Real Variables: Solutions to Homework 9

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1 Real Variables: Solutions to Homework 9 Theodore D Drivas November, 20 xercise 0 Chapter 8, # : For complex-valued, measurable f, f = f + if 2 with f i real-valued and measurable, we have f = f + i f 2 Prove that f is finite if and only if f is finite, and f f Proof Firstly, we know by Theorem (52) that if f is measurable, f L if and only if f L so both integrals or either finite or infinite Now, for some θ R, write f = f eiθ Then, calculate f = e iθ f = e iθ f = Re e iθ f = Re ( e iθ f ) e iθ f = f Here we have use the fact that the integral is real to proceed from the first line to the second line and we used the fact that e iθ = to get the last equality xercise 02 Chapter 8, # 2: Prove the converse of Holder s inequality for p = and Show also that for real-valued f / L p (), there exists a function g L p (), /p + /p =, such that fg / L () Proof First for p = Then it is clear that for all g L with g, f sup fg For the other direction, take g = sgnf, then g L and g f = f = fg But this gives us the other direction for the inequality and therefore we have proven what we set out to prove

2 Now for p =, the for all g L with g, then we have f sup fg Again, for the other direction, there are 3 cases, f = 0, f finite and f = For f = 0 it is immediate For f finite, suppose without loss of generality f = Define set := {x : f(x) > with n N} The measure n n > 0 for all n and define function g such that { strictly positive with g n (x) = n g n (x) = x n 0 otherwise Then f g n = f g n n Then for all g L with g, then f = sup ( ) g n = n n n f g = sup fg For the last case, repeat the above argument with the set F := {x : f(x) > n with n N} and the result follows xercise 03 Chapter 8, # 3: Prove theorems (82) and (83) Show that Minkowski s inequality for series fails when p < We will simplify the statements in the text by proving the result for functions f and g Theorem 04 Holder s Inequality for sequences Proof We will prove Holder s inequality for sequences by employing the more general statement for functions from the text The proof the the statement (Theorem 86) can be found in the text) Take the f and g given in the theorem to be step functions defined in the following way: { a k x (k, k] for k N f(x) = Then it is clear that g(x) = 0 otherwise { b k x (k, k] for k N 0 otherwise fg = a k b k, k= f L p = a k l p and g L p = b k l p Thus the theorem is proven by substitution into Theorem 86 2

3 Theorem 05 Minkowski s inequality for sequences Proof Minkowski s inequality is proven in precisely an analogous way to the way we proved Holder s inequality by taking f and g given in the theorem to be step functions defined in the same way and then applying Theorem 83 However, feeling like this was not enough work, here is an alternate proof of it /q = /p Now we know by Holder that Let also ( ) /p ( ) /q a i a i + b i p/q a i p a i + b i p ( ) /p ( ) /q b i a i + b i p/q b i p a i + b i p Because p = + p/q, we know that a i + b i p = a i + b i a i + b i p/q and putting together the two above statements, we have ( ) /p ( ) /p ( ) /q a i + b i p/q a i p + b i p a i + b i p but /p + /q = so ( ) /p ( ) /p ( ) /p a i + b i p a i p + b i p This is Minkowski s inequality To see that Minkowski s inequality does not hold for p <, Take a = (, 0, 0, )) and b = (0,, 9, 0, ) xercise 06 Chapter 8, # 4: Let f and g be real-valued, and let < p < Prove that equality holds in the inequality fg f p g p if and only if fg has constant sign ae and f p is a multiple of g p ae Proof First the easier direction Let f p = c g q Then ( f p g q = = c /p ( = c /p ( f p ) /p ( ) q g q q g q ) 3 g q ) /q ( g q ) /q

4 But f g = c /p g q g = c /p g q so we have the desired result For the other direction, normalize the functions so that F = f f p and G = g g q Then the p norms of F and G are Supposing Holder s equality holds, then F G = = p F p + q But F G so F G < p F p + q Gp by Youngs inequality But then ( p F p + ) q Gq F G = 0 so p F p + q Gq F G = 0 ae This happens if and only if F p = G q ae so With that we are done f p = f p g q g q G q xercise 07 Chapter 8, # 5: For < p < and 0 < <, define N p [f] = ( ) /p f p Prove that if p < p 2, then N p [f] N p2 [f] Prove also that N p [f + g] N p [f] + N p [g], fg N p[f]n p [g], /p + /p =, and that lim p N p [f] = f Proof If p < p 2 <, let q = p 2 p we find that x = q q = p 2 p 2 p > Then, solving for the conjugate exponent q + x = > Now we estimate with Holder s inequality: f p p = ( ( = f p = f p q f p 2 f p ) /q ( x ) /x ) p /p 2 p 2 p p 2 4

5 Taking both sides to the power /p yields: ( ) /p2 f p = f p 2 p 2 p p p 2 But so N p [f] = ( ) /p f p = f p, /p N p [f] = f /p p ( ) /p2 = f p 2 p 2 = N p2 [f] Now, by Minkowski, N p [f + g] = /p f + g p /p ( f p + g p ) = N p [f] + N p [g] And, by Holder, fg f /p p g p /p = N p [f]n p [g] Finally, as finite and non-zero, lim N p[f] = lim p p f /p p = lim f p = f p With this we have completed the problem xercise 08 Chapter 8, # 6: Prove the generalization of Holder s inequality If k /p i = /r, p i, r, then f f k r f p f k pk Proof We proceed by induction considering first the case n = 2 f f 2 r r = f f 2 r f r p /r f 2 r p2 /r ( ) r/pr ( ) r/p2 = f r(p /r) f 2 r(p 2/) = f r p f 2 r p 2 5

6 Here the first inequality was by Holder Suppose now that this holds true for n For the case n, we apply our result fo n = 2 with now by replacing f 2 with n j=2 f j and p 2 replaced by p = ( n j=2 /p j), then n f j = f j= n j=2 Now apply the n equality and we are finished f j f p n f j p xercise 09 Chapter 8, # 9: If f is real-valued and measurable on, define its essential infimum on by ess inf f = sup{α : {x : f(x) < α} = 0} If f 0, show that ess inf f = (ess sup /f) Proof It is the greatest number α such that f(x) α except for on a subset of measure zero j=2 ess inf f = sup{α : {x : f(x) < α} = 0} = inf{α : {x : f(x) < α } = 0} = inf{α : {x : f(x) > α} = 0} = (ess sup /f) 6