M38C Prctice for the finl Let f L ([, ]) Prove tht ( /p f dm) p = exp p log f dm where, by definition, exp( ) = To simplify the problem, you my ssume log f L ([, ]) Hint: rewrite the left hnd side in form to which you cn pply L Hopitl s rule Solution Use L Hopitl s rule: ( /p ( f dm) p = exp (/p) log p p ( = exp = exp ( p p log ( f p dm p log( f ) f p dm f p dm Use the Dominted Convergence Theorem to complete the exercise )) f p dm ) 2 Let φ be differentible Lipschitz function on R If f is integrble on [, b] show tht the function Ψ(t) defined by is differentible Solution Ψ(t + h) Ψ(t) h Ψ(t) = = φ(tx)f(x) dx ) (φ((t + h)x) φ(tx)) f(x) dx h By ssumption there is constnt M > such tht (φ((t + h)x) φ(tx)) h M x Mb So the integrnd is dominted by M b f(x) Lebesgue s Dominted Convergence Theorem implies Ψ (t) = ( d φ(tx))f(x) dx dt 3 Suppose E R is mesurble nd E = E + for every nturl number n Show n tht either m(e) = or m(e c ) = Hint: Fix number N nd let F (x) = m(e [N, x]) (for x > N) Show tht F (x + ɛ) F (x ɛ) = F (y + ɛ) F (y ɛ)
whenever N + ɛ < x < y Wht does this imply bout F (x)? Wht does the Lebesgue Differentition Theorem pplied to χ E sy bout F? (This exercise cn be used to show tht if G R is proper subgroup then m(g) = ) Solution Becuse E = E +/n, F (x+ɛ) F (x ɛ) = F (x+q/n+ɛ) F (x+q/n ɛ) for ny integer q > Since we cn choose q, n rbitrry nd since F is continuous, it follows tht F ( ( ) ( ) ( ) x + n) F x n = F y + n F y n whenever N + < x < y n So F (x) = c for some constnt c Note tht F m(e (x h, x + h)) (x) = h 2h The Lebesgue Differentition Theorem pplied to χ E sys tht for e x, χ E (x) = h m(e (x h, x + h)) 2h = F (x) Since F is constnt e we must hve either F = e in which cse m(e) = or F = e in which cse m(e c ) = 4 Let f L ([, ]) Wht is n Solution There is constnt c > such tht log ( + e nf(x)) dx? (/n) log ( + e nf(x)) f(x) + c So Lebesgue s Dominted Convergence Theorem pplies We observe tht So 5 Compute n (/n) log ( + e nf(x)) f + (x) log ( + e nf(x)) dx = x dx f + dx Hint: Use Fubini s Theorem nd the reltion /x = e xt dt for x > Solution π/2 x dx = = = = = e xt dt dx e xt dt dx e xt dx dt [ ] n cos(x) + t e xt dt t 2 [ (cos(n) + t sin(n)) e nt ] dt t 2
By the dominted convergence theorem this is 6 Stte nd prove Egoroff s Theorem + t 2 dt = rctn(t)] = π/2 7 Let T : [, ] [, ] be n ergodic mesure-preserving Borel trnsformtion This mens tht m(t (E)) = m(e) for ny Borel E [, ] nd if f L 2 ([, ]) stisfies f T = f then f is constnt e Let Ω = {f T f + c : f L ([, ]), c C} () Show Ω is dense in L 2 ([, ]) Hint: becuse Ω is subspce, it suffices to show tht if v L 2 ([, ]) is orthogonl to every element of Ω then v = (b) Show tht for every f L 2 ([, ]) if f n is the function f n = n + n f T i then {f n } n= converges in L 2 ([, ]) to the constnt f dm Hint: first show this is true if f Ω (This is known s von Neumnn s men ergodic theorem) 8 Recll tht outer mesure is defined on R by m (E) = inf n i= m(i i) where the infimum is over ll collections {I i } n i= of intervls tht cover E nd n N { } From this definition, prove tht outer mesure is sub-dditive Tht is: show m ( i E i ) i m (E i ) for ny E, E 2, R 9 Let µ be finite Borel mesure on [, ] Define Prove: i= f(x) = µ([, x)) () µ is bsolutely continuous to Lebesgue mesure if nd only if f is bsolutely continuous (b) µ is singulr to Lebesgue mesure if nd only if f = e Hints: µ = µ c +µ sing where µ c is bsolutely continuous to Lebesgue mesure nd µ sing is singulr to Lebesgue mesure Also there is Rdon-Nikodym derivtive g = dµc dm It my be esier to prove the contrpositives For exmple, by showing for (b), if f then µ is not singulr to Lebesgue Find closed subset C of l 2 (N) tht does not hve n element with smllest norm In other words there does not exist v C with v = inf{ w : w C} Hint: If {e n } n N re the bsis vectors nd c n s re numbers then there is n exmple of the form {c n e n } n= Let C be the set of ll x l 2 (N) such tht x i /i for ll i Show tht C is compct On the other hnd, if D is the set of ll x l 2 (N) with x i then D is noncompct Why?
