Math 320: Real Analysis MWF 1pm, Campion Hall 302 Homework 9: OPTIONAL Please write neatly, and in complete sentences when possible.

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1 Math 320: Real Analysis MWF pm, Campion Hall 302 Homework 9: OPTIONAL Please write neatly, and in complete sentences when possible. Do the following problems from the book: 5.2., 5.2.2, 5.2.3, 5.2.5, 5.2.8, 5.3., 5.3.3, Solution Since f and g are differentiable at c, we have f f() f(c) (c), and g g() g(c) (c). c c c c By the algebraic properties of functional its, we have f (c) + g f() f(c) g() g(c) (c) + c c c (f + g)() (f + g)(c) c c c (f() + g()) (f(c) + g(c)) c Since the latter eists, f + g is differentiable, and (f + g) (c) = f (c) + g (c). Similarly, we have kf (c) = k c f() f(c) c (kf)() (kf)(c), c c so kf is differentiable and (kf) (c) = kf (c). c kf() kf(c) c Solution 5.2.2(a). Eploring the definition of the derivative, for any c 0 we have c c c c c c c c c = c, 2 where we ve used the algebraic it properties to evaluate the it of /(c) as approaches c. Since the it of the difference quotient eists, f () = / 2. Solution 5.2.2(b). Given two differentiable functions u and v, we may interpret the quotient function u/v as the product of u and /v. Since /v = f v, where f() = / as above, we may interpret u/v as the product of u and f v, whose derivative we compute with the

2 2 product rule: ( u ) (c) = (u (f v)) (c) = u (c) (f v)(c) + u(c) (f v) (c). v Using the chain rule, we have (f v) = (f v) v, and the computation of the derivative f () above now gives ( u ) (c) = u (c) v v(c) + u(c) v(c) 2 v (c) = u (c) v(c) u(c)v (c) v(c) 2 as desired. = u (c)v(c) u(c)v (c) v(c) 2, Solution 5.2.2(c). We manipulate the difference quotient directly: c u() u(c) v() v(c) c c u()v(c) u(c)v() ( c)v()v(c) u()v(c) u(c)v(c) + u(c)v(c) u(c)v() c ( c)v()v(c) u() u(c) c ( c)v() u(c) v() v(c) v(c) ( c)v(). As we assume that u and v are each differentiable at c and v(c) 0, we have that v is also continuous at c, so that c v() = v(c), and we obtain c u() u(c) v() v(c) c = u (c) v(c) u(c)v (c) v(c) 2 Solution Let f() be defined by 2 if Q, f() = 0 if R \ Q. We claim that f is differentiable at 0: f() f(0) 0 0 = u (c)v(c) u(c)v (c) v(c) 2. 0 f(). If { n } is any sequence converging to 0, then we have 0 f( n ) 2 n, so that 0 f(n) n n for all n. By the squeeze theorem, f( n )/ n

3 must also converge to 0, and we conclude that f() 0 = 0. On the other hand, f is not continuous at any 0, so it is certainly not differentiable: Since both the rationals and the irrationals are dense in R, we may choose sequences of irrationals {α n } and rationals {β n }, each converging to. On the other hand, f(α n ) = 0 of all n, while f(β n ) converges to 2. Solution 5.2.5(a). Since it will be necessary to choose non-integer values of a in our search for functions g a with specified properties, it is important to point out that there are some restrictions on what noninteger values of a we may choose. For instance, if a is a dyadic rational (e.g. 3/2) then g a will not be a real-valued function for negative values of. On the other hand, there are many values of a for which g a is perfectly fine, for eample rational values of a with odd denominators. Whenever we refer below to a non-integer value of a, we mean such a value. The definition of the derivative at = 0, if it eists, is ( g a(0) g a () g a (0) a sin ) a sin. The latter eists and equals 0 as long as a >, by the squeeze theorem: For all 0, we have a a sin a, and 0 a 0 a = 0 since a >. Thus g a () is differentiable at = 0 as long as a >. As for 0, the chain rule implies that ( ) g a() = a sin = a a sin a 2 cos so that for a > we have a a sin g a() a 2 cos ( ) if 0, = 0 if = 0., 3

