利用拉氏中值定理与夹逼准则求解函数表达式一例
作者：互联网
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
\[\left| g\left( x \right) \cdot f\left( x \right) +\lambda g'\left( x \right) \right|\leqslant \left| g\left( x \right) \right| \]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
\[h'\left( x \right) =e^{\frac{1}{\lambda}F\left( x \right)}\left( \frac{1}{\lambda}f\left( x \right) \cdot g\left( x \right) +g'\left( x \right) \right) \]得到
\[\begin{align*} \left| h'\left( x \right) \right|&=e^{\frac{1}{\lambda}F\left( x \right)}\left| \frac{1}{\lambda}f\left( x \right) \cdot g\left( x \right) +g'\left( x \right) \right| \\ &\leqslant e^{\frac{1}{\lambda}F\left( x \right)}\cdot \frac{1}{\lambda}\left| g\left( x \right) \right| \\ &=\frac{1}{\lambda}\left| h\left( x \right) \right| \end{align*} \]设\(\displaystyle a\leqslant x\leqslant a+\frac{\lambda}{2}\)
则有
\[\begin{align*} \left| h\left( x \right) \right|&=\left| h\left( x \right) -h\left( a \right) \right| \\ &=\left| h'\left( \xi _1 \right) \right|\left( x-a \right) \\ &\leqslant \frac{1}{\lambda}\left| h\left( \xi _1 \right) \right|\left( x-a \right) \\ &\leqslant \frac{1}{2}\left| h\left( \xi _1 \right) \right| \\ &\leqslant \cdots \\ &\leqslant \frac{1}{2^n}\left| h\left( \xi _n \right) \right|\rightarrow 0\left( n\rightarrow \infty \right) \end{align*} \]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\(\displaystyle f\left( x \right) \equiv 0\).
标签：right,frac,leqslant,拉氏,中值,lambda,displaystyle,表达式,left 来源： https://www.cnblogs.com/xuke0721/p/14933228.html