EXAMINATION SOLUTIONS

Transcription

1 1 Marks & (a) If f is continuous on (, ), then xf is continuously differentiable 5 on (,) and F = (xf)/x as well (as the quotient of two continuously differentiable functions with the denominator never vanishing). We have (xf) = F +xf on the one hand and (xf) = f by the fundamental theorem of analysis (which holds true here since f is continuous). So xf = f F. (b) We have (F p ) (x) = pf p 1 (x)f (x). Integration by parts gives 7 F p (x)dx = p F p 1 (x)f (x)xdx+[xf p (x)]. Now, since f is compactly supported on (,), there exists a, b such that < a < b < and f(t) = for all t a or t b. So, F(x) = for all x such that < x a and F(x) = C/x where C = f(t)dt for all x b. Therefore, lim x xf p (x) = and lim x xf p (x) = = lim x C p /x p 1 =, since p > 1. If follows that F p (x)dx = p F p 1 (x)f (x)xdx, as requested. (c) By inserting this into the previous equality, we get 3 F p (x)dx = p F p 1 (x)f(x)dx+p F p (x)dx So F p (x)dx = p F p 1 (x)f(x)dx. p 1 (d) Applying Hölder s inequality to the right hand side with exponent p on f 5 and q on F p 1 (with = 1), we get p q F p (x)dx p ( F q(p 1) (x)dx) 1/q ( f p (x)dx) 1/p p 1 But q(p 1) = p. Thus, ( F p (x)dx) 1 1 q p ( f p (x)dx) 1/p p 1 But again 1 1 = 1 and we are left with F q p p p f p 1 p. 1

2 2 Marks & (a) If f L p (,), then, f (,x) L p (,x), where (,x) denotes 3 the restriction to (,x). Now, we have: x f(t)dt f px 1/q < with = 1, thanks to Hölder s inequality. p q Therefore, F(x) exists and is finite for all x (,). (b) Take a sequence (f n ) n 1 as suggested and define 4 F n (x) = 1 x f x n(t)dt. We may assume that F n L p (,) and F n p p f p 1 n p. Now, the right hand side converges to p f p 1 p as f n f p, So, taking the lim inf of the left-hand side and the limit of the right hand side, and owing to the fact that the lim inf preserves inequalities, we have liminf n F n p p f p 1 p. (c) Since f n f a.e. in (,), thanks to Fatou s Lemma (that may be 5 applied since all considered functions are nonnegative), we have, 1 x f(t)dt = 1 x liminf x x n f n (t)dt liminf n ( 1 x f x n(t)dt). So by the monotony of the function s s p on (,), we have: 1 x x f(t)dt p liminf n 1 x x f n(t)dt p = liminf n 1 x x f n(t)dt p. (d) Applying Fatou s Lemma again, we get: liminf n ( 1 f x n(t)dt p dx 5 liminf n 1 x f x n(t)dt p dx = liminf n F n p p. x It follows that F p liminf n F n p p p 1 p 3 and the result is proved. 2

3 3 Marks & (a) If u V then u = u n = 1 (nu n n). 4 By Cauchy-Schwarz, we get u 2 ( 1 )( n 2 n2 u 2 n) unseen = K n2 u 2 n with K = 1 <. n 2 (b) a(u,v) u v, so a is continuous on V. Indeed, we know from 6 the course that continuity of a bilinear form on a normed vector space unseen is equivalent to the existence of a constant C > such that a(u,v) C u v. Here this property is satisfied with C = 1. Now, since n 1 in the sums, applying the previous results, we get: a(u,u) = n2 u 2 n 1 3 n2 u 2 n u2 n + 1 3K u2 min{ 1, 1 3 3K n= (1+n2 )u 2 n = min{ 1, 1 3 3K } u 2. Therefore, a is coercive on V. (c) Since (f v) f v f v, the mapping v V (f v) R is a 5 continuous linear form on V and its norm is less that f. We can apply Lax-Milgram s theorem and get the existence and uniqueness of u and the existence of a constant C > such that u C f. Thus, the map T is a continuous linear map from l 2 to V and therefore, 2 to h 1. (d) Since J is compact (result assumed), T = J T is compact itself. 3 Indeed, T maps bounded sets onto bounded sets (by its continuity) and J maps bounded sets to sets whose closure is compact. 3

4 4 Marks & (a) Since Tf = λf V and λ we have that f V. 5 We have λ > since T is positive. unseen u = Tf = λf satisfies a(λf,v) = (f v), v V. i.e. λ n2 f n v n = n= f nv n, v V. Rearranging, we get: f v + (λn2 1)f n v n =, v V, and since v V, we have v = v n. So finally ((λn2 1)f n +f )v n =, v V. By taking v n = v = 1 and v k = for all k, n, we form an element v in V which we can use in the above identity and finally get (λn 2 1)f n +f =, for all n 1. (b) If λ = 1 then λm 2 1 = and thus f m 2 = (apply the above 5 equation for n = m). Then, λn 2 1 for all n m unseen and thus f n =, for all n m. So, the only non-zero term of f is f m But f V so = n= f n = f m. So f = and λ is not an eigenvalue of T. (c) If λ A, then λn 2 1, n 1. Then, f n = f, n 1. 5 λn 2 1 f V if and only if = n= f n = f (1 1 ). unseen λn 2 1 We need f otherwise f =. So λ is an eigenvalue iff (4) is true. (d) The eigenspace corresponding to λ m is one-dimensional since 2 all the f n s for n 1 are uniquely determined by f. (e) {e (m) } m 1 forms a Hilbert basis of l 2 by the spectral theorem for compact 3 self-adjoint operators on Hilbert spaces. 4

5 5 Marks & (a) Take v = u k in the variational formulation. One directly gets the reque- 4 sted identity (the energy identity). Using Cauchy-Schwarz, we get unseen u k 2 (f u k ) f u k. Thus u k f. By the Banach-Alaoglu theorem owing to the fact that a Hilbert space is its own dual, there exists a subsequence (still denoted by (u k ) k N ) such that u k u. (b) We test (5) with v n = (u k ) n for n N and v n = otherwise. 6 Since v n is zero but for a finite number of terms, v h 1. This gives unseen the equality to the left. We now test with v n = n 2 (u k ) n for n N and v n = otherwise. Again, v h 1. We get ε N k n4 (u k ) n 2 + N n= n2 (u k ) n 2 = N n= n2 (u k ) n f n. We deduce using Cauchy-Schwarz: N n= n2 (u k ) n 2 ( N n= n2 (u k ) n 2 ) 1/2 ( N n= n2 f n 2 ) 1/2 which gives the inequality to the right. (c) We have (u k f u k ) = n= ((uk ) n f n )(u k ) n 6 N n= ((uk ) n f n )(u k ) n + n=n+1 ((uk ) n f n )(u k ) n ). unseen Using Part (b), we have N n= ((uk ) n f n )(u k ) n ε N k n= n2 f n 2 N 2 ε N k n= f n 2 N 2 ε k f 2. We have n=n+1 ((uk ) n f n )(u k ) n n=n+1 (uk ) n 2 + +( n=n+1 (uk ) n 2 ) 1/2 ( n=n+1 f n 2 ) 1/2. But n=n+1 (uk ) n 2 n=n+1 f n 2 as assumed. By collecting the estimates of the two sums, we get the requested inequality. (d) For given δ >, there exists N large enough such that 4 2 n=n+1 f n 2 δ. Then, (u k f u k ) ε k N 2 f 2 +δ and thus unseen lim (u k f u k ) δ, δ >, which implies lim (u k f u k ) =. 5