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Trying to Prove Uniform Convergence: Analysis II

  1. Jan 22, 2012 #1
    1. The problem statement, all variables and given/known data

    I have a solution to the following problem. I feel it is somewhat questionable though

    If fn converges uniformly to f, i.e. fn[itex]\rightarrow[/itex]f as n[itex]\rightarrow[/itex]∞ and
    gn converges uniformly to g, i.e. gn[itex]\rightarrow[/itex]f as n[itex]\rightarrow[/itex]∞ ,

    Prove that fngn [itex]\rightarrow[/itex] fg

    The domain is ℝ for all functions.

    3. The attempt at a solution

    If f and g are uniformly continuous, then for any ε>0, there is an N for which n[itex]\geq[/itex]N implies | fn - f | < ε. The same for g.

    This is my idea:

    For any given ε >0, find N such that for n>N, |fn-f| < ε1, and |gn-f| < ε1. This ε1 should be made very small, such that:

    |fngn - fg| = | (f+ε1)(g+ε1) - fg|
    = | f*g + f*ε1 + g*ε1 + (ε1)2 - fg|
    The ε1 terms can be made very small as needed, such that:
    = | f*g + [STRIKE]f*ε1[/STRIKE] +[STRIKE] g*ε1[/STRIKE] + [STRIKE](ε1)2[/STRIKE] - fg| < ε

    Agree, or disagree? Secondly, is there a shorter, simpler method? Thanks.
    --Abraham
     
    Last edited: Jan 22, 2012
  2. jcsd
  3. Jan 22, 2012 #2

    LCKurtz

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    I probably should refrain from answering this because I have to leave momentarily. But my suggestion is that you need to state the question more carefully to begin with. I think you are confusing the ideas of uniform continuity and uniform convergence. You haven't told us what ##f_n\rightarrow f## means. Pointwise convergence? Uniform convergence? Any information about the domain? Problems like this are frequently addressed by adding and subtracting terms:$$|f_ng_n-fg|
    =|f_ng_n-f_ng+f_ng-fg|$$ and working with that.
     
  4. Jan 22, 2012 #3
    Sorry, I wrote the problem incorrectly. I meant to write:

    fn converges uniformly to f, i.e. fn→f.

    I don't know why I wrote "uniformly continuous" instead.....

    I see what you mean though, adding and subtracting quantities. I'll start with that.
     
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