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Uniform convergence problem

  1. May 6, 2010 #1
    let f is a continuous, real-valued function on [a,b]

    then, for any e, there exist a polygonal function p such that

    sup|f(x)-p(x)|<e

    using uniform convergence, this might be shown... but i cannot figure it out...
     
  2. jcsd
  3. May 6, 2010 #2

    LCKurtz

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    Use the fact that a continuous function on [a,b] is uniformly continuous.
     
  4. May 6, 2010 #3
    how does it have to d with polygonal function?
     
  5. May 7, 2010 #4
    This is a basic rough sketch of the argument ...there are a few gaps that need to be filled in by yourself to be rigorous...

    Since [a,b] is a closed and bounded interval, note that f(x) is uniformly continous on this interval.

    So given any epsilon > 0 , then there must exist a delta such that
    | f(x) - f(y) | < (epsilon/2) whenever |x-y| < delta. This is simply the definition of uniform convergence.

    Now choose an appropriate partition for the interval [a,b] such that |x(n+1) - x(n)| < delta for all n.

    Now by the triangle inequality |p(x)-f(x)|<= |p(x)-p(x(n))|+|p(x(n))-f(x(n))|+|f(x(n))-f(x)| =
    |p(x)-p(x(n))|+|f(x(n))-f(x)|<= epsilon, since |x-x(n)|<delta.
     
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