ia2sp10fes

# Solution the new series can be written in the form z

• Notes
• 8

This preview shows pages 1–3. Sign up to view the full content.

Solution. The new series can be written in the form z X n =0 a n ( z 2 ) n . It clearly converges at z if and only if the original series converges at z 2 . In particular, it converges for | z 2 | < 2; i.e., for | z | < 2, diverges for | z 2 | > 2; i.e., for | z | > 2. The radius is 2. 3. Let { a n } be a sequence of positive numbers such that n =1 a n < . Prove: There exists a sequence { c n } of positive numbers diverging to such that n =1 a n c n < . Solution. This is too nice an exercise to be spoiled already by a solution. Only two people selected it in the final and I am afraid that their “solutions” are totally wrong. Very wrong. Fundamentally wrong. Everybody still has a chance to work on it.

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

4. Let F be an equicontinuous set of real valued functions on the interval [ a, b ]. That is, each element of F is a function from [ a, b ] to R and for every ² > 0 there is δ such that if x, y [ a, b ] and | x - y | < δ , then | f ( x ) - f ( y ) | < ² . Assume -∞ < a < b < . Prove: If the sequence of real numbers { f ( a ) : f ∈ F} is bounded, then the set F is uniformly bounded. Solution. I did this in class! (Maybe slurring over some details.) The hypothesis is that F is equicontinuous and { f ( a ) : f ∈ F} is bounded; let M > 0 be such that | f ( a ) | ≤ M for all f ∈ F . Let δ be the δ that works for ² = 1 in the definition of equicontinuity; that is, let δ > 0 be such that if x, y [ a, b ], then | f ( x ) - f ( y ) | < 1. Let n N be such that ( b - a ) /n < δ and let x 0 = a, x 1 = a + b - a n , . . . , x k = a + k b - a n , . . . , x n = x k = a + n b - a n = b. Notice x k - x k - 1 = ( b - a ) /n < δ for k = 1 , 2 , . . . , n . Let x [ a, b ], f ∈ F . There is k ∈ { 1 , . . . , n } such that x [ x k - 1 , x k ], then | x - x k - 1 | < δ and we have: | f ( x ) | ≤ | f ( x ) - f ( a ) | + | f ( a ) | = | f ( a ) | + fl fl fl fl fl fl f ( x ) - f ( x k - 1 ) + k - 1 X j =1 [ f ( x j ) - f ( x j - 1 )] fl fl fl fl fl fl . If k = 1, we interpret 0 j =1 as 0; the sum does not appear. The sum, of course, telescopes to f ( x k - 1 ) - f ( x 0 ) = f ( x k - 1 ) - f ( a ). Thus | f ( x ) | ≤ | f ( a ) | + | f ( x ) - f ( x k - 1 ) | + k - 1 X j =1 | f ( x j ) - f ( x j - 1 ) | ≤ M + 1 + k - 1 X j =1 1 = M + k M + n. Since x [ a, b ] is arbitrary, we proved | f ( x ) | ≤ M + n for all x [ a, b ], f ∈ F . The sequence is uniformly bounded my M + n . 5. If f, g : R R define f g if and only if f = g a.e. Prove: is an equivalence relation. Solution. We have to see f f , f g g f and f g, g h f h , for all functions f, g, h : R R . Since the empty set is a null set, it is clear that f f for all f : R R . Assume f, g : R R and f g . Then the set E = { x : f ( x ) 6 = g ( x ) } is a null set; this is of course the same set as { x : g ( x ) 6 = f ( x ) } , thus g = f a.e. and g f . Alternatively, the definition of f = g a.e. is symmetric in f, g , thus f = g a.e. if and only if g = f a.e.. In any case f g g f . Finally assume f g, g h . The sets E = { x : f ( x ) 6 = g ( x ) } , F = { x : g ( x ) 6 = h ( x ) } are then null sets. Clearly f ( x ) 6 = h ( x ) implies that either x E or x F , thus { x : f ( x ) 6 = h ( x ) E F , thus a null set. Thus f = h a,e. and f h .
This is the end of the preview. Sign up to access the rest of the document.
• Spring '11
• Speinklo
• Empty set, Open set, Topological space, Dominated convergence theorem, Lebesgue integration, Non-measurable set

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern