(1) If $X_n\rightarrow X$ and $Y_n\rightarrow Y$ both in $L^p$, then $$X_n\pm Y_n\rightarrow X\pm Y\mbox{ in }L^p;$$
(2) If $X_n\rightarrow X$ in $L^p$ and $Y_n\rightarrow Y$ in $L^q$, where $p>1$ and $1/p+1/q=1$, then $$X_nY_n\rightarrow XY\mbox{ in }L^1.$$
$\bullet$ Proof.
(1) First, we introduce an inequality.
[Inequality]
If $X\geq0$ and $Y\geq0$, $p\geq0$, then $$\mathscr{E}\{(X+Y)^p\}\leq2^p\{\mathscr{E}(X^p)+\mathscr{E}(Y^p)\}.$$If $p>1$, the factor $2^p$ may be replaced by $2^{p-1}$. If $0\leq p\leq1$, it may be replaced by $1$.
Back to the proof. $X_n\rightarrow X$ and $Y_n\rightarrow Y$ both in $L^p\implies$ $$\begin{array}{c}\underset{n\rightarrow\infty}{\lim}\mathscr{E}|X_n-X|^p=0,\\ \underset{n\rightarrow\infty}{\lim}\mathscr{E}|Y_n-Y|^p=0.\end{array}$$Hence, $$\begin{array}{rl}\mathscr{E}\left|(X_n+Y_n)-(X+Y)\right|^p & = \mathscr{E}\left|(X_n-X)+(Y_n-Y)\right|^p\\ &\leq\max{\{1,2^p\}}\mathscr{E}\left(\left|X_n-X\right|^p+\left|Y_n-Y\right|^p\right)\\ &\rightarrow0\mbox{ as }n\rightarrow\infty.\end{array}$$ Similarly, by replacing $Y'_n=-Y_n$ and $Y'=-Y$, we have $$X_n\pm Y_n\rightarrow X\pm Y\mbox{ in }L^p.$$
(2) $$\begin{array}{c}X_n\overset{L^p}{\rightarrow}X\implies\underset{n\rightarrow\infty}{\lim}\mathscr{E}|X_n-X|^p=0,\\ Y_n\overset{L^q}{\rightarrow}Y\implies\underset{n\rightarrow\infty}{\lim}\mathscr{E}|Y_n-Y|^q=0.\end{array}$$Hence by H$\ddot{o}$lder's inequality, we have $$\begin{array}{rl}\mathscr{E}\left|X_nY_n-XY\right|&=\mathscr{E}\left|X_nY_n-X_nY+X_nY-XY\right|\\ &=\mathscr{E}\left|X_n(Y_n-Y)+Y(X_n-X)\right|\\ &\leq\left(\mathscr{E}\left|X_n\right|^p\right)^{\frac{1}{p}}\left(\mathscr{E}\left|Y_n-Y\right|^q\right)^{\frac{1}{q}}+\left(\mathscr{E}\left|X_n-X\right|^p\right)^{\frac{1}{p}}\left(\mathscr{E}\left|Y\right|^q\right)^{\frac{1}{q}}\\ &\rightarrow0\mbox{ as }n\rightarrow\infty.\end{array}$$We have $$X_nY_n\rightarrow XY\mbox{ in }L^1.$$
$\Box$
沒有留言:
張貼留言