Problem set 4, Real Analysis I, Spring, 215. (18) Let f be a measurable finite-valued function on [, 1], and suppose f(x) f(y) is integrable on [, 1] [, 1]. Show that f is integrable on [, 1]. [Hint: Show that if p is a real or complex number, and f(x) p is integrable over [, 1], then f(x) is integrable.] Solution: Part (i) of Fubini s Theorem 3.1 implies that for almost every y [, 1], the function f(x) f(y) is integrable as a function of x on [, 1]. In other words, for such a y, f(x) f(y) dx <. But then by the Triangle Inequality f(x) dx f(x) f(y) dx+ f(y) dx = f(x) f(y) dx+ f(y) <, and so f is integrable. (19) Suppose f is integrable on R d. For each α >, let E α = {x : f(x) α}. Prove that f(x) dx = m(e α ) dα. R d [Hint: Consider Corollary 3.8.] Solution: Corollary 3.8(ii) shows that for A = {(x, y) R d R : y f(x) }, f(x) dx = m(a). R d Now m(a) = χ A. R d+1 Then we can apply either Fubini s Theorem or Tonelli s Theorem to show χ A = χ A (x, α) dx dα R d+1 R d = χ Eα (x) dx dα R d = 1 m(e α ) dα.
(6) (a) Let I be an interval, and let φ: I R be a convex function, and let a, b I with a < b. Assume there is a λ (, 1) so that φ(λb + (1 λ)a) = λφ(b) + (1 λ)φ(a). Show the restriction of φ to [a, b] is linear. Hint: Draw a picture. The graph of φ on [a, b] must lie below the secant line l. Then for c = λb + (1 λ)a, let d (a, c). If (d, φ(d)) l, then draw a secant line to produce a contradiction. Do the same for d (c, b). Solution: Let c = λb + (1 λ)a. We assume that φ(c) = λφ(b) + (1 λ)φ(a). Let d (a, c). We prove by contradiction the (d, φ(d)) l. Define l(x) to be the linear function through (a, φ(a)) and (b, φ(b)). By the convexity of φ, we know φ(d) l(d). So to proceed with the proof by contradiction, we may assume φ(d) < l(d). Now again since φ is convex, if we define l(x) to the the linear function between (d, f(d)) and (b, f(b)), then for all x (d, b), φ(x) l(x). Note also l(d) = φ(d) < l(d), while l(b) = φ(b) = l(b). Since l l is linear, we find that l(x) l(x) < for all x < b. In particular, we see that l(c) < l(c) = φ(c), and c (d, b). This provides a contradiction, and so we see that φ(d) = l(d). There is a remaining case of d (c, b) to rule out. This proceeds as above by contradiction, where instead we consider the line from (a, φ(a)) to (d, φ(d)) to derive the contradiction. (b) Show that a convex function φ is strictly convex if and only if its graph contains no line segments. Solution: We prove each direction by using the contrapositive. If the graph of φ contains a line segment, it is obviously not strictly convex (consider the endpoint of the line segment and any point in the interior of the segment). On the other hand, if φ is convex but not strictly convex, then by definition there is an [a, b] in the domain and a c (a, b) so that φ(c) = l(c), where l is defined as above. Part (a) then shows that φ is linear on [a, b], and thus the graph contains a line segment. 2
(7) Let (, M, µ) be a measure space. Let φ be a strictly convex function on an interval I, and assume the range of a measurable function g : R is contained in I. Formulate and prove a version of the equality case of Jensen s Inequality. Solution: The correct statement is Theorem 1 (Jensen s Inequality, The Case of Equality). Let (, M, µ) be a probability measure space. Let g be an integrable function on with range in an interval I R. Let φ : I R be strictly convex. Then if ( ) φ g dµ = φ g dµ, g is constant almost everywhere (in other words, there is a constant C I so that g = C almost everywhere). The proof is as follows (we modify the proof of Jensen s Inequality in the notes): Let α = g dµ. Then, as in the proof in the notes, we find that either g = α for almost all x (in which case we are done), or α I the interior