Solutions to home assignments (sketches)

Transcription

1 Matematiska Istitutioe Peter Kumli 26th May 2004 TMA401 Fuctioal Aalysis MAN670 Applied Fuctioal Aalysis 4th quarter 2003/2004 All documet cocerig the course ca be foud o the course home page: Solutios to home assigmets (sketches) Problem 1: Let Y be a fiite-dimesioal subspace of a ormed space X. Show that Y is closed. Solutio: It is eough to show that if (y ) =1 is a sequece i Y covergig to some elemet y i X the y Y. So assume that (y ) =1 is a sequece i Y coverig i X ad call the limit elemet y. Fix a basis e 1, e 2,..., e i Y. Every elemet y ca be writte i the form y = Σ k=1α () k e k. Moreover all orms o fiite-dimesioal spaces, here we cosider the space Y with the iduced orm from X, are equivalet ad we see that z = Σ k=1 α k, where z = Σ k=1 α ke k, defies a orm. This implies that (α () k ) =1, k = 1, 2,...,, are Cauchy sequeces i (if Y is a complex ormed space) ad hece coverges. Call the limits α k, k = 1, 2,...,. Set ỹ = Σ k=1 α ke k. This implies that y ỹ i Y ad sice y y i X we have y = ỹ Y. This proves that Y is closed. Problem 2: Show that l 1 (as a vector space) is a subspace of l 2. Is this subspace closed i l 2 with the l 2 -orm? Solutio: Let = (x 1, x 2,...) l 1. Sice Σ k=1 x k 2 (Σ k=1 x k ) 2 holds true for every positive iteger we obtai l 2 l 1 < ad l 2. Moreover sice l 1 is a vector space we see that l 1 is a subspace of l 2. To see that l 1 is ot closed i l 2 cosider for istace the sequece ( ) =1 where = (1, 1 2, 1 3,..., 1, 0, 0,...). Here l 1 for all ad i l 2 where = (1, 1 2, 1 3,...). This is clear sice as. l 2 = (Σ k=+1 1 k 2 ) Problem 3: Let X be a ormed space. Show that X is fiitedimesioal if ad oly if every closed ad bouded set i X is compact. Solutio: Will be give later sice it is quite log, but ot very difficult, to prove usig Riesz lemma i oe directio.

2 Fuctioal Aalysis sid. 2 av 5 Problem 4: Set X = l 2 with the l 2-orm ad defie the mappigs T 1, T 2 by T 1 (x 1, x 2, x 3,..., x,...) = (x 1, 1 2 x 2, 1 3 x 3,..., 1 x,...) ad T 2 (x 1, x 2, x 3,..., x,...) = (x 1, x 2 2, x 3 3,..., x,...) for (x 1, x 2, x 3,..., x,...) l 2. Is T 1 a liear mappig? Is T 2 a liear mappig? Is T 1 cotiuous at ay poit i l 2? Is T 2 cotiuous at ay poit i l 2? Calculate sup{ T (x 1, x 2, x 3,..., x,...) l 2 : (x 1, x 2, x 3,..., x,...) l 2 r} for all r > 0 for both T equal to T 1 ad to T 2. Explai the differece. Solutio: It is easily see that T 1 is a liear mappig ad that T 2 is ot a liear mappig o l 2. It is also easy to see that the operator orm of T 1 is equal to 1 ad that sup{ T 1 ( ) : < r} = r for every r > 0. Sice T 1 is a bouded liear mappig it is also cotiuous at every l 2. It remais to treat the mappig T 2. First we see that T 2 ( ) l 2 for every l 2. To see this we fix a = (x 1, x 2, x 3,... ) l 2. From (Σ =1 x 2 ) 1/2 < it follows that there exists a iteger N such that x 1 for all N. This implies that ad hece l 2. We easily see that T 2 (x 1, x 2, x 3,...) l 2 = (x 1, x 2 2, x 3 3,...) l 2 { iequality} (x 1, x 2 2,..., x N 1 N 1, 0, 0,...) l 2 + (0, 0,..., 0, xn N, x N+1 N+1,...) l 2 (x 1, x 2 2,..., x N 1 N 1, 0, 0,...) l 2 } {{ } < sup{ T 2 ( ) : < r} = + (0, 0,..., 0, x N, x N+1,...) l 2 < < { r 0 r 1 1 < r. Here the last statemet follows e.g. from the observatio that ad lettig. r + 1 T 2 (0, 0,...,, 0,...) = (0, 0,..., ( r + 1 }{{ 2 } 2 ), 0,...) positio positio Fially we observe that T 2 is cotiuous at every l 2. To prove this fix a l 2 ad a sequece ( k) k=1 i l 2 such that k i l 2. Set = (x 1, x 2, x 3,...) ad Sice k = (x (k) 1, x(k) 2, x(k) 3 l 2 there exists a iteger N such that,...), k = 1, 2,.... x < 1 2 for all N, ad sice k i l 2 there exists a iteger K such that k l 2 < 1 4 k K. This implies that x (k) x (k) x + x < = 3 4

