[再寄小读者之数学篇](2014-06-18 积分、微分不等式)

 

设 $f$ 为 $[0,1]$ 上的连续正函数, 且 $\dps{f^2(t)\leq 1+2\int_0^t f(s)\rd s}$. 证明: $f(t)\leq 1+t$.

 

证明: 设 $\dps{F(t)=\int_0^t f(s)\rd s}$, 则 $F(0)=0$, 且 $$\beex \bea F'^2(t)&\leq 1+2F(t),\\ \cfrac{\rd F(t)}{\sqrt{1+2F(t)}}&\leq \rd t,\\ \sqrt{1+2F(t)}-\sqrt{1+2F(0)}&\leq t,\\ \sqrt{1+2F(t)}&\leq 1+t,\\ 2F(t)&\leq (1+t)^2-1=2t+t^2,\\ f^2(t)&\leq 1+2F(t)=1+2t+t^2=(1+t)^2,\\ f(t)&\leq 1+t. \eea \eeex$$