Ngô Quốc Anh

Tháng Mười 30, 2009

A characteristic of essentially bounded functions

Chuyên mục: Các Bài Tập Nhỏ, Giải Tích 6 (MA5205), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 11:53

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.

No Comments Yet »

Chưa có phản hồi.

RSS các phản hồi của bài viết này. URL TrackBack

Để lại phản hồi

Blog at WordPress.com.