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Homework Help: Hard limit problem

  1. Nov 6, 2004 #1
    This problem was given in one of these math contests my school had and that no one was able to prove it (in extreem detail).
    Problem: Suppose that f is a function with the property that
    lim f(x) exists for all real c

    Define a new function g this way:
    g(x) = lim f(t)

    Prove in hideous detail that g is continuous everywhere, (i.e. continuous at every real number).

    As far as i can tell you, i m not at this level of proving such a problem, but it would be helpful if someone could explain it to me on doing this problem. As it would help me understand my class related topics better. This is was a contest question, so I dont expect everyone being able to do it, as even my teacher said it is difficult.
  2. jcsd
  3. Nov 6, 2004 #2
    You probably mean t->x and not x->t in the definition of g(x) right?

    Start from the consideration of the limit of the function at an arbitrary point in the real set. Is it defined? Can it help you to generalize?


    [ps--my 400th post :rofl:]
  4. Nov 6, 2004 #3
    yes sorry, its t->x
  5. Nov 6, 2004 #4


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    Let D be the domain of f. Define a function j : [itex]\mathbb{R} \to \mathbb{R}[/itex] by:

    [tex]j(x)=\left\{\begin{array}{cc}f(x)&\mbox{ if }
    x\in D\\ \lim _{t \to x}f(t) & \mbox{ if } x \notin D\end{array}\right.[/tex]

    j is defined on all [itex]\mathbb{R}[/itex], now let's show it's continuous. If [itex]x \in D[/itex], then there is some open interval around x, [itex]I = (x - \delta, x + \delta[/itex] for some [itex]\delta > 0[/itex] such that [itex]I \subseteq D[/itex]. If no such interval existed, then you could find no interval around x such that f(c) for all c in that interval was defined, so you could never show that the limit of f at c existed. This contradiction assures us that such an interval exists. Now note that this interval is in D, so f is defined for each point in this interval. We know that f must be continuous on this interval, because if it weren't, then the only kind of discontinuity we could have, given that f is defined on I, is a "jump" discontinuity as opposed to a "hole" discontinuity. But if that's true, then the limit could not be defined at the jump, which contradicts the hypothesis that f has limit everywhere. So, since j is f on D, and f is continuous on D, j is continuous there.

    Now, it's easy to see that [itex]\mathbb{R} - D[/itex] is a set of disjoint points. If it contained any non-trivial intervals, then the limit of f would not be defined on any of those intervals. Essentially, f is a continuous function that had some holes punched out of it, and j is f with those holes filled back in. j is continuous on [itex]A = \mathbb{R} - D[/itex] if the limit of j(x) as x approaches some a in A is equal to j(a). The limit as x approaches a of j(x) is the same as the limit as x approaches a for f(x), since a in an open set which is in D (except for a), and so we have something like [itex]0 < |x - a| < \delta[/itex] where "[itex]\delta[/itex]" is like (half of) the size of our interval. We know such a [itex]\delta[/itex] exists because we just showed that A consisted of disjoint points. For these x, j behaves like f, so the limit as x approaches a for j(x) is the same for f(x), which is, by definition, also j(a), so j is continuous at a, and clearly, for all a in A. So j is continuous.

    Anyways, having proved that f is continuous on D, we have that for all d in D, [itex]f(d) = \lim _{x \to d}f(x)[/itex], so we can rewrite j as g, and since j is continuous, so is g.

    I did this in a rush, so it might be confusing, redundant, and mistaken, so ask if something seems off.
  6. Nov 7, 2004 #5
    aaaahhh i see, so even though the function has holes it is still continuous, since the holes are removeable. And no matter how small it is, as you said half of delta, it will have holes in there as well, but again they are removeable, thus the function is continuous. Makes perfect sense to me.
  7. Nov 7, 2004 #6
    Also i forgot to mention this earlier, but our teacher said this question is this book by Spivak, Third Edition, page 119, 16(d). I haven't checked it as i have no resources to find this book but i understand the problem now. so no worries... thanks again
  8. Nov 7, 2004 #7


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    Well f need not be continuous, but we want to prove that g is continuous. g doesn't have holes in it, f might, and g is just f with those holes filled in. As for the rest, I'm not sure exactly which delta you're referring to, or what it is you're really saying.

    Basically, what the proof did was showed that a function that has a limit everywhere will look, "at best" like a continuous function, and "at worst" a continuous function with holes in it, and not a function with gaps in it (like some function that isn't defined over the entire interval (0, 0.12) for example), nor a function with "jumps" in it (like the function H(x) = 0 for negative x, H(x) = 1 for non-negative x, i.e. H(0) = 1, the limit of H(x) as x approaches 0 from the left is 0, and the limit of H(x) as x approaches 0 from the right is not 0, it's 1).

    If f had gaps in it, let's say it wasn't defined on the interval (0, 0.12). Then the limit at x = 0.06 can't exist. Why? Well when we evaluate a limit, we want to find a bunch of points "around" x for which the function at those points evaluate to some number "close" to some number L, and so L would be the limit of f as x approaches 0.06. But no only will the points "around" x not evaluate close to some L, they won't evaluate at all, so there can be no limit at 0.06, but the problem says that there is a limit everywhere, so that can't be the case.

    Obviously, if there were "jumps" then the left limit would not equal the right limit, and so the limit would not exist there at all, again, a contradiction. Therefore, the only types of discontinuities f can have, if it has any at all, are single disjoint holes (no gaps or jumps).

    Actually, I was in a rush and didn't consider another possibility. It can still be proven, so maybe I'll try again and be more rigorous this time.
  9. Nov 7, 2004 #8


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    The Spivak book wouldn't happen to be called "Calculus on Manifolds" would it?

    EDIT: Thinking about it, I think I did cover all the possibilities essentially, although some of what I said may have been wrong, the overall ideas should hold.
    Last edited: Nov 7, 2004
  10. Nov 7, 2004 #9
    ummm, my notes just say calculus on it, and then the date of 1994.
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