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Prove Uniformly Continuous

  1. Dec 2, 2007 #1
    Let [tex]f(x) = \frac{x}{x-1}[/tex]. Prove [tex]f(x)[/tex] is uniformly cont. on the interval [tex][1.5,\infty)[/tex]
  2. jcsd
  3. Dec 2, 2007 #2
    We have a homework forum for this.

    To prove that something does not have discontinuities on a given interval, it suffices to show that all of the function's discontinuities lie outside of the interval. Find the discontinuity(s) for this function then show that they are not in the given interval.
  4. Dec 2, 2007 #3
    to demonstrate uniform continuity on a noncompact set requires more than showing it is continuous at that interval.

    But here is a trick you can use. limf(x) exists as x goes to infinity, right? Now use the fact that the terms will be as close as you want them to be for some [y, infty), and then for [1.5,y] use the theorem that continuous functions on compact sets are uniformly continuous. Choose an appropriate delta.

    IN general if you have a function that is continuous on an infinite interval and the limit exists as x goes to plus or minus infinity, the function is uniformly continuous there.

    btw, when i say the limit exists i mean that it is finite.
  5. Dec 3, 2007 #4
    Can't we use the derivative?

    Never mind. I missed the key word, uniformly, here.
    Last edited: Dec 3, 2007
  6. Dec 3, 2007 #5
    Hint: What can you say about f'(x). You can prove this in almost one line of math.
  7. Dec 3, 2007 #6
    Uh, [tex]\frac{-1}{(x-2)^2}[/tex] ... I don't see how f'(x) helps in showing f(x) is uniformly continuous. We know every function that is integrable is continuous, but we can't say the same with differentiability. Mind helping?
  8. Dec 3, 2007 #7
    I think perhaps the hint may be too subtle. Honestly, I'm trying to avoid giving the answer as this seems like an analysis h.w. problem. Think 'bounded,' and look up 'Lipschitz.' At least this way you're doing some of your own "research" on the problem.

    P.S. Unless I'm mistaken, your derivative is incorrect if we are to assume your problem statement is correct. Check your denominator.

    [tex]f(x) = \frac{x}{x-1} = \frac{x-1}{x-1} + \frac{1}{x-1} = 1 + (x-1)^{-1} \implies
    f'(x) = -\frac{1}{(x-1)^2}[/tex]
  9. Dec 3, 2007 #8
    Indeed, I meant 1. Typo. I'll look for "Lipschitz" although it doesn't sound familiar.
  10. Dec 3, 2007 #9
    If Lipschitz doesn't ring a bell, the mean value theorem should.
  11. Dec 3, 2007 #10
    Well, looking in my notes, to show [tex]f(x) = \frac{1}{x}[/tex] is unif. continuous on [tex]A = [a, \infty)[/tex] for [tex]a > 0[/tex], they prove it with:

    We note [tex]x,y \in A[/tex], then:

    [tex]|f(x) - f(y)| = \left|\frac{y-x}{xy}\right| \leq \frac{|y-x|}{a^2}[/tex]

    Here, if we have that [tex]\epsilon > 0[/tex] and if [tex]|x-y| < a^2\epsilon[/tex], then:

    [tex]|f(x) - f(y)| < \epsilon[/tex]

    Hence f is unif. continuous.

    ...I see no use of derivative. I tried manipulating mine to be similar to the above one, but I'm not even enitrely sure of how the above proof works. Sorry to be a pain.
  12. Dec 3, 2007 #11
    First, let's see what they do:

    [tex]f(x)-f(x) = \frac{1}{x} - \frac{1}{y} = \frac{y-x}{xy} = (y-x) \cdot \frac{1}{xy}[/tex]

    Notice that

    [tex]f(x) = \frac{1}{x} \le \frac{1}{a}[/tex]

    The rest follows immediately.

    There is an easier way to do that same problem using the mean value theorem. The method above works, but it doesn't generalize easily. Here's the general method (and I'll only sketch the proof).

    Your function is clearly continuous on [tex][1.5,\infty)[/tex]. Apply the mean value theorem:

    [tex]\frac{f(y)-f(x)}{y-x} = f'(c) \text{ where } c \in (x,y) [/tex]

    Toss in absolute values on both sides.

