問：\(f(x):R \rightarrow R\)， 3階可導，且\(f(x),f^{\prime\prime\prime}(x)\)有界，證明\(f^\prime(x),f^{\prime\prime}(x)\)有界。

證：\(\forall x_0\in R\), 根據帶Lagrange餘項的Taylor公式，將\(f(x_0+1)\)和\(f(x_0-1)\)在\(x=x_0\)處展開，得到：

\[\begin{align} f(x_0+1)&=f(x_0)+f^\prime(x_0)+\frac{1}{2}f^{\prime\prime}(x_0)+\frac{1}{6}f^{\prime\prime\prime}(x_0+\theta_1),\quad \theta_1:=\theta_1(x_0)\in [0,1]\\ f(x_0-1)&=f(x_0)-f^\prime(x_0)+\frac{1}{2}f^{\prime\prime}(x_0)-\frac{1}{6}f^{\prime\prime\prime}(x_0-\theta_2),\quad \theta_2:=\theta_2(x_0)\in [0,1] \end{align} \]

因此 \(f^\prime(x_0)\)

和 \(f^{\prime\prime}(x_0)\) 可表示為：

\[\begin{align} f^\prime(x_0)&=\frac{1}{2}[f(x_0+1)-f(x_0-1)-\frac{1}{6}f^{\prime\prime\prime}(x_0+\theta_1)-\frac{1}{6}f^{\prime\prime\prime}(x_0-\theta_2)]\\ f^{\prime\prime}(x_0)&=f(x_0+1)+f(x_0-1)-2f(x_0)-\frac{1}{6}f^{\prime\prime\prime}(x_0+\theta_1)+\frac{1}{6}f^{\prime\prime\prime}(x_0-\theta_2) \end{align} \]

由於 \(\forall x\in R, f(x),f^{\prime\prime\prime}(x)<0\)

，故\(f(x_0+1),f(x_0-1),f(x_0),f^{\prime\prime\prime}(x_0+\theta_1),f^{\prime\prime\prime}(x_0-\theta_2)<\infty\)，於是

\[f^\prime(x_0),f^{\prime\prime}(x_0)<\infty \]

根據\(x_0\)的任意性，\(f^\prime(x),f^{\prime\prime}(x)<\infty,\forall x\in R\)。