[家里蹲大学数学杂志]第442期一个积分不等式
设 $f$ 在 $[a,b]$ 上连续可微且 $f(a)=0$. 试证: $$ex int_a^b |f'(x)|^2 d xgeq frac{2}{(b-a)^2}int_a^b |f(x)|^2 d x. eex$$
证明: $$eex ea f(x)&=int_a^x f'(t) d t,\ |f(x)|^2&leq int_a^x 1^2 d tcdot int_a^x |f'(t)|^2 d t leq (a-x)int_a^b |f'(t)|^2 d t,\ int_a^b |f(x)|^2 d x &leq int_a^b |f'(t)|^2 d tcdot int_a^b (x-a) d x =frac{(b-a)^2}{2} int_a^b |f'(t)|^2 d t. eea eeex$$