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Thread: continuity: problem 64 in practice book

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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)| <= 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.
    Last edited by alamps3; 04-01-2008 at 04:45 PM. Reason: typo

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    how bout f(x) = sqrt(x)

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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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    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].
    Last edited by oldmathguy; 12-18-2008 at 11:15 PM.

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    The version above is now correct.
    Last edited by oldmathguy; 12-17-2008 at 06:20 PM.

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    Excuse me, What is this "Practice Book"? Could you please give me a link to download?

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