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let $(X, \| \|$) be the normed linear space of bounded uniformly continuous real valued functions defined on $R$. I need to prove that X is complete (under sup norm).

My attempt:

Let {$f_n$} be a Cauchy sequence in X converging to $f$ in the completion of X.

I am checking the uniform continuity of f below.

$|f(x)-f(y)|$ = $ |f(x) - f_n(x) + f_n(x) - f_n(y) + f_n(y) - f(y)|\;$ $\leq\; |f(x) - f_n(x)|+ |f_n(x) - f_n(y)| + |f_n(y) - f(y)|$

By choosing an x in X, I get an $N_1(x,\epsilon/3)$ such that $|f(x) - f_n(x)| \leq \epsilon/3$ $\forall$ n $\geq$ $N_1$. However, for the next two terms I don't know how to deal with them. I can't fix either y nor n because both are interdependent. I want to use the fact that $f_n$-s are bounded somehow into it.

I can prove that f is bounded on X using $|f(x)| \leq |f(x) - f_n(x)| + |f_n(x)|$ once I get uniform convergence of $f_n$.

If I could get help in proving that $f_n$ is uniformly convergent on $f$, then it would be great.

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First, $f_n$ Cauchy in $||\cdot||_{\infty}$ norm implies that for every $x\in \mathbb{R}$ $f_n(x)$ is Cauchy, so by completeness of $\mathbb{R}$ $f_n\to f$ pointwise.

Next as $f_n$ is Cauchy given $\varepsilon>0$ there is $N$ such that $m,n\ge N$ implies $||f_m-f_n||_{\infty}<\varepsilon/3$. Taking limit as $m\to\infty$ we get

$$ ||f-f_n||_{\infty}\le\varepsilon/3 $$

so $f_n\to f$ uniformly.


You also need $f$ to be uniformly continuous. For this note that since $f_n$ is uniformly continuous for all $n$, there is $\delta>0$ such that $$|x-y|<\delta \implies |f_N(x)-f_N(y)|<\varepsilon/3 $$

So if $|x-y|<\delta$

$$ |f(x)-f(y)|\le |f(x)-f_N(x)| + |f_N(x)-f_N(y)| + |f(y)-f_N(y)| <\varepsilon/3+\varepsilon/3+\varepsilon/3=\varepsilon $$

Note that you get $|f(x)-f_N(x)|<\varepsilon/3$ and $|f(x)-f_N(y)|<\varepsilon/3$ from uniform convergence (recall how we chose $N$)

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  • $\begingroup$ Thank you for your answer....a small clarification...is it correct to use $ \| .\|_{\infty}$ on $ \| f - f_n \|_{\infty} < \epsilon/3$ when we don't know yet if $f$ is in our normed space X. Could you justify (for my sake of understanding), "Taking limit as $m\to\infty, \| f - f_n \|_{\infty} < \epsilon/3$"? $\endgroup$ – user166305 Feb 4 '16 at 19:21
  • $\begingroup$ May be in this case, we use the same norm on higher spaces too, like continuous functions on $R$. But in general if I am in a normed space about whose completeness I am unaware, how can i use the same norm on $f$? $\endgroup$ – user166305 Feb 4 '16 at 19:30
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    $\begingroup$ @user166305 the norm itself can be defined for arbitrary $f$, it's just that such "norm" of $f$ might not be finite. For instacne here $||\cdot||_{\infty}$ is just a supremum of the function over domain, and this can even be defined for some functions like $f=1_{\mathbb{Q}}$ which take is $1$ on rationals and $0$ otherwise (of course then $||f||_{\infty}=1$. We can take limit inside the norm since norm is continuous function. $\endgroup$ – user160738 Feb 4 '16 at 20:21
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    $\begingroup$ @user166305 if you are unsure, consider this: for all $n,m\ge N$, we know $|f_n(x)-f_m(x)|< \varepsilon/3$ for every $x\in \mathbb{R}$. Take limit on both sides as $m\to \infty$, then since $|\cdot|$ is continuous function, we have $|f_n(x)-f(x)|\le\varepsilon/3$, for all $x$. So this means $||f_n-f||_{\infty}\le\varepsilon/3$ $\endgroup$ – user160738 Feb 4 '16 at 20:23
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The sequence $f_n$ is uniformly convergent because of the norm you have chosen: $$\Vert f_n-f\Vert=\sup_{x}|f_n(x)-f(x)|.$$ Thus, for any $y$, $|f_n(y)-f(y)|\leq \sup_{x}|f_n(x)-f(x)|=\Vert f_n-f\Vert.$ So, if you insist that $\Vert f_n-f\Vert<\varepsilon/3,$ then $|f_n(y)-f(y)|<\varepsilon/3$ for every $y$.

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