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Set of continuous bounded functions.

  1. Dec 25, 2007 #1
    1. The problem statement, all variables and given/known data

    This is not a homework question. I am solving this from the lecture notes that one of my friend's has got from last year.

    If C(X) denotes a set of continuous bounded functions with domain X, then if X= [0,1] and fn(x) = x^n. Does the sequence of functions {fn} closed , bounded and compact? If not then why.

    2. Relevant equations

    In a compact metric space , every sequence has a convergent subsequence.

    3. The attempt at a solution

    My confusion is the following.

    For x that belongs to [0,1), the sequence fn(x) -> - as n-> infinity. And
    for x=1 , fn(x) ->1 as n -> infinity.

    In this case, obviously fn(x) is bounded as for all n and all x, |fn(x)| <= 1. It is also closed. But, in the first place , how come is fn(x) continuous at x=1. Because for any n, for x->1, fn(x) ->1 from 0. Hence, we can choose an epsilon = 1/2 such that for delta = 0.5, we have |fn(x) - fn(1)| > epsilon.

    Hence, fn(x) indicates a set of functions that is continuous only at X= [0,1) and not in [0,1]. Then, how come is fn(x) = x^n belong to C(X) , where X = [0,1].

    Please explain.
  2. jcsd
  3. Dec 26, 2007 #2


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    Why is it "also closed"?

    I don't understand what you mean by "for x->1, fn(x)->1 from 0". If you simply mean that fn(x) ranges from 0 to 1 as x goes from 0 to 1, yes, that is true. But that has nothing to do with continuity at x= 1. Here fn(x)= x^n and that is, of course, continuous for all n. "If epsilon= 1/2, then |fn(x)- fn(1)|>1/2 for delta= 1/2" is not relevant. To show that the function is not continous at 1 you must show that |fn(x)- fn(1)| is not less than epsilon for any delta- you can't just choose delta to be 1/2. Here [itex|f_n(x)- f_n(1)|= |x^n- 1|= |x-1||x^{n-1}+ x^{n-1}+ \cdot\cdot\cdot+ 1|[/itex]. If x< 2 (delta< 1) then [itex]x^{n-1}+ x^{n-1}+ \cdot\cdot\cdot+ 1|< 2^{n-1}[/itex] so it is sufficient to take delta< epsilon/[itex]2^{n-1}[/itex].

    I would have thought that you learned that x^n is continuous back in Calculus 1. Every function in this sequence, that is, every power of x, is continuous for all x and so on [0, 1].

    That says nothing, of course, about the limit. The limit of the sequence is, as you said, f(x)= 0 of x in [0, 1), f(x)= 1 for x= 1. This sequence in C(X) does NOT converge to a function in C(X) and so is NOT closed nor is it compact.

    By the way, is it clear to you that saying the functions are bounded is not the same as saying this set of functions is bounded? That depends on what metric you are using. The standard metric on C(X) is the "uniform metric"- distance from f to g is max|f(x)- g(x)| where the maximum is taken over all x in [0, 1]. You can, in fact, show that since all of the functions in the sequence are bounded on [0,1] the set is bounded in that metric.
  4. Dec 26, 2007 #3
    It forms a closed sphere with radius =1. Also, the limit points of the sequence of functions is either 0 or 1 and both are contained in the range of fn(x).

    Yes, I stand corrected. I realized that the sequence of functions are all continuous. It is just the f(x) to which the sequence tends to which is not continuous at x=1. Because
    f(x) =0 for 0<=x<1 and f(x) =1 for x=1. Hence if we choose epsilon = 0.5,
    |f(x) - f(1)| > epsilon, for any delta = |x-1|.

    Now, you are being unclear. The sequence, I understand converges to either 0 or 1 pointwise. I also understand that f(x), by itself is not part of C(X) because it is not continuous at x=1. But, why is the sequence of functions not closed? The range forms a closed subset in R ( the usual).

