3.5. Convergence in \(L^p\)

Definition

A sequence \(\{f_n:n\ge 1\}\) of \(L^p\) functions, \(0<p<\infty\) is said to converge in \(L^p\) to a measurable function \(f\) (written \(f_n\stackrel{L^p}\rightarrow f\)) if \[ \int |f_n-f|^p d\mu\rightarrow 0\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ as }n\rightarrow \infty. \] In special case, if \(X_n\) and \(X\) are random variables \(X_n\stackrel{L^p}\rightarrow X\), \[ \int |X_n-X|^p d\mu\rightarrow 0\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ as }n\rightarrow \infty, \] then we sometimes say that \(X_n\) converges to \(X\) in \(p\)th mean.

For \(1\le p<\infty\), \(f_n\stackrel{L^p}\rightarrow f\) is equivalent to \(||f_n-f||_p\rightarrow 0\), and this is also the definition for the case \(p=\infty\),

i.e., a sequence \(\{f_n,n\ge 1\}\) of \(L^\infty\) functions is said to converge in \(L^\infty\) to a measurable function \(f\) if \(||f_n-f||_\infty\rightarrow 0\) as \(n\rightarrow \infty\).

Proposition

If \(f_n\stackrel{L^p}\rightarrow f\) form some \(0<p\le \infty\), then \(f\in L^p\).

• 증명 idea:

\[ |f|^p=|(f_n-f)-f_n|^p\le (|f_n-f|+|f_n|)^p \le 2^p(|f_n-f|^p+|f_n|^p). \]

매우 중요

Theorem

If \(f_n\stackrel{L^p}\rightarrow f\) form some \(0<p\le \infty\), then \(f_n\stackrel{\mu}\rightarrow f\).

• 증명 : 마코프 부등식을 이용한다.

\[ \mu(|f_n-f|>\epsilon)=\mu(|f_n-f|^p>\epsilon^p)\le \frac{1}{\epsilon^p}\int |f_n-f|^pd\mu\rightarrow 0\mbox{ }\mbox{ }\mbox{ as }n\rightarrow \infty \mbox{ }\mbox{ }\mbox{ for all }\epsilon>0 . \]

중요

Theorem(Riesz-Fisher)

For \(0<p\le \infty\), the space \(L^p\) is complete:

a sequence of \(L^p\) function converges in \(L^p\) if and only if the sequence is Cauchy in \(L^p\).

• (\(\implies\)): \[ |f_m-f_n|^p=|f_m-f+f-f_n|^p=|(f_m-f)-(f_n-f)|^p\le(|f_m-f|+|f_n-f|)^p\le 2^p(|f_m-f|^p+|f_n-f|^p). \]

• (\(\Longleftarrow\)): Sequence \(\{f_n\}\)이 Cauchy in \(L^p\)라고 하자. 마코프 부등식에 의해 \[ \mu(|f_m-f_n|>\epsilon)=\mu(|f_m-f_n|^p>\epsilon^p)\le \frac{1}{\epsilon^p}\int |f_m-f_n|^p d\mu \] Thus, \(\{f_n\}\)은 Cauchy in measure이다. 또한 Cauchy in measure는 convergence in measure와 동치이므로 \(f_n\stackrel{\mu}\rightarrow f\)인 \(f\)가 존재한다. 그러므로 \(f_{n_k}\rightarrow f\) a.e. 를 만족하는 subsequence \(\{f_{n_k}\}\)가 존재한다. 마지막으로 Fatou’s Lemma에 의해 \[\begin{eqnarray*} \int |f_n-f|^pd\mu&=&\int \lim_{n\rightarrow \infty}|f_n-f_{n_k}|^pd\mu \\ &\le& \liminf_{k}\int |f_n-f_{n_k}|^pd\mu\\ &\le& \limsup_{k}\int |f_n-f_{n_k}|^pd\mu \\ &=&\sup_{m\ge n}\int |f_m-f_n|^pd\mu\rightarrow 0 \mbox{ }\mbox{ }\mbox{ }\mbox{ as }n\rightarrow \infty\mbox{ }\mbox{ (Cauchy in }L^p\mbox{)}. \end{eqnarray*}\]

정리

\(\{f_n\}\in L^p\)에 대해 \(f_n\stackrel{L^p}\rightarrow f\)라면

1. \(f\in L^p\);

2. \(f_n\stackrel{\mu}\rightarrow f\) (\(L^p\) convergence \(\implies\) convergence in measure);

3. Convergenge in \(L^p\) \(\iff\) Cauchy in \(L^p\) (\(L^p\) space is complete).

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