Analytical Methods Exam: Preparatory Exercises


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1 Anlyticl Methods Exm: Preprtory Exercises Question. Wht does it men tht (X, F, µ) is mesure spce? Show tht µ is monotone, tht is: if E F re mesurble sets then µ(e) µ(f). Question. Discuss if ech of the following sttement holds true or flse: i) L ([, b]) L ([, b]). ii) L ([, b]) L ([, b]). iii) L () L (). iv) L () L (). Justify your nswer providing quick proof if true or counterexmple if flse. Question. Let H be Hilbert spce. Describe the Grm Schmidt lgorithm to determine n orthonorml system from system of linerly independent vectors ( f n ) n N. Compute in detils the first three vectors obtined by, x, x,... in the cse H = L ([, ]) endowed with usul sclr product. Exercise. Stte precisely continuous nd derivble dependence by prmeters for integrls like f (ξ) = φ(ξ, x) dx, E, ξ I. Let E f (ξ) := sin(ξ x) x( + x ) dx. i) Show tht f is well defined for every ξ nd it is continuous. ii) Show tht f is lso derivble, compute f nd connect this to some Fourier Trnsform. iii) Determine f. Exercise. On X := C ([, ]) let s onsider the supnorm, the totl vrition norm f v := f + f f + nd the C norm f := f + f. i) Show tht there exist c, C > such tht f c f v C f. f (x) dx, ii) By using sequences like f k (x) := c k sin(kπx) or g k (x) = c k x k with c k > show tht do not exists constnts m, M > such tht f v m f, f M f v.
2 Question. Wht does it men to sy tht " mesure µ is continuous from bove"? Under which conditions this property holds true? Is probbility mesure lwys continuous from bove? Question. Wht does it men tht two norms re equivlent? Define precisely. Assume tht X is Bnch spce under certin norm. Prove tht it is lso Bnch spce for ny equivlent norm. Question. Wht is the multiplictionderivtion dulity in the FT? Stte precise property nd prove it in the cse of multipliction by x nd d dx. Exercise. Let H = L ([, ]) endowed with usul sclr product f, g = f (x)g(x) dx. i) Describe, in generl, the Grm Schmidt lgorithm to compute n orthonorml system by system of linerly independent vectors ( f n ) n N. ii) Let U be the subspce of H generted by functions x, x, x 4. Determine n orthonorml bse for U. iii) Determine (if ny) the element of U tht best pproximtes in H. Exercise. Let f (x) = ( x ) χ [,] (x). i) Compute the Fourier Trnsform of f. Is f L ()? Is f L ()? ii) Let u = u(t, x) be the solution of the following PDE: u t (t, x) = tu xx (t, x), t, x, u(, x) = f (x), x, Let v(t, ξ) := u(t, )(ξ). Determine v(t, ξ). Wht cn be sid bout u?
3 . Prtil Exm December 8 Question. Let ( f n ) L([, b]). Which of the following sttements hold true? L i) If f L n then f n. L ii) If f L n then f n. L iii) If f L n then f n. Justify your nswer by proving the conclusion if true or disproving by counterexmple if flse. Question. Stte precisely the derivtion theorem for integrls depending by prmeters. Consider the identity + e xy dy = x, for every x >. Show tht the previous identity cn be derived n times, for every n N, n, nd use this to compute + x n e x dx. Question. Give the definition of Fourier Trnsform for function f L ( d ). Let T be d d orthogonl mtrix (tht is T T = I). Prove tht if f (T x) = f (x) for every x d then f (Tξ) = f (ξ) for every ξ d. Exercise. Let H be rel Hilbert spce, φ, ψ H two linerly independent vectors, U = φ {φ : } nd, similrly, V = ψ. i) Determine the orthogonl projections Π U nd Π V. ii) Determine Π U+V. iii) In generl, wht is the reltion between Π U+V nd Π U + Π V? Under which condition on φ, ψ is it true tht Π U+V = Π U + Π V? iv) Let f be fixed. Discuss how big cn be Π U+V f (Π U f + Π V f ). Exercise. Let f (x) := (+x ). i) Use multiplictionderivtion dulity to compute explicitly f ( ) (hint: x f (x) = x...) ii) Use i) to determine f. Justify crefully. iii) Use f to compute + ( + x ) dx, + sin x ( + x ) dx.
