Ngô Quốc Anh

October 30, 2009

A characteristic of essentially bounded functions

In this topic, we prove the following statement

Statement: Let $(X,\mathcal B, m)$ be a probability space. Let $h \in L^2(m)$. Then $h$ is essentially bounded iff $h \cdot f \in L^2(m)$ for all $f \in L^2(m)$.

Proof. If $h$ is bounded, then by using the Holder inequality one has

$\displaystyle\int_X {{{\left| {h \cdot f} \right|}^2}dm}\leq \underbrace {\sqrt {\int_X {{{\left| h \right|}^2}dm} } }_{ \leqslant c}\sqrt {\int_X {{{\left| f \right|}^2}dm} }<+\infty$

for all $f \in L^2(m)$. Conversely, we suppose $h$ is such that $h \cdot f \in L^2(m)$ whenever $f \in L^2(m)$. Let

$\displaystyle X_n = \{ x \in X : n-1 \leq |h(x)| < n\}, \quad \forall n>0$.

Then $\{X_n\}_1^\infty$ partitions $X$. Let

$\displaystyle f\left( x \right) =\sum\limits_{n = 1}^\infty{\frac{1}{{n\sqrt {m\left( {{X_n}} \right)} }}{\chi_{{X_n}}}\left( x \right)} ,$

where it is understood that the $n$-term is omiited if $m(X_n)=0$. Then

$\displaystyle\int_X {{{\left| f \right|}^2}dm}=\int_X {{{\left({\sum\limits_{n = 1}^\infty {\frac{1}{{n\sqrt {m\left( {{X_n}}\right)} }}{\chi _{{X_n}}}\left( x \right)} } \right)}^2}dm}\leq \sum\limits_{n = 1}^\infty{\frac{1}{{{n^2}}}}<\infty$

which implies $f \in L^2(m)$. Since

$\displaystyle\int_X {{{\left| {hf} \right|}^2}dm}=\sum\limits_{n \in F} {\int_{{X_n}} {{{\left| {hf} \right|}^2}dm}}\geq\sum\limits_{n \in F}{{{\left( {\frac{{n - 1}}{n}}\right)}^2}}$

where $F = \left\{ {n:m\left( {{X_n}} \right) \ne 0} \right\}$. Sincc $h \cdot f \in L^2(m)$ we have that $F$ is finite and therefore $h$ is essentially bounded.