment #7 1. Let r, s, t be positive real numbers such that r + s + t = 1. Suppose that f, g, h are nonnegative measurable functions on a measure space (X, S, µ). Prove that Z r s t r s t f g h dµ ≤ kfk1kgk1khk1. X s t Proof . Let u := g s+t h s+t . Then f rgsht = f rus+t. By H¨older’s inequality we have Z Z rZ s+t r s+t r 1/r s+t 1/(s+t) r s+t f u dµ ≤ (f ) dµ (u ) dµ = kfk1kuk1 . X X X For the same reason, we have Z Z s t s t s+t s+t s+t s+t kuk1 = u dµ = g h dµ ≤ kgk1 khk1 . X X Consequently, Z r s t r s+t r s t f g h dµ ≤ kfk1kuk1 ≤ kfk1kgk1khk1. X 2. Let Cc(IR) be the linear space of all continuous functions on IR with compact support. (a) For 1 ≤ p < ∞, prove that Cc(IR) is dense in Lp(IR). Proof . Let X be the closure of Cc(IR) in Lp(IR). Then X is a linear space. We wish to show X = Lp(IR). First, if I = (a, b) with −∞ < a < b < ∞, then χI ∈ X. Indeed, for n ∈ IN, let gn the function on IR given by gn(x) := 1 for x ∈ [a + 1/n, b − 1/n], gn(x) := 0 for x ∈ IR \ (a, b), gn(x) := n(x − a) for a ≤ x ≤ a + 1/n and gn(x) := n(b − x) for b − 1/n ≤ x ≤ b. Then gn ∈ Cc(IR), 0 ≤ gn(x) ≤ χI (x) and limn→∞ gn(x) = χI (x) for all x ∈ IR. By the Lebesgue dominated convergence theorem, we have Z p lim |gn − χI | dλ = 0. n→∞ IR Hence, χI ∈ X. Second, if G is a nonempty open subset of IR with λ(G) < ∞, then χG ∈ X. Indeed, if J is the union of finitely many disjoint open intervals I1,...,Im, ∞ then χJ = χI1 + ··· + χIm ∈ X. In general, G can be expressed as G = ∪n=1Jn, where Jn ⊆ Jn+1 for all n ∈ IN and each Jn is the union of finitely many disjoint open intervals. By the Lebesgue dominated convergence theorem, we have Z p lim |χJn − χG| dλ = 0. n→∞ IR 1 Hence, χG ∈ X. Third, if E is a Lebesgue measurable subset of IR with λ(E) < ∞, then χE ∈ X. Indeed, there exists a sequence (Gn)n=1,2,... of open sets such that ∞ ∞ λ(G1) < ∞, Gn+1 ⊆ Gn for all n ∈ IN, E ⊆ ∩n=1Gn, and λ(∩n=1Gn \ E) = 0. We have χGn ≤ χG1 for every n ∈ IN and limn→∞ χGn (x) = χE(x) for almost every x ∈ IR. By the Lebesgue dominated convergence theorem, we have Z p lim |χGn − χE| dλ = 0. n→∞ IR Pm Hence, χE ∈ X. Fourth, if h = j=1 cjχEj is a simple function with λ(Ej) < ∞ for all j ∈ {1, . , m}, then h ∈ X. Finally, if f ∈ Lp(IR), then there exists a sequence (hn)n=1,2,... of simple functions such that |hn| ≤ |f| for every n ∈ IN and limn→∞ hn(x) = f(x) for all x ∈ IR. By the Lebesgue dominated convergence theorem, we have Z p lim |hn − f| dλ = 0. n→∞ IR Hence, f ∈ X. This shows that Cc(IR) is dense in Lp(IR). (b) Show that Cc(IR) is not dense in L∞(IR). Proof . Let Y be the closure of Cc(IR) in L∞(IR). We show that Y ⊆ C0(IR). Indeed, for f ∈ Y and ε > 0, there exists some g ∈ Cc(IR) such that kf − gk∞ < ε. Since g ∈ Cc(IR), there exists some A > 0 such that g(x) = 0 for all x ∈ IR \ [−A, A]. Then |f(x)| < ε whenever |x| > A. This verifies Y ⊆ C0(IR). The constant function 1 lies in L∞(IR) but not in C0(IR). Therefore, Cc(IR) is not dense in L∞(IR). 3. Let C[0, 1] be the linear space of continuous functions on [0, 1], and let C1[0, 1] be the linear space of all continuous functions on [0, 1] having continuous derivatives. (a) Prove that C[0, 1] with the norm k·k∞ is a Banach space. Proof . It suffices to show that the space C[0, 1] with the norm k·k∞ is complete. Let (fn)n=1,2,... be a Cauchy sequence in C[0, 1] equipped with the norm k·k∞. For given ε > 0, there exists a positive integer N such that kfm −fnk∞ ≤ ε whenever m, n ≥ N. It follows that |fm(x)−fn(x)| ≤ ε for all x ∈ [0, 1]. Hence, (fn(x))n=1,2,... is a Cauchy sequence in IR. Let f(x) := limn→∞ fn(x). Consequently, for n ≥ N we have |f(x) − fn(x)| = lim |fm(x) − fn(x)| ≤ ε ∀ x ∈ [0, 1]. m→∞ This shows that f ∈ L∞[0, 1] and limn→∞ kfn −fk∞ = 0. Furthermore, for x0 ∈ [0, 1], there exists some δ > 0 such that |fN (x)−fN (x0)| < ε for all x ∈ (x0 −δ, x0 +δ)∩[0, 1]. Consequently, for x ∈ (x0 − δ, x0 + δ) ∩ [0, 1] we have |f(x) − f(x0)| ≤ |f(x) − fN (x)| + |fN (x) − fN (x0)| + |fN (x0) − f(x)| < 3ε. Therefore f ∈ C[0, 1]. 