# Ngô Quốc Anh

## November 30, 2009

### A property of the essentially bounded function 2

Filed under: Các Bài Tập Nhỏ, Giải Tích 6 (MA5205) — Ngô Quốc Anh @ 22:52

This topic is a companion to the following topic. In this topic, we consider the case when $E$ is the whole space, i.e. $E = \mathbb R^n$. We also add an extra function $g$ to $a_n$. To be precise, we have

Question. Suppose $g>0$ on $\mathbb R^n$ is in $L^1(\mathbb R^n)$ in Lebesgue sense. Let $f \in L^\infty(\mathbb R^n)$ such that $\| f\|_\infty > 0$. Define

$\displaystyle {a_n} = \int_{\mathbb R^n} {{{\left| f \right|}^ng}}$

for $n=1,2,3,...$ Show that

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \frac{{{a_{n + 1}}}} {{{a_n}}} = {\left\| f \right\|_\infty }$.

Solution. For any $\alpha$ with $0<\alpha < \|f\|_\infty$, let

$\displaystyle{E_\alpha } = \left\{ {x \in E: f\left( x \right) \geqslant \alpha } \right\}$

and

$\displaystyle {F_\alpha } = E\backslash {E_\alpha }$

then $\infty> |E_\alpha|>0$. Clearly when $\alpha$ is sufficiently closed to $\|f\|_\infty$, $\int_{E_\alpha}g>0$. For any $k \in \mathbb N$ ($k$ can be zero), note that

$\displaystyle\int_{{E_\alpha }} {{{\left| f \right|}^n}g} \geqslant {\alpha ^n}\int_{{E_\alpha }} g$

and

$\displaystyle\int_{{F_\alpha }} {{{\left| f \right|}^{n + k}}g} \leqslant \left\| f \right\|_\infty ^k\int_{{F_\alpha }} {{{\left| f \right|}^n}g}$.

Then

$\displaystyle\frac{{\int_{{F_\alpha }} {{{\left| f \right|}^{n + k}}g} }}{{\int_{{E_\alpha }} {{{\left| f \right|}^n}g} }} \leqslant \frac{{\left\| f \right\|_\infty ^k\int_{{F_\alpha }} {{{\left| f \right|}^n}g} }}{{{\alpha ^n}\int_{{E_\alpha }} g }} = \frac{{\left\| f \right\|_\infty ^k}}{{\int_{{E_\alpha }} g }}\int_{{F_\alpha }} {{{\left| {\frac{f}{\alpha }} \right|}^n}g}$.

By the Dominated Convergence Theorem, one gets

$\displaystyle 0 \leqslant \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{\int_{{F_\alpha }} {{{\left| f \right|}^{n + k}}g} }}{{\int_{{E_\alpha }} {{{\left| f \right|}^n}g} }}} \right) \leqslant \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{\left\| f \right\|_\infty ^k}}{{\int_{{E_\alpha }} g }}\int_{{F_\alpha }} {{{\left| {\frac{f}{\alpha }} \right|}^n}g} } \right) = 0$.

Hence

$\displaystyle\begin{gathered}\mathop {\lim \inf }\limits_{n \to \infty } \left( {\frac{{\int\limits_E {{{\left| f \right|}^{n + 1}}g} }}{{\int\limits_E {{{\left| f \right|}^n}g} }}} \right) \geqslant \mathop {\lim \inf }\limits_{n \to \infty } \left( {\frac{{\int\limits_{{F_\alpha }} {{{\left| f \right|}^{n + 1}}g} + \int\limits_{{E_\alpha }} {{{\left| f \right|}^{n + 1}}g} }}{{\int\limits_{{E_\alpha }} {{{\left| f \right|}^n}g} + \int\limits_{{F_\alpha }} {{{\left| f \right|}^n}g} }}} \right) \hfill \\\qquad \geqslant \mathop {\lim \inf }\limits_{n \to \infty } \left( {\frac{{\int\limits_{{F_\alpha }} {{{\left| f \right|}^{n + 1}}g} + \alpha \int\limits_{{E_\alpha }} {{{\left| f \right|}^n}g} }}{{\int\limits_{{E_\alpha }} {{{\left| f \right|}^n}g} + \int\limits_{{F_\alpha }} {{{\left| f \right|}^n}g} }}} \right) \hfill \\\qquad = \mathop {\lim \inf }\limits_{n \to \infty } \left( {\frac{{\int\limits_{{F_\alpha }} {{{\left| f \right|}^{n + 1}}g} }}{{\int\limits_{{E_\alpha }} {{{\left| f \right|}^n}g} }} + \alpha } \right)/\left( {1 + \frac{{\int\limits_{{F_\alpha }} {{{\left| f \right|}^n}g} }}{{\int\limits_{{E_\alpha }} {{{\left| f \right|}^n}g} }}} \right) \hfill \\ \qquad = \alpha . \hfill \\ \end{gathered}$

Letting $\alpha \nearrow {\left\| f \right\|_\infty }$, we get that

$\displaystyle\mathop{\lim }\limits_{n\to\infty }\left({\frac{{\int_{E}{{{\left| f\right|}^{n+1}g}}}}{{\int_{E}{{{\left| f\right|}^{n}g}}}}}\right) ={\left\| f\right\|_\infty }$.

As an application, if we put $a_0 = 1$, then from

$\displaystyle {a_{n + 1}} = \frac{{{a_1}}} {{{a_0}}}.\frac{{{a_2}}} {{{a_1}}} \cdots\frac{{{a_{n + 1}}}} {{{a_n}}}$

we deduce that

$\displaystyle\mathop{\lim }\limits_{n\to\infty }\sqrt[n]{{{a_{n}}}}=\mathop{\lim }\limits_{n\to\infty }\frac{{{a_{n+1}}}}{{{a_{n}}}}={\left\| f\right\|_\infty }$.

In other words,

$\displaystyle\mathop{\lim }\limits_{n\to\infty }{\left({\int_{E}{{{\left| f\right|}^{n}g}}}\right)^{\frac{1}{n}}}={\left\| f\right\|_\infty }$.

1. Your application may be proved with another way. Here is what I did:

$\displaystyle {a_n} = \int_{\mathbb R^n} {{{\left| f \right|}^ng}}\leq \|f\|_\infty^n\|g\|_1$,

hence

$a_n^{1/n}\leq \|f\|_\infty\|g\|_1^{1/n}\to\|f\|_\infty$

as $n\to\infty$. Take any $\epsilon > 0$, there exists a measurable $E$ such that $m(E) > 0$ and that $f(x)\geq \|f\|_\infty-\epsilon$ on $E$. So

$a_n^{1/n}\geq \left(\int_E|f|^n gdx\right)^{\frac{1}{n}}\geq (\|f\|_\infty-\epsilon)\cdot \left(\int_E gdx\right)^{1/n}\to \|f\|_\infty-\epsilon$

as $n\to\infty$. So I have proved

$\|f\|_\infty\geq \limsup a_n^{1/n}\geq \liminf a_n^{1/n}\geq \|f\|_\infty-\epsilon$,

for any $\epsilon > 0$. Thus, $\lim a_n^{1/n}=\|f\|_\infty$.

Comment by mrstoke — April 29, 2012 @ 21:13

• Thanks Dr. Hung for this approach and for your interest in my blog.

Comment by Ngô Quốc Anh — April 29, 2012 @ 21:18