alamps3 Posted April 1, 2008 Share Posted April 1, 2008 64) Suppose that f is a continuous real-valued function defined on the closed interval [0,1]. Which of the following must be true? I. There is a constant C>0 st |f(x) - f(y)| II. There is a constant D>0 st |f(x) - f(y)| III. There is a constant E>0 st |f(x) - f(y)| answer is I and II only. Can someone provide a counterexample to 3 to show why it doesn't have to be true. Quote Link to comment Share on other sites More sharing options...
alamps3 Posted April 1, 2008 Author Share Posted April 1, 2008 how bout f(x) = sqrt(x) Quote Link to comment Share on other sites More sharing options...
RiverMyst Posted April 2, 2008 Share Posted April 2, 2008 I is true because f(x) is continuous on a compact set, hence it is totally bounded below and above. II is true because it is the definition of epsilon-delta continuity. III is FALSE. Here is why. f(x) is continuous on a compact set hence it is uniformly continuous. However, it is well known that uniform continuity does not imply Lipschitz continuity, which is condition III, i.e. f(x) = sqrt(x) Quote Link to comment Share on other sites More sharing options...
oldmathguy Posted December 17, 2008 Share Posted December 17, 2008 Here's a practical way to view a counterexample. It's a problem about slopes of pieces of the graph since (f(x)-f(y) average rate of change (f(x)-f(y))/(x-y) must be Draw a jagged continuous graph f(x)which is piecewise-linear ( made up of line segments) with infinitely many larger and larger slopes. If you can imagine this, mark option III false and go on to the next problem. If you want a rigorous definition of a function try this: f(0)=f(1) = 0, between 0 and 1/2 it has slope 1 going up to (1/4,1/4) and slope -1 down to (1/2, 0), from x=1/2 to x=3/4, it goes up to height 2/8 with slope 2 to the half-way point at x=5/8 and down with slope -2 to (3/4,0) Continue in this way, with slopes n and -n between points (1- 1/2^{n-1}, 0) and (1 - 1/2^n, 0), achieving maximal height n/2^{n+1} halfway between these two points. Notice that the limit of n/2^{n+1} as n--> infinity is 0, so the function is continuous at x=1. Then given any E there is an integer n>E and a piece of the graph with slope n. So it won't be true that the slope( f(x)-f(y))/(x-y) x and y in [0, 1]. Quote Link to comment Share on other sites More sharing options...
oldmathguy Posted December 17, 2008 Share Posted December 17, 2008 The version above is now correct. Quote Link to comment Share on other sites More sharing options...
amintwist Posted March 25, 2009 Share Posted March 25, 2009 Excuse me, What is this "Practice Book"? Could you please give me a link to download? Quote Link to comment Share on other sites More sharing options...
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