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Homework Help: Continuity and liimit of functions

  1. Dec 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Suppose [tex]f_n : [0, 1]\rightarrow R[/tex] is continuous and lim[tex]_{n \rightarrow \infty}f_n(x)[/tex] exists for each x in [0,1]. Denote the limit by [tex]f(x)[/tex].

    Is f necessarily continuous?

    2. Relevant equations
    We know by Arzela-Ascoli theorem:
    If [tex]f_n: [a,b] \rightarrow R[/tex] is continuous, and [tex]f_n[/tex] converges to [tex]f [/tex]uniformly, then [tex]f[/tex] is continuous.

    3. The attempt at a solution
    Question: Does the fact of knowing
    give us insight to declare that [tex]f_n[/tex] converges to [tex]f[/tex] uniformly- and thus satisfying Arzela-Ascoli's theorem?


    Jeffrey Levesque
    Last edited: Dec 12, 2009
  2. jcsd
  3. Dec 12, 2009 #2


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    Well, no. They only gave you that the fn are continuous and converge pointwise. I think they want you find an example of a sequence of continuous functions that are pointwise convergent but don't have an continuous limit. Can you think of one?
  4. Dec 12, 2009 #3
    Can someone provide some insight for me as to what the following means:

    And how I could use this fact to construct my justification for whether f is necessarily continuous?
  5. Dec 12, 2009 #4


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    f isn't necessarily continuous. Face it. It says fn converges at each point. That's not enough to prove f is continuous.
  6. Dec 13, 2009 #5


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    Dick is suggesting that you find a counter-example. Taking fn to be piecewise linear will suffice.
  7. Dec 13, 2009 #6


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    The Arzela-Ascoli theorem asserts something about a sequence of equicontinuous functions. This has little to do with your question [or you have seen a different version of A-A].

    Just construct a counter-example, i.e. a sequence of continuous functions (f_n)_n which converges pointwise to some discontinuous function. (Every book that introduces the concept of 'uniform convergence' will have such a counter-example, so you probably have encountered one already.)
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