Math 104: Homework 7 solutions

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1 Math 04: Homework 7 solutions. (a) The derivative of f () = is f () = 2 which is unbounded as 0. Since f () is continuous on [0, ], it is uniformly continous on this interval by Theorem 9.2. Hence for all ɛ > 0, there eists δ > 0 such that for all, y [0, ] and y < δ, then f () f (y) < ɛ. Since this property is still satisfied if and y are chosen from (0, ], then f () is uniformly continuous on (0, ] also. (b) Choose a ɛ > 0, and consider, y [, ). Then f () f (y) = y = ( + y y) + y = y + y y 2 where the final line makes use of the inequality for. Hence for y < δ where δ = 2ɛ, then f () f (y) < ɛ and thus f is uniformly continuous on [, ). 2. (a) Choose ɛ > 0. Then since g is uniformly continuous, there eists a δ > 0, such that for all a, b R with a b < δ, g(a) g(b) < ɛ. Similarly, since f is uniformly continuous, there eists κ such that for all, y S with y < κ, f () f (y) < δ. Hence, by equating a = f () and b = g(y), it can be seen that for all y < κ, where, y S, g( f ()) g( f (y)) < ɛ and thus g f is uniformly continuous.

2 (b) Choose ɛ > 0. Then there eists δ > 0 such that for all, y S where y < δ, then f () f (y) < ɛ 2 and there eists δ 2 > 0 such that for all, y S where y < δ 2, then g() g(y) < ɛ 2. Now consider any, y S satisfying y < δ, where δ = min{δ, δ 2 }. Then ( f + g)() ( f + g)(y) = ( f () f (y)) (g() g(y)) f () f (y) + g() g(y) < ɛ 2 + ɛ 2 = ɛ. Hence f + g is uniformly continuous on S. (c) Consider f () = on R. Then for any ɛ > 0, it can be seen that for any, y R satisfying y < δ where δ = ɛ, then f () f (y) < ɛ. Hence f is uniformly continuous on R. Now consider f () = and g() =. Then the multiplication is h() = f () g() = 2. To show that h is not continuous, pick ɛ =, and consider any δ > 0. If = δ + 2 δ and y = δ, then y = 2 δ but ( h() h(y) = δ + δ ) 2 2 = + δ2 4 > Hence there does not eist a δ > 0 such that for all y < δ, h() h(y) <, so h is not continous on R. 3. (a) Figure shows a graph of the function f () = ( ) ( 2) 2. (b) Consider a sequence (a n ) with terms in (2, 3) which converges to 2. Then any term can be written at 2 + λ for some λ (0, ), and f (2 + λ) = δ 2 (( + λ)λ 2 > 2λ 2 Consider any M > 0. Then there eists an N N such that a n < 2 + / 2M for all n > N. Then f (a n ) > M for all n > N, and thus lim 2 + f () =. Similar arguments show that lim 2 = lim + = lim f () =

3 f () Figure : A graph of the function f () = ( ) ( 2) 2. (c) By Theorem 20.0, a limit at a point is well defined if and only if the positive and negative limits are equal. Hence lim 2 f () =, and lim f () is undefined. 4. (a) Suppose that f () f 2 () for all (a, b), but L > L 2. Then L = L 2 + for some > 0. Now consider the sequence c n = a + (b a)/(2n) which converges to a. Since lim a + f () = L, there eists an N such that n > N implies f (c n ) L < 2 and hence f (c n ) L > /2, so that f (c n ) > L /2. Similarly there eists an N 2 such that n > N 2 implies f 2 (c n ) L 2 < 2 and hence f 2 (c n ) L 2 < /2, so that f 2 (c n ) < L 2 + /2. Bu since = L L 2, then L /2 = L 2 + /2, and hence f (c n ) < f 2 (c n ) which is a contradiction. Hence L L 2. (b) Consider f () = 0, and f 2 () =. Then for all (0, ), f () < f 2 (). However lim 0 + f () = 0 and lim 0 + f 2 () = 0, so L = L 2.