3 for all N, k K. Now fix a ɛ > 0. We see that Fuctioal Aalysis sid. 3 av 5 where T 2 ( k) T 2 ( ) l 2 = (Σ =1 (x (k) ) (x ) 2 ) 1/2 { iequality} (Σ N 1 =1 (x(k) ) (x ) 2 ) 1/2 + (Σ =N (x (k) ) (x ) 2 ) 1/2 (x (k) ) (x ) x (k) x Σ 1 x (k) x ( 3 4 ) 1 l=0 x(k) l x 1 l for all N ad k K. However sup + ( 3 4 ) 1 C <. This fially implies that T 2 ( k) T 2 ( ) l 2 (Σ N 1 =1 (x(k) ) (x ) 2 ) 1/2 0 sice k i l 2 because it implies that x (k) x,k for all. +C (Σ =N x (k) x 2 ) 1/2 0 sice k i l 2 Note that we have prove that both T 1 ad T 2 are cotiuous at every poit. However, cotiuity for a oliear mappig does ot imply that it maps bouded sets oto bouded sets while this is true for every liear mappig. Problem 5: Let X be a Baach space ad let T B(X, X), = 1, 2, 3,... Assume that lim T x exists for every x X. Show that T B(X, X) where T is defied by for x X. T x = lim T x Solutio: Clearly T is a liear mappig o X sice T (αx + βy) = lim T (αx + βy) = lim (αt (x) + βt (y)) = = α lim T (x) + β lim T (y) = αt (x) + βt (y) for every x, y X ad all scalars α, β. Moreover sice for all x X the sequece (T (x)) =1 coverges, ad hece is bouded, we coclude from the Baach-Steihaus Theorem that the sequece ( T ) =1 is bouded. This yields T (x) = lim T (x) = lim T (x) (sup T ) x, for all x X. This shows that T is a bouded liear mappig o X. Problem 6: Let T : H H be a compact liear operator o a Hilbert space H. Show that I +T is compact if ad oly if H is fiite-dimesioal. Here I deotes the idetity operator o H. Solutio: If H is a ifiite-dimesioal Hilbert space there exists a ON-sequece (e ) =1 i H. Here e 0 ad so T (e ) 0 sice T is compact. From this we see that the sequece ((I + T )(e )) =1 ca ot have ay coverget subsequece sice for m 2 = e e m (I + T )(e ) (I + T )(e m ) + T (e ) + T (e m ) ad so (I + T )(e ) (I + T )(e m ) 2 T (e ) T (e m ) 2 as, m, m. O the other had, if H is fiite-dimesioal Hilbert space the I is a compact operator ad so I + T is compact. Problem 7: Set T f(x) = π 0 cos(x y)f(y) dy, 0 x π. Fid the orm of T where T is regarded as a operator o L 2 ([0, π]).

