(云南大学). 已知 $$ex 0leq fin C[0,infty),quad int_0^infty frac{1}{f^2(x)} d x<infty. eex$$ 试证: $$ex vlm{A}frac{1}{A^2}int_0^A f^2(x) d x=infty. eex$$
证明: 由 Cauchy 收敛准则, $$ex forall M>0, exists X>0,st Ageq 2X a int_X^Afrac{1}{f^2(x)} d x<frac{1}{4M}. eex$$ 又由 Cauchy-Schwarz 不等式, $$eex ea (A-X)^2&=sex{int_X^A f(x)cdot frac{1}{f(x)} d x}^2\ &leq int_X^A f^2(x) d xcdot int_X^Afrac{1}{f^2(x)} d x\ &<frac{1}{4M} int_X^A f^2(x) d x. eea eeex$$ 于是 $$ex frac{1}{A^2}int_X^A f^2(x) d x>4Mfrac{(A-X)^2}{A^2}geq M. eex$$ 综上, $$ex forall M>0, exists X,st Ageq 2X a frac{1}{A^2}int_0^Af^2(x) d x geq frac{1}{A^2}int_X^A f^2(x) d xgeq M. eex$$ 故有结论.