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Uniform convegence and continuity

  1. Nov 16, 2009 #1
    If we know that some function f(x) converges uniformly on some interval does that imply that it is also continuous on the interval?
  2. jcsd
  3. Nov 16, 2009 #2


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    it does not make sense to speak of the convergence of a function.
  4. Nov 16, 2009 #3
    Sorry, say we have a series [tex]f(x)= \Sigma n \cos(nx) e^{-n^2 x} [/tex] and know that is converges uniformly on some interval [tex] [a, \infty) [/tex] could we then conclude that it was continuous for all x in [tex] [a, \infty) [/tex] ?

    I know there is a theorem that says that in order for there to be uniform convergence, f(x) and f (where [tex] f(x) \rightarrow f [/tex]) must be continuous. But does the theorem work the other way?

    If not, how would I find for which x [tex] f(x) [/tex] is continuous?
  5. Nov 16, 2009 #4


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    If you have a sequence of functions which are continuous and converge uniformly to a function, then that function must be continuous. In your example they are referring to the partial sums as the sequence of functions that are converging uniformly.

    Without speaking of the continuity of the sequence of functions nothing about continuity can be said, for example if fn(x) = (1-1/n)*X(x) where X is the indicator function of the rationals between 0 and 1, fn converges uniformly to the indicator function of the rationals between 0 and 1 but obviously none of these functions are continuous. As another example if fn(x) = X(x)/n then fn converges uniformly to 0 which is continuous, but none of the fns are continuous themselves
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