of I. By Proposition 5 in the notes, we my choose l(x) = φ(α) + m(x α) φ(x) for all x I. Therefore, ( ) φ g dµ φ(α) + m(g α) dµ = φ(α) = φ g dµ. Let l be a linear function whose graph is a supporting line to the graph of φ at α (this is guaranteed by Proposition 5 in the notes on L p Spaces and Convexity. In particular, the convexity of φ implies φ(c) l(c) for all c I. Moreover, the strict convexity of φ and Problem 6b above imply that φ(c) > l(c) for all c I \ {α}. Write l(c) = φ(α) + m(c α). Then compute (φ g l g) dµ = φ g [φ(α) + m(g α)] dµ = φ g φ(α) m g dµ + mα ( ) = φ g φ g dµ = 3
Here the last inequality is by our assumption of equality. Now on, φ g l g is a nonnegative measurable function with integral. Therefore, φ g l g = almost everywhere. But φ g = l g if and only if g = α. Thus g = α almost everywhere. (8) Let (, M, µ) be a measure space. Let 1 < s < r <. Assume f L s () L r () and that f L s. Define for p [s, r], φ(p) = f p Lp. Prove that log φ is convex on [s, r]. Recall there are two parts to this. (a) For a, b [s, r] and λ (, 1), prove log φ(λb + (1 λ)a) λ log φ(b) + (1 λ) log φ(a). Hint: Apply Hölder s Inequality to the integral φ(p) = f γ f p γ dµ for a judicious choice of γ. Solution: Let s a < b r and let p (a, b). For u, v (1, ) conjugate exponents (so that u 1 + v 1 = 1), compute for γ (1, p) using Hölder s Inequality φ(p) = f γ f p γ dµ = f γ L u f p γ L v ( ) u 1 ( ( f γ ) u dµ ( f p γ ) v dµ = φ(uγ) u 1 φ(v(p γ)) v 1, log φ(p) u 1 log φ(uγ) + v 1 log φ(v(p γ)). u = b a p a, ) v 1 To convert this inequality into the convexity inequality, note a, b, p are fixed, and we would like to set uγ = b, v(p γ) = a, u 1 = λ, v 1 = 1 λ, and p = λb + (1 λ)a. Together with u 1 + v 1 = 1, these equations form a linear system for u 1, v 1, γ, λ which can be solved to find v = b a b p, λ = p a b(p a), γ = b a b a. We can check that a < p < b implies u, v (1, ), λ (, 1), γ (1, p), as is necessary. (b) Prove log φ is continuous at the endpoints r, s. Hint: In order to prove the continuity of log φ at the endpoints s, r, use the convexity inequality to show 4
lim sup p r + log φ(p) log φ(r). Then use Fatou s Lemma to conclude that lim p r + log φ(p) = log φ(r). Solution: The convexity inequality implies that for p = λr + (1 λ)s, λ (, 1), that log φ(p) λ log φ(r) + (1 λ) log φ(s). Taking λ +, we find lim sup log φ(p) log φ(s). p s + Similarly, we may take λ 1 to find lim sup log φ(b) log φ(r). p r For the opposite inequalities, let λ n (, 1) so that λ n. Then for p n = λ n r + (1 λ n )s, we have f pn is a sequence of nonnegative measurable functions which converge everywhere to f s as n. Then by Fatou s Lemma, φ(s) = f s dµ lim inf f pn dµ = lim inf φ(p n). n n Since the sequence p n s + is arbitrary, we find Thus lim inf p s + φ(p) φ(s). lim sup φ(p) φ(s) lim inf φ(p), p s + p s + and so the limit exists and must equal φ(s). Similar considerations work for the other endpoint as p r. Thus φ is continuous on [s, r] and so is convex. (9) Let = {p 1, p 2 } equipped with the counting measure. The set of real-valued functions on can be identified with R 2. (a) Describe the L 1, L 2 and L norms on as norms on R 2. Solution: A function on is described by its values on p 1, p 2. So we describe each function x as an ordered pair (x 1, x 2 ) = (x(p 1 ), x(p 2 )) R 2. The integral with respect to the counting measure is just the sum x 1 + x 2. There are no nonempty sets of measure zero, and so the essential supremum is just the supremum. 5
