[家里蹲大学数学杂志]第412期积分与极限

(云南大学). 已知 $$\bex 0\leq f\in C[0,\infty),\quad \int_0^\infty \frac{1}{f^2(x)}\rd x<\infty. \eex$$ 试证: $$\bex \vlm{A}\frac{1}{A^2}\int_0^A f^2(x)\rd x=\infty. \eex$$

证明: 由 Cauchy 收敛准则, $$\bex \forall\ M>0,\ \exists\ X>0,\st A\geq 2X\ra \int_X^A\frac{1}{f^2(x)}\rd x<\frac{1}{4M}. \eex$$ 又由 Cauchy-Schwarz 不等式, $$\beex \bea (A-X)^2&=\sex{\int_X^A f(x)\cdot \frac{1}{f(x)}\rd x}^2\\ &\leq \int_X^A f^2(x)\rd x\cdot \int_X^A\frac{1}{f^2(x)}\rd x\\ &<\frac{1}{4M} \int_X^A f^2(x)\rd x. \eea \eeex$$ 于是 $$\bex \frac{1}{A^2}\int_X^A f^2(x)\rd x>4M\frac{(A-X)^2}{A^2}\geq M. \eex$$ 综上, $$\bex \forall\ M>0,\ \exists\ X,\st A\geq 2X\ra \frac{1}{A^2}\int_0^Af^2(x)\rd x \geq \frac{1}{A^2}\int_X^A f^2(x)\rd x\geq M. \eex$$ 故有结论.