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Let S be any subset of $\mathbb{R^n}$. Let $C_b(S)$ denote the vector space of all bounded continuous functions on S. For $f \in C(S)$, define $\| f \|_\infty = \sup_{x \in S} |f(x)|$

When is this a norm of the vector space of all continuous functions on S?

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The presentation here is confusing. Why do you mention what the notation $C_b(S)$ means? Note that by definition of norm, $\|f\|_\infty$ must be a real number for all such $f$. When is it true that for all $f\in C(S)$, $\sup\limits_{x\in S}|f(x)|<\infty$? – Jonas Meyer Feb 6 '13 at 3:33
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Like Jonas mentioned, your question seems incomplete, since you define $C_{b}(S)$, yet make no use of it. Here are some facts to perhaps answer the question you intended:

(1) For continuous functions with compact support, $||f||_{\infty}$ is in fact equivalent to the ''sup norm' as you defined in your question. Note that $||f||_{L^{\infty}(S,\mathscr{S},\mu)}:=\inf\limits_{\alpha\in\mathbb{R}}\mu\left(\{x:|f(x)|>\alpha\}\right)=0.$ In other words, it is the smallest positive real number $\alpha$ whereby the set on which $|f|(x)>\alpha$ has measure zero (we call this the essential upper bound, since as far as integration is concerned, what $f$ does on a set of measure zero can be ignored). The previous assertion now follows from the observation that compact support and continuity means $f$ achieves its (finite) optimal values on its support.

(2) When the space of continuous functions with compact support, $C_{B}(S)$ as you refer to it, is endowed with the $L^{p}$ norm ($1\leq p<\infty$), then unlike for the usual $L^{p}$ spaces, $||\cdot||$ defines an actual metric without having to first quotient out almost everywhere equal functions: this is because if two continuous functions differ at a point, then they necessarily (because of continuity) differ on an open set, and such an open set has positive measure since it also contains a k-cell. But $C_{B}(S)$ is not complete, for it is an open and dense sub(space!) within $L^{p}$, which is complete (this is, when combined with Lusin's theorem, a simple consequence of step functions also being a dense and open subspace within $L^{p}$). Recall that complete metric spaces are closed and can never be open and dense within a complete metric space, for the metric completion of such a subspace is unique upto isomorphic embedding. Anyway, the point is, with the $||\cdot||_{L^{p}}$ metric, the completion of $C_{B}(S)$ is in fact $L^{p}$ (note $p=\infty$ is excluded from this point!).

(3) Recall that the ordinary Riemann integral is defined only for continuous functions with compact support (boundedness being implied). Then (2) then implies that the Lebesgue integral is the proper construction for extending the Riemann integral! Compare this to the usual direct proof.

(4) For the case $p=\infty$, the completion of $C_{B}(S)$ is not $L^{\infty}$!. It is in fact the space of continuous functions that are in some sense "locally compactly supported." Precisely, the completion of $C_{B}(S)$ consists of those continuous functions defined on $S$ which vanish at infinity: for any $\epsilon>0$ there exists a compact set $K\subset\text{supp}f$ whereby $|f|(x)<\epsilon$ for $x\in K^{c}$. The proof of this is just a direct application of usual metric space techniques. Despite this, going back to your original question, this does not mean $|\cdot|_{\infty}$ doesn't define a metric on $C_{B}(S)$, it of course does. Just that its completion is not in relation to the $L^{\infty}$ space as it is for the case $1<p<\infty$ (note that the cases $0<p<1$ are not treated since, among other things, the $L^{p}$ norm here only defines a quasimetric).

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I thought the question was probably trying to get at what the characterization is of subspaces of $\mathbb R^n$ such that every continuous $\mathbb R$ valued function on the subspace is bounded, i.e., pseudocompactness, which for subspaces of metric spaces is equivalent to compactness. – Jonas Meyer Feb 6 '13 at 17:29
Honestly, the question is from my textbook and i think it has a typo, since it refers to two cases of continuous functions, thanks for your response, i shall consult with my professor to see if it indeed a typo – bobdylan Feb 6 '13 at 18:36
Well there is another case, which is just the bounded continuous functions defined on an arbitrary metric space. This space is complete in and of itself with respect to the "sup norm" metric, which is just the $L^{\infty}$ norm. Note carefully though the distinction between this particular space of continuous functions and the ones mentioned above. – Taylor Martin Feb 6 '13 at 18:39

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