【第五章 习题15】
设 a ∈ R 1 a \in R^{1} a∈R1， f f f是 ( a , + ∞ ) (a, + \infty) (a,+∞)上的二次可微实函数， M 0 , M 1 , M 2 M_{0},M_{1},M_{2} M0,M1,M2分别是 ∣ f ( x ) ∣ , ∣ f ′ ( x ) ∣ , ∣ f ′ ′ ( x ) ∣ \left| f(x) \right|,\left| f^{'}(x) \right|,\left| f^{''}(x) \right| ∣f(x)∣, f′(x) , f′′(x) 在 ( a , + ∞ ) (a, + \infty) (a,+∞)上的最小上界。试证
M 1 2 ≤ 4 M 0 M 2 M_{1}^{2} \leq 4M_{0}M_{2} M12≤4M0M2
【证明】
- 根据泰勒定理，可知存在某个 ξ ∈ ( x , x + 2 h ) \xi \in (x,x + 2h) ξ∈(x,x+2h)，使得
f ′ ( x ) = 1 2 h [ f ( x + 2 h ) − f ( x ) ] − h f ′ ′ ( ξ ) f^{'}(x) = \frac{1}{2h}\left\lbrack f(x + 2h) - f(x) \right\rbrack - hf^{''}(\xi) f′(x)=2h1[f(x+2h)−f(x)]−hf′′(ξ)
所以
∣ f ′ ( x ) ∣ ≤ ∣ 1 2 h [ f ( x + 2 h ) − f ( x ) ] ∣ + ∣ h f ′ ′ ( ξ ) ∣ \left| f^{'}(x) \right| \leq \left| \frac{1}{2h}\left\lbrack f(x + 2h) - f(x) \right\rbrack \right| + \left| hf^{''}(\xi) \right| f′(x) ≤ 2h1[f(x+2h)−f(x)] + hf′′(ξ)
≤ 1 2 h ( ∣ f ( x + 2 h ) ∣ + ∣ f ( x ) ∣ ) + h ∣ f ′ ′ ( ξ ) ∣ \leq \frac{1}{2h}\left( \left| f(x + 2h) \right| + \left| f(x) \right| \right) + h\left| f^{''}(\xi) \right| ≤2h1(∣f(x+2h)∣+∣f(x)∣)+h f′′(ξ)
根据 M 0 , M 1 , M 2 M_{0},M_{1},M_{2} M0,M1,M2的定义，可知
∣ f ′ ( x ) ∣ ≤ 1 2 h ( M 0 + M 0 ) + h M 2 = h M 2 + M 0 h \left| f^{'}(x) \right| \leq \frac{1}{2h}\left( M_{0} + M_{0} \right) + hM_{2} = hM_{2} + \frac{M_{0}}{h} f′(x) ≤2h1(M0+M0)+hM2=hM2+hM0
由于 h h h是一个自由的变量，不受 M 0 , M 1 , M 2 M_{0},M_{1},M_{2} M0,M1,M2影响，所以当
h = M 0 M 2 h = \sqrt{\frac{M_{0}}{M_{2}}} h=M2M0
有
∣ f ′ ( x ) ∣ ≤ 2 M 0 M 2 \left| f^{'}(x) \right| \leq 2\sqrt{M_{0}M_{2}} f′(x) ≤2M0M2
所以
M 1 2 ≤ 4 M 0 M 2 M_{1}^{2} \leq 4M_{0}M_{2} M12≤4M0M2
- 根据 f ( x ) f(x) f(x)的定义，可以知道
f ( x ) = { 2 x 2 − 1 − 1 < x < 0 1 − 2 x 2 + 1 0 ≤ x < + ∞ f(x) = \left\{ \begin{matrix} 2x^{2} - 1 & - 1 < x < 0 \\ 1 - \frac{2}{x^{2} + 1} & 0 \leq x < + \infty \end{matrix} \right.\ f(x)={2x2−11−x2+12−1<x<00≤x<+∞
f ′ ( x ) = { 4 x − 1 < x < 0 4 x ( x 2 + 1 ) 2 0 ≤ x < + ∞ f^{'}(x) = \left\{ \begin{matrix} 4x & - 1 < x < 0 \\ \frac{4x}{\left( x^{2} + 1 \right)^{2}} & 0 \leq x < + \infty \end{matrix} \right.\ f′(x)={4x(x2+1)24x−1<x<00≤x<+∞
f ′ ′ ( x ) = { 4 − 1 < x < 0 4 − 12 x 2 ( x 2 + 1 ) 3 0 ≤ x < + ∞ f^{''}(x) = \left\{ \begin{matrix} 4 & - 1 < x < 0 \\ \frac{4 - 12x^{2}}{\left( x^{2} + 1 \right)^{3}} & 0 \leq x < + \infty \end{matrix} \right.\ f′′(x)={4(x2+1)34−12x2−1<x<00≤x<+∞
可以验证 M 0 = 1 , M 1 = 4 , M 2 = 4 M_{0} = 1,M_{1} = 4,M_{2} = 4 M0=1,M1=4,M2=4，于是
M 1 2 = 4 M 0 M 2 M_{1}^{2} = 4M_{0}M_{2} M12=4M0M2
