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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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up vote 3 down vote accepted

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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But this bound only works almost everywhere. Sure, you can find a representative of $f$ from same equivalence class for which this works, but in general $f$ can be unbounded. Right? – T. Eskin Aug 6 '12 at 13:33
@ThomasE. the order structure on $L^\infty$ (as a Riesz space) is $f \leq g$ if and only if $g - f \geq 0$ almost everywhere. This is true in every $L^p$ with $0 \leq p \leq \infty$. Pointwise ordering wouldn't make sense for the reason you note. – t.b. Aug 6 '12 at 13:34
@t.b. Thanks. I overlooked that part and it makes perfect sense now. – T. Eskin Aug 6 '12 at 13:37

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