Iowa State University. Instructor: Alex Roitershtein Summer Homework #5. Solutions

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1 Math 50 Iowa State University Introduction to Real Analysis Department of Mathematics Instructor: Alex Roitershtein Summer 205 Homework #5 Solutions. Let α and c be real numbers, c > 0, and f is defined on [, ] by { x fx = α sinx c if x 0 0 if x = 0. Find all the values of the parameters α and c such that: a f is continuous. b f 0 exists. c f is continuous. d f 0 exists. Solution: First, notice that the only problematic point is 0. Everywhere else, the function is differentiable infinitely many times. Thus: a Since whenever lim x 0 fx exists we have lim fx = lim x 0 x 0 x α+c sinxc x c then lim x 0 fx = 0 if and only if α > c. = lim x 0 x α+c, b We have: f fx 0 = lim x 0 x = lim x α +c sinxc = lim x α +c, x 0 x c x 0 and hence f 0 exists if and only if α c. In fact, { 0 if α > c f 0 = if α = c. c Away from zero we have: f x = αx α sinx c + cx α+c cosx c = x α +c α sinxc + c cosx c, 2 x c and hence { 0 if α > c lim f x = x 0 α + c = if α = c and c 0. It then follows from that f is continuous if and only if either α > c or α = c and c 0.

2 d It follows from and 2 that when α > c, f x = x α 2+c α sinxc + c cosx c x c exists if and only if α 2 c. When α = c and c 0, f [ sinx c x = lim α + c ] cosx c x 0 x x c = use the L hospital rule lim x 0 x 2c α 6 c 2 exists if and only if c /2 note that α 6 + c 2 = c 6 + c 2 = +2c Let a R be a given real number. Suppose that a function f : R R satisfies and Prove that f a does not exists. f x < 0 < f x if x < a, f x > 0 > f x if x > a. Hint: To see informally the reason why f a doesn t exist, draw the graph of, for example, { x a 2 if x < a, fx = if x a. +x a The monotonicity of fx on a, and, a dictates f a = 0 if the derivative exists. But f a = 0 means that a is a local maximum of f, and hence f 0 < 0. This is impossible since f is decreasing and positive on x < a. To make a formal proof from this argument, use the mean value theorem fx fa = f y x a x and the fact that the derivative is monotone on each side of a, to deduce that if f a exists then f a = lim f y x = sup f x = inf f x = 0. x a x>a x<a Use then the mean value theorem for the derivative f x f a x a assertion f a = 0 contradicts the assumptions. = f z x to check that the Solution: Since 0 < f x if x < a, f x is increasing for x < a. Since 0 > f x if x > a, 2

3 f x is decreasing for x > a. To get a contradiction, assume that f a exists and is equal to b. Then fx fa b = lim. x x a By the mean value theorem, for any x 0, we have fx fa x a = f y x, for some y x within the interval containing a and x as its endpoints. Since f x is monotone and bounded on the both half-lines x > a as well as x < a, this implies that f a = 0 = sup f x = inf f x. x<a x>a But then, the mean value theorem implies that for any x > a there exists z x a, x such that f z x = f x f a x a = f x x a > 0, in contradiction to the assumption that f z < 0 for any z > a. The calculation formally confirms the heuristic suggested by the picture see the Hint that the problem with the existence of the derivative is caused by the assumption that f is concave for x > a. 3. a Let gy = arcsin y 3. Compute g y for y 0,. b Suppose that f : R R is differentiable and that there exists n N such that ftx = t n fx for any t > 0 and x R. Show that xf x = nfx for all x R. Solution: a gy = arcsin y 3 and hence by the definition of the arcsin function, gy [ π 2, π 2 ]. In calculations below we occassioanllly use the fact that this imples cos gy 0. First method: Use a table of derivatives or the inverse function theorem to verify that The chain rule therefore implies: arcsin x = arcsin y 3 = y 3 x 2. y3 2 = 3y2 y 6. 3

