f(x)在[0,1]上具有二阶导数,|f(x)|

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证明：把f(x)在x=c处泰勒展开得
f(0)=f(c)-f'(c)*c+f''(m)*c²/2
f(1)=f(c)+f'(c)*(1-c)+f''(n)*(1-c)²/2
两式相减,得
f(1)-f(0)=f`(c)+f''(n)*(1-c)²/2-f''(m)*c²/2
f'(c)=f(1)-f(0)+f''(m)*c²/2-f''(n)*(1-c)²/2
所以
|f'(c)|≤|f(1)|+|f(0)|+|f''(m)|*c²/2+|f''(n)|*(1-c)²/2
因为|f(x)|≤a,所以|f(0)≤a |f(1)|≤a,同样|f``(x)|≤b,所以|f``(m)|≤b | f``(n)|≤b
所以f(c)≤a+a+b/2*[c²+(1-c)²]
又0

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