设 $f(x)$ 在 $\bbR$ 上连续, 又 $$\bex \phi(x)=f(x)\int_0^x f(t)\rd t \eex$$ 单调递减. 证明: $f\equiv 0$.
证明: 设 $$\bex g(x)=\cfrac{\sez{\int_0^x f(t)\rd t}^2}{2}, \eex$$ 则 $g'(x)=\phi(x)$ 递减, 而 $$\bex g'(x)\sedd{\ba{ll} \geq g'(0)=0,&x<0,\\ \leq g'(0)=0,&x>0; \ea} \eex$$ 进一步, $$\bex g(x)\sedd{\ba{ll} \leq g(0)=0,&x<0,\\ \leq g(0)=0,&x>0. \ea} \eex$$ 如此, $g(x)\leq 0$, $$\bex \int_0^x f(t)\rd t=0,\quad \forall\ x, \eex$$ $$\bex f(x)=\sez{\int_0^x f(t)\rd t}'=0,\quad \forall\ x. \eex$$
