设f'(x)∈C[a,b],f(a)=f(b)=0,证明|f(x)|≤1/2∫(a,b)|f'(x)|dx

设f'(x)∈C[a,b],f(a)=f(b)=0,证明|f(x)|≤1/2∫(a,b)|f'(x)|dx
设f'(x)∈C[a,b],f(a)=f(b)=0,证明
|f(x)|≤1/2∫(a,b)|f'(x)|dx

设f'(x)∈C[a,b],f(a)=f(b)=0,证明|f(x)|≤1/2∫(a,b)|f'(x)|dx
f(a)=f(b)=0,所以|f(x)|在【a,b】上的最大值肯定在区间中取到,并且这点的导数为零.设为c 所以有|f(x)|≤f(c)=1/2（∫(a,c)f'(x)dx-∫(c,b)f'(x)dx）≤1/2（∫(a,c)|f'(x)|dx+∫(c,b)|f'(x)|dx）=1/2∫(a,b)|f'(x)|dx