Math 316, Intro to Analysis Final project. of of functions. YOUR NAME GOES HERE. I HAVE ADDED SOME PROMPTS IN CAPS. THESE SHOULD NOT APPEAR IN YOUR FINAL WRITEUP. INSTEAD REPLACE THEM WITH SOME ACTUAL ANAL- YSIS. FEEL FREE TO ALSO CORECT ANY TYPOS YOU FIND. You have proven that if you improve the concept of convergence to uniform convergence, then the limit of a of continuous functions is continuous. This project will have you prove that instead of powering up the notion of convergence, you can improve the notion of continuity and get the limit to be continuous.

Definition 1. A sequence of functions (fn : U → R) is called equicontinuous if for all  > 0 and all x ∈ U there is a δ > 0 such that for all n ∈ N and all y ∈ U if |x − y| < δ then |fn(x) − fn(y)| < . Basically, this definition says that we may allow δ to depend on x, but δ cannot depend on n. The same choice of δ must be used for each fn, independent of n.

Claim 2. Let (fn : R → R) be given by  0 if x < n f (x) = n x − n if n ≤ x

Then (fn) is equicontinuous on R. Proof. YOUR PROOF GOES HERE. YOUR PROOF MIGHT START WITH “CONSIDER ANY  > 0 AND ANY x ∈ U. LET δ =??? AND CONSIDER ANY n ∈ N.”  The whole point of this packet is that everything that happens for uniformly convergent sequences of continuous functions holds for pointwise convergent equicontinuous sequences of functions.

Theorem 3. Let (fn : U → R) be an equicontinuous sequence of . Let f : U → R be its pointwise limit. Then f : U → R is continuous. Proof. YOUR PROOF GOES HERE. THINK ABOUT HOW WE PROVED THAT THE UNIFORM OF CONTINUOUS FUNCTIONS IS CONTINUOUS.  We can add the prefix “equi-” to other notions of continuity:

Definition 4. A sequence of functions (fn : U → R) is called uniformly equicontinuous if for all  > 0 there is a δ such that for all n ∈ N and all x, y ∈ U if |x − y| < δ then |fn(x) − fn(y)| < . Importantly, we get good results out of this:

Theorem 5. Let (fn : U → R) be an uniformly equicontinuous sequence of functions. Let f : U → R be its pointwise limit. Then f is uniformly continuous. Proof. YOUR PROOF GOES HERE. MAKE SURE THAT YOUR DELTA DEPENDS ONLY ON EPSILON.  1 2

An important payoff of the fact that uniform convergence preserves continuity was the continuity of power . The notion of equicontinuity also recovers the continuity of .

Theorem 6. Let a0, a1, a2,... be a sequence of real numbers. Let 1/k λ = lim sup |ak| k→∞ and   1/λ if 0 < λ < ∞ R = ∞ if λ = 0  0 if λ = ∞ n X k (1) Then for every L with 0 < L < R, the sequence of partial sums Sn(x) = ak · x is k=0 uniformly equicontinuous on [−L, L], ∞ X k (2) ak · x diverges if |x| > R, k=0 (3) The sequence of partial sum Sn(x) converges pointwise on (−R,R). Using the preceding theorems on this page, we may prove the following corollary: ∞ X k Corollary 7. Let S(x) = ak · x be a power series with R. k=0 • If 0 < R < ∞ then (−R,R) ⊆ Dom(f) ⊆ [−R,R] and f is continuous on (−R,R). • If R = ∞ then S has domain R and S in continuous on all of R. ∞ X k • If R = 0 then ak · x has domain {0} and f(0) = a0 on that domain. k=0 Proof of Corollary 7. YOUR PROOF GOES HERE CONSIDER LOOKING AT THE NOTES ON POWER SERIES FOR INSPIRATION.  Proof of Theorem 6. We begin our proof of Claim (1) with the following lemma. 1/k Lemma 8. Let (ak) be a sequence of real numbers. Let λ = lim sup ak . Then for every 1 number L with 0 < L < λ , ∞ X k−1 |ak| · k · L k=0 converges. Proof of Lemma 8. YOUR PROOF GOES HERE CONSIDER TAKING ADVANTAGE OF THE ROOT TEST, JUST AS WE DID IN THE NOTE PACKET ABOUT POWER SE- RIES.  We now prove Claim (1) of Theorem 6. YOUR PROOF GOES HERE, MAYBE TAKE ADVANTAGE OF LEMMA 8, IF YOU DON’T USE THE LEMMA, THEN YOU DON’T NEED TO INCLUDE IT IN YOUR WRITEUP. Next we prove Claim (2) of Theorem 6. YOUR PROOF GOES HERE. THIS ONE IS EASIER. NOTICE THAT THIS CLAIM IS IDENTICAL TO ONE MADE IN THE NOTES ON POWER SERIES. 3

Finally we prove Claim (3) of Theorem 6. YOUR PROOF GOES HERE. THIS ONE IS EASIER THAN THE ANALOGOUS RE- SULT FROM THE NOTES ON POWER SERIES, AS YOU ONLY NEED .