4 4 Answers nd Solutions to Exercises. 4. First Set Question. For the Definition, see notes. To prove tht µ is monotone: just notice tht if E F, then F = E (F\E), F\E F (being this σ lgebr) whence µ(f) = µ(e) + µ(f\e) µ(e). Question. i) L ([, b]) L ([, b]): true, by Hölder (or Cuchy Schwrz) inequlity, f = b f dx = b ii) L ([, b]) L ([, b]): flse, f (x) = x L \L. iii) L () L (): flse f (x) = + x L \L. iv) L () L (): flse f (x) = x(+x ) L \L. ( b ) / ( b / f dx f dx dx) = b f. Question. Given ( f n ) n orthonorml system generted by this is defined s f n n j= f n, e j e n e = f f, e n = f n n j= f n, e j e. n Let f, f = x, f = x nd let s compute e, e, e. Notice tht f = dx = so e. Now, f, e = so e = f f, where f = x dx =, by which e = Therefore x, e = e = x x, x = x dx =, x =, x, e = x. Finlly x dx =. (x ) dx = = 5, e = 5 ( x ).
5 Exercise. i) We hve first to check tht the integrl is well defined for ny ξ, tht is the function sin(ξ x) f (ξ, x) := x(x +) L ([, + [) w.r.t. x for every ξ. Notice tht if ξ, f (ξ, x) = ξ sin(ξx) ξx + x ξ + x, becuse of the inequlity sin t t. In prticulr, f (ξ, ) is bounded by L function for every ξ (if ξ = this is trivil being f (, x) ). ii) To discuss the existence of g (ξ) it is nturl to pply the derivtion under integrl sign. Formlly g (ξ) = ξ f (ξ, x) dx = ξ f (ξ, x) dx. To this im we need of course the existence of πx cos(πξx) ξ f (ξ, x) = x(x = π cos(πξx) + ) x +, nd n integrble bound of it independent by the prmeter ξ, nd indeed ξ f (ξ, x) = cos(ξx) x + x + L (). Therefore the derivtion under integrl sign pplies nd we obtin g cos(ξ x) (ξ) = π x + dx. By Euler formul cos(ξx) = ei ξ x +e iξ x hence g (ξ) = ( ) + x eiξ x dx + + x e iξ x dx = + ( ξ/π) + + (ξ/π) = e ξ. iii) For ξ >, ξ η g(ξ) = g() + g (η) dη = + e η dη = [ ] e η η=ξ = e ξ η= For ξ <, being g ( ξ) = g (ξ) nd g() =, g must be odd, whence g(ξ) = sgn(ξ) e ξ, ξ., ξ. 5 Exercise. i) Trivilly, f f + f = f v for every f X, therefore c =. Let s prove tht f v C f for some C. Notice tht f v = f + f dx f + f dx = f + f = f. ii) Let f k (x) = sin(kπx). Then f k = mx x [,] sin(kπx) = mx y [,kπ] sin y = while f k v = f + f k (x) dx = + kπ cos(kπx) dx = + kπ cos(y) dy +, k +,
6 6 whence it is impossible tht f v C f otherwise f k v C. Now, by considering g k (x) = x k we hve g k = g k + g k where therefore g k = + k. On the other hnd, g k = mx x [,] xk =, g k = mx x [,] k xk = k, g k v = g k + g k = + k x k dx =, thus it cnnot exists constnt M such tht f M f v for every f otherwise, tking f = g k, + k C for every k. 5. Second Set Question. A mesure is continuous from bove if µ(e n ) µ(e) whenever (E n ) F, E n E (tht is E n E n+, E = n E n ). In generl this property is flse. A sufficient condition is tht µ(e ) < +. In prticulr, for probbility mesures it is lwys true. Question. Two norms nd on X re equivlent if c, C > : c f f C f, f X. Let s ssume tht X be Bnch spce respect nd let n equivlent norm. We clim X is Bnch, tht is complete, respect to. To prove this, let ( f n ) X be Cuchy sequence in the norm, ε >, N : f n f m ε, n, m N. By equivlence f n f m Cε, n, m N, whence ( f n ) is Cuchy respect to. Being X complete respect to this norm, f n f X nd becuse the two norms re equivlent, we hve lso f n f, tht is X is complete under. Question. The multiplictionderivtion dulity for the FT sys tht (iπξ) ξ f = x ( iπx f ). This follows by combining the following two well known properties: i) if f, x f L then x f = (iπξ) f ; ii) if f, x f L then ξ f = iπx f. Be these, (iπξ) ξ f ii) = (iπξ) iπx f i) = x ( iπx f ).
