问题：设\(\displaystyle f\left( x \right)\)，\(\displaystyle g\left( x \right)\)在\(\displaystyle\left[ a,b \right]\)上连续，在\(\displaystyle\left( a,b \right)\)上可微，且\(\displaystyle g\left( a \right) =0\)，若有实数\(\displaystyle\lambda \ne 0\)，使得
成立，试证：\(\displaystyle g\left( x \right) =0\).
过程如下：不妨设\(\displaystyle\lambda >0\)
记\(\displaystyle F‘\left( x \right) =f\left( x \right)\)，令\(\displaystyle h\left( x \right) =e^{\frac{1}{\lambda}F\left( x \right)}\cdot g\left( x \right)\)
则有
得到
设\(\displaystyle a\leqslant x\leqslant a+\frac{\lambda}{2}\)
则有
其中用到了\(\text{Lagrange}\)中值定理.
若\(\displaystyle a+\frac{\lambda}{2}>b\)，则可直接说明\(\displaystyle h\left( x \right) =0,x\in \left[ a,b \right]\)，进而得到\(\displaystyle g\left( x \right) =0,x\in \left[ a,b \right]\).
若\(\displaystyle a+\frac{\lambda}{2}\leqslant b\)，则可利用同样的办法逐段对区间\(\displaystyle\left[ a+\frac{\lambda}{2},a+\frac{2\lambda}{2} \right] ,\left[ a+\frac{2\lambda}{2},a+\frac{3\lambda}{2} \right] ,\cdots\)进行研究，进而可得\(\displaystyle g\left( x \right) =0,x\in \left[ a,b \right]\).
事实上，若\(\displaystyle f\left( x \right)\)在\(\displaystyle\left[ 0,+\infty \right)\)上可微，\(\displaystyle f\left( 0 \right) =0\)，\(\displaystyle A>0\)，且有\(\displaystyle\left| f‘\left( x \right) \right|\leqslant A\left| f\left( x \right) \right|\)在\(\displaystyle\left[ 0,+\infty \right)\)上恒成立，则\(\displaystyle f\left( x \right) \equiv 0\).