    [tex] \left| \frac{f(y)-f(x)}{y-x}\right| = |f'(c)|
    \implies |f(y) - f(x)| = |f'(c)| \cdot |y-x|[/tex]

    Again, what do you property do notice about the derivative on the interval [tex][1.5,\infty)[/tex]? To answer this question, think about what you would need to make the rest of this sketch work? Once you figure this out, you should be able where the derivative "appears" in your example problem (it has to do with the [tex]1/a^2[/tex]).
  13. Dec 3, 2007 #12
    This is false, there are functions that are discontinuous that are still integrable. Also if a function is differentiable it is necessarily continuous.

    Although it is possible to show f(x) is uniformly continuous on [1.5, infinity) by using the fact that f(x) goes to 1 as x goes to infinity as i previously mentioned, if you have the theorems of differentiation it is much easier to do using this theorem:

    If f is differentiable on a set E and there exists a real number M so that M>|f'| for all x in E, then f is uniformly continuous on the set E.

    This theorem is easy to prove and rs1n has given a big hint.

    There is an interesting converse to the above theorem.

    A twice differentiable function from R to R whose second derivative is bounded is uniformly continuous if and only if it's first derivative is bounded. However, to proof this is not very easy.
  14. Dec 4, 2007 #13
    We know that the derivative is always negative for values > 1.

    So we have:

    [tex]f(y) - f(x) \rightarrow \frac{\frac{y}{y-1} - \frac{x}{x-1}}{y-x} = \frac{-1}{(x-1)(y-1)}[/tex]

    ... Not sure what this quite does... we know this equals |f'(c)|*|y-x| from what you showed me. f'(c) will have to be 1.5 or greater, and thus this is negative. Hence this product will be positive assuming y > x...
    Last edited: Dec 4, 2007
  15. Dec 4, 2007 #14
    if delta=epsilon/M where M is a bound for |f'(x)|, what happens?
  16. Dec 4, 2007 #15


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    That's exactly backwards! If a function is differentiable, then it is continuous. An integrable function may not be continuous.

    The point in the orginal response was that if a function is continuous on a closed interval, then it is uniformly continuous there. [1.5, [itex]\infty[/itex]) is closed and x/(x-1) is continuous on it.
  17. Dec 4, 2007 #16
    You mean compact interval, otherwise take y=x^2 on [0, infinity), which is closed and f is continuous there, but not uniformly.

    So that argument is not satisfactory.
    Last edited: Dec 4, 2007
  18. Dec 4, 2007 #17
    Either you meant:

    [tex]\frac{f(y) - f(x)}{y-x} = \frac{\frac{y}{y-1} - \frac{x}{x-1}}{y-x} = \frac{-1}{(x-1)(y-1)}[/tex]

    or you meant:

    [tex]f(y) - f(x) = \frac{y}{y-1} - \frac{x}{x-1} = -\frac{y-x}{(x-1)(y-1)}[/tex]

    You can use the example from your book to finish this. Take the absolute value of both sides in the either 'corrected' equation and rewrite to get:

    [tex]|f(y)-f(x)| = |y-x| \cdot \left| \frac{1}{x-1}\right| \cdot \left| \frac{1}{y-1}\right|[/tex]

    Question: Can you place a bound on [tex]\frac{1}{x-1}[/tex] and [tex]\frac{1}{y-1}[/tex]? If so what is the bound? (Refer to the example in the book and try to figure out how they obtained [tex]1/a^2[/tex]. Then work from there.

    If you wish to know the more general method (one that is much more useful later on), take a look at what I wrote earlier. Examine [tex]f'(c)[/tex] (where [tex]c\in [1.5,\infty)[/tex]). Since you eventually would like to have

    [tex]|f(y)-f(x)|< \epsilon[/tex]​

    and you already have

    [tex]|f(y)-f(x)| = |f'(c)| \cdot |y-x|[/tex],​

    the latter equation should have raised the question: can I place a bound on

    [tex]|f'(c)| = \frac{1}{(c-1)^2}[/tex]​

    where [tex]c\in [1.5,\infty)[/tex]?

    If so, then you can choose [tex]\delta[/tex] in [tex]|y-x|<\delta[/tex] based on the bound so that your desired result

    [tex]|f(y)-f(x)|< \epsilon[/tex]​

    follows immediately.
    Last edited: Dec 4, 2007
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