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Continuous functions on intervals

  1. Mar 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose that f : ℝ→ℝ is continuous on ℝ and that lim f =0 as x→ -∞ and lim f =0 as x→∞. Prove that f is bounded on ℝ and attains either a maximum or minimum on ℝ. Give an example to show both a maximum and a minimum need not be attained.


    3. The attempt at a solution

    The example i plan on using is f(x) = 0. But i am not sure which theorem i am supposed to use. I am thinking the maximum-minimum theorem but i am unsure how to start this without a closed bounded interval. Can anyone help me start?
     
  2. jcsd
  3. Mar 19, 2012 #2
    Usually for stuff like this, I start of by doing a proof by contradiction. Now, I very rarely turn in a proof-by-contradiction (I'm always afraid I've messed something up), but this usually gives insight. So, I'd start off by assuming that f is unbounded at some x. Then see how this messes with the contnuity of the limits.

    As for an example, I think that you need to come up with one that has either a max or a min, not a function that has neither. For this, I suggest thinking of something like the Bell Curve.
     
  4. Mar 19, 2012 #3
    That is one of my biggest hangups, I never know when to use a proof by contradiction, direct proof, or a proof by induction. I almost always use the direct method because it is where i am most comfortable.

    Is there a general rule of thumb or hint in a proof that generally tells us what type of proof we should try first?
     
  5. Mar 19, 2012 #4
    Well, its mostly just a matter of experience. The solutions to some problems are pretty evident from the beginning and the direct method just sort of "pops" out, you know? But for ones like this, it always serves me to try to find a contradiction. This usually tells me enough about the problem, and I am usually able to "get" whatever needs to be "gotten" to understand the solution to the problem. Once I have done that, I can usually figure out something that works. For example, I might use contraposition instead of proof by contradiction.
     
  6. Mar 19, 2012 #5

    jgens

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    I am going to disagree that trying for a contradiction with this problem is a good way to go, especially since the continuity on [itex]\mathbb{R}[/itex] can be relaxed with the result still holding. I would suggest trying to turn the problem from one dealing with an unbounded interval, to one dealing with a bounded one. The limiting behavior of the function provides all the information you need to do this. The idea of turning a problem dealing with something unbounded or infinite into something that is bounded or finite is a common theme in mathematics, so I think that getting practice with that idea is helpful.
     
  7. Mar 19, 2012 #6
    Does this proof by contradiction make sense then?

    Suppose that f : ℝ→ℝ is continuous on ℝ and that lim f =0 as x→ -∞ and lim f =0 as x→∞ where f(x) is unbounded on ℝ and attains either a maximum ot minimum on ℝ. Since f(x) is unbounded on ℝ then if given M>0, there exists Xm in ℝ where |f(Xm)| > M, Now WLOG, supppose that f(x) has a maximum on ℝ then there exists s :={max f(x) for all x in ℝ} where s ≥ f(x) for all x in ℝ. However Xm is in R so S≥ |f(Xm)|>M for all x in ℝ, meaning f(Xm) is bounded on ℝ. [Contradiction]

    Therefore f(x) is bounded on ℝ.

    --
    I am still not sure what kind of example to use though...
     
  8. Mar 19, 2012 #7

    jgens

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    Not at all. You have shown that if you assume that f is both unbounded above and bounded above, then you obtain a contradiction, but that is hardly a surprise.
     
  9. Mar 19, 2012 #8
    Well, that rained on my parade. I was quite proud of that :(
     
  10. Mar 19, 2012 #9

    jgens

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    Well you should know you have a problem when you write something like "assume [itex]f[/itex] is unbounded and is either bounded above or below". You should also realize you have a problem when you do not utilize the limiting hypotheses in the statement of the problem.

    The problem with doing this problem by contradiction is that the continuity on all of [itex]\mathbb{R}[/itex] is actually not necessary (i.e. it can be relaxed a little). So it is just easier to use the limiting behavior of the function to turn the question into one dealing with bounded domains.
     
  11. Mar 19, 2012 #10
    Yes, I should have been more precise. What I meant was that it is good to think about what might happen if f was un bounded at a point. I didn't say it explicitly, but I was thinking of putting a closed interval around the point at which f was unbounded. I'm just suggesting doing this as a way to wrap ones head around the problem. Like I said though, a contradiction is probably not the best proof to turn in.

    As for relaxing the continuity on [itex]\mathbb{R}[/itex] I'm a little confused about what you mean. For example [itex]\frac{1}{x}[/itex] is continuous everywhere except 0 and it's limit is 0 as x goes to positive and negative infinity, but it is not bounded. So what can the continuity condition be relaxed to?
     
  12. Mar 19, 2012 #11
    Can't it be relaxed to (0,1]
     
  13. Mar 19, 2012 #12
    Do you mean like saying only jump discontinuities are allowed?
     
  14. Mar 19, 2012 #13

    jgens

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    The end behavior of the function means the function is bounded on [itex](-\infty,-M) \cup (M,\infty)[/itex] for some [itex]M \in \mathbb{R}[/itex] and then the continuity of [itex]f[/itex] on [itex][-M,M][/itex] gives the boundedness on [itex][-M,M][/itex]. The boundedness due to the end behavior though is not dependent on the continuity, so [itex]f[/itex] can actually be discontinuous for sufficiently large values of [itex]x[/itex].

    Now the way of getting around relaxing the continuity restriction means you essentially need to give the problem away (since you need to be careful that you have boundedness on [itex](-\infty,-M) \cup (M,\infty)[/itex] and continuity on [itex][-M,M][/itex]), but the point is that the condition can be relaxed.

    That works too. The point is that you can get all sorts of discontinuities with [itex]f[/itex] and still have boundedness hold. The other problem is that unboundedness does not necessarily contradict continuity. It can contradict the end behavior condition. So there are essentially two types of behaviors that you need to check if you are doing a proof by contradiction.
     
  15. Mar 20, 2012 #14
    So is the limiting condition telling me that 0 is a two sided limit and is thus the max/min of the equation since both are going to that value?
     
  16. Mar 20, 2012 #15
    To turn the problem into using a bounded interval do i state something like "there exists an Interval I in R such that i is a closed, bounded interval from [a,b]"?
     
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