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Let $(S,\mathcal A, \mu)$ be a measure space and consider the Riesz space $L^\infty=L^\infty(S,\mathcal A, \mu)$ (under point-wise ordering). Let $1_X$ denote the indicator function on $S$ (which is contained $L^\infty$).

Given an arbitrary $f\in L^\infty$, is it possible to find a $\alpha\in\mathbf{R}$ such that $-\alpha\cdot1_S\le f\le \alpha\cdot 1_S$ The $\cdot$-symbol denotes the point-wise multiplication, that is $\alpha\cdot 1_S=\alpha\cdot 1_S(x)$ for all $x\in S$.

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As $f \in L^\infty$, we have $|f| \le \|f\|_\infty$ almost everywhere. If we let $\alpha = \|f\|_\infty$, we are done.

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    @t.b. Thanks. I overlooked that part and it makes perfect sense now.2012-08-06