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Homework Help: Sequence of functions, uniform convergence

  1. Nov 12, 2008 #1
    1. The problem statement, all variables and given/known data
    {f_n} is a sequence of continuous functions on E=[a,b] that converges uniformly on E. for each x in E, set g(x)=sup{f_n(x)}. Prove that g is continuous on E

    2. Relevant equations

    3. The attempt at a solution
    I've got an idea to prove it:
    Let e>0 be given, there is a N such that
    for all n>N and all x in E.
    Let K_n={x|f_n(x)>=e+f(x)}, n=1,2,3....,N.
    each K_n is closed since f_n is continuous.
    Let [tex]A=\cup^{N}_{1} K_{i}[/tex] then A is closed.
    g(x) is continuous on A (since there are only N functions to deal with. And I find it tedious to prove?)
    Let B=E-A.
    Then f(x)<=g(x)<f(x)+e, for any x in B.
    let p be an interior point in B.
    if g(p)=f_k(p), for some k>N.
    since f_k(p) is continuous, there is a neighborhood of p such that if q is in it and also in B.
    hence g(x) is continuous at p.
    if g(p)=f(p).
    |g(p)-g(q)|<=|f(p)-f(q)|+|f(q)-g(q)|=2e, which implies
    g(x) is continuous at p.
    It is analogous to say that g(x) is continuous if p is the boundary point of B.

    It seems okay. But I'm not satisfied with it, it looks so lenthy and not well organized and I believe there is some other proof much better than this one. Any hint?
  2. jcsd
  3. Nov 12, 2008 #2


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    I don't think you quite have it down.

    Imagine if your functions were just fk(x) = 1-1/k on the whole interval. Then fk converges to 1 uniformly, and clearly A is empty as fn < f everywhere for all n. Then B is the whole interval, yet g(x) =/= fk(x) for any x,k combination. I think this is fixed by a better definition of Kn (probably what you wrote isn't what you really intended)
  4. Nov 12, 2008 #3
    I think the case that g(x)=f(x) is included in the last few lines...check it and tell me if anything goes wrong.Thanks!
  5. Nov 12, 2008 #4


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    Whoops.... I missed the word 'if'. Pretty important
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