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Let $\left(X,M,\mu\right)$ be a measure space, $f:X\to\mathbb{C}$ a measurable function and $z_0\in R_f$ where $R_f$ is the essential range of $f$, that is $$R_{f}=\left\{ z\in\mathbb{C}\vert\forall\epsilon>0\quad\mu\left(\left\{ x\vert\left|f\left(x\right)-z\right|<\epsilon\right\} >0\right)\right\} $$

Claim

If $\mu\left(\left\{ x\vert\left|f\left(x\right)\right|>\left|z_{0}\right|\right\} \right)>0$ then there exists a $z\in R_f$ such that $\left|z\right|>\left|z_{0}\right|$.

Discussion

What I believe the claim means, very intuitively, is that if I $f$ gets a significant amount of values "bigger" than $z_0$ then there is a point $z$ in which $f$ gets a significant amount of values.

The claim sounds true but I can't seem to prove it - I suspect my lack of familiarity with the complex plane. Or perhaps I am missing something elementary?

To provide context, I require this claim when I attempt to prove that for $f\in L^\infty$, $\left\Vert f\right\Vert _{\infty}\leq\max_{z\in R_{f}}\left(\left|z\right|\right)$.

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  • $\begingroup$ Also, I wasn't sure how to phrase the title -- if anyone has a better idea please feel free to edit it. $\endgroup$ – 8l2s Jan 11 '17 at 11:27
  • $\begingroup$ I don't feel why this should be true, what's your intuition ? $\{z, |z|>|z_0|\}$ is unbounded, so it needn't have some accumulation point. $\endgroup$ – Gabriel Romon Jan 11 '17 at 12:10

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