日经题

2020-10-01 本文已影响0人洛玖言

函数 $f(x)$ 在 $[0,1]$ 上一阶可导, $f(0)=f(1)=0$ . 证明
$\int_{0}^1f^2(x)\text{d}x\leqslant c\int_0^{1}f'^2(x)\text{d}x.$
其中 $c$ 分别为 $\dfrac{1}{2},\;\dfrac{1}{4},\;\dfrac{1}{8}$

Sol:
因为 $f(0)=0$
因此 $f(x)=\displaystyle\int_0^xf'(x)\text{d}x$ .
$\begin{aligned} f^2(x)=&\Bigg(\int_0^xf'(t)\text{d}t\Bigg)^2\\ \leqslant& x\int_0^{x}f'^2(t)\text{d}t\quad(Cauchy-Schwarz)\\ \leqslant& x\int_0^1f'^2(t)\text{d}t\quad(x\in(0,1))\\ \end{aligned}$
因此
$\begin{aligned} \int_0^1f^2(x)\text{d}x\leqslant&\int_0^1x\text{d}x\cdot\int_0^1f'^2(t)\text{d}t=\dfrac{1}{2}\int_0^1f'^2(t)\text{d}t \end{aligned}$
由此证明了 $c=\dfrac{1}{2}$ 的情况.

又有
$\begin{aligned} f^2(x)=\Bigg(\int_x^1f'(t)\text{d}t\Bigg)^2\leqslant(1-x)\int_x^1f'^2(t)\text{d}t\leqslant(1-x)\int_0^1f'^2(t)\text{d}t\quad(x\in(0,1)) \end{aligned}$

因此
$\begin{aligned} \int_0^1f^2(x)\text{d}x=&\int_0^{\frac{1}{2}}f^2(x)\text{d}x+\int_{\frac{1}{2}}^1f^2(x)\text{d}x\\ \leqslant&\int_0^{\frac{1}{2}}x\text{d}x\cdot\int_0^{1}f'^2(t)\text{d}t+\int_{\frac{1}{2}}^1(1-x)\text{d}x\cdot\int_0^{1}f'^2(t)\text{d}t\\ =&\dfrac{1}{8}\int_0^{1}f'^2(t)\text{d}t+\dfrac{1}{8}\int_0^{1}f'^2(t)\text{d}t=\dfrac{1}{4}\int_0^{1}f'^2(t)\text{d}t \end{aligned}$
由此证明了 $c=\dfrac{1}{4}$ 的情况. 微调上面的两个不等式得
$\displaystyle f^2(x)\leqslant x\int_0^{\frac{1}{2}}f'^2(t)\text{d}t\quad(x\in(0,\frac{1}{2})),\quad\quad$
$\displaystyle f^2(x)\leqslant (1-x)\int_{\frac{1}{2}}^1f'^2(t)\text{d}t\quad(x\in(\frac{1}{2},1))$
由此得
$\begin{aligned} \int_0^1f^2(x)\text{d}x=&\int_0^{\frac{1}{2}}f^2(x)\text{d}x+\int_{\frac{1}{2}}^1f^2(x)\text{d}x\\ \leqslant&\int_0^{\frac{1}{2}}x\text{d}x\cdot\int_0^{\frac{1}{2}}f'^2(t)\text{d}t+\int_{\frac{1}{2}}^1(1-x)\text{d}x\cdot\int_\frac{1}{2}^{1}f'^2(t)\text{d}t\\ =&\dfrac{1}{8}\int_0^{\frac{1}{2}}f'^2(t)\text{d}t+\dfrac{1}{8}\int_{\frac{1}{2}}^{1}f'^2(t)\text{d}t\\ =&\dfrac{1}{8}\int_0^{1}f'^2(t)\text{d}t \end{aligned}$
由此证明了 $c=\dfrac{1}{8}$ 的情况.

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