Chapter 4 29 Let f: → satisfy |f(x) – f(y)| ≤ |x-y|2. Prove that f is constant. |f(x) – f(y)| ≤ |x-y|2 ⇒ |f(x) - f(y)||x-y| ≤ |x – y| ⇒ xyxyxfyf−≤−− )()( Then 0|)('|lim)()(lim ≤⇒−≤−−→→xfxyxyxfyfxyxy This implies f’(x) = 0 ∀x, which means that f is constant. 37 Prove the following intermediate value theorem for derivatives: If f is differentiable at all points of [a,b], and if f’(a) and f’(b) have opposite signs, then there is a point c ∈ (a,b) such that f’(c) = 0. Suppose f’(x) < 0 < f’(b). f is continuous on [a,b] since it is differentiable there. By the minimum-maximum theorem it has a minimum somewhere in [a,b]. Let c be this minimum, then c ∫ a since f(x) < f(a) if x > a and c ∫ b since f(x) < f(b) for x < b. Therefore, a < x < b which implies that f’(c) = 0by the intermediate value theorem for continuous functions. Similarly, for f’(x) > 0 > f’(b). Let c be the minimum, then c ∫ a since f(x) > f(a) if x < a and c ∫ b since f(x) > f(b) for x < b. Therefore, a > x > b which implies that f’(c) = 0 by the intermediate value theorem for continuous functions. 38 A real-valued function defined on (a,b) is called convex when the following inequality holds for x,y in (a,b) and t ∈ [0,1]: f(tx + (1 –t)t) ≤ tf(x)+(1-t)f(y). If f has a continuous second derivative and f’’ > 0, show that f is convex. Let g(t) = f(tx + (1-t)y) and h(t) = tf(x) + (1-t)f(x). Then we get g(0) = f(y) = h(0) and g(1) = f(x) = h(1). Then g – h has a local minimum or a local maximum in [0,1]. If g-h has a local maximum at t ∈ (0,1) then (g-h)’’(t) < 0. But g = f’’(t)(x-y)2 which is a contradiction to the assumption. Therefore, h-h has a local minimum in (0,1) and cannot have another extrema. Then (g-h)(t) < (g-h)(0) = 0 for all t ∈ (0,1) and g(t) < h(t) ∀t ∈ (0,1). ⇒ f(tx + (1 –t)t) ≤ tf(x)+(1-t)f(y), ∀t ∈ (0,1). 42 For x > 0, define L(x) = ∫xdtt1)/1( . Prove the following, using the definition. a) L is increasing in x. If a > b > 0, then L(a) – L(b) = 0/1/1/111>=−∫∫∫abbatdttdttdt since 1t > 0 on [b,a].b) L(xy) = L(x) + L(y) )()()()(1/1dtdu xt u Let ./1)(/1/1/1)(1111yLxLxyLyLduuxduuxtdtxtdtxLtdttdttdtxyLyyxyxxyxxyxxxy+=⇒===⇒=⇒==+==∫∫∫∫∫∫∫ c) L’(x)=1/x By the fundamental theorem of calculus, L’(x) = 1/x. d) L(1) = 0 ∫=110)( dxxf , for any x. Therefore, ∫=110)/1( dtt . e) Properties c and d uniquely determine L. What is L? L(x) = ln(x). From c, ln’(x) = 1x ⇒ ddx ln(x) + C = 1x . And from d we know that C = 0 since L(1) = ln(1) = 0. 45 Prove the following second mean value theorem. Let f and g be defined on [a,b] with g continuous, f ≥ 0, and f integrable. Then there is a point x0 ∈ (a,b) such that ∫∫=babadxxfxgdxxgxf )()()()(0. Let m = inf(g([a,b])) and M = sup(g([a,b])). Then ∫∫∫≤≤bababadxxfMdxxgxfdxxfm )()()()( since m = inf(g([a,b])) and M = sup(g([a,b])) and ∫∫=babadxxmfdxxfm )()( and ∫∫=babadxxMfdxxfM )()( . Since ∫badxxft )( depends on t, by the intermediate value theorem, there is c ∈ [m,M] such that ∫∫=babadxxfcdxxgxf )()()( . And then there must exist a d such that g(x0) = c, also by the intermediate value theorem. ⇒ ∫∫=babadxxfxgdxxgxf
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