TERENCE TAO S AN EPSILON OF ROOM CHAPTER 3 EXERCISES. 1. Exercise 1.3.1

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1 TRNC TAO S AN PSILON OF ROOM CHAPTR 3 RCISS KLLR VANDBOGRT 1. xercise We merely consider the inclusion f f, viewed as an element of L p (, χ, µ), where all nonmeasurable subnull sets are given measure 0. Linearity is trivial. Surjectivity is also immediate, since the preimage is merely the function f itself. It remains to prove injectivity, so assume f 0. Then, Supp(f) A for some null set A. Integrating over A, we find that f = 0 a.e, which proves injectivity. 2. xercise (i). Note that whenever p < 1 and x 1, x x p. Then, 1 = f f + g + g f + g ( f ) p ( g + f + g f + g = f p + g p ( f + g ) p ) p = ( f + g ) p f p + g p Combining this with the triangle inequality and integrating, f + g p p f p p + g p p Date: September 17,

2 2 KLLR VANDBOGRT (ii). Note first that x x p is concave for p < 1. By homogeneity, it is of no loss of generality to assume that f p + g p = 1, so that for F p = G p = 1, f = (1 θ)f, g = θg, θ (0, 1) So that by concavity, (1 θ)f + θg p (1 θ) F p + θ G p Integrating yields f + g p 1 = f p + g p As asserted. (iii). We may again assume f p + g p = 1, so that for F p = G p = 1, By part (i), f = (1 θ)f, g = θg, θ (0, 1) f + g p ( (1 θ) p + θ p) 1/p Since x x 1/p is convex, ( (1 θ) p + θ p) 1/p (1 θ) (1 θ) 1 1/p + θ θ 1 1/p = (1 θ) 2 1/p + θ 2 1/p Optimizing in θ, we find that the minimum is achieved for θ = 1/2, implying ( 1 ) 2 1/p ( 1 2 1/p f + g p + 2 2) = 2 1/p 1 Which yields our optimal constant as 2 1/p 1, as asserted.

3 TRNC TAO S AN PSILON OF ROOM CHAPTR 3 RCISS 3 (iv). Note that by strict convexity/concavity, equality holds if and only if g = cf, c R. In the p = 1 case, we merely require that f and g always have the same sign (that is, fg 0). 3. xercise Suppose first that is a norm. If B denotes our unit ball, let x, y B, t (0, 1): (1 t)x + ty (1 t) x + t y 1 t + t = 1 So that B is convex. Conversely, we merely use contraposition. Then there exist points x and y B such that the triangle inequality does not hold. By homogeneity, we may assume that x = (1 t)x, y = ty for x, y B and that x + y = 1. If we consider the line segment through x and y, we see that for θ = t, (1 t)x + ty = x + y > x + y = 1 So that B is not convex, whence the result. 4. xercise [ Define A 0 := [1, ], A n := Supp( f ), we see: Yielding σ-finiteness. Supp(f) = 1 n+1, 1 n ]. Noting that Supp(f) = f 1 (A n ) n=0 5. xercise If f = 0, f 0 trivially. Assume now that f 0. For sufficiently small ɛ > 0, consider S ɛ := {x f(x) f ɛ}

4 4 KLLR VANDBOGRT By the previous problem, we may assume without loss of generality that µ(s ɛ ) <. Then, Taking the limit inferior, ( f p ( f ɛ) p dµ S ɛ = ( f ɛ)µ(s ɛ ) 1/p lim inf p f p f ɛ ) 1/p As ɛ > 0, is arbitrary, lim inf p f. Now, as f L p 0 L, Hölder s inequality yields for all p > p 0 : Letting p, p p 0 p f p f f p 0p p 0 lim sup f p f p Combining with the reverse inequality, we deduce lim p f p = f As asserted. Suppose now that f / L. Then, for every n > 0, there exists a set n and ɛ > 0 such that µ( n ) ɛ > 0 and f(x) > n for every x T ɛ. We see: Letting p, nɛ 1/p = nµ( n ) 1/p ( ) 1/p < f p dµ f p n n < lim inf p f p for all integers n. Hence, letting n, lim p f p =.

