因为f(x)在[0,1]上二阶可导,所以f(x)可以展开成一阶泰勒公式:
f(x)=f(a)+f'(a)*(x-a)+f''(c)*[(x-a)^2]/2!,其中0<=a<=1,c介于x和a之间
①令x=0
f(0)=f(a)-f'(a)*a+f''(c1)*(a^2)/2,其中0<=c1<=a
②令x=1
f(1)=f(a)+f'(a)*(1-a)+f''(c2)*[(1-a)^2]/2,其中a<=c2<=1
下式减上式,得:
f(1)-f(0)=f'(a)+(1/2)*[f''(c2)*(1-a)^2-f''(c1)*a^2]
因为f(0)=f(1)
所以f'(a)=(1/2)*[f''(c1)*a^2-f''(c2)*(1-a)^2]
因为|f''(x)|<=2,所以|f''(c1)|<=2,且|f''(c2)|<=2
|f'(a)|=(1/2)*|f''(c1)*a^2-f''(c2)*(1-a)^2|
<=(1/2)*[|f''(c1)*a^2|+|f''(c2)*(1-a)^2|]
<=(1/2)*[2a^2+2(1-a)^2]
=2a^2-2a+1
=2(a-1/2)^2+1/2
因为0<=a<=1,所以2(a-1/2)^2+1/2<=1
即|f'(a)|<=1
证毕
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