Math630-7 - Real Analysis - Math 630 Homework Set #7 -...

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Unformatted text preview: Real Analysis - Math 630 Homework Set #7 - Chapter 5 by Bobby Rohde 10-19-00 Problem 1 Let f be the function defined by f 0 0 and f x Find D f 0 , D f 0 , D f 0 , and D f 0 . D f0 x sin 1 x for x 0. D f 0 = lim h0 fh can find a value of D f0 h h0 1 h which gives f0 = lim h sin h 1 h = lim sin h0 1 h = 1, since for arbitrarily small positive h we sine its maximal value. D f 0 = lim h0 fh h 1 h f0 = lim h0 h sin h 1 h = lim sin h0 1 h = -1, since for arbitrarily small positive h we can find a value of D f0 which gives sine its minimal value. D f 0 = lim h0 f0 f h 1 h h = lim h0 h sin h 1 h = lim sin h0 1 h = 1, since for arbitrarily small positive h we can find a value of D f0 which gives sine its maximal value. D f 0 = lim h0 f0 f h 1 h h = lim h0 h sin h 1 h = lim sin h0 1 h = -1, since for arbitrarily small positive h we can find a value of which gives sine its minimal value. MATH630-7.nb 2 Problem 2 a) Show that D Proof D h0 fx = D fx lim = fxh fx fx 0 = fx h fx h0 h = fx h0 fx lim = fxh fx h inf inf h00 sup inf h h0 lim h00 fxh fx = h = inf sup fx fx h = D f x . QED. b) If g x = f Proof Df x = D g x . QED x , then D g x = D f lim f x f h xh x = lim h0 gx gx h h h0 = h0 lim gx gx h h = lim g x h h0 gx h = MATH630-7.nb 3 Problem 3 a) If f is continuous on [a, b] and assummes a local maximum at c then D fc D fc fch fc h (a, b), 0 D fc D fc , since lim sups are neccesarily greater Proof Well D f c = lim h0 D f c = lim h0 fch fc h than lim infs. Similarly D f c D f c . So we only really need to establish that 0 D f c . Since f c is a local maximum an open interval I a, b such that c D fc I and f c is the maximum of f over I. WLOG we may consider h to be sufficiently small such that c h and c h I. Consider D f c = lim h0 fch fc h , since c h fch fc h I, f c h fc h0 fc h fc h 0. , Furthermore since h > 0, we know that 0. By definition lim g h h 00 inf sup g but as h 0 , the supremums form a nonincreasing sequence, and hence so long as we permit ± as acceptable limits, lim g h must exist. h0 Thus we have that lim h0 fch fc h fc 0, since each fch fc h < 0, and the limit exists. Similarly D f c extended reals, fch f has f c h c h h h0 f so lim f c h c h 0. Hence D f h0 lim > 0 over I, and the limit must exist over the c 0 D f c . QED. b) What if f has a local maximum at a or b? Answer If f is maximal at a or b then the two derivates which are still inside the interval must exist and obey the above inequality, however the other two may not even exist. Thus with a < b and f a a local maximum we have that D f a D fa 0, and if f b is a local max then 0 D fb D f b. MATH630-7.nb 4 Problem 4 Prove: If f is continuous on a, b and one of its derivates (say D ) is everywhere nonnegative on a, b , then f is nondecreasing on a, b ; i.e. fx f y for x y. Proof Lemma - For a function g x such that D g x > 0, the above property holds. Suppose x y, such that g x g y . Since g is continuous we may apply the normal knowledge of continuous functions to conclude that either g has a local maximum value or g is always decreasing. If g were always decreasing then we know that D g x = lim g x hh g x < 0 since gx h gx 0, x. Which is a contradiction, thus g has a local max. h0 However by problem 3, D g c 0, where c is the location of the local max, but D g x which gives a contradiction. Thus we know that g is nondecreasing. Consider f x lim gx h gx h 0, x gx x, where g x is defined as in the lemma. h0 fxh fx h gx h gx h Therefore D f x = lim h0 = lim h0 gx h xh h gx x = lim h0 gx h gx h h h = = lim h0 0. So if f is an arbitrary function of the type given in the problem then we know that > 0, fx x, is a nondecreasing function. Suppose x y such that f x f y , yet we know that x fy y. So this means that f x f y 0 and f x f y y x . Let = fx f x f y , then > 0, and y x . However we may choose = y x 2 > 0. Which give that *(y-x) = 2 . Which is a contradiction. Hence f x must be nondecreasing. MATH630-7.nb 5 Now all we need to show is that D f x 0 iff the other derivates are 0. However we know that nondecreasing is equivalent to D f x 0, and it follows immediately that f x h f x and f x f x h are each 0 for h > 0. However this means that the terms for which we are considering the limit of in each derivate must be always 0. So by definition all the derivates are 0 if D f x 0. D f x and D f x Now we need only look at the other direction. Clearly for both D f x D f x , so if respectively D f x and D f x are 0 than the other two will be. So in order to complete the other direction we only need that D f x 0 implies that the function is nondecreasing. However if we replace D f x with D f x and the related definitions in the above two steps its easy to see that the remainder of the proof will preceed verbatim. Hence D f x 0 implies that f x is nondecreasing. This completes the proof. QED. ...
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This note was uploaded on 03/23/2010 for the course MATH 515 taught by Professor Staff during the Spring '08 term at Iowa State.

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