问题及解答
1
设 $0 < x_0\leqslant 1$. 注意到
\[
\Bigl((x-x_0)f'(x)-f(x)\Bigr)'=f'(x)+(x-x_0)f''(x)-f'(x)=(x-x_0)f''(x),
\]
因此,
\[
\begin{split}
\int_0^1 (x-x_0)f''(x)dx&=\Bigl[(x-x_0)f'(x)-f(x)\Bigr]\biggr|_0^1\\
&=\bigl[(1-x_0)f'(1)-f(1)\bigr]-\bigl[(0-x_0)f'(0)-f(0)\bigr]\\
&=x_0.
\end{split}
\]
因此, 由 Hölder 不等式
\[
\begin{split}
x_0^2=\biggl(\int_0^1 (x-x_0)f''(x)dx\biggr)^2 &\leqslant\int_0^1 (x-x_0)^2 dx\cdot\int_0^1(f''(x))^2 dx\\
&=\frac{1}{3}(x-x_0)^3\biggl|_0^1 \cdot\int_0^1(f''(x))^2 dx\\
&=\frac{(1-x_0)^3+x_0^3}{3}\cdot\int_0^1(f''(x))^2 dx\\
\end{split}
\]
这推出
\[
\begin{split}
\int_0^1(f''(x))^2 dx&\geqslant\frac{3x_0^2}{(1-x_0)^3+x_0^3}\\
&=\frac{3x_0^2}{(1-3x_0+3x_0^2-x_0^3)+x_0^3}\\
&=\frac{3}{\frac{1}{x_0^2}-\frac{3}{x_0}+3}\\
&=\frac{3}{(\frac{1}{x_0}-\frac{3}{2})^2+\frac{3}{4}}\\
\end{split}
\]
注意到 $x_0$ 是某个取定的数, 特别的, 我们取 $x_0=\frac{2}{3}$(可以在上述过程中将 $x_0$ 统一替换为 $\frac{2}{3}$), 则可得
\[
\int_0^1(f''(x))^2 dx\geqslant\frac{3}{\frac{3}{4}}=4.
\]
Remark:
This proof is provided by Shouwen FANG (方守文).