4 4 When a < 2, this will be unbounded in a neighborhood of = 0: a a 2 g a(/2πn) = a sin(2πn) cos(2πn) = (2πn) 2 a. 2πn 2πn If a < 2, the latter approaches as n. Solution 5.2.5(b). The epression above for g a() defines a continuous function at = 0 as long as a > 2, using the squeeze theorem again: a a a 2 a a sin a 2 cos a a + a 2. When a > 2, each of a a a 2 and a a + a 2 approach 0, and g a() is continuous. On the other hand, if a 3, then g a() will be not differentiable at 0: ( g a() g a(0) a a sin ) a 2 cos a a 2 sin a 3 cos 0 When a 3 this last it doesn t eist: Consider n = /πn, in which case we have an a 2 sin n a 3 cos = (πn) 3 a cos(πn) = ( ) n+ (πn) 3 a. n Thus g a is not differentiable at = 0 for a 3. n Solution 5.2.5(c). For 0, we can use the chain and product rules to compute g a() = a(a ) a 2 sin a a 3 cos (a 2) a 3 cos a 4 sin = (a(a ) a 2 a 4 ) sin 2(a ) a 3 cos As for = 0, the computation above shows that g a is differentiable at = 0, with value 0, as long as a > 3. If a 4 however, g a() doesn t approach 0 as approaches 0, so that g a is not continuous at = 0.

5 Solution 5.2.8(a). If g = f, for a differentiable function f on an interval (a, b), then by Darbou s Theorem g has the intermediate value property. If g is not constant, there are values, 2 (a, b) so that g( ) g( 2 ). Since irrationals are dense in R, there is an irrational value α between g( ) and g( 2 ). Since g has the intermediate value property, there is a c (a, b) so that g(c) = α, so that g takes on an irrational value. 5 Solution 5.2.8(b). This is false: Consider f() = g 2 () +, where 2 g 2 () is defined as in Eercise Then f is differentiable everywhere, and f () = 2 sin ( ( ) cos ) + if 0, 2 if = 0. 2 Thus f () > 0, but on the other hand f (/2πn) = + 2 = 2. Since the sequence {/2πn} approaches 0 as n goes to, in any ɛ- neighborhood of 0 there will be poiints where f () < 0. Solution 5.2.8(c). This is true: Suppose on the contrary that L > f (0), and let ɛ = (L f (0))/3 > 0. Then, since 0 f () = L, there eists a δ > 0 so that f () L < ɛ for 0 < < δ. But then f () doesn t satisfy the intermediate value property, contradicting Darbou s Theorem: For any 0 in the interval [ δ, δ], we have f () (L ɛ, L+ɛ), and moreover L f (0) = 3ɛ. Thus, while f (0)+2ɛ is between f (δ) and f (0), there eists no element of [ δ, δ] taking this value. The case L < f (0) is similar. Solution 5.2.8(d). This is true: Suppose on the contrary that f (0) does not eist, i.e. the it f(h) f(0) h 0 h does not eist. In that case, there eists a sequence {h n }, converging to 0, but so that the sequence {(f(h n ) f(0))/h n } does not converge. On the other hand, for each h n, by the Mean Value Theorem there is, so that the sequence {f (k n )} does not converge. On the other hand, since k n is between 0 and h n, we have a k n between h n and 0 so that f (k n ) = f(hn) f(0) h n h n k n h n.

6 6 By the squeeze theorem {k n } converges to 0 as well. Thus {f (k n )} does not converge, but {k n } converges to 0. This violates the assumption that 0 f () = L. This contradiction implies that f (0) eists, and by the solution to 5.2.8(c) we have f (0) = L. Solution Since f is continuous on [a, b], by the preservation of compact sets (Theorem 4.4.2) the image f ([a, b]) is compact, and in particular bounded in absolute value. Thus there eists an M > 0 so that f () M for all [a, b]. For any pair of distinct points, y [a, b], by the Mean Value Theorem there is a c between and y so that In particular, we have f (c) = f() f(y). y f() f(y) y M, so that f is Lipschitz. Solution 5.3.3(a). Consider the function g() = h(). Since h is differentiable on [0, 3], it is continuous on [0, 3], so that g is continuous as well. By assumption, g(0) = and g(3) =. By the Intermediate Value Theorem, there is a value d [0, 3] so that g(d) = 0, i.e. h(d) = d. Solution 5.3.3(b). By the Mean Value Theorem, there is a c [0, 3] so that h (c) = (h(3) h(0))/3 = /3. Solution 5.3.3(c). By Darbou s Theorem, derivatives of differentiable functions on intervals have the Intermediate Value Property. By the Mean Value Theorem, there is a c [0, 3] so that h (c ) = (h(3) h())/2 = 0. Since /4 is between 0 and /3, the Intermediate Value Property implies that there is a c [0, 3] so that h (c ) = /4. Solution Suppose on the contrary that f is differentiable on an interval (a, b) so that f () for all (a, b), but so that f has two fied points and y in (a, b). In that case, by the Mean Value

7 Theorem, there would be a c between and y so that f (c) = f() f(y) y violating the assumption that f (c). = y y =,