4 Fuctioal Aalysis sid. 4 av 5 Solutio: (sketch) T is a self-adjoit (k(x, y) = cos(x y) satisfies k(x, y) = k(y, x)) compact (k L 2 ([0, π] [0, π])) liear operator o the Hilbert space L 2 ([0, π]). Hece T = sup λ eigevalue λ. It is a easy exercise to calculate the eigevalues to T. Problem 8: Prove the existece ad uiqueess of solutio to the followig boudary value problem: { 4u (x) = x + u(x), 0 x 1 u(0) 2u(1) = u (0) 2u (1) = 0, u C 2 ([0, 1]). Solutio: Stadard problem. Calculatios are omitted. Problem 9: Let (x ) =1 be a bouded sequece i a separable Hilbert space H. Show that there exists a subsequece (x k ) k=1 ad a x H such that What happes if H is ot separable? x k x. Solutio: (sketch) Assume that H is a separable Hilbert space ad that (e k ) k=1 is a ON-basis. Applyig a diagoal sequece -argumet as i Theorem we obtai a subsequece (x p ) =1 of (x ) =1 such that x p, e k coverges for all k = 1, 2,... Call the limits α k. Set M = sup x. Here M < by the hypothesis. Note that Σ k=1 x p, e k 2 = x p M 2 by Parseval s formula ad lettig we coclude Now fix a arbitrary x H. We obtai ad so x p, x = x p, Σ k=1 x, e k e k = Σ k=1 α k 2 M 2. = Σ k=1 x, e k x p, e k = Σ k=1 x, e k α k + Σ k=1 x, e k ( x p, e k α k ) x p, x Σ k=1 x, e k α k Σ N k=1 x, e k ( x p, e k α k ) +Σ k=n+1 x, e k ( x p, e k α k ). For fixt N the first term o the RHS teds to 0 as while the secod term ca be estimated from above, usig the Cauchy-Schwartz iequality, by (Σ k=n+1 x, e k 2 ) 1 2 2M. which teds to 0 as N.Hece we have that x p, x coverges as. (We also see that x p Σ k=1 α ke k ) Fially, if H is ot separable cosider let H deote the closure of the liear spa of the set {x : = 1, 2,...}. The H is a Hilbert space cotaiig all x. Moreover H is separable (possibly fiite-dimesioal) sice a ON-basis ca be costructed by the Gram-Schmidt process applied to the sequece (x ) =1. By the costructio above we have a subsequece (x p ) =1 that coverges weakly o H. Now x p, x coverges for every x H as sice every x ca be decomposed as y + z, y H ad z H ad x p, z = 0. Problem 10: Let T : H H be a compact positive self-adjoit operator o a Hilbert space H. Moreover assume that T 2. Give a estimate 1 for 1 better tha the trivial estimate T 2 3T + I 11. T 2 3T + I.

5 Fuctioal Aalysis sid. 5 av 5 Solutio: (sketch) Applyig the Hilbert-Schmidt theorem we have a ON-sequece (e ) =1 of eigevectors correspodig to the o-zero eigevalues (λ ) =1 to T such that T S = 0, where S = Spa{e : = 1, 2,...}. Moreover we kow that 0 < λ 2 for all. This yields (T 2 3T +I)x 2 = Σ =1(λ 2 3λ +1) x, e e 2 = Σ =1 λ 2 3λ +1 2 x, e 2 ( 5 4 x )2 for all x Spa{e : = 1, 2,...}. Here we have used Parseval s formula together with max 0 x 2 x2 3x + 1 = 5 4. For z S we get (T 2 3T + I)(z) = z ad hece (T 2 3T + I)(z) = z. Fially if x H the x = y + z, where y Spa{e : = 1, 2,...} ad z S = Spa{e : = 1, 2,...}, we get We coclude that T 2 3T + I 5 4. (T 2 3T + I)(x) 2 = (T 2 3T + I)(y + z) 2 = = (T 2 3T + I)(y) 2 + (T 2 3T + I)(z) 2 ( 5 4 )2 y 2 + z 2 ( 5 4 )2 ( y 2 + z 2 ) = ( 5 4 )2 x 2. The writte exam will take place i the V-buildig o May 29.