Math 320: Real Analysis MWF pm, Campion Hall 302 Homework 8 Solutions Please write neatly, and in complete sentences when possible. Do the following problems from the book: 4.3.5, 4.3.7, 4.3.8, 4.3.9, 4.4., 4.4.2, 4.4.9, 4.5.3 Problem G. Use the Intermediate Value Theorem to show that every polynomial of odd degree has a root in R. (Recall that a polynomial p(x) = a n x n +... + a 0, with a n 0, has degree n.) Problem H. () Give an example of a continuous function on (, ) that is not bounded. (2) Give an example of a uniformly continuous function on (, ) that is not bounded. (3) Suppose that f is uniformly continuous on (, ). Prove that f is bounded. Problem I. Let t(x) be defined by if x = p Q in lowest terms. q q t(x) = 0 if x R \ Q. () Show that t is not continuous at x Q. (2) Show that t is continuous at x R \ Q. Solution 4.3.5. Suppose g is continuous at x = c and g(c) 0. Let ɛ = g(c). By continuity, there is a δ > 0 so that x c < δ implies that g(x) g(c) < ɛ. For such x, by the triangle inequality, g(x) g(c) g(c) g(x), so that g(c) g(x) < ɛ. Rearranging, we find g(x) > g(c) ɛ = 0, so that g(x) 0 for any x c < δ. This implies that f/g is defined on V δ (c), an open interval containing c. Solution 4.3.7. Let c be a limit point of K, so that there exists a sequence {x n } K \ {c} converging to c. By continuity of h, we have h(c) = lim n h(x n ). Since x n K, we have h(x n ) = 0 for each n, so that h(c) = 0, and c K. Since K contains its limit points, K is closed.
2 Solution 4.3.8(a). Suppose f : R R is a continuous function with f(x) = 0 for all x Q. Given c R, since Q is dense in R, there exists a sequence {r n } Q converging to c. By continuity, f(c) = lim n f(r n ) = 0. Thus f is identically 0 on R. Solution 4.3.8(b). If f and g are not both assumed to be continuous, then f and g may be distinct functions of R, even if f and g are equal on Q. For example, we may have f(x) = 0 for all x, and g equal to Dirichlet s function (called g as well on p. 00). On the other hand, if f and g are both continuous, then so is f g. The latter is a continuous function that vanishes on Q by assumption, so that it is identically zero by part (a). Solution 4.3.9(a). Given ɛ > 0 and x R, let δ = ɛ/c. Then, for any x y < δ, we have f(x) f(y) c x y < c δ = ɛ. (Note that we ve actually proved uniform continuity of f on R). Solution 4.3.9(b). Let C = y 2 y. We claim first that y n+ y n c n C. This is easily seen by induction: y n+ y n = f(y n ) f(y n ) c y n y n, and the latter is less than or equal to c c n C = c n+ C by the inductive hypothesis. Notice now that this allows us to estimate y n y for any n: n n y n y = (y k+ y k ) y k+ y k k= k= n n c k C = C c n. k= The last sum can be computed explicitly using the formula for a geometric series, and we obtain k= y n y C cn c C c.
(Note that we have used 0 < c < in the last inequality, since we needed that c n is positive, i.e. c n <.) Finally, we observe that f k (y l ) = y k+l, for any k, l N. m > n, we have y m y n = f m (y ) f n (y ) = f n (y m n ) f n (y ) c n y m n y c n C c. 3 Thus, for Since 0 < c <, we have lim n c n = 0 (see example 2.5.3 on p. 56). Thus, for any ɛ > 0, there is an N so that for all n N we have c n < ɛ( c)/c. For any pair m, n N with m > n N, we have y m y n c n C c < ɛ( c) C so that {y n } is a Cauchy sequence, as desired. C c = ɛ, Solution 4.3.9(c). Consider evaluating the function f along the sequence {y n } obtained above. Since f is continuous on R, we have ( ) f(y) = f lim y n = lim f(y n ) = lim y n+ = y, n n n so that y is a fixed point of f. This fixed point is unique: Suppose that y R were another fixed point of f, and note that f(y) f(y ) c y y by the contraction property of f. Since y and y are both fixed points of f, we have y y = f(y) f(y ) c y y. If y y, then y y 0, and we may cancel it from both sides above and obtain c, contradicting the assumption that c <. Thus y is the unique fixed point of f. Solution 4.3.9(d). The argument in part (b) did not rely on the value of y, so that we may perform the above argument with y = x, for any x R. We conclude that the sequence of iterates {x, f(x), f(f(x)),...} is Cauchy, and hence converges, say to y. The continuity argument in part (c) applies again, so that f(y ) = y, and y is a fixed point of f. By the unicity demonstrated in part (c), we conclude that y = y, so that the sequence of iterates {x, f(x), f(f(x)),...} approaches y, for any choice of x R.
