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Proving Completeness of a function space

  1. Dec 24, 2009 #1
    Let [tex]F = \left\{f : [0, \infty) \rightarrow R, norm(f) = \sup_{x \in [0,\infty)} \frac{|f(x)|}{x^{2} + 1} < \infty\right\}[/tex]

    Is F complete , under the given norm ?

    My approach was to look at the pointwise limit of an arbitrary Cauchy sequence, but I am not able to prove that it converges in the metric induced by the norm.

    Thank you and Merry Christmas !
     
  2. jcsd
  3. Dec 24, 2009 #2
    Hi , is the following correct ?
    (an outline of the proof )
    Given an arbitrary Cauchy sequence (f_{n})
    we have that
    [tex] \forall \epsilon > 0, \exists n_{\epsilon} \leq m < n \, s.t \sup_{0 \leq x < \infty} \frac{|f_{n}(x) - f_{m}(x)|}{x^{2} + 1} < \epsilon [/tex]

    [tex] g_{n}(x) = f_{n}(x)/(x^{2} + 1) [/tex]

    this means that [tex] \sup_{ 0 \leq x \infy} |g_{n}(x) - g_{m}(x)| < \epsilon [/tex] for m, n as above

    so that [tex] g_{n}(x) \rightarrow g(x) , \forall x, uniformly [/tex]
    which means that [tex] \sup_{0 \leq x < \infty} |g_{n}(x) - g(x)| \rightarrow 0 [/tex]

    with [tex] f(x) = g(x)(x^{2} + 1) [/tex]

    [tex] \sup_{0 \leq x < \infty} |\frac{f_{n}(x) - f(x)}{x^{2} + 1}| \rightarrow 0 [/tex]
    [tex] \forall x, \frac{|f(x)|}{x^{2} + 1} \leq \frac{|f_{n}(x) - f(x)|}{x^{2} + 1} + \frac{|f_{n}(x)|}{x^{2} + 1} [/tex] which should give that f \in F ?
    Thank you for your time !
     
    Last edited: Dec 24, 2009
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