# Thread: continuity: problem 64 in practice book

1. ## continuity: problem 64 in practice book

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)| <= C for all x and y in [0,1]
II. There is a constant D>0 st |f(x) - f(y)| <= 1 for all x and y in [0,1] that satisfy |x-y\<=D
III. There is a constant E>0 st |f(x) - f(y)| <= E|x-y| for all x, y in [0,1]

answer is I and II only.

Can someone provide a counterexample to 3 to show why it doesn't have to be true.  Reply With Quote

2. how bout f(x) = sqrt(x)  Reply With Quote

3. 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)  Reply With Quote

4. 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)<E(x-y) means that the
average rate of change (f(x)-f(y))/(x-y) must be <E. So do this:
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) < E for all
x and y in [0, 1].  Reply With Quote

5. The version above is now correct.  Reply With Quote  Reply With Quote