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 }.

2 Comments »

  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


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