7 Exercise. i) In generl, given system ( f n ) of linerly independent vectors, n orthonorml system generted by this cn be computed in the following wy: e := f f, e n := f n j = n f n, e j e j f n j = n f n, e j e j. ii) Let U = Spn(x, x, x 4 ) = {x + bx + cx 4 :, b, c }. To determine n orthonorml bse for U let s pply the Grm Schmidt lgorithm. We set f = x, f = x, f = x 4 nd e = f f. becuse f = x dx =, = e = x. Now, e = f f,e e.... We hve f, e = Finlly, being f, e = x 4 = x4 5 7 x... f = x x dx =, whence e = f f, where x 4 dx = 5 5, = e = x. 5 x dx =, f, e = x 4 x dx = 5 7 = 7, we hve e = f f,e e.... Becuse x4 5 ( 7 x = x 4 5 ) 7 x dx = = , ( by which e = x x). iii) U is closed becuse finite dimensionl, hence the best pproximtion of in U exists nd it is the orthogonl projection of on U tht is Π U =, e j e j. We hve, e = x dx =,, e = 5 x dx = j= 5 =, 7 Therefore, e = ( x x) dx = Π U = 5 x 5 ( 5 ) 5 7 = 4 5. (x 4 57 ) x = ( ) 5 5 x4 + + x.
8 8 Exercise. i) Let f (x) = ( x ) χ [,] (x). Clerly f L (). For ξ, f (ξ) = f (x)e iπξ x dx = ( x )e iπξ x dx = sin(πξ) πξ = sin(πξ) πξ = sin(πξ) πξ ( [ x e iπ ξ x iπξ { ( e iπ ξ iπξ e iπ ξ iπξ ] x= + e iπ ξ x x= iπξ ) + [ ] x= e iπ ξ x ( iπξ) ( xe iπξ x dx + xe iπξ x dx dx + [ ] x e iπ ξ x x= iπξ ) e iπ ξ x x= iπξ dx x= [ ] x= e iπ ξ x ( iπξ) = 4π ξ ( e iπξ (e iπξ ) ) = π ξ ( cos(πξ)), while f () = f (x) dx = cos t ( x ) dx =. eclling tht for t = by definition nd conclude tht f (ξ) = cos(πξ) π ξ, ξ. x= } t t we my ssume cos t t = Notice tht f L L. Indeed: s ξ we hve f (ξ), whence f, f re both integrble t ξ =. At, f (ξ) = C, this mens tht f, f re both integrble t ±. π ξ ξ ii) Let u = u(t, x) be the solution of the following PDE: Let v(t, ξ) := u(t, )(ξ). Then Deriving under integrl sign t v(t, ξ) = According to the eqution, Furthermore, therefore v solves the eqution u t (t, x) = tu xx (t, x), t, x, u(, x) = f (x), x, t v(t, ξ) = t u(t, )(ξ) = t u(t, x)e iπξ x dx. t u(t, x)e iπξ x dx = t u(t, )(ξ). t v(t, ξ) = t xx (t, )(ξ) = t(iπξ) u(t, )(ξ) = 4π ξ tv(t, ξ). v(, ξ) = u(, )(ξ) = ( ) χ[,] ( )(ξ) = cos(πξ) π ξ, t v(t, ξ) = 4π ξ tv(t, ξ), v(, ξ) = cos(πξ) π ξ. )
9 9 The differentil eqution is n ODE respect to t, nd it cn be esily solved: writing tht is v (t) = 4π ξ tv(t), v(t) = v()e π ξ t, v(t, ξ) = cos(πξ) π ξ e π ξ t. Now, becuse v is clerly n L function, we my obtin u by inverting the FT, u(t, x) = v(t, )( x). 6. Prtil Exm December 8 Question. i) True: becuse of the Cuchy Schwrz inequlity, b ( b ) / ( b / f L = f dx f dx) = b f L, by which we deduce tht L is stronger thn L. It s stndrd fct tht by this it follows tht if ( f n ) converges in the L norm it converges in L norm s well. ii) True: just notice tht f (x) f L.e., then f L = b f dx b f L dx = (b ) f L, by which, gin, the conclusion follows being L stronger thn L. iii) Flse: tke for instnce [, b] = [, ] nd f n = n / χ [, n ]. Then /n f n L = f n dx = n / dx = /n n /, but f n = n 4/ L In prticulr, ( f n ) cnnot be convergent in L otherwise it should be bounded. dx = n / +. Question. For the sttement see notes. Let F(x) := + f (x, y) dx with f (x, y) = e xy. Clerly f (x, ) L ([, + [) for every x >, x f (x, y) = ye xy exists for every x, y (hence for every x, y ], + ]) nd x f (x, y) = y e xy. Now, if x [r, + [ with r > fixed, x f (x, y) ye ry L ([, + [). We my pply the derivtion theorem on [r, + [ to conclude tht F (x) = + ye xy dy, x r. Becuse r > is rbitrry, the previous holds for every x >. Iterting this procedure we hve n x F(x) = + ( y) n e xy dy.