4 Second method: Observe that sin gy = y 3. Differentiate this identity to obtain Conclude that g y = g y cos gy = 3y 2. 3y2 cos gy = 3y 2 sin 2 gy = 3y 2 y6.. Third method: First, observe that sin gy = y 3 and hence 3 sin gy = y. This implies that gy = f y, where fx = 3 sin x. Next, by the chain rule, f x = 3 sin x 2/3 cos x = 3 sin x 2/3 sin 2 x. Thus, if fx = y = 3 sin x, then f x = 3 sin x 2/3 sin 2 x = 3 y 2 y 6. It follows from the inverse function theorem, that g y = f y = f x = 3y2 y 6. b Differentiate both sides of the identity ftx = t n fx with respect to t. This yields: Now plug in t = into the last identity. xf tx = nt n f x. 4. Solve Exercise 80 in Chapter 2 of the textbook. Solution: a Suppose that the graph G f subsets of M R. Say, = {p, y M R : y = fp} is a union of two clopen H = {p, y A R : y = fp} and F = {p, y B R : y = fp}, 4

5 where A and B are two disjoint subsets of M whose union is entire M. We will next show that if f is continuous, then A and B are clopen in M, and hence M is a disconnected set in contrary to the assumption. To this end, observe that if p n A and lim n p n = p M, then, since f is continuous, lim n fp n = fp. Since p n, fp n H and H is a closed set, p, fp H. Hence p A. Thus A is a closed set. Similarly, B is closed. In particular, no point of A is a cluster point of B. Thus all points of A are its interior points, and hence A is open. b The result is actually true if M R. Therefore, to construct a counterexample, one has to use higher-dimensional domains. Consider, for instance, f : R 2 R defined as follows: { y if x 0 and y 0, fx, y = 0 otherwise. The graph of this function is connected, but the function is discontinuous, for example, at 0,. Indeed, f0, = but f ε, = 0 for any ε > 0. Notice that two parts of the graph corresponding, respectively, to the arguments in the first quadrant and the rest of R 2 meet at the origin, and hence the graph is connected. We remark in passing that it is not hard to show that if f : M R is continuous and G f is connected, then M must be connected. c Suppose that M is path-connected. Take any two points p, q M. Let h : [a, b] M be a path connecting p M and q M. That is, h is continuous, p = ha, and q = fb. Define g : [a, b] M by setting gx = f hx and F : [a, b] M R by setting F x = hx, gx. Then F : [a, b] M R is a path which connects p, fp and q, fq in the graph of the functions f. Since p, fp and q, fq are arbitrary points in the graph, the graph is path-connected. d Recall that a function F x = F x, F 2 x : [a, b] M R is continuous if and only if both its components F : [a, b] M and F 2 : [a, b] R are continuous. Thus, if F is a path joining p, fp and q, fq in the graph of a continuous function f : M R, then F : [a, b] R is a path joining p and q in M. Therefore, M is a path-connected set if the graph of f is path-connected. However, f doesn t have to be continuous if both M and G f are path-connected. To show this, one may exploit the same counterexample as in part b. 5. Consider a function f : R R. Suppose that i f is continuous for x 0. ii f x exists for x > 0. iii f0 = 0. 5

6 iv f is monotonically increasing. Let Prove that g is monotonically increasing. gx = fx x, x > 0. Solution: We have Hence, it suffices to show that g x = xf x fx x 2. xf x fx 0 for all x > 0. 3 To this end, observe that by the mean value, for any x R there exists y x 0, x such that Hence f y x = fx f0 x 0 = fx x xf x fx = x f x f y x, and 3 follows from the monotonicity of the derivative condition iv in the statement of the problem. 6. Let X be the set of all bounded real-valued functions on a non-empty set S. a For x, y X set dx, y = sup xt yt. Show that d is a metric on X. b For x X, let fx = inf xt and gx = sup xt. Show that f and g are uniformly continuous functions from X, d to R. Solution: a Notice that, since functions in X are bounded, dx, y < + for any x, y X. Furthermore, dx, y 0 and dx, y = 0 if and only if xt = yt for all t S. Clearly, dx, y = dy, x. Finally, for any t S, and hence xt zt xt yt + yt zt dx, y + dy, z, dx, z = sup xt zt dx, y + dy, z. 6

7 b First, observe that for any x, y X we have [ ] inf xt = inf xt yt +yt inf [xt yt] + inf yt. 4 Indeed, for any s S, inf[xt yt] + inf yt [xs ys] + ys = xs, showing that inf [xt yt] + inf yt is a lower bound for {xt : t S}. It follows from 4 that Similarly, fx fy = inf xt inf = sup yt inf[xt yt] [yt xt] sup yt xt. fy fx sup yt xt. Combining these two inequalities, we obtain fy fx sup yt xt = dx, y. Thus, for any ε > 0, fy fx ε whenever dx, y ε. Since gx = f x, the claim about g follows directly from its counterpart for f. 7