# Ngô Quốc Anh

## July 31, 2009

### A property of the essentially bounded function

Let $E$ be a subset of $\mathbb R^n$ with $|E| < \infty$ in Lebesgue sense. Suppose $f \in L^\infty(E)$ and $\| f\|_\infty > 0$. Set

$\displaystyle {a_n} = \int_E {{{\left| f \right|}^n}}$

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 $|E_\alpha|>0$. For any $k \in \mathbb N$, by the Dominated Convergence Theorem,

$\displaystyle\mathop{\lim }\limits_{n\to\infty }\left({\dfrac{{\int_{{F_\alpha }}{{{\left| f\right|}^{n+k}}}}}{{\int_{{E_\alpha }}{{{\left| f\right|}^{n}}}}}}\right)\leqslant\underbrace{\mathop{\lim }\limits_{n\to\infty }\frac{1}{{\left|{{E_\alpha }}\right|}}\int_{{F_\alpha }}{{{\left|{\frac{f}{\alpha }}\right|}^{n}}\left\| f\right\|_\infty^{k}}}_{0}$.

Hence

$\displaystyle\mathop{\lim\inf }\limits_{n\to\infty }\left({\frac{{\int_{E}{{{\left| f\right|}^{n+1}}}}}{{\int_{E}{{{\left| f\right|}^{n}}}}}}\right)\geqslant\mathop{\lim\inf }\limits_{n\to\infty }\left({\frac{{\alpha\int_{{E_\alpha }}{{{\left| f\right|}^{n}}}+\int_{{F_\alpha }}{{{\left| f\right|}^{n+1}}}}}{{\int_{{E_\alpha }}{{{\left| f\right|}^{n}}}+\int_{{F_\alpha }}{{{\left| f\right|}^{n}}}}}}\right) =\alpha$.

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

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

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,

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