4 Solution 4.4.(a). We ve seen in class that g(x) = x is continuous at each c R. By the Algebraic Continuity Theorem, lim f(x) = lim g(x) g(x) g(x) = g(c) g(c) g(c) = x c x c c3, so that f is continuous at c R, for each c R. Solution 4.4.(b). Let x n = n + and y n n = n for each n N, so that x n y n = /n, which converges to 0 as n. We have ( f(x n ) f(y n ) = n + 3 n n) 3 = 3n + 3 n + 3, n 3 for all n. By Theorem 4.4.6, f cannot be uniformly continuous on R. Solution 4.4.(c). Let A be a bounded subset of R, so that there is an R > 0 with x R for all x A. Note that we have x 3 y 3 = x y x 2 + xy + y 2 x y ( x 2 + x y + y 2). For x, y A, we thus have x 3 y 3 x y (R 2 + R 2 + R 2 ) = 3R 2 x y. Given ɛ > 0, let δ = ɛ. Then, for any x, y A with x y δ, we 3R 2 have f(x) f(y) = x 3 y 3 3R 2 x y < 3R 2 ɛ 3R = ɛ, 2 so that f is uniformly continuous on A. Solution 4.4.2. Note that we have x 2 y 2 = y 2 x 2 x 2 y 2 For x, y [, ), we have x + y x 2 y 2 = x y x + y x 2 y 2. = x + y x 2 y 2 = xy 2 + x 2 y 2. Thus, for any ɛ > 0, we let δ = ɛ/2. Then for x y < δ we have x x + y 2 y 2 = x y 2 x y < 2δ = ɛ, x 2 y 2 so that /x 2 is uniformly continuous on [, ).
On the other hand, consider x n = /n, while y n = /(n + ). Then we have x n y n = n n + = n 2 + n, which goes to 0 as n. Meanwhile, we have x 2 n yn 2 = n2 (n + ) 2 = 2n +. By Theorem 4.4.6, /x 2 is not uniformly continuous on (0, ]. Solution 4.4.9(a). Given ɛ > 0, let δ = ɛ/m. Then, for any x y < δ, we have f(x) f(y) M x y < M δ = ɛ, so that f is uniformly continuous on A. Solution 4.4.9(b). Not every uniformly continuous function is Lipschitz: Consider f(x) = x on [0, ]. As explained in example 4.3.8, f is continuous on [0, ]. By Theorem 4.4.8, f is uniformly continuous on [0, ]. On the other hand, f is not Lipschitz. This can be seen graphically, since the slope of the tangent line to the graph of f is equal to f (x) = /(2 x), which approaches as x 0. More precisely, suppose that f were Lipschitz, with constant M > 0. Namely, for any x, y [0, ] we would have M x y x y x y = ( x + y)( x y) =. x + y By choosing x and y distinct and very small, we will find that M for all x, y [0, ] is impossible. Namely, consider the points x+ y x n = /n 2 and y n = /(4n 2 ) in [0, ], for each n N. Then we have M + = 2n/3, n 2n for all n. Since 2n/3 as n, this is impossible, and we conclude that f is not Lipscitz on [0, ], though it is uniformly continuous. Solution 4.5.3. There is no continuous function f on R with range f(r) = Q, since the continuous image of a connected set is connected. More precisely, suppose that f was such a function. Since R satisfies the interval property, it is connected by Theorem 3.4.7. By Theorem 4.5.2 we must have that f(r) = Q is connected. On the other hand, 5
6 Q doesn t satisfy the interval property: The real number c = 2 is in [0, 0], and 0, 0 Q, but c / Q. By Theorem 3.4.7 again, Q is not connected, a contradiction. Solution G. Let p(x) = a n x n + a n x n +... + a x + a 0, with n odd and a n 0. Consider the sequence {p(k)/k n }. Since p(x) x n = a n + a n x + a n 2 x 2 +... + a x n + a 0 x n, it is evident that the sequence {p(k)/k n } approaches a n as k goes to. This means that, for k large enough, p(k)/k n a n < a n, so that p(k) a n k n < a n k n and p(k) and a n k n have the same sign. Similarly, {p( k)/( k) n } also approaches a n, so that, again for k large enough, p( k) and a n ( k) n have the same sign. Since n is odd, we have a n ( k) n = a n k n, and thus (for k large enough) p(k) and p( k) have opposite signs. In particular, 0 is between the values of p at k and k. By the Intermediate Value Theorem, there is a value c ( k, k) so that p(c) = 0, as desired. Solution H(). Let f(x) =. Then f((n )/n) = n, so that f is x not bounded on (, ). Solution H(2). Let f(x) = x. This function is evidently uniformly continuous: For ɛ > 0, let δ = ɛ. Then x y < δ implies that f(x) f(y) = x y < ɛ. On the other hand, f is also not bounded: For any M > 0, f( M ) = M, which is not bounded in absolute value by M. Solution H(3). Choose ɛ = > 0, so that by the uniform continuity of f, there is a δ > 0 so that x y < δ implies that f(x) f(y) <, for any x, y (, ). For each integer n Z, let x n = δn/2, so that x n x n = δ/2 < δ. Choose x > 0, and let N = 2x/δ. (Note that a indicates the greatest integer less than a). Then x N = δn/2 x, and x N + δ/2 = x N+ = δ(n + )/2 > x, so that x [x N, x N + δ/2). Note that this implies that x x N < δ/2. We may now use the uniform continuity
of f to deduce f(0) f(x) = f(0) f(x ) + f(x ) f(x 2 ) +... + f(x N ) f(x N ) + f(x N ) f(x) f(0) f(x ) + f(x ) f(x 2 ) +... < N +. + f(x N ) f(x N ) + f(x N ) f(x) Finally, since N < δx/2 < δ/2, we conclude that f(0) f(x) < δ 2 +. The triangle inequality now implies that f(x) δ 2 + + f(0). Thus f(x) is bounded in absolute value by δ/2 + + f(0), for all x > 0. The case x < 0 is almost identical ( N should be replaced with the least integer greater than δx/2), and we conclude that f is bounded. 7 Solution I(). Since R \ Q is dense in R, for any c Q there is a sequence {x n } R \ Q converging to c. Since f(x n ) = 0 for all n N, the limit lim n f(x n ) = 0, but on the other hand f(c) 0. Thus lim f(x n ) f(c), and f is not continuous at c. Solution I(2). For a fixed ɛ > 0, consider the set F ɛ = {x : t(x) ɛ}. Evidently, F ɛ Q. Given x F ɛ written in lowest terms x = p/q, t(x) ɛ implies that q /ɛ, so that the elements of F ɛ all have denominators bounded above by /ɛ. We show first that, for any ɛ > 0 and any interval [a, b], the intersection F ɛ [a, b] is finite. Suppose x = p/q F ɛ [a, b], and note that the above argument implies that there are only finitely many possibilities for q (namely, integers less than /ɛ). Then a p q b so that aq p bq. For each q, there are finitely many integers p in the interval [aq, bq], so that there are only finitely many choices for p and q so that x = p/q F ɛ [a, b]. This implies that F ɛ has no limit points: If there were a limit point c of F ɛ, then the interval [c, c + ] must intersect infinitely many points of F ɛ, which we know is impossible by the above argument. We
conclude that F ɛ is closed. Finally, this implies that t is continuous at x 0 R \ Q: Since F ɛ is closed, R \ F ɛ is open, and x 0 R \ Q R \ F ɛ implies that there is some δ > 0 so that V δ (x 0 ) (R \ F ɛ ). For any x V δ (x 0 ), we have x / F ɛ, so that t(x) < ɛ. Since t(x) 0 for all x R, this means t(x) < ɛ, for any x such that x x 0 < δ, and t is continuous at any irrational point.