(Yunnan University). Known $$\bex 0\leq f\in c[0,\infty), \quad \int_0^\infty \frac{1}{f^2 (x)}\rd x<\infty. \eex$$ test: $$\bex \vlm{a}\frac{1}{a^2}\int_0^a f^2 (x) \rd x=\infty. \eex$$
Proof: by Cauchy convergence Criterion, $$\bex \forall\ m>0,\ \exists\ x>0,\st a\geq 2x\ra \int_x^a\frac{1}{f^2 (X)}\rd x<\frac{1}{4m}. \eex$$ again by Cauchy-schwarz inequality, $$\beex \bea (a-x) ^2&=\sex{\int_x^a f (x) \cdot \frac{1}{f (x)}\rd x}^2\\ &\leq \int_X ^a f^2 (x) \rd X\cdot \int_x^a\frac{1}{f^2 (x)}\rd x\\ &<\frac{1}{4m} \int_x^a f^2 (x) \rd x. \EEA \eeex$$ so $$\bex \fr Ac{1}{a^2}\int_x^a f^2 (X) \rd x>4m\frac{(a-x) ^2}{a^2}\geq M. \eex$$ integrated, $$\bex \forall\ m>0,\ \exists\ X,\st A\geq 2X \ra \frac{1}{a^2}\int_0^af^2 (x) \rd x \geq \frac{1}{a^2}\int_x^a f^2 (x) \rd X\geq m. \eex$$ there is a conclusion.
[Home Squat University Mathematics magazine] NO. 412 period integral and limit