This exercise appeared in the last analysis qualifier

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

View Full Document Right Arrow Icon
(This exercise appeared in the last analysis qualifier. You have the advantage over the people who took the exam of knowing that it has to do with something we covered recently.) Solution. Claim that the sequence { f n } is equicontinuous and bounded. In fact, given x, y [0 , 1] we have that | f n ( x ) - f n ( y ) | ≤ | x - y | by the Mean Value Theorem, thus given ² > 0 we can take δ = ² to conclude that | f n ( x ) - f n ( y ) | < ² for all x, y [0 , 1] with | x - y | < δ , n = 1 , 2 , 3 , . . . . This establishes equicontinuity. For boundedness, we can use that f n ( x ) = f n (0) + Z x 0 f 0 n ( t ) dt, 3
Image of page 3

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

View Full Document Right Arrow Icon
hence | f n ( x ) | ≤ | f n (0) | + Z x 0 | f 0 n ( t ) | dt 1 + Z 0 x dt = 1 + x 2 for all n N , x [0 , 1]. The claim is established. By the Theorem of Arzela-Ascoli, it follows that { f n } has a uniformly convergent subsequence { f n k } . Since this sequence also converges pointwise, the limit must be f . As the uniform limit of a sequence of continuous functions, f must be continuous. The rest of the exercise is a bit ambiguous. I think I copied it from a non final version of the qualifier without checking too carefully. “Must the sequence converge?” can be interpreted in at least two ways: (a) Given all the other hypothesis, is the hypothesis lim n →∞ f n ( x ) = f ( x ) for all x [0 , 1] redundant; is it needed? Interpreted in this way the answer is no; the sequence does not have to converge; counterexamples provided upon request. (b) Must the full sequence converge uniformly ? The answer in this case is yes, and I made this an exercise Homework 5. In either case, the answer to the last question is yes. I will only grade your proof of the continuity of f (or lack thereof). 4. Let [ a, b ] be a closed and bounded interval and let E ⊂ C ([ a, b ]). Assume that E is equicontinuous and bounded. Define f : [ a, b ] R by: if a x b , then f ( x ) = sup { g ( x ) : g ∈ E} . Prove: f is continuous. Solution. Because E is bounded, sup { g ( x ) : g ∈ E} < for all x [ a, b ]; f is a well defined, real valued, function on [ a, b ]. By equicontinuity, given ² > 0 there is δ > 0 such that if x, y [ a, b ] satisfy | x - y | < δ , then | g ( x ) - g ( y ) | < ²/ 2 for all for all g E . Now let x, y [ a, b ] and assume | x - y | < δ . By definition of f ( x ) as the sup of the set of all values g ( x ), g ∈ E , there is g ∈ E such that f ( x ) - ² 2 < g ( x ) , thus f ( x ) < g ( x ) + ² / 2 .
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern