Math 1270 Honors Fall, 2008 Background Material on Uniform Convergence

Transcription

1 Math 27 Honors Fall, 28 Background Material on Uniform Convergence Uniform convergence is discussed in Bartle and Sherbert s book Introduction to Real Analysis, which was the tet last year for 42 and 45. This topic is in chapter 8. It is my understanding that this material was covered last spring in 45, but not in 42. But I think that everyone would bene t from reviewing it. Here I will give the basic de nitions, the main theorems that we will use, and some eamples. For proofs please look at Bartle and Sherbert ( B&S ). In referring to functions I will sometimes write f, and sometimes f (). Strictly speaking, the second notation is incorrect, because f () means the value of f at a certain number, rather than the function itself. (See chapter of B&S.) However we are often sloppy about this, especially when we wish to indicate what letter we will be using for elements of the domain of the function. Suppose that ff n g is a sequence of real valued functions de ned on the interval [; ], and that g is also such a function. We say that ff n g converges to g if, for each in [; ], n! f n () = g () : Eamples: (i) f n () = for. Then ff n ng converges to the function g () = on the entire interval [; ]. Here is a graph indicating this: y I plotted ; 2 ; 3 ; 4 ; 5 and.

2 (ii) f n () = n for. Then ff n ()g converges to the following function g : if < g () = if = : Here is the same sort of graph, again for the cases n = ; 2; 3; 4; 5; : y n What you see is that n! 2 =, and 99 n n! = ; but the rst sequence converges much faster than the second. The other thing to notice is that in this case, all of the functions f n are continuous, but the it g is not continuous. And here is another thing to look at: Z n d = n+ n + j = n + : Hence, n! R f n () d = : Now we can ask about the integral of g: Recall the graph of g: 2

3 y (Notice the single point at (; ).) Now we ask: What is R g () d? We have to be careful. Is R g () d even de ned? You may need to look back to chapter 7 of B&S. What you nd is that yes, the integral is de ned, and g () d = : We therefore have the following satisfying fact: R Z n! f n () d = Z g () d = Z f n () d n! For this sequence of functions, we can either integrate rst and then take the it, or take the it and then integrate. We get the same answer either way. 3

4 (iii) Let f n () = n 2 e n () Once again I plot several, this time for n = ; 2; 3; 5; 7; 8; 9; : y It s not so clear what is going on. From the formula () we see that f () = : Suppose > : Then n! n2 e n = n! n 2 = en n! 2n e n by L hôpital s rule, where we have considered n to be the variable. derivative of the numerator and denominator with respect to n.) (We took the But the answer is still not obvious, so we use L hôpital s rule again, and get 2 n! = ; because the numerator is constant and the term e n goes to in nity. 2 e n So,.for every. f n () = (2) n! But, what about those graphs, which seem to show the functions growing? To see what is happening, we can calculate the maimum of each of the functions f n. 4

5 We nd that the maimum occurs at = n. And, f n = n 2 n n e = n e : (3) So the maima go to in nity. Can you see how (2) and (3) are both possible? If we take a ed, then f n ()! as n!. But if we take a di erent for each n (that is, = ), then we get equation (3). n Now let s do the integral. I won t do the details. We get Z n 2 e n d = ne n e n We should remember that n! ne n = : (Or use L hôpital s rule once again,) Therefore, n! Z f n () d = 6= since the it inside the integral is. Z f n () d; n! This turns out to be bad news for di erential equations. We don t want this kind of thing happening! We will have to get around it somehow. De nition: Suppose that f is continuos on [; ]. We de ne the norm of f to be jjfjj = ma jf ()j : In making this de nition, we are using Theorem (The maimum-minimum theorem) from B&S, page 3. That is how we know this maimum eists. The theorem is being applied to jfj. De nition De nition: We say that a sequence of continuous functions ff n g on [; ] converges uniformly to the function g if jjf n gjj = : n! The de nition in B&S (8..4, pg 23) uses " and K ("). I think the de nition with norms is easier to understand. Lemma 8..8 shows that they are really the same thing. 5

6 You should try to convince yourself that the sequence in eample (i) above converges uniformly to its it function ; but that the sequences in the other two eamples do not converge uniformly. For eample, in eample (iii), f n () = n 2 e n, and g () = (since we showed that ff n ()g converges to for each, but we saw that jjf n gjj = n e ; so f n does not converge uniformly to g. In our rigorous study of di erential equations, we will need the following two Theorems. The eamples above illustrate why uniform convergence is needed in these theorems. Theorem 2 If ff n g converges uniformly to g on [; ], then g [; ]. is continuous on This is Theorem 8.2.2, pg. 234 of B&S. Theorem 3 If ff n g is a sequence of functions which are all Riemann integrable on [; ], and ff n g converges uniformly on [; ] to a function g, then g is Riemann integrable on [; ], and This is Theorem Z n! f n () d = Z g () d: 6