MATH 104: INTRODUCTORY ANALYSIS SPRING 2008/09 PROBLEM SET 8 SOLUTIONS

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1 MATH 04: INTRODUCTORY ANALYSIS SPRING 008/09 PROBLEM SET 8 SOLUTIONS. Let f : R R be continuous periodic with period, i.e. f(x + ) = f(x) for all x R. Prove the following: (a) f is bounded above below attains its maximum minimum; Solution. Define a function g : [0, ] R, g(x) = f(x) (the function g is called the restriction of f to [0, ] is often denoted by f [0,]. Since f is continuous on R, g is continuous on [0, ] by the extreme value theorem, g attains its maximum minimum on [0, ]. Since f is periodic with period, for any x R, so max f(x) = max g(x), x R x [0,] f(x) = g(x x ) min f(x) = min g(x), x R x [0,] so f attains its maximum minimum in [0, ]. (b) f is uniformly continuous on R; Solution. Define a function h : [0, ] R, h(x) = f(x) Since h is continuous on a closed bounded interval [0, ], it is uniformly continuous on [0, ]. Let ε > 0. Then there exists δ > 0 such that h(u) h(v) < ε for all u, v [0, ], u v < δ. If x, y R x y < δ, then there exist u, v [0, ] such that u v < δ x u, y v Z. So f(x) = h(u) f(y) = h(v) f(x) f(y) = h(u) h(v) < ε. Hence f is uniformly continuous on R. (c) there exists x 0 R such that f(x 0 + π) = f(x 0 ). Solution. Define a function e : R R, e(x) = f(x + π) f(x). Then e is continuous on R since f is. By part (a), f attains its minimum maximum at some a, b R. So In other words, f(a) f(a + π) f(b + π) f(b). e(b) 0 e(a) 0, so there exists x 0 between a b such that e(x 0 ) = 0, i.e. f(x 0 + π) = f(x 0 ).. Suppose the interval I is contained in the interior of dom(f). Define the set of difference quotient f(x) f(y) D := x y }. x, y I, x y Prove that if D is bounded then f is uniformly continuous on I. Solution. If D is bounded, then there exists m, M R such that for all x, y I x y, Date: May 5, 009 (Version.0). m f(x) f(y) x y M.

2 Let K := max m, M } +. Then for all x, y I x y, f(x) f(y) x y < K. Given ε > 0, let δ = ε/k. Then for all x, y I such that x y < δ, we have So f is uniformly continuous on I. f(x) f(y) < K x y < Kδ = ε. 3. Define the following functions f, g : R R, f(x) = 4 n if x = for some n N, n 0 otherwise, g(x) = n+ if x = for some n N, n 0 otherwise. Are f g differentiable at x = 0? Prove your answers. Solution. f is differentiable at x = 0 since 4 n 0 = n if h = for some n N, n h 0 otherwise, h if h = for some n N, = n 0 otherwise. So for all h R, 0 h. h For given ε > 0. Let δ = ε. So when h < δ, 0 h h < ε. Hence = 0 so f (0) exists ( is equal to 0). g is not differentiable at 0 since if we choose a n = / n b n = /3 n, we have g(a n ) = a n g(b n ) = 0 for all n N; so so does not exist. g(a n ) g(0) = n a n g(h) g(0) g(b n ) g(0) = 0 n b n 4. For each of the following functions, find f (x) for each x dom(f) or show that it does not exist.

3 (a) f a : R R, f a (x) = x 3. f a (x) = f a(x) = Note that f a is continuous at 0. Also x 3 if x 0, x 3 if x < 0, 3x if x > 0, 3x if x < 0. f a (h) f a (0) h 3 0 = = = 0 + f a (h) f a (0) h 3 0 = = ) = 0. h 0 ( h Hence f a (h) f a (0) exists f a(0) = 0. In other words, (b) f b : R R, f b (x) = e x. Also f b (x) = f a(x) = 3x x. e x if x 0, e x if x < 0, f b (x) = e x if x > 0, e x if x < 0. f b (h) f b (0) e h e 0 = f b (h) f b (0) Hence f b (0) does not exist. (c) f c : R R, f c (x) = x. = h 0 e h e 0 h f c (x) = dh e h dh eh x if x 0, x if x < 0, = e h h=0 = h=0 = e h h=0 =. h=0 Also f c(x) = x x if x > 0, if x < 0. f c (h) f c (0) h 0 = = h = + h 0 + f c (h) f c (0) h 0 = = =. h 0 h 3

4 Hence f c(0) does not exist. (d) f d : R\0} R, f d (x) = log x. so f d (x) = log x if x > 0, log( x) if x < 0, for all x 0. f d (x) = x 5. For each of the following functions, find g (x) for each x dom(g) or show that it does not exist. (a) g a : R R, 4x 4 if x <, g a (x) = x if x. Solution. (b) g b : R R, arctan x if x, g b (x) = π 4 sgn x + x if x >. Recall that the sign function sgn x =, 0, + for x negative, zero, positive respectively. Solution. Observe that sgn x is a constant function on x > x < so g b (x) = + x if x <, if x >. It remains to check whether g b () g b ( ) exist. Note that g b is continuous at. We have g b (x) g b () = x + x x + π 4 + x π 4 x = g b (x) g b () arctan x arctan = x x x x dx arctan x = x= + x = x=. b () = /. We also have g b (x) g b ( ) arctan x arctan( ) = x + x ( ) x + x ( ) dx arctan x = x= + x = x= b ( ) does not exist. (c) g c : R R, g b (x) g b () π 4 = + x x x ( ) x x + g c (x) = π 4 x e x if x, e if x >. = +. 4

5 Solution. Clearly, g c(x) = x( x )e x if x <, 0 if x >. It remains to check whether g c() g c( ) exist. Note that g c is continuous at. Also, we have g c (x) g c () e = e x + x x + x = 0, g c (x) g c () x e x e = x x x x dx x e x = 0. x= c() = 0. Since g c is an even function, i.e. g c (x) = g c ( x) for all x R, we have that g c( ) = 0 too. (d) g d : R R, arctan if x 0, g d (x) = x π if x = 0. Solution. Clearly g d (x) = + x if x > 0, + x if x < 0. It remains to check whether g d (0) exist. We have d (0) does not exist. g d (h) g d (0) arctan h = π θ π = θ (π/) tan θ = θ (π/) (θ π g d (h) g d (0) arctan( h = ) π θ π = θ (π/) tan θ = θ (π/) (θ π ) tan θ = ) tan θ =. 5