Advanced Real Analysis

Transcription

1 McGill Uiversity December 26 Faculty of Sciece Fial Exam Advaced Real Aalysis Math 564 December 9, 26 Time: 2PM - 5PM Examier: Dr. J. Galkowski Associate Examier: Prof. D. Jakobso INSTRUCTIONS. Please write your aswers clearly i the exam booklets provided. 2. You may quote ay result/theorem see i the lectures or i the assigmets without provig it. 3. This a CLOSED BOOK exam. Notes, crib sheets, textbooks or ay other aids are NOT permitted. 4. Traslatio dictioary is permitted. Other dictioaries are ot permitted. 5. Calculators are ot permitted. This exam comprises the cover page ad four pages of questios umbered Q to Q6.

2 Math 564 Fial Exam Page 2 December 26. Let f L ([, ]), f >. Let α = f(x) dx. Show that α + lim = f. α Proof. We may assume f = without loss of geerality. Let A = {x f > ɛ}, B = {x ɛ f 2ɛ} ad C = {x f < 2ɛ}. The Now, f + f A f + + B f + A f + B f + ( 2ɛ) µ(c) ( 2ɛ) + ( 2ɛ) µ(c) A f A f µ(a)( ɛ). Therefore, f + f ( 2ɛ) sice µ(a) >. Now, sice ɛ > is arbitrary, + ( 2ɛ) µ(c) ( ɛ) µ(a) ( 2ɛ) lim if α + α f. But, it is easy to see that ad hece α + f α lim sup α + α f. 2. If µ is a positive measure, each f L (µ) defies a multiplicatio operator M f o L p (µ) ito L p (µ), such that M f (g) = fg. (a) Prove that M f f. Proof. Observe that fg p f p g p = f p g p. Therefore, M f (g) p f g p.

3 Math 564 Fial Exam Page 3 December 26 (b) For which measures µ ad p [, ) is it true that M f = f for all f L. Proof. Suppose µ is semifiite. The fix ɛ > ad let B = {x f f ɛ}. The there exists A B with < µ(a) <. So, let g = A µ(a) /p. The g p = ad fg ( f ɛ) A µ(a) /p So that fg p ( f ɛ). I particular, sice ɛ > is arbitrary, M f = f. O the other had, if µ is ot semifiite, the there exists A with µ(a) = ot cotaiig ay subsets with fiite but positive measure. Let f = A L. The observe that for g L p, g = a.e. o A ad hece gf = a.e. ad i particular M f (g) p = for all g L p, so M f = as a operator o L p. (c) For which f L does M f map L p (µ) oto L p (µ), p <? (Hit: Use the ope mappig theorem.) Proof. First observe that M f (g ) = M f (g 2 ) implies f(g g 2 ) = a.e. Hece g = g 2 a.e. o f. Suppose that there exists A with > µ(a) > so that f A. The clearly A / M f L p but A L p ad hece M f is ot oto. Therefore, if f A = the µ(a) = or µ(a) =. I particular, {f = } has o subsets with fiite but positive measure. The if g L p, g has g = a.e. o f =. I particular, this implies M f is ijective. Now, if M f is surjective, by the ope mappig theorem it has a bouded iverse. Clearly, (M f ) = M /f, so M /f is bouded o L p. Notice that if there exists A µ( f < ɛ) with < µ(a) <, the A L p ad M /f ( A ) p ɛ A p. I particular, M /f ɛ. So, sice M /f is bouded, there exists ɛ > so that { f < ɛ} cotais o subsets with fiite, ozero measure. Now, suppose there exists ɛ > so that { f < ɛ} cotais o subsets with fiite, ozero measure ad let g L p. The g/f ɛ g sice g = a.e. o { f < ɛ} ad hece g/f L p. I particular, M f (g/f) = g ad hece M f is oto. 3. Give a measurable fuctio f : X C where (X, µ) is a σ-fiite measure space, defie a fuctio F f : R + R + (the distributio fuctio of f) by F f := µ{x f(x) > t}. (a) Prove that for < p <, ad f measurable, f p p = pt p F f (t)dt.