5 TRNC TAO S AN PSILON OF ROOM CHAPTR 3 RCISS 5 Define d(f, g) := f g. 6. xercise Homogeneity: d(cf, cg) = cf cg = c f g = c d(f, g) Triangle Inequality: d(f, h) = f h f g + g h = d(f, g) + d(g, h) Separation: d(f, g) = 0 f g = 0 f = g Symmetry: d(f, g) = f g = g f = d(g, f) Translation Invariance: d(f + h, g + h) = (f + h) (g + h) = f g = d(f, g) Conversely, suppose we have a translation invariant homogeneous metric d : V V [0, ]. Define f := d(f, 0). This choice is clearly unique, since any definition with respect to a nonzero basepoint loses homogeneity. It remains only to show the triangle inequality: f + g = d(f + g, 0) = d(f, g) d(f, 0) + d(0, g) = d(f, 0) + d(g, 0) = f + g So that defines a unique norm.

6 6 KLLR VANDBOGRT 7. xercise Assume first that V is complete. Given an absolutely convergent series n=1 f n, the sequence of partial sums s N is Cauchy. But f n f n n=1 So that N n=1 f n is bounded by s N. As the s N are Cauchy, we deduce that N n=1 f n is Cauchy. By completeness, this sequence converges, so that n=1 f n exists. Conversely, suppose that any absolutely convergent sum converges conditionally. Let f n be a Cauchy sequence, and extract a subsequence f nk such that n=1 f nk+1 f nk < 1 2 k for k N. Then, obviously k=1 f n k+1 f nk converges. By assumption, this implies k=1 f n k+1 f nk converges as well. But this sum in telescoping with limit lim k f nk f n1, so we deduce that f nk f for some f. It remains to show that f n f, but as f n is Cauchy: f n f f n f nk + f nk f 0 as n, k, so that f n f, implying that every Cauchy sequence is convergent, so V is complete. 8. xercise If f L, f is essentially bounded. Hence there exists a sequence of simple functions increasing to f (by the simple approximation theorem), and density is trivial. Note that the space of simple functions with finite measure is not dense in L. To see this, merely consider the characteristic function χ R. This has χ R = 1. Choosing a sequence

7 TRNC TAO S AN PSILON OF ROOM CHAPTR 3 RCISS 7 of simple functions s n with finite support, we see that for every n, s n vanishes outside of some sufficiently large set, so that χ R s n = 1 for all n. Therefore, this set cannot possibly be dense. 9. xercise numerate the generators of our σ algebra Ω by B = { 1, 2,... }. One immediately sees that { finite χ n n B} is dense in the space of characteristic function. Since Q is also dense in the reals, we see that S := span Q {χ n n B} is dense in the space of simple functions with finite measure support, denoted S 0, which in turn is dense in L p. But this implies S = S 0 and S 0 = L p From which we immediately see that S is dense in L p as well. But S is countable, so we deduce that L p is separable. For p =, consider the family C := {χ [ r,r] r R + } Then for any two distinct x, y C, x y = 1. But then L is certainly not separable, as no sequence of distinct elements could ever converge to an element of C.

8 8 KLLR VANDBOGRT 10. xercise Young s inequality is established by concavity, and the case of equality for strict concavity will occur precisely when a p = cb q. By homogeneity, we may assume f p = g q = c = 1. Then, Yielding the result. f(x)g(x) = 1 p f(x) p + 1 q g(x) q = fg 1 = 1 = f p g q Let f L p, q < p. Then ( f q 11. xercise ) 1/q 1/p ( ) 1/p 1dµ f p dµ = µ() 1/q 1/p f p So f L q. constant. By the previous problem, equality occurs when f is a 12. xercise Since q > p, we may apply the reverse H ólder inequality to find f p f q µ() 1/p 1/q By assumption, µ() m for every in our σ-algebra, so we rearrange the above inequality to find f q m 1/q 1/p f p So that f L q if f L p. quality holds again for f const.

9 TRNC TAO S AN PSILON OF ROOM CHAPTR 3 RCISS 9 We have: f(x) p dµ = ( 13. xercise f pθ f (1 θ)p d]mu = f pθ p 1 f p(1 θ) p 0 Taking pth roots in the above, whence the result. By Hölder s, ( ) ) f pθ p pθ ( 1 p1 pθ ( dµ ) ) f (1 θ)p p p(1 θ) 0 p (1 θ)p dµ 0 f p f θ p 1 f 1 θ p xercise so that f p p µ() 1 p/p 0 f p p 0 lim sup f p p µ() For the reverse inequality, lim inf f p dµ lim inf f p dµ lim inf f p dµ = dµ = µ() Hence we conclude that lim f p p = µ(). (Fatou s)