MATH 421/510 Assignment 5 Suggested Solutions April 2018 1. Given any point t0 2 [0; 2π], show using the uniform boundedness principle that there exists a continuous 2π-periodic function whose Fourier series diverges at t0. We sketched a proof of this result in class. Fill in the details. Proof. Consider the N-th Dirichlet kernel N X int DN (t) := e : n=−N It is a fact that the partial sum sequence SN f(t) = (DN ∗ f)(t), where the integral defining the convolution is normalized by a factor 1=2π: 1 Z π SN f(t) = DN (t − s)f(s)ds: 2π −π We show this in several steps, using contradiction. Suppose for all f 2 C[−π; π], we have SN f(t0) ! f(t0). Then: (a) The mapping lN : f 7! SN f(t0) is linear and bounded from C[−π; π] := (C[−π; π]; k·k1) to C, with a supN klN k ≤ C < 1. This is a result of the uniform boundedness principle. (b) We show that this implies that SN is bounded from C[−π; π] to C[−π; π], with the bound independent of N. Indeed, given f 2 C[−π; π], suppose jSN fj attains its maximum at t1. Consider the translated function g(t) := f(t + t1 − t0), which has kgk1 = kfk1 and SN g(t0) = SN f(t1). Hence kSN fk1 = jSN f(t1)j = jSN g(t0)j ≤ Ckgk1 = Ckfk1: (c) We state a special case of the Young's convolution theorem: Theorem 1. Let (X; µ) be a measure space, and g be a measurable function. The convolution operator T : f 7! f ∗ g is bounded from L1 to L1 if and only 1 if g 2 L . Moreover, kT k 1 1 = kgk . L !L L1 Now in our situation, supN kSN k < 1 implies that supN kDN k1 < 1. 1 (d) Lastly, we show the above cannot happen. Direct computation shows that 1 sin N + 2 x DN (x) = 1 : sin 2 x 1 1 By considering the integral over jxj 2 [kπ=(N + 2 ); (k + 1)π=(N + 2 )] for each k, we see kDN k1 is bounded below by a constant times the first N terms of the harmonic series. Letting N ! 1, we have kDN k1 ! 1, contradiction to the conclusion above. 2. In class, we introduced the concept of a locally convex space, whose topology is generated by a family of seminorms. When is such a topology equivalent to a metric topology? A norm topology? Note: If X is locally convex, it separates points by definition taught in class. (a) We claim such a topology is a metric topology if and only if it is generated by a countable family of seminorms. 1 Proof. (\ (= ") Let fpigi=1 be the countable family of seminorms that gener- ates a topology on X. Then we define a metric by 1 X pi(x − y) d(x; y) := 2−i : 1 + p (x − y) i=1 i It is direct to check that d is a metric. Indeed, d(x; y) ≥ 0, and if d(x; y) = 0, then pi(x − y) = 0 for all i. Since fpig separates points, we have x = y. Symmetry is trivial. For the triangle inequality, refer to the following question: https://math.stackexchange.com/questions/309198/if-d-is-a-metric- then-d-1d-is-also-a-metric. 1 It remains to show d generates the same topology as fpigi=1 does. By transla- tion invariance, it suffice to consider their neighbourhood bases at 0: Bd(") := fx 2 X : d(x; 0) < "g; n \ Bi("i) := fx 2 X : pi(x) < "i 81 ≤ i ≤ ng: i=1 P1 −i • Given " > 0, take N such that i=N+1 2 < "=2. Take "i := "=2 for all 1 ≤ i ≤ N. Thus if pi(x) < "=2 for all 1 ≤ i ≤ N, we have N 1 X " X " " d(x; 0) ≤ 2−i + 2−i < + = ": 2 2 2 i=1 i=N+1 This implies that N \ Bi("i) ⊆ Bd("): i=1 2 • On the other hand, given "i > 0 for i = 1; 2; : : : ; n, then d(x; 0) < " := minf"i : 1 = 1; 2; : : : ; ng implies that pi(x) < "i. Hence n \ Bd(") ⊆ Bi("i): i=1 Therefore they generate the same topology. 