4 Math 564 Fial Exam Page 4 December 26 Proof. Cosider f (x) > f(x) p dµ(x) = pt p dtdµ(x) = pt p f(x) >t dµ(x)dt = pt p F f (t)dt where i the third lie we apply the Toelli theorem. (b) Prove Chebyshev s iequality: for f L p (X, µ), we have ( ) p f p F f (t). t Proof. Observe that by mootoicity of measure, F f (t) is mootoe decreasig ad hece T T f p p = pt p F f (t)dt pt p F f (t)dt F f (T ) pt p dt = F f (T )T p. (c) Prove that if f L (µ), the lim sup t tf f (t) =. Proof. Observe that > f = F f (t)dt t tf f (t)dt. Now, if T F f (T ) > ɛ, the for tf f (t) > ɛ/2 for t [T/2, T ]. I particular, there exists T so that T F f (T ) > ɛ, the we may assume T > 2T ad hece a cotradictio. t tf f (t)dt ɛ/2 log(t ) log(t /2) = ɛ/2 log(2) =, (d) Fid a fuctio f ad measure space (X, µ) so that lim sup tf f (t) =, but f / L (µ). Proof. Let (X, µ) = ([, ], m) ad f =. The f / L x log x (m) but lettig x = t /(log t), t(log t) = x log x log(t(log t)) = + t log log t log t < t

5 Math 564 Fial Exam Page 5 December 26 for t large eough. Hece, { f > t} { < x < t log t } ad m( f > t) t log t. 4. Let f be a bouded liear fuctioal o a subspace M of a Hilbert space H. Prove that f has a uique orm preservig extesio to a bouded liear fuctioal o H ad that this extesio vaishes o M. Proof. We have that f(x) f M x, x M. Therefore, by the Hah-Baach theorem, there exists a extesio F of f to all of H with F (x) F x for all x H. I particular, the this extesio has F = f sice we may take x M with x = ad F (x ) = f(x ) f. Suppose that there exists y M with y, g. z = sg(f (y))y. The cosider F (w + t f z) f + t f F (z) The let w = sg(f (x ))x ad ad So, w + t f z 2 = + t 2 f 2 z 2. F (w + t f z) + t 2 f 2 z f + t F (z) 2 + t 2 f 2 z 2 ad hece, there exists t > so that F (w + t f z) > f, + t 2 f 2 z 2 a cotradictio. Therefore F vaishes o M. Now, suppose that F is aother orm preservig extesio of f. The F also vaishes o M. For y M, cotiuity of F ad F imply with y y, F (y) = lim f(y ) = F (y ). Notice also that M M. Suppose that w M. The suppose y M with {y } M y y. The y, w = lim y, w =.

6 Math 564 Fial Exam Page 6 December 26 So, w M ad M = M. Therefore, if x H, the x = y + z with y M ad z M. Thus, F (x) = F (y + z) = F (y) + F (z) = lim(y ) = F (y) = F (y) + F (z) = F (x) ad hece F = F 5. Suppose that {Λ } is a sequece of bouded liear trasformatios from a ormed liear space X to a Baach space Y, suppose that Λ M < for all ad suppose that there is a dese set E X such that {Λ x} coverges for each x E. Prove that {Λ x} coverges for each x X. Proof. Fix y X ad ɛ >. Let x E such that x y < ɛ. The Λ y Λ m y = (Λ Λ m )x + (Λ Λ m )(y x) (Λ Λ m )x + (Λ Λ m )(y x) (Λ Λ m )x + 2M y x < (Λ Λ m )x + 2Mɛ Now, sice x E, there exists N > large eough so that for, m > N, ad i particular, for, m > N, (Λ Λ m )x < ɛ Λ y Λ m y < (2M + )ɛ. So, sice ɛ > was arbitrary, {Λ y} is Cauchy ad hece coverges. 6. Suppose that {x i } C has α i x i coverges for every {α i } C with α i. Prove that xi <. Proof. Defie the maps T : l C by The, sice α i x i coverges Hece, sice l T (α, α 2,... ) = α i x i. i= sup T (α) < is a Baach space, by the priciple of uiform boudedess, sup T <

7 Math 564 Fial Exam Page 7 December 26 Now, lettig α = (sg(x ), sg(x 2 ),..., sg(x ),,... ). The α l = ad T (α ) = x i i= I particular, Therefore, sice sup T <, T x i. i= x i = sup x i <. i=