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continuity: problem 64 in practice book


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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.

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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)

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  • 8 months later...

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].

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  • 3 months later...

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