1 (\ =) ") Note that fBd(1=n)gn=1 forms a neighbourhood base at 0. For each n, there is Bi;n("i;n); i = 1; 2;:::;Kn such that K 1 \n B 0; ⊇ B (" ): d n i;n i;n i=1 1 Relabel the countable collection fpi;n : 1 ≤ i ≤ Kn; n 2 Ng as fpjgj=1. Then 1 fpjgj=1 generates the metric topology d. (b) We claim such a topology is a norm topology if and only if it is generated by a finite collection of seminorms. N Proof. (\ (= ") Let P := fpigi=1 be the finite collection of seminorms that generates a topology on X. Then we define a norm by kxk := maxfpi(x); i = 1; 2;:::;Ng: It is direct to check that k·k is a norm. Indeed, kxk ≥ 0, and if kxk = 0, then pi(x) = 0 for all i. Since fpig separates points, we have x = 0. Homogeneity and the triangle inequality follows from the corresponding prop- erties of the seminorms. N It remains to show k·k generates the same topology as fpigi=1 does. But this is similar and easier than the countable case. (\ =) ") This is trivial. 3. Let (X; Ω; µ) be a σ-finite measure space, 1 ≤ p < 1. Suppose that K : X ×X ! F is an Ω × Ω-measurable function such that for f 2 Lp(µ) and almost every x 2 X, the function K(x; ·)f(·) 2 L1(µ) and Z Kf(x) = K(x; y)f(y)dµ(y) defines an element Kf 2 Lp(µ). Show that K is a bounded operator on Lp(µ). Proof. We first prove a lemma: Lemma 1. Let (X; Ω; µ) be a measure space, and f be a measurable function. Sup- R p p0 pose X fg converges absolutely for every f 2 L , 1 ≤ p ≤ 1. Then g 2 L , where p0 is the dual exponent of p. 3 Proof of the Lemma. Suppose, towards contradiction, that g2 = Lp0 . By duality, this p R n is to say that there is a sequence fn 2 L with kfnkp = 1 such that X fng > 4 . R n In particular, X jfngj > 4 . P1 −n P1 −n Now we define f := n=1 2 jfnj. We have kfkp ≤ n=1 2 kfnkp = 1, by the triangle inequality. However, we see that Z 1 Z 1 X −n X −n n jfgj ≥ 2 jfngj > 2 4 = 1; X n=1 X n=1 so fg2 = L1, a contradiction. Hence g 2 Lp0 . By the lemma, K(x; ·) 2 Lp0 for a.e. x 2 X. By H¨older'sinequality, f 7! Kf(x) is a bounded linear functional on Lp for a.e. x 2 X. We will use the closed graph theorem to show that K is bounded on Lp. Let p p p fn ! f in L , Kfn ! g in L . Since f 7! Kf(x) is continuous on L for a.e. x, Kfn(x) !Kf(x) a.e. By uniqueness of limits, we have g = Kf(x), which completes the proof. 4. (a) Show that the weak topology on X is the weakest topology for which all l 2 X∗ is continuous. Proof. We take the definition of weak topology on X as the topology generated by the seminorms p(x) := jl(x)j over l 2 X∗. Recall that a linear functional l : X ! F is continuous if and only if there exists finitely many seminorms pi, 1 ≤ i ≤ n, and a constant C such that for all x 2 X, n X jl(x)j ≤ C pi(x): i=1 Now we take C = 1 and a single p1 = jlj to finish the proof. On the other hand, given any topology on X such that each l 2 X∗ is contin- uous. Since taking modulus on the scalar field is continuous, we see that each x 7! p(x) = jl(x)j is continuous. Hence the weak topology is weaker than any topology such that each l 2 X∗ is continuous. Lastly, by taking intersection of all such topologies, we see that the weak topology on X is unique, so it is indeed the weakest topology such that each l 2 X∗ is continuous. (b) Show that the weak-star topology is the smallest topology on X∗ such that for each x 2 X, the map l 7! l(x) is continuous. Proof. We take the definition of weak-star topology on X∗ as the topology generated by the seminorms qx(l) := jl(x)j over x 2 X. Recall that a linear functional q : X∗ ! F is continuous if and only if there exists finitely many seminorms qxi , 1 ≤ i ≤ n, and a constant C such that for all l 2 X∗, n n X X jq(l)j ≤ C qxi (l) = C jl(xi)j: i=1 i=1 4 Now for each x 2 X, q(l) = qx(l) is the mapping l 7! jl(x)j. We take C = 1 and a single x1 = x to finish the proof. On the other hand, given any topology on X∗ such that for each x 2 X, l 7! l(x) is continuous. Since taking modulus on the scalar field is continuous, we see that each l 7! qx(l) = jl(x)j is continuous.