$\newcommand{\esssup}{\mathrm{ess\,sup}}$$\newcommand{\essrng}{\mathrm{ess\,range}}$ I am trying to prove that $\esssup(f) = \sup(\essrng(f))$, where we define $$ \esssup(f) = \inf \{b \in \mathbb{R}_+ : \mu(f^{-1}((b, \infty))) = 0\} $$ and similarly $$ \essrng(f) = \{w \in \mathbb{R}_+ : \mu(f^{-1}(B(w, \epsilon))) > 0\}. $$ Actually, I already showed that $\esssup(f) \leq \sup(\essrng(f))$, but I have not been able to prove the other direction. The answer on this related question hasn't been useful for me to prove the desired reverse inequality.

If you could give me a hint to prove it, I'll be really grateful.

  • $\begingroup$ Hi. How is this different from the question you asked here on May 22? math.stackexchange.com/questions/3686917/… $\endgroup$ – Alonso Delfín Jul 2 '20 at 23:19
  • $\begingroup$ Does this answer your question? Equivalence of definitions for $L_\infty$ norm via essential range and essential supremum $\endgroup$ – Alonso Delfín Jul 2 '20 at 23:21
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    $\begingroup$ Oh, es la misma pregunta de hecho, Alonso jaja, sólo que en ella nunca respondieron, así que, ya la eliminé para evitar mal entendidos, (espero no te moleste que te responda en español ya que vi que eres de la CDMX) y ahora ya tengo una desigualdad resuelta, la que menciono. $\endgroup$ – dante arroyo Jul 2 '20 at 23:36
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    $\begingroup$ En esa que mencionas pues, de hecho no resuelve mi duda, ya que sólo dicen que es fácil de probar, y la verdad, no logro ver como. $\endgroup$ – dante arroyo Jul 2 '20 at 23:37
  • $\begingroup$ Hola Dante. Ya veo, gracias por aclarar. Voy a agregar lo que me comentas a tu pregunta para que no te la cierren. $\endgroup$ – Alonso Delfín Jul 2 '20 at 23:54

$\newcommand{\esssup}{\mathrm{ess\,sup}}$$\newcommand{\essrng}{\mathrm{ess\,range}}$ You want to prove that $\esssup(f) \geq \sup (\essrng(f))$. For this, it's enough to show that if $w \in \essrng(f)$, then $\esssup(f) \geq |w|$.

Hint: You can do this by contradiction: Assume there is $w_0 \in \essrng(f)$ such that $\esssup(f)<|w_0|$ and try to find a contradiction by playing around with the definitions of $\esssup$ and $\essrng$.

Solution: (Try not to look at this until you've tried to solve it by yourself)

I am assuming that the measure space here is called $X$. By definition of essential supremum we have $|f(x)| \leq \esssup(f)$ a.e. $[\mu]$, so$$ \mu\big( \left\{ x \in X : |f(x)| > \esssup(f) \right\} \big) =0$$Take any $w_0 \in \Bbb{C}$ and suppose that $|w_0| > \esssup(f)$. Then, there is $\varepsilon_0$ such that $|w_0| - \varepsilon_0 >\esssup(f)$. Therefore, if $x \in X$ is such that $|f(x)-w_0|< \varepsilon_0$ the reverse triangle inequality implies that $|f(x)|>|w_0|-\varepsilon_0 > \esssup(f)$. Hence $$\{x \in X : |f(x)-w_0|< \varepsilon_0 \} \subset \{ x \in X : |f(x)| >\esssup(f)\}$$However, if $w_0 \in \essrng(f)$,$$0 < \mu\big( \{x \in X : |f(x)-w_0|< \varepsilon_0 \} \big) \leq \mu\big( \{ x \in X : |f(x)| > \esssup(f) \} \big)=0$$a contradiction! Then we must have $|w| \leq \esssup(f)$ for any $w \in \essrng(f)$, as desired.

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    $\begingroup$ Oh, thank you so much, I haven't seen the $ess range$ in that form, it's really helpful, =) $\endgroup$ – dante arroyo Jul 6 '20 at 17:36
  • $\begingroup$ De nada Dante! Me da gusto que te fue útil la respuesta. $\endgroup$ – Alonso Delfín Jul 7 '20 at 1:10

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