2 Define mesure on the unit sphere S n in R n s follows For E S n, let Ē be the set of ll x R n such tht x nd if x then x E Define σ(e) := m(ē)n x Show tht this defines mesure, denoted by σ, on S n Moreover prove tht for ny mesurble X R n, m(x) = χ X (rv)r n dm(r) dσ(v) S n Hint: first show tht the formul holds whenever there is n open set A S n nd rdii r < r 2 such tht X = {rv : r < r < r 2, v A} 3 If µ nd ν re signed mesures on R then we define their convolution µ ν in the following wy For ny set E R, let E 2 = {(x, y) R 2 : x + y E} Then define µ ν(e) := µ ν(e 2 ) We sy mesure µ is purely tomic if there is countble set C R such tht µ(r C) = We sy µ is continuous if µ({x}) = for every x R Prove: () if µ, ν re purely tomic then µ ν is purely tomic (b) if µ is continuous then µ ν is continuous (c) if µ << m then µ ν << m (d) Are there mesures µ, ν tht re singulr to Lebesgue mesure such tht µ ν << m? Hints: for (), purely tomic mesure is liner combintion of Dirc mesures ( Dirc mesure is mesure supported on single point) The convolution is bi-liner, so it suffices to show tht convolution of Dirc mesures is Dirc mesure) For (b,c) you my ssume µ is sigm-finite For (d) think bout mesures concentrted on the Cntor set 4 Let m denote Lebesgue mesure on [, ] nd c denote counting mesure on [, ] In other words, c(e) = E, the crdinlity of E Let = {(x, x) : x [, ]} [, ] 2 denote the digonl Compute the integrls χ (x, y) dm(x) dc(y) nd χ (x, y) dc(x) dm(y) Note they re not equl Does this violte Fubini s Theorem? Does it violte Tonelli s Theorem? If not, why? 5 Identify the circle T with [ π, π) Define the Fourier coefficents of f L (T) by ˆf(n) = 2π π π f(x)e ixn dx Recll tht f is odd if f(x) = f( x) nd it is even if f(x) = f( x)
() Show tht f is odd if nd only if ˆf(n) = ˆf( n) for ll n (b) Show tht f is even if nd only if ˆf(n) = ˆf( n) for ll n 6 Suppose f L (T) nd f is continuous t x Also suppose tht n= ˆf(n)e inx converges t x Show tht f(x) = n= ˆf(n)e inx Hint: recll tht the Fejér kernels F N form n pproximtion to the identity **My pologies; I don t think this one hs solution Insted, show tht F N f(x) x s N (A more difficult result is tht if f is Lipschitz in neighborhood of x then N ˆf(n)e n= N inx converges to f(x) s N For tht result, use tht D N (x) = sin((n+/2)x) sin(x/2)