设 $f(x)$ 是 $[0,+\infty)$ 上的有界可积函数, 证明
\[\liminf_{t\rightarrow\infty}f(t)\leqslant\liminf_{t\rightarrow\infty}\frac{1}{t}\int_{0}^{t}f(s)\mathrm{d}s.\]
Remark: $f(x)$ 在 $[0,\infty)$ 上可积即可.
Keywords: 上极限($\limsup$)、下极限($\liminf$)
1
记 $T(t)=\inf_{\tau\geqslant t}f(\tau)$, 对 $\tau\geqslant t$,
\[
\begin{split}
\frac{1}{\tau}\int_{0}^{\tau}f(s)\mathrm{d}s&=\frac{1}{\tau}\biggl[\int_{0}^{t}f(s)\mathrm{d}s+\int_{t}^{\tau}f(s)\mathrm{d}s\biggr]\\
&\geqslant\frac{1}{\tau}\int_{0}^{t}f(s)\mathrm{d}s+\frac{1}{\tau}\int_{t}^{\tau}T(t)\mathrm{d}s\\
&=\frac{1}{\tau}\int_{0}^{t}f(s)\mathrm{d}s+\frac{1}{\tau}(\tau-t)\cdot T(t)
\end{split}
\]
从而
\[
\begin{split}
\inf_{\tau\geqslant t}\frac{1}{\tau}\int_{0}^{\tau}f(s)\mathrm{d}s&\geqslant\inf_{\tau\geqslant t}\biggl(\frac{1}{\tau}\int_{0}^{t}f(s)\mathrm{d}s+\frac{1}{\tau}(\tau-t)\cdot T(t)\biggr)\\
&\geqslant\inf_{\tau\geqslant t}\frac{1}{\tau}\int_{0}^{t}f(s)\mathrm{d}s+\inf_{\tau\geqslant t}\frac{1}{\tau}(\tau-t)\cdot T(t)\\
&=0+T(t)=T(t).
\end{split}
\]
即有
\[
\inf_{\tau\geqslant t}f(\tau)\leqslant\inf_{\tau\geqslant t}\frac{1}{\tau}\int_{0}^{\tau}f(s)\mathrm{d}s.
\]
两边取极限 $t\rightarrow\infty$, 得
\[
\lim_{t\rightarrow\infty}\inf_{\tau\geqslant t}f(\tau)\leqslant\lim_{t\rightarrow\infty}\inf_{\tau\geqslant t}\frac{1}{\tau}\int_{0}^{\tau}f(s)\mathrm{d}s,
\]
此即
\[
\liminf_{t\rightarrow\infty}f(t)\leqslant\liminf_{t\rightarrow\infty}\frac{1}{t}\int_{0}^{t}f(s)\mathrm{d}s.
\]