{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

rudin-ch7

# rudin-ch7 - Homework 10 — Chapter 7 of Rudin Revised...

This preview shows pages 1–5. Sign up to view the full content.

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

View Full Document

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Homework 10 — Chapter 7 of Rudin Revised version of solutions by Aja Johnson December 2, 2005 1 Theorem. Every uniformly convergent sequence of bounded functions is uni- formly bounded. Proof. Let (fn) be a uniformly convergent sequence of bounded functions on a space X. Pick an integer N such that if m, n 2 N, then lfn(:L') — fm < 1 for all a: E X. This N is guaranteed to exist by the Cauchy criterion for uniform convergence. By hypothesis, each f” is bounded. By deﬁnition this means that for each n E N there exists an M,, E R such that 3 M,, for all :17 E X. Thus, if n 2 N, then lfn(\$)| = lfn(\$) — .f'N(III) + fN(1L')|Slfn(\$)-—fN(w)l+|fN(w)l < 1 + MN. Let M = max{1 + MN,M1,M2,...,MN_1}. Then|fn(a:)| g M for all n E N and alleX. D 2 Theorem. If and (gn) converge uniformly on a set E, then (fn + gn) converges uniformly on E. Proof. Let 6 > 0 and write f, g for the uniform limits of (fn), (gn) respectively. Apply the deﬁnition of uniform convergence to choose N1, N2 6 N, such that if n 2 N1,N2, then — < and — < 5, respectively, for all a: E E. Take N : max{N1,N2}. Suppose n 2 N and :17 E E. Then we have lfn(:r) +gne) — (re) + mm : lfn(:r) — f(:r)| + lgn(m) — you < 3+; as desired. D Recall that if h : E —> R is a bounded function, then we write z: sup{|h(:17)| : a: E (If more than one domain is being considered, we write to avoid ambiguity, but this is not usually necessary.) This notation will be used below. It saves us from writing “for all .r in E” over and over again. Theorem. If and (gn) are sequences of bounded functions that converge uniformly on a set E, then converges uniformly on E. Proof. Let 6 > 0 and write f, g for the uniform limits of the given sequences of functions. Apply Exercise 1 to get a single constant M satisfying full 3 M and g M for all n E J. It follows from (even pointwise) convergence that ||f g M and g M as well. Now apply the uniform convergence hypothesis to ﬁnd N E J so that ||fn — f|| < 2ME+1 and ~g|| < 2ME+1 for all n 2 N. Let n 2 N. Then 2M6 llfngn - fgll S llfnllllgn ~ 9H + llgllllfn — f|| < < 6- _2M+1 By definition, this means (fngn) converges uniformly to f g as desired. D 3 Question. Construct sequences and (gn) which converge uniformly on some set E, but does not converge uniformly on E. Answer. Let fn(as) = gn(:c) = a: + % and = g(:L') : gt. Fix 6 > 0. Choose N > 1. Ifn Z N, then for all x E E, we have 6 lfn(:r) — f(:v)| = g < Thus converges uniformly to f. Likewise (gn) converges uniformly to g. It remains to show that (fngn) does not converge uniformly. If it were to converge to some function uniformly that function would be f : m2.N0te that fn(x)gn(x) = + %)2 = + 279” + Pick 6 = 1. Let N be given. Take n = a: = N. Then . 2 1 2N2 1 |fn(~’r)gn(m) — f(:r)g(w)l = + f + 71—. — w2| = > 1. Thus (fngn) does not converge uniformly. 5 Theorem. Let 1 0 A a: < n+1 Mm) = sinzg ,5? s :1: 3% O i < Then converges to a continuous function, but not uniformly. Proof. Let 6 > 0. Fix a: E R. Let N > i E N. (When a: g 0, we take N = 1.) If n 2 N, then ‘ 0| = 0 < 6. Thus for each a: 6 R, the numerical sequence ( converges to 0. Now we show that (fn) does not converge uniformly. If it were to converge 1 to some function uniformly that function would be 0. Pick 6 = 5. Let N be given. Take n 2 N and a: = Then 2 1 sin2(N + —)7r =1> 2 lumen: g Thus does not converge uniformly. D Theorem. Pointwise absolute convergence for a series of functions does not imply uniform convergence. Proof. By deﬁnition, (pointwise or uniform) convergence of the series 2 fn is equivalent to that type of convergence for its sequence of partial sums. Fix a: E R. As shown above, there is an integer N (depending on 1r) so that = 0 for all n 2 N. But this means sn(:r) = SN(:L') for all n 2 N, whence lim,H00 3,,(33) = sN(:L'), and the series Z fn(:r) converges to sN(:r) by deﬁnition. Since fn(:r) Z 0 for all n E N and all a: E E, this convergence is absolute by default. Suppose Z fn converged uniformly to some function 8. Then the sequence = (8,, — sn_1) would converge uniformly to s — s = 0. Since the ﬁrst part of this problem showed that does not converge uniformly, we conclude that the series Z fn cannot converge uniformly either. D 7 Theorem. For n E J and a: E R, put 2 Then converges