[题目]设$p>1,f\in C(0,+\infty)$，且$\int_0^\infty|f(t)|^pdt$收敛，证明：

\left\{\int_0^\infty\left[\frac{1}{x}\int_0^x|f(t)|dt\right]^pdx\right\}^{\frac{1}{p}}\leq\frac{p}{p-1}\left(\int_0^\infty|f(t)|^pdt\right)^{\frac{1}{p}}.

<script id="MathJax-Element-1" type="math/tex; mode=display">\left\{\int_0^\infty\left[\frac{1}{x}\int_0^x|f(t)|dt\right]^pdx\right\}^{\frac{1}{p}}\leq\frac{p}{p-1}\left(\int_0^\infty|f(t)|^pdt\right)^{\frac{1}{p}}.</script>

[证明]我们首先证明：\begin{equation}\label{e1}\int_0^N\chi_{\{\int_0^x|f(t)|dt\geq x\}}dx\leq\int_0^N|f(x)|\chi_{\{\int_0^x|f(t)|dt\geq x\}}dx,\forall N>0.\end{equation}

<script id="MathJax-Element-2" type="math/tex; mode=display">\begin{equation}\label{e1}\int_0^N\chi_{\{\int_0^x|f(t)|dt\geq x\}}dx\leq\int_0^N|f(x)|\chi_{\{\int_0^x|f(t)|dt\geq x\}}dx,\forall N>0.\end{equation}</script>

事实上，命$F(x)=\int_0^x(|f(t)|-1)dt$。我们取一列$t_m\in\{x\in[0,N]:F(x)\geq 0\}$,$t_m\rightarrow \mathop{\rm sup}\{x\in[0,N]:F(x)\geq 0\}$，则$\{x\in[0,t_m]:F(x)<0\}$是一开集。设它的构成区间为$\{(a_j,b_j):j\geq 1\}$。由连续性可知$F(a_j)=F(b_j)=0$.对于固定的自然数$m,n$,我们有：

\begin{equation}\label{e2}\int_{[0,t_m]\setminus\bigcup_{j=1}^n(a_j,b_j)}F‘(x)dx=F(t_m)-F(0)-\sum_{j=1}^n(F(b_j)-F(a_j))=F(t_m)\geq 0.\end{equation}

<script id="MathJax-Element-3" type="math/tex; mode=display">\begin{equation}\label{e2}\int_{[0,t_m]\setminus\bigcup_{j=1}^n(a_j,b_j)}F‘(x)dx=F(t_m)-F(0)-\sum_{j=1}^n(F(b_j)-F(a_j))=F(t_m)\geq 0.\end{equation}</script>

由$F‘=|f|-1\in L^1[0,t_m]$以及控制收敛定理,对$\eqref{e2}<script id="MathJax-Element-4" type="math/tex">\eqref{e2}</script>$命$n\rightarrow\infty$，有

\begin{equation}\label{e3}\int_0^{t_m}\chi_{\{\int_0^x|f(t)|dt\geq x\}}dx\leq\int_0^{t_m}|f(x)|\chi_{\{\int_0^x|f(t)|dt\geq x\}}dx.\end{equation}

<script id="MathJax-Element-5" type="math/tex; mode=display">\begin{equation}\label{e3}\int_0^{t_m}\chi_{\{\int_0^x|f(t)|dt\geq x\}}dx\leq\int_0^{t_m}|f(x)|\chi_{\{\int_0^x|f(t)|dt\geq x\}}dx.\end{equation}</script>

再令$m\rightarrow\infty$,由单调收敛定理，便得$\eqref{e1}<script id="MathJax-Element-6" type="math/tex">\eqref{e1}</script>$.下面来证明本题：

\begin{equation}\begin{split}\int_0^N\left[\frac{1}{x}\int_0^x|f(t)|dt\right]^pdx&=\int_0^N\int_0^{\frac{1}{x}\int_0^x|f(t)|dt}ps^{p-1}dsdx\\ &=\int_0^\infty ps^{p-1}ds\int_0^N\chi_{\{\frac{1}{x}\int_0^x|f(t)|dt\geq s\}}dx\\&=\int_0^\infty ps^{p-1}ds\int_0^N\chi_{\{\frac{1}{x}\int_0^x\frac{|f(t)|}{s}dt\geq 1\}}dx\\ &\leq \int_0^\infty ps^{p-1}ds\int_0^N\frac{|f(x)|}{s}\chi_{\{\frac{1}{x}\int_0^x|f(t)|dt\geq s\}}dx\\&=p\int_0^N|f(x)|dx\int_0^{\frac{1}{x}\int_0^x|f(t)|dt}s^{p-2}ds\\&=\frac{p}{p-1}\int_0^N|f(x)|\left[\frac{1}{x}\int_0^x|f(t)|dt\right]^{p-1}dx\\ &\leq\frac{p}{p-1}\left(\int_0^N|f(x)|^pdx\right)^{\frac{1}{p}}\left\{\int_0^N\left[\frac{1}{x}\int_0^x|f(t)|dt\right]^pdx\right\}^{\frac{p-1}{p}}.\end{split}\end{equation}

<script id="MathJax-Element-7" type="math/tex; mode=display">\begin{equation}\begin{split}\int_0^N\left[\frac{1}{x}\int_0^x|f(t)|dt\right]^pdx&=\int_0^N\int_0^{\frac{1}{x}\int_0^x|f(t)|dt}ps^{p-1}dsdx\\ &=\int_0^\infty ps^{p-1}ds\int_0^N\chi_{\{\frac{1}{x}\int_0^x|f(t)|dt\geq s\}}dx\\&=\int_0^\infty ps^{p-1}ds\int_0^N\chi_{\{\frac{1}{x}\int_0^x\frac{|f(t)|}{s}dt\geq 1\}}dx\\ &\leq \int_0^\infty ps^{p-1}ds\int_0^N\frac{|f(x)|}{s}\chi_{\{\frac{1}{x}\int_0^x|f(t)|dt\geq s\}}dx\\&=p\int_0^N|f(x)|dx\int_0^{\frac{1}{x}\int_0^x|f(t)|dt}s^{p-2}ds\\&=\frac{p}{p-1}\int_0^N|f(x)|\left[\frac{1}{x}\int_0^x|f(t)|dt\right]^{p-1}dx\\ &\leq\frac{p}{p-1}\left(\int_0^N|f(x)|^pdx\right)^{\frac{1}{p}}\left\{\int_0^N\left[\frac{1}{x}\int_0^x|f(t)|dt\right]^pdx\right\}^{\frac{p-1}{p}}.\end{split}\end{equation}</script>

移项便得：

\begin{equation}\label{e5}\left\{\int_0^N\left[\frac{1}{x}\int_0^x|f(t)|dt\right]^pdx\right\}^{\frac{1}{p}}\leq\frac{p}{p-1}\left(\int_0^N|f(t)|^pdt\right)^{\frac{1}{p}}.\end{equation}

<script id="MathJax-Element-8" type="math/tex; mode=display">\begin{equation}\label{e5}\left\{\int_0^N\left[\frac{1}{x}\int_0^x|f(t)|dt\right]^pdx\right\}^{\frac{1}{p}}\leq\frac{p}{p-1}\left(\int_0^N|f(t)|^pdt\right)^{\frac{1}{p}}.\end{equation}</script>

再命$N\rightarrow\infty$即得结论。