    Yes, that is the metric I am using || f-g|| , where ||f|| = sup{|f(x)|}
  5. Dec 26, 2007 #4
    If you can find a sequence {x_n} in a subspace of a metric space that converges to a point x, that implies that x is a limit point of the subspace (because it's a limit point of the subset {x_n}). Since f_n converges to an f not in the space, the space must fail to be closed. In particular, the sequence {f_n} has f as a limit point, and so it fails to contain all its limit points. Hence it cannot be closed.
    Last edited: Dec 26, 2007
  6. Dec 27, 2007 #5
    There seems to be some confusion on this question.. for me the confusion started with the question -- it sounds strange to ask if a sequence {a_n} is closed/compact, though bounded makes sense.

    The question is whether or not the set A = {a_n: n in N} is closed/bounded/compact. But it is closed: because it doesn't have any limit points in C(X). You have to associate some subspace analogy, say the space X = (0,1). A space is always open/closed in of itself.

    The other confusion is the difference between pointwise and uniform limit. In the uniform metric, a_n -> a if and only if the functions a_n -> a uniformly. So a pointwise limit is not necessarily a limit in the metric space.

    There was no confusion on whether or not it is bounded. BUT: it's not compact, because the set A is infinite and has no limit point! Chapter 2/3 of baby Rudin are good to keep in mind!
  7. Dec 27, 2007 #6
    More clarification...
    What I mean is that if you take a metric space X = (5,7), say, the arguments above look like the subset A = (6,7) is not closed in X because 7 is a limit point of A. But that is of course wrong because 7 is not in X either!
  8. Dec 27, 2007 #7
    I know what that closed set have all its limit points. You don't need to explain that part to me. By the way, I think you should read rudinreader's comments. They are right on dot!
  9. Dec 27, 2007 #8
    This is exactly what I had on my mind. Thanks for clarifying the same. It is very helpful. I understood the part of boundedness and the fact that the set of functions {x^n} is also not compact as there is a sequence {fn} which has limit point {f} that is not in C(X). The only confusion was about the set being closed/open. But you made that clear.
  10. Dec 27, 2007 #9
    You always have to be careful how you word things.. because even in this case (6,7) is not closed in X but [6,7) is.

    But even so, this analogy doesn't even apply to this case... Let f(x) be the pointwise limit of the f_n(x) defined above. As is known, it does not converge uniformly - thus f is not a limit of f_n in any metric space...

    I.e. if B(X) is the bounded functions with uniform metric, then f in B(X), but f_n -> f does not hold (with respect to metric) because the convergence is not uniform... In any case, f is not a limit of f_n in any metric space with uniform metric.

    Edit: As is probably clear by now I think the point of this problem is that you can't take the Heine-Borel theorem (compact <=> closed and bounded) for granted outside of R^k. http://en.wikipedia.org/wiki/Heine-Borel_theorem
    Last edited: Dec 27, 2007
  11. Dec 28, 2007 #10

    Yes, the Heine-Borel theorem and compactness is clear to me. I do have a question though about the limit of f_n "not being" a limit in any metric space. This part is not clear
    to me. Thanks again though!
  12. Dec 28, 2007 #11
    Again let f_n = x^n on [0,1], and f(x) is the pointwise limit (f(x) = 0, x < 1, f(1) = 1).
    Then f_n -> f pointwise, but not uniform.

    So in any metric space with uniform metric, we do not have f_n -> f as a limit of a sequence. Just look at why: d(f_n,f) = 1 for all n, so we do not have d(f_n,f) -> 0 - which is required for a limiting sequence.

    B([0,1]) = {the set of all bounded functions on [0,1]} is a metric space (check yhis claim) under the metric d(f,g) = sup|f(x)-g(x)|. In this case, you have f_n in B(X) for all n, and also f in B(X). But because the convergence is not uniform, you do not have d(f_n,f) -> 0.

    Unfortunately I don't have any topology books with me, so I can't discuss anything about the product topology and it's relation to pointwise convergence. But any such topology is not the same as the uniform metric situation above. Can look here: http://en.wikipedia.org/wiki/Topology_of_pointwise_convergence
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