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Consider the subset S = $L^\infty(\mathbb{R}) \cap L^2(\mathbb{R})$ of $L^\infty(\mathbb{R})$. We know that $L^\infty(\mathbb{R})$ is a Banach space and complete. Is this subset $S$ complete under the $\|\cdot\|_\infty$ distance metric? If yes, where can I find a proof for it?

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No. Since $S$ is a subset of the complete space $L^\infty$, it will be complete if and only if it is closed. But this is not true since any continuous function of compact support is in $S$, and the closure in the $L^\infty$-norm of the continuous functions of compact support is the set $C_0$ of continuous functions vanishing at infinity. It's fairly simple to furnish such a function which is not in $L^2$. For example, $f(x) = \frac{1}{(1 + |x|)^{1/2}}$ will do the trick.

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Funny, now you've added (independently I guess) the same example of a missing limit I gave. In any case, we gave different ways of approaching it. I like this answer. – Jonas Meyer Jul 1 '12 at 5:24
@JonasMeyer Yes, your answer came while I was editing. Rest assured I was not trying to steal your work; after all, this is the standard example of a non-integrable function vanishing at infinity. Incidentally, I like your answer too, its a little bit more constructive. – user12014 Jul 1 '12 at 5:30
Remark: An unclosed subspace is incomplete regardless of whether the containing space is complete. – Jonas Meyer Jul 1 '12 at 5:38

If $f_n(x)=\chi_{[-n,n]}(x)\cdot\dfrac{1}{\sqrt{1+|x|}}$, then $(f_n)$ is a Cauchy sequence with no limit.

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How does $1\cdot\chi_{[-n,n]} \rightarrow 1$ in $L^\infty$? The $\infty$ - norm of the difference is always equal to one. Note that $L^\infty$ is not separable. Are you talking about those $f$ which vanish at $\infty$? – user20266 Jul 1 '12 at 5:59
@Thomas: I had initially been thinking of $f$ which vanish at $\infty$, hence the example I gave, then I apparently stopped thinking. Thanks for pointing out the errors, which I will now remove. – Jonas Meyer Jul 1 '12 at 6:03

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