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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...
    [tex]\frac{1}{[x-1]^2+1}[/tex]
    [tex]\frac{1}{x^2-2x+2}[/tex]
    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

    HallsofIvy

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