Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $x\in S$.

share|improve this question

1 Answer 1

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.

share|improve this answer
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? –  Thomas E. 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. –  Thomas E. Aug 6 '12 at 13:37

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.