題目

Problem

Suppose $f^2(x)$ , $g^2(x)$ , and $f(x)g(x)$ are all integrable functions in the interval $[0,1]$ .

(a) (10 pts) Please prove the inequality

\left(\int_0^1 f^2(x)\,dx\right)\left(\int_0^1 g^2(x)\,dx\right)\ge \left(\int_0^1 f(x)g(x)\,dx\right)^2.

(Hint: $(tf(x)+g(x))^2\ge 0$ for all $x\in[0,1]$ and $t\in\mathbb{R}$ .)

(b) (10 pts) Please show that

\int_0^1 \frac{1}{1+x^n}\,dx\ge \frac{n+1}{n+2}

for all $n>0$ . (Hint: use (a).)

解答

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(a) 對任意 $t\in\mathbb R$ ，由提示可得

\begin{align*} 0\le&\,\int_0^1\bigl(tf(x)+g(x)\bigr)^2\,\mathrm{d}x\\[4mm] =&\,At^2+2Bt+C, \end{align*}

其中

A=\int_0^1f^2(x)\,\mathrm{d}x,\qquad B=\int_0^1f(x)g(x)\,\mathrm{d}x,\qquad C=\int_0^1g^2(x)\,\mathrm{d}x.

此二次式對所有 $t$ 非負，故其判別式不大於零：

4B^2-4AC\le0.

因此

\boxed{\left(\int_0^1f^2\,\mathrm{d}x\right)\left(\int_0^1g^2\,\mathrm{d}x\right) \ge\left(\int_0^1fg\,\mathrm{d}x\right)^2}.

(b) 在 (a) 取

f(x)=\frac1{\sqrt{1+x^n}},\qquad g(x)=\sqrt{1+x^n}.

則 $f(x)g(x)=1$ ，並且

\int_0^1g^2(x)\,\mathrm{d}x =\int_0^1(1+x^n)\,\mathrm{d}x =1+\frac1{n+1}=\frac{n+2}{n+1}.

由 (a)，

\left(\int_0^1\frac1{1+x^n}\,\mathrm{d}x\right)\frac{n+2}{n+1}\ge1,

故

\boxed{\int_0^1\frac1{1+x^n}\,\mathrm{d}x\ge\frac{n+1}{n+2}}.