Proving Completeness of a function space

1. Dec 24, 2009

vineethbs

Let $$F = \left\{f : [0, \infty) \rightarrow R, norm(f) = \sup_{x \in [0,\infty)} \frac{|f(x)|}{x^{2} + 1} < \infty\right\}$$

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. Dec 24, 2009

vineethbs

Hi , is the following correct ?
(an outline of the proof )
Given an arbitrary Cauchy sequence (f_{n})
we have that
$$\forall \epsilon > 0, \exists n_{\epsilon} \leq m < n \, s.t \sup_{0 \leq x < \infty} \frac{|f_{n}(x) - f_{m}(x)|}{x^{2} + 1} < \epsilon$$

$$g_{n}(x) = f_{n}(x)/(x^{2} + 1)$$

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

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

with $$f(x) = g(x)(x^{2} + 1)$$

$$\sup_{0 \leq x < \infty} |\frac{f_{n}(x) - f(x)}{x^{2} + 1}| \rightarrow 0$$
$$\forall x, \frac{|f(x)|}{x^{2} + 1} \leq \frac{|f_{n}(x) - f(x)|}{x^{2} + 1} + \frac{|f_{n}(x)|}{x^{2} + 1}$$ which should give that f \in F ?
Thank you for your time !

Last edited: Dec 24, 2009