4 5. (a) At =, the series becomes a n. In order for this series to converge, then lim a n = 0. However, if the sequence (a n ) has infinitely many non-zero integers, then there does not eist an N such that n > N implies a n 0 < /2. Hence the series does not converge at =, so the radius of convergence must be less than or equal to. (b) Suppose that lim sup a n = a > 0. Then there eist infinitely many terms a nk such that a nk > a/2. Now consider the sequence with terms a n /n. This has a subsequence a nk, which satisfies ( a nk /n k a ) /nk > 2 and as n k, a nk /n k. Hence, lim sup an /n. 6. (a) For a fied value of [0, ), lim n n = lim n n = 0. and hence the sequence of functions converges pointwise to f () = 0. (b) Choose ɛ > 0. Then let N = ɛ. If n > N, then for all [0, ], f n () f () = n n < ɛ. and hence f n f uniformly on [0, ]. (c) Pick ɛ =. To prove that f n does not tend to f uniformly on [0, ), it must be shown that there does not eist an N such that n > N implies f n () f () < for all [0, ). However, for any n, if = n, then f n () f () = n n 0 =. Hence for all n there eists an such that f n () f (), so f n does not converge uniformly to f. 7. (a) For = 0, lim f 0 n() = lim n n = 0. For a fied value of (0, ), lim f n() = lim n n n + n 2 = lim n Hence on the interval [0, ), f n converges pointwise to { 0 if = 0, f () = if > 0. n + = 2 2 =.

5 y y = f () y = f 3 () y = f 6 () y = f 0 () y = f 20 () y = f 50 () y = f () Figure 2: Graphs of the function f n () = n/( + n 2 ) for several values of n, as well as its pointwise limit f () = 2. (b) To show that f n does not converge to f uniformly on [0, ], consider = n for f n : f n () f () = + n n = n n n + The fraction n/(n + ) is smaller than for all n N. Hence for n 2, and = /n, f n () f () >. Hence there does not eist an N N such that n > N implies f n () f () < for all [0, ]. (c) For [, ), f n () f () = n + n 2 = ( + n 2 ) < n. Thus for any ɛ > 0, if N = ɛ, then n > N implies that f n () f () < ɛ for all [, ), and hence f n f uniformly on this interval.

6 8. Suppose that f (I) is open for any open interval I, but that f is not monotonic. Then there would eist some interval [a, b] over which it is non-monotonic. Suppose f (a) = f (b). Then if it is non-monotonic, it is non-constant so there eists an (a, b) such that f () = f (a). Consider f ([a, b]): by Corollary 8.3, the set must be an interval, and by Theorem 8. it must be bounded and attain its bounds, thus being some closed interval [c, d]. If f () > f (a), then d > f (a), in which case f ((a, b)) must still contain d, and thus this set is not open since d is non an interior point. If f () < f (b), then c < f (a), in which case f ((a, b) must still contain c, and thus this set is not open. Either possibility leads to a contradiction. Now suppose that f (a) < f (b). Then, by the argument above, f ([a, b]) must be a closed interval [c, d]. If d > f (b), then d is still in f ((a, b)), and thus this set is not open. Hence, since d f (b), then the only remaining possibility is d = f (b), and hence f () f (b) for all [a, b]. Similarly, it can be shown that c = f (a), and hence f (a) f () for all [a, b]. Now consider the interval [a, ]. If f (a) = f (), this leads to a contradiction following the same argument above. Hence f (a) < f (), and thus, by applying the argument above, f (y) f () for all y [a, b]. Since and y are chosen arbitrarily, it has been shown that for all, y [a, b] where y, then f () f (y), showing that the function non-decreasing, and hence monotonic. If f (a) > f (b), the above arguments can be applied to f, to show that f is non-increasing, and hence monotonic also. Either possibility violates the original assumption the f is non-monotonic. All possibilities lead to a contradiction, so the original assumption must be false. Hence if f (I) is open for any open interval I, then f is monotonic.