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 }.

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