测试数学公式效果
问题. Let be \(f:[0,1]\to[0,1]\) a continuous and bijective function, such that: \(f(0)=0\). Then the following inequality holds:
\[\left( \alpha +2 \right) \int_0^1{x^{\alpha}\left( f\left( x \right) +f^{-1}\left( x \right) \right)\text{d}x \leqslant 2},\forall \alpha \geqslant 0.
\]
证明. 由于 \(f:[0,1]\to[0,1]\) 是双射，且 \(f(0)=0\)，故 \(f(x)\) 在 \([0,1]\) 上单调递增，\(f(1)=1\). 令 \(g(x)=\int_0^x (f(t)+f^{-1}(t))\text{d}t\). 由 Young 不等式得 \(g(x)\geqslant x^2\)，当且仅当 \(f(x)=x\) 时取等号，从而 \(g(1)=1\).
\[\begin{aligned}
\int_0^1{x^{\alpha}\left( f\left( x \right) +f^{-1}\left( x \right) \right) \mathrm{d}x}&=\int_0^1{x^{\alpha}\mathrm{d}g\left( x \right)}\\
&=x^{\alpha}g\left( x \right) \mid_{0}^{1}-\alpha \int_0^1{x^{\alpha -1}g(x)\mathrm{d}x}\\
&\leqslant 1-\alpha \int_0^1{x^{\alpha +1}\mathrm{d}x}=\frac{2}{\alpha +2}
\end{aligned}
\]
从而
\[\left( \alpha +2 \right) \int_0^1{x^{\alpha}\left( f\left( x \right) +f^{-1}\left( x \right) \right)\text{d}x \leqslant 2},
\]
当且仅当 \(f(x)=x\) 时取到等号.
问题. 设 \(f\) 在 \([0,1]\) 上 \(n\) 阶连续可微，\(f\left(\frac{1}{2}\right)=0\)，\(f^{(i)}\left(\frac{1}{2}\right)=0\)，\(i=1,2,\cdots,n\). 证明
\[\left( \int_0^1{f\left( x \right) \mathrm{d}x} \right) ^2\leqslant \frac{1}{\left( 2n+1 \right) 4^n\left( n! \right) ^2}\int_0^1{\left( f^{\left( n \right)}\left( x \right) \right) ^2\mathrm{d}x}.
\]
证明. 将 \(f(x)\) 在 \(x=\frac{1}{2}\) 处用带积分余项的 Taylor 公式展开到 \(x^{n-1}\)，由题设得到
\[f\left( x \right) =r_{n-1}\left( x \right) =\int_{1/2}^x{\frac{\left( x-t \right) ^{n-1}}{(n-1)!}f^{\left( n \right)}\left( t \right) \mathrm{d}t}.
\]
也即
\[\begin{align*}
\int_0^1{f\left( x \right) \mathrm{d}x}&=\int_0^1{\int_{1/2}^x{\frac{\left( x-t \right) ^{n-1}}{\left( n-1 \right) !}f^{\left( n \right)}\left( t \right) \mathrm{d}t\mathrm{d}x}}\\
&=\int_0^{1/2}{\int_{1/2}^x{\frac{\left( x-t \right) ^{n-1}}{\left( n-1 \right) !}f^{\left( n \right)}\left( t \right) \mathrm{d}t\mathrm{d}x}}+\int_{1/2}^1{\int_{1/2}^x{\frac{\left( x-t \right) ^{n-1}}{\left( n-1 \right) !}f^{\left( n \right)}\left( t \right) \mathrm{d}t\mathrm{d}x}}
\end{align*}
\]
一方面，
\[\begin{align*}
\int_0^{1/2}{\int_{1/2}^x{\frac{\left( x-t \right) ^{n-1}}{\left( n-1 \right) !}f^{\left( n \right)}\left( t \right) \mathrm{d}t\mathrm{d}x}}&=-\int_0^{1/2}{\int_0^t{\frac{\left( x-t \right) ^{n-1}}{\left( n-1 \right) !}f^{\left( n \right)}\left( t \right) \mathrm{d}x\mathrm{d}t}}\\
&=-\frac{\left( -1 \right) ^n}{n!}\int_0^{1/2}{t^nf^{\left( n \right)}\left( t \right) \mathrm{d}t}
\end{align*}
\]
另一方面，
\[\begin{align*}
\int_{1/2}^1{\int_{1/2}^x{\frac{\left( x-t \right) ^{n-1}}{\left( n-1 \right) !}f^{\left( n \right)}\left( t \right) \mathrm{d}t\mathrm{d}x}}&=\int_{1/2}^1{\int_t^1{\frac{\left( x-t \right) ^{n-1}}{\left( n-1 \right) !}f^{\left( n \right)}\left( t \right) \mathrm{d}x\mathrm{d}t}}\\
&=\frac{1}{n!}\int_{1/2}^1{\left( 1-t \right) ^{n}f^{\left( n \right)}\left( t \right) \mathrm{d}t}
\end{align*}
\]
从而由 \(\int_a^b f(x)\text{d}x\leqslant \int_a^b |f(x)|\text{d}x\)，
\[\begin{align*}
\int_0^1{f\left( x \right) \mathrm{d}x}&=\frac{1}{n!}\left( \left( -1 \right) ^n\int_0^{1/2}{t^nf^{\left( n \right)}\left( t \right) \mathrm{d}t}+\int_{1/2}^1{\left( 1-t \right) ^nf^{\left( n \right)}\left( t \right) \mathrm{d}t} \right) \\
&\leqslant \frac{1}{n!}\int_0^1{\left( \min \left\{ t,1-t \right\} \right) ^n\left| f^{\left( n \right)}\left( t \right) \right|\mathrm{d}t}
\end{align*}
\]
再由 Cauchy-Schwarz 不等式，
\[\left( \int_0^1{f\left( x \right) \mathrm{d}x} \right) ^2\leqslant \frac{1}{\left( n! \right) ^2}\int_0^1{\left( \min \left\{ t,1-t \right\} \right) ^{2n}\mathrm{d}t}\int_0^1{\left( f^{\left( n \right)}\left( t \right) \right) ^2\mathrm{d}t}
\]
而 \(\int_0^1{\left( \min \left\{ t,1-t \right\} \right) ^{2n}\mathrm{d}t}=2\int_0^{1/2}{t^{2n}\mathrm{d}x}=\frac{1}{\left( 2n+1 \right) 4^n}\)，代入上式即得
\[\left( \int_0^1{f\left( x \right) \mathrm{d}x} \right) ^2\leqslant \frac{1}{\left( 2n+1 \right) 4^n\left( n! \right) ^2}\int_0^1{\left( f^{\left( n \right)}\left( x \right) \right) ^2\mathrm{d}x}.
\]
（1）Markdown 编辑器中输入多行公式：https://zhuanlan.zhihu.com/p/95674430 ，如果不想公式有编号，可以在“$$”内使用 aligned 或者 align* 环境
（2）&emph; 这样的符号似乎并不能在文章中编译出来
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