- 对于向量值函数， M 1 2 ≤ 4 M 0 M 2 M_{1}^{2} \leq 4M_{0}M_{2} M12≤4M0M2仍然成立
设
f ( x ) = ( f 1 ( x ) , f 2 ( x ) … f n ( x ) ) \mathbf{f}(x) = \left( f_{1}(x),f_{2}(x)\ldots f_{n}(x) \right) f(x)=(f1(x),f2(x)…fn(x))
任取一点 x 0 x_{0} x0，设
F ( x ) = f ′ ( x 0 ) ⋅ f ( x ) F(x) = \mathbf{f}^{\mathbf{'}}\left( x_{0} \right) \cdot \mathbf{f}(x) F(x)=f′(x0)⋅f(x)
此时，设
M 0 = sup ∣ f ( x ) ∣ M_{0} = \sup{|\mathbf{f}(x)|} M0=sup∣f(x)∣
M 1 = sup ∣ f ′ ( x ) ∣ M_{1} = \sup{|\mathbf{f}^{'}(x)|} M1=sup∣f′(x)∣
M 2 = sup ∣ f ′ ′ ( x ) ∣ M_{2} = \sup{|\mathbf{f}^{''}(x)|} M2=sup∣f′′(x)∣
所以
F ′ ( x ) = f ′ ( x 0 ) ⋅ f ′ ( x ) F^{'}(x) = \mathbf{f}^{\mathbf{'}}\left( x_{0} \right) \cdot \mathbf{f}^{\mathbf{'}}(x) F′(x)=f′(x0)⋅f′(x)
F ′ ′ ( x ) = f ′ ( x 0 ) ⋅ f ′ ′ ( x ) F^{''}(x) = \mathbf{f}^{\mathbf{'}}\left( x_{0} \right) \cdot \mathbf{f}^{\mathbf{'}\mathbf{'}}(x) F′′(x)=f′(x0)⋅f′′(x)
根据柯西施瓦茨不等式
sup ∣ F ( x ) ∣ ≤ M 0 × ∣ f ′ ( x 0 ) ∣ \sup{|F(x)|} \leq M_{0} \times \left| \mathbf{f}^{\mathbf{'}}\left( x_{0} \right) \right| sup∣F(x)∣≤M0× f′(x0)
sup ∣ F ′ ′ ( x ) ∣ ≤ M 2 × ∣ f ′ ( x 0 ) ∣ \sup{|F^{''}(x)|} \leq M_{2} \times \left| \mathbf{f}^{\mathbf{'}}\left( x_{0} \right) \right| sup∣F′′(x)∣≤M2× f′(x0)
由于 F ( x ) F(x) F(x)为实函数，根据泰勒定理，可知存在某个 ξ ∈ ( x 0 , x 0 + 2 h ) \xi \in \left( x_{0},x_{0} + 2h \right) ξ∈(x0,x0+2h)，使得
F ′ ( x 0 ) = 1 2 h [ F ( x 0 + 2 h ) − F ( x 0 ) ] − h F ′ ′ ( ξ ) F^{'}\left( x_{0} \right) = \frac{1}{2h}\left\lbrack F\left( x_{0} + 2h \right) - F\left( x_{0} \right) \right\rbrack - hF^{''}(\xi) F′(x0)=2h1[F(x0+2h)−F(x0)]−hF′′(ξ)
根据前面的结论
∣ F ′ ( x 0 ) ∣ ≤ 2 × sup ∣ F ( x ) ∣ × sup ∣ F ′ ′ ( x ) ∣ \left| F^{'}\left( x_{0} \right) \right| \leq 2 \times \sqrt{\sup\left| F(x) \right| \times \sup\left| F^{''}(x) \right|} F′(x0) ≤2×sup∣F(x)∣×sup∣F′′(x)∣
从而
∣ f ′ ( x 0 ) ⋅ f ′ ( x 0 ) ∣ ≤ 2 × M 0 M 2 × ∣ f ′ ( x 0 ) ∣ \left| \mathbf{f}^{\mathbf{'}}\left( x_{0} \right) \cdot \mathbf{f}^{\mathbf{'}}\left( x_{0} \right) \right| \leq 2 \times \sqrt{M_{0}M_{2}} \times \left| \mathbf{f}^{\mathbf{'}}\left( x_{0} \right) \right| f′(x0)⋅f′(x0) ≤2×M0M2× f′(x0)
∣ f ′ ( x 0 ) ∣ ≤ 2 M 0 M 2 \left| \mathbf{f}^{\mathbf{'}}\left( x_{0} \right) \right| \leq 2\sqrt{M_{0}M_{2}} f′(x0) ≤2M0M2
由于 x 0 x_{0} x0是任意的，所以
M 1 ≤ 2 M 0 M 2 M_{1} \leq 2\sqrt{M_{0}M_{2}} M1≤2M0M2
M 1 2 ≤ 4 M 0 M 2 M_{1}^{2} \leq 4M_{0}M_{2} M12≤4M0M2
- 当向量值函数的所有分量为上边的例子函数的常数倍时，
M 1 2 = 4 M 0 M 2 M_{1}^{2} = 4M_{0}M_{2} M12=4M0M2
可以成立