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Convergence on R->R

  1. Mar 5, 2009 #1
    Let fn : R [tex]\rightarrow[/tex] R be the function
    fn= [tex]\frac{1}{n^3 [x-(1/n)]^2+1}[/tex]

    Let f : R [tex]\rightarrow[/tex] R be the zero function.
    a. Show that fn(x) [tex]\rightarrow[/tex] f(x) for each x [tex]\in[/tex] R
    b. Show that fn does not converge uniformly to f. (This shows that the converse of Theorem 21.6 does not hold; the limit function f may be continuous even though the convergence is not uniform.)

    a. i'm not sure...
    is f(x) equivalent to f1(x)?
    If it is.... then... the function would be...
    but i'm not sure how to show fn(x) [tex]\rightarrow[/tex] f(x) for each x [tex]\in[/tex] R
    b. Theorem 21.6 states,
    " let fn: X[tex]\rightarrow[/tex] Y be a sequence of continuous functions from the topological space X to the metric space Y. If (fn) converges uniformly to f, then f is continuous. "

    The converse of this is...
    "If f is continuous, then (fn) converges uniformly to f. "

    i don't know how to prove a function is not convergent.

    Can someone help me?
    Thank You
  2. jcsd
  3. Mar 5, 2009 #2
    ad a) I would say, just multiply out the brackets in the denominator and look at what happens if you take [tex]\lim_{n\rightarrow \infty} f_{n}(x) [/tex] for fixed [tex] x [/tex].
    ad b) Uniform convergence means, that [tex] \sup_{x\in \mathbb{R}} |f_{n}(x)-f(x)| \rightarrow 0[/tex]. Now try to choose your x such that you see, that this does not happen (you should see what to do by looking at the demoninator).
  4. Mar 5, 2009 #3


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