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If $f(x)$ is a continuous function on all of $\mathbb R$, with the property that $\sup_{x\in\mathbb R}|f(x)|\leq 1$. If this is the case, how I can test if the sup is attained or not? (i.e., if there exists at least $x_{o}\in \mathbb R$ such that $|f(x_{o})|\geq |f(x)|, \forall x\in \mathbb R$).

Should we have something like $\lim_{x\to\pm\infty}|f(x)|=0$, or something else?

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With $f(x)=1-e^{-x^2}$ we can see that the $\sup$ is not attained. However, under the condition $\lim_{|x|\to +\infty}f(x)=0$, it is reached. Two cases: $f$ is identically $0$ (hence it's obvious) or not. In this case, $2s:=\sup_{x\in \Bbb R}|f(x)|>0$. You can find $R$ such that if $|x|\geq R$ then $|f(x)|\leq s$. Hence, by continuity, the supremum is reached somewhere in the compact $[-R,R]$.

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    @Catherine I fact I didn't understand well the question. I thought first that the question was: if the limit at infinity is$0$then $\sup|f|$ is attained. In fact you asked the converse, which doesn't seem to be true, for example with $f$ constant equal to $1$.2012-06-17
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If,e.g., you somehow know that

$\sup_\mathbb{R}|f|> \lim_{x\rightarrow\infty}\sup_{|y|>|x|} |f(y)|$

you can conclude that there must be some bounded closed interval $I$ such that $\sup_I|f| =\sup_{\mathbb{R}}|f|$ and $f$ will attain it's supremum. In general you will run into problems (see Davide's answer for an example ;-).