6 x L 1 = x L 2 = ( x = x 1 + x 2, x 2 ) 1 2 = ( x1 2 + x 2 2 ) 1 2, x L = sup{ x 1, x 2 } = max( x 1, x 2 ). (b) Let f, g be functions in L 2 (), and assume f + g L 2 = f L 2 + g L 2. Show directly by using the geometry of R 2 that there are nonnegative constants α, β, not both zero, so that αf = βg. Solution: The L 2 norm is just the usual Euclidean norm on R 2. Let (, ) denote the corresponding inner product. Compute f + g 2 L 2 = ( f L 2 + g L 2)2, (f + g, f + g) = f 2 + 2 f g + g 2, f 2 + 2(f, g) + g 2 = f 2 + 2 f g + g 2, (f, g) = f g. Now for f, g both not zero, we know that (f, g) = f g cos θ, where θ is the angle between f and g as vectors in R 2. Thus in this case, our equality is just that cos θ = 1, which is equivalent to f, g being colinear and pointing in the same direction. This is exactly what we are to prove. (The remaining cases in which f or g or both are zero follow easily.) (c) Describe all pairs of functions f, g on so that f +g L 1 = f L 1 + g L 1. Do the same for L. Solution: f + g L 1 = f L 1 + g L 1 is equivalent to (1) f 1 + g 1 + f 2 + g 2 = f 1 + f 2 + g 1 + g 2. The Triangle Inequality in R implies f j +g j f j + g j for j = 1, 2. Thus (1) is equivalent to both these inequalities being equalities, and so f 1 + g 1 = f 1 + g 1, f 2 + g 2 = f 2 + g 2. This is true if and only if there are nonnegative constants α 1, β 1, not both zero, so that α 1 f 1 = β 1 g 1, and in addition there are two other nonnegative constants α 2, β 2, not both zero, so that α 2 f 2 = β 2 g 2. Roughly this means that for f 1, g 1 must have the same sign (with a zero value being
treated as either positive or negative), and the same must also be true for f 2, g 2. Note this is a much weaker condition than for L 2, as for example f = (3, 5), g = (1, ) satisfy f + g L 1 = f L 1 + g L 1. For the case of L, the case of equality happens if and only if both of the pairs f 1 +f 2, g 1 +g 2 and f 1 f 2, g 1 g 2 have the same sign (in the sense above; zero is considered to have the same sign as any real number, positive or negative). There is a slick way of seeing this, by considering the map Φ: R 2 R 2 given by Φ(x 1, x 2 ) = (x 1 + x 2, x 1 x 2 ). Then we can express f L 1 = Φ f L and f L = 2 Φ f L 1. Thus we can relate the condition for L to that of L 1 via Φ. Otherwise, it is possible to analyze the case of L by considering various cases. First of all note f + g L = f L + g L is equivalent to (2) max( f 1 + g 1, f 2 + g 2 ) = max( f 1, f 2 ) + max( g 1, g 2 ) We begin by considering a few cases. First of all, if f 1 > f 2 and g 1 > g 2, then we note f 2 + g 2 f 2 + g 2 < f 1 + f 2, and so the max on the left-hand side of (2) must be f 1 +g 1. So we are left with f 1 +g 1 = f 1 + g 1 in this case. This is only true if f 1, g 1 have the This analysis holds in these cases: f 1 > f 2, g 1 > g 2 together imply f 1, g 1 have the f 1 = f 2, g 1 > g 2 together imply f 1, g 1 have the f 1 > f 2, g 1 = g 2 together imply f 1, g 1 have the The same analysis leads to the following cases: f 2 > f 1, g 2 > g 1 together imply f 2, g 2 have the f 2 = f 1, g 2 > g 1 together imply f 2, g 2 have the f 2 > f 1, g 2 = g 1 together imply f 2, g 2 have the There are a few cases remaining. First of all if f 1 > f 2 and g 2 > g 1, then the right-hand side of (2) is f 1 + g 2, while for the left-hand side, f 1 +g 1 f 1 + g 1 < f 1 + g 2, and similarly f 2 +g 2 f 2 + g 2 < f 1 + g 2. This shows 7
there is no solution to (2) in this case. We list another similar case as well f 1 > f 2, g 2 > g 1 together imply there is no solution to (2). f 2 > f 1, g 1 > g 2 together imply there is no solution to (2). Finally, there is the remaining case of f 1 = f 2, g 1 = g 2. In this case, (2) implies we may have f 1 + g 1 = f 1 + g 1 or f 2 + g 2 = f 2 + g 2 (or both). This corresponds to f 1 = f 2, g 1 = g 2 together imply f 1, g 1 have the same sign or f 2, g 2 have the same sign (or both). It is not difficult to translate these cases into the conditions on f g, f + g mentioned above. 8