uniformly to a function f, and = lim is correct ifzc 7f 0, but false n—>OO ifzv = 0. Proof. We ﬁrst show that the sequence converges uniformly to f E 0, i.e., that limn_>00 = 0. One can get a reasonable upper bound for by separately bounding fn(:r) for .v close to, and far away from, 0, but it is easier to use calculus. A direct computation shows that the only positive critical point of f” is at % and that in fact f,’, > 0 on the interval (0, %) while ﬂ, < 0 0n (%, 00). In view of the mean value theorem, this means that 0 g f S f = % for all a: Z 0. Since all these functions are odd = — for all n and all 1r), we conclude that lim : lim = O, as desired. The statements concerning limnnoo are also direct computations. D 9 Theorem. Let be a sequence of continuous functions which converges uni- formly to a function f on a set E. Then lim = for every sequence 77,—}00 of points 33,, E E such that xn —> a, and a E E. Proof. Let 6 > 0. Apply the uniform convergence hypothesis to choose N1 6 N such that llfn — f|| < for all n 2 N. Theorem 7.12 tells us f is continuous. Then by deﬁnition a 6 > 0 can be chosen such that |f — f (a)| < 6 whenever |y—a| < 6. Apply the deﬁnition of convergence to the numerical sequence to choose N2 6 N such that Irrn»a| < 6 whenever n 2 N2. Take N : max{N1, N2}. Suppose n 2 N. Then lfn(:rn) — f(a)| : vim) — f(:vn)l + um) — f(a)| < 3+ 3 = 6. Thus lim = f(a). D TL—>OO Question. Is the converse true? Answer. No. Let E = Z and = 1 if a: = n and = 0 otherwise. All functions from Z to R are continuous, so each f” is continuous. Also, it is clear that converges to 0 pointwise. It remains to show that does not converge uniformly. If it were to converge to some function uniformly that function would be f E 0. Pick 6 = 1 Let N be given. Taken = a: = N. Then = fN(N) = 1 > %, i. so does not converge uniformly. 11 Theorem. Suppose that and are deﬁned on E, has uniformly bounded partial sums, 9,, —> 0 uniformly on E, and Z 92 Z 93 Z for every :17 E E. Then Z fngn converges uniformly on E. Proof. This is a matter of adapting the proof of Dirictlet’s Theorem 3.42 by replacing “a, b” by “ f, g” and sprinkling in the term “uniformly”. D 15 Question. Suppose f is a real continuous function on R, = f(nt) for n = 1,2,3, . --, and is equicontinuous on [0,1]. What conclusion can you draw about f? Answer. f must be constant on the interval [0, 00). Proof. The condition is sufﬁcient because the equicontinuity hypothesis has no bearing on the values of f for a: < 0. (This was pointed out by Ms. Ulrich.) To establish necessity, ﬁx t > 0. Let 6 > 0, and take 6 > 0 to be the corresponding number guaranteed by equicontinuity. Choose n > %. Then % < 6 so |f(t) — f(0)| = — f(0)| < e. The arbitrariness of 6 means f(t) = f(0) as desired. D 16 Theorem. Suppose that is an equicontinuous sequence of functions on a compact set K, and converges pointivise on K. Then converges uniformly on K. The following argument is similar to the last few paragraphs in the proof of the Arzela-Ascoli Theorem 7.25. One can also use the result of that theorem to establish the assertion of this problem. Proof. (outline) Write f for the pointwise limit of (fn). Use the hypotheses in conjunction with the inequality lf(\$) - f(y)l S lf(\$) - fn(:v)| + lfn(:v) — fn(y)l + |fn(y) - f(y)l to establish uniform continuity of f. Now, let 6 > 0 and take 6 > 0 to be the corresponding number guaranteed by equicontinuity. Temporarily ﬁx a E K 7 and apply the inequality lfn(\$) - f(\$)l S lfn(\$) — fn(a)| + |fn(a) — f(a)| + |f(a) — f(\$)l to ﬁnd a positive integer Ma such that — < 6 whenever n 2 N and a: 6 N5 (a). Finally, apply compactness to ﬁnd ﬁnitely many points a1,--- ,am so that the neighborhoods N5(a1),--- ,N5(am) cover K. Take M : max{Mai} and check that — < e for all n 2 M and all a: E K. D 18 Theorem. Let be a uniformly bounded sequence of functions which are Riemann-integrable on [a, b], and put = fn(t) dt for a g a: g h. Then there emists a subsequence which converges uniformly on [a, b]. Proof. Choose a number M so that g M for each n E J. For each :17 E [a, b] and n E J, we have 3 g - a)M 3(1) — a)M, so the are uniformly (and hence pointwise) bounded. Moreover7 for each :3, y E [a, b], and n E J 7 we have 9 an(y) ~ Fn(m)l = I/ fnl s Mly —x|, establishing equicontinuity of the family { The proof is completed by ap- pealing to the Arzela—Ascoli Theorem 7.25. D ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 5

rudin-ch7 - Homework 10 — Chapter 7 of Rudin Revised...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online