10 On the other hnd ( ) x n F(x) = x n x = n x x = n x Setting x = we finlly obtin x = ( ) x n x = ( ) x n x 4 =... = ( ) n n!x n. + ( ) n y n e y dy = ( ) n n!, + y n e y dy = n! Question. We hve f (Tξ) = d d d f (x)e iπξ T x dx. Setting y = T x, or x = T y, dx = det T dy = dy becuse T is orthogonl, therefore f (Tξ) = f (T y)e iπξ y dy = f (y)e iπξ y dy = f (ξ). d d Exercise. i) Let e := φ φ. Then U = e (evident) nd Π U f = f, e e. Indeed, Π U f U nd f Π U f, e = f, e f, e e, e = ( f, e f, e ) =. Similrly, setting ê := ψ ψ we hve V = ê nd Π V f = f, ê ê. ii) Being U + V = e + ê, to determine the orthogonl projection on U + V we need n orthonorml bse. In generl, e, ê re not orthogonl, so we my replce ê by Therefore iii) Set δ := ê ê, e e. Then By ii), ẽ = ê ê, e e ê ê, e e. Π U+V f = f, e e + f, ẽ ẽ. δ = ê + ê, e e ê, ê, e e = + ê, e ê, e = ê, e. Π U+V f = f, e e + δ f, ẽ ( ê ê, e e ). Notice tht f, ẽ = δ f, ê ê, e e = ( ) f, ê ê, e f, e δ
11 therefore Π U+V f = f, e e + δ ( f, ê ê, e f, e ) (ê ê, e e ) = f, e e + δ ( f, ê ê f, ê ê, e e f, e e, ê ê + ê, e f, e e ) = Π U f + δ ( ΠV f Π U Π V f Π V Π U f + ê, e Π U f ) = ( ) + ê,e Π δ U f + Π δ V f (Π δ U Π V f + Π V Π U f ) = δ (Π U f + Π V f (Π U Π V f + Π V Π U f )). We hve Π U+V = Π U +Π V if nd only if φ ψ. Indeed: if φ ψ then e, ê =, so δ =. Furthermore Π U (Π V f ) = Π V (Π U f ) =. Therefore Π U+V = Π U + Π V. Vice vers: ssume Π U+V = Π U + Π V. Then ) ( δ Π U f + δ Π UΠ V f = ( δ ) Π V f δ Π V Π U f. Becuse U nd V re linerly independent, the previous identity cn be true if nd only if both sides re. But then (( Π U ) δ f + ) δ Π V f =, tht is ( δ ) f + δ Π V f U or f δ ( f Π V f ) U. Now, if f V, being f Π V f = we obtin f U for every f V, tht is V U. But this is possible if nd only if φ ψ. iv) It is cler tht the gp Π U+V f (Π U f + Π V f ) is mximum when U = V, in such cse Π U+V = Π U so Π U+V f (Π U f + Π V f ) = Π U f Π U f = Π U f = f for f U. Exercise. i) We hve x x f (x) = ( + x ) = x + x, so f ( )(ξ) = x + (ξ) = (iπξ) + (ξ) = iπ ξe π ξ. ii) Now, reclling tht f (ξ) = iπ f ( )(ξ) = iπ f ( )(ξ) = (iπ)(iπ ξe π ξ ) = π ξe π ξ. In prticulr f is primitive of π ξe π ξ. Let s determine this. Becuse of the modulus, we distinguish ξ by ξ. In the first cse, ) ) f (ξ) = π ξe πξ dξ + c = π (ξe πξ e πξ dξ + c = π (ξe πξ + e πξ + c π
12 In the second cse f (ξ) = π ξe πξ dξ + c = π (ξe πξ ) ) e πξ dξ + c = π (ξe πξ eπξ + c π To determine c, c we notice tht, becuse f L (), ccording to L Lemm, f (ξ) s ξ +. In prticulr we get esily tht c = c =. The conclusion is ξ, = π ( ) ξe πξ + e π ξ π, ( f (ξ) = ξ, = π ( ) = πe π ξ π ξ + ). ξe πξ eπ ξ π iii) We cn esily reduce the two integrls to suitble Fourier integrls f (ξ) = ( + x ) e iπξ x dx. Indeed: in the first cse we hve + ( + x ) dx = About the second, reclling tht sin x = ei x e i x, we hve + sin x (+x ) dx = i ( + i e i x dx + + (+x ) = f ( ) ( π = πe + ) = π e. ( + x ) dx = f () = π 4. e i x dx ) = (+x ) (+x ) e ix dx
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