ia2sp10e1

B 10 bonus points prove that f n has a subsequence f

Info iconThis preview shows pages 3–4. Sign up to view the full content.

View Full Document Right Arrow Icon
(b) ? (10 bonus points) Prove that { f n } has a subsequence { f n k } converging pointwise to some f : R R such that for every closed and bounded interval [ a,b ] the sequence of restrictions { ( f n k ) | [ a,b ] } converges uniformly to f | [ a,b ] . Solution. For this part, one needed to assume that the sequence was uniformly bounded. I’ll give credit for pointing this out. Even more credit for realizing that it was needed, adding it as a hypothesis, and proving the result. I assume from now on equicontinuity and uniform boundedness. This time Arzela-Ascoli only does 70% or so of the job. I’ll merely sketch a proof, but filling in the details should be easy (in a perfect world?) By Arzela Ascoli, the sequence has a subsequence converging uniformly to some continuous g 1 on the interval [ - 1 , 1]. This subsequence has in turn a subsequence converging uniformly to some g 2 in [ - 2 , 2]. Necessarily, the restriction of g 2 to [ - 1 , 1] must be g 1 . The subsequence of the subsequence now has in turn a subsequence converging uniformly to some continuous g 3 on the interval [ - 3 , 3 , ]. And so forth. The functions g m : [ - m,m ] to R satisfy g m | [ - k,k ] = g k if k m and this allows us to define f : R R by f ( x ) = g m ( x ) if x [ - m,m ]; f is then clearly (?) continuous. Returning to our subsequence of a subsequence of a subsequence. .., the diagonal subsequence; that is the sequence whose n -th element is the n -th element of the n -th subsequence, is a subsequence of the original subsequence converging uniformly to g m ; i.e., to f , on every interval [ - m,m ]. Since every [ a,b ] is contained in some (and many) interval [ - m,m ], this diagonal subsequence converges uniformly to f on all closed and bounded intervals. 5. (20 points) Let [ a,b ] be a closed and bounded interval and assume f : [ a,b ] R is continuous and such that Z b a x 3 k f ( x ) dx = 0 for k = 0 , 1 , 2 ,... . Prove: f ( x ) = 0 for all x [ a,b ]. Solution. It is immediate that if p ( x ) = n k =0 a k x 3 k , then Z b a p ( x ) f ( x ) dx = 0. Let A be the set of all polynomials of this type, restricted to the interval [ a,b ]. That is, g ∈ A iff g = p | [ a,b ] where p is a polynomials involving only powers that 3
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
are multiples of 3. It is immediate that A is a subalgebra of C ([ a,b ]). It separates points; if x 6 = y , then x 3 6 = y 3 . Since 1 ∈ A , for every x [ a,b ] there is g ∈ A with g ( x ) 6 = 0. Thus A satisfies all hypotheses of Stone-Weierstrass, hence is dense in C ([ a,b ]). In particular, there is a sequence { g n } in C ([ a,b ]) converging uniformly to f . It is then very easy, immediate perhaps, that g n f converges uniformly to f 2 . We then have Z b a f ( x ) 2 dx = lim n →∞ Z b a g n ( x ) f ( x ) dx = 0 . (Certain steps that were detailed in the solution of a homework exercise were skipped). Since f 2 0 and continuous, we conclude that f = 0. 4
Background image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page3 / 4

b 10 bonus points Prove that f n has a subsequence f n k...

This preview shows document pages 3 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online