2 1 (b) Show that C [0, 1] with the norm k·k∞ is not complete. Proof . Let 0 if 0 ≤ x ≤ 1/2, f(x) := x − 1/2 if 1/2 ≤ x ≤ 1. Moreover, for n = 1, 2,..., let 0 if 0 ≤ x ≤ 1/2, 2 fn(x) := n(x − 1/2) 2 − n(x − 1/2) if 1/2 ≤ x ≤ 1/2 + 1/n, x − 1/2 if 1/2 + 1/n ≤ x ≤ 1. 1 1 Then fn ∈ C [0, 1] for each n. But limn→∞ kfn − fk∞ = 0 and f∈ / C [0, 1]. 1 0 1 (c) For f ∈ C [0, 1], let kfk := kfk∞ + kf k∞. Show that C [0, 1], equipped with this norm, is a Banach space. 1 0 Proof . It suffices to show that the space C [0, 1] with the norm kfk := kfk∞+kf k∞ is 0 complete. Let (fn)n=1,2,... be a Cauchy sequence in this space. Let gn := fn for n ∈ IN. Then both (fn)n=1,2,... and (gn)n=1,2,... are Cauchy sequences in C[0, 1] with the norm k·k∞. By part (a), there exist f and g in C[0, 1] such that limn→∞ kfn − fk∞ = 0 and limn→∞ kgn − gk∞ = 0. By the fundamental theorem of calculus, we have Z x fn(x) = fn(0) + gn(t) dt, x ∈ [0, 1]. 0 Letting n → ∞ in the above equation, we obtain Z x f(x) = f(0) + g(t) dt, x ∈ [0, 1]. 0 0 It follows that g = f . Therefore, limn→∞ kfn − fk = 0. 4. Prove the following two statements: (a) If M is a closed subspace of a Hilbert space H, then (M ⊥)⊥ = M. Proof . If x ∈ M and y ∈ M ⊥, then hx, yi = 0. Hence, x ∈ (M ⊥)⊥. If x∈ / M, then x = y + z for some y ∈ M ⊥ and z ∈ M. Since x∈ / M, we must have y 6= 0. Consequently, hx, yi = hy, yi 6= 0. Hence, x∈ / (M ⊥)⊥. This shows that (M ⊥)⊥ = M. ⊥ ⊥ ⊥ (b) If M1 and M2 are subspaces of a Hilbert space H, then (M1 +M2) = M1 ∩M2 . ⊥ ⊥ Proof . We have M1 ⊆ M1 + M2 and M2 ⊆ M1 + M2. Hence, (M1 + M2) ⊆ M1 and ⊥ ⊥ ⊥ ⊥ ⊥ (M1 + M2) ⊆ M2 . Consequently, (M1 + M2) ⊆ M1 ∩ M2 . On the other hand, if ⊥ ⊥ x ∈ M1 ∩ M2 , then hx, y1i = 0 for every y1 ∈ M1 and hx, y2i = 0 for every y2 ∈ M2. It follows that hx, y1 + y2i = 0. But every element y ∈ M1 + M2 can be represented ⊥ as y = y1 + y2. Hence, x ∈ (M1 + M2) . This completes the proof of the identity ⊥ ⊥ ⊥ (M1 + M2) = M1 ∩ M2 . 3 5. Let H be the linear space of all absolutely continuous functions f on [0, 1] such that 0 kf kL2(0,1) < ∞. For f, g ∈ H, let Z 1 Z 1 hf, gi := f(x)g(x) dx + f 0(x)g0(x) dx. 0 0 (a) Prove that H with this inner product becomes a Hilbert space. Proof . For f ∈ H, we have hf, fi ≥ 0, and hf, fi = 0 if and only if f = 0. Moreover, hf, gi = hg, fi. If f, g, h ∈ H and a, b ∈ IR, then haf + bg, hi = ahf, hi + bhg, hi. It remains to show that H with the norm induced by this inner product is complete. 0 Let (fn)n=1,2,... be a Cauchy sequence in H. For each n ∈ IN, let gn := fn. Then both (fn)n=1,2... and (gn)n=1,2,... are Cauchy sequences in L2[0, 1]. Since L2[0, 1] is complete, there exist f and g in L2[0, 1] such that limn→∞ kfn − fk2 = 0 and limn→∞ kgn − gk2 = 0. Since each fn is absolutely continuous on [0, 1], by the funda- mental theorem of calculus, we have Z x fn(x) = fn(0) + gn(t) dt, x ∈ [0, 1]. 0 Letting n → ∞ in the above equation, we obtain Z x f(x) = f(0) + g(t) dt, x ∈ [0, 1]. 0 0 It follows that g = f . Therefore, limn→∞ kfn − fkH = 0. (b) Let φ(f) := f(0) for f ∈ H. Prove that φ ∈ H∗. Proof . Clearly, φ is a linear functional on H. For f ∈ H, let g be the constant function on [0, 1] given by g(x) := f(0) for x ∈ [0, 1].