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函数的凸性

1.设\(f\)是一个下凸函数，且满足\(\lim_\limits{x \to-\infty} f(x) = -\infty\)

证明必有 \(\lim_\limits{x \to+\infty} f(x) = +\infty\)

证明：假设 \(\exists M > 0,\forall X > 0,s.t. x > X, f(x) < M\),由下凸函数定义得：

当\(\lambda = \frac{1}{2},y = -x时, f(\frac{x+y}{2}） \le \frac{1}{2}f(x) +\frac{1}{2}f(y)\).

即\(f(0) \le \frac{1}{2}f(x) + \frac{1}{2}f(-x)\)

\(\lim_\limits{x \to+\infty}f(0)\) \(\le\) \(\lim_\limits{x \to+\infty} \frac{1}{2} f(x)\) \(+\) \(\lim_\limits{x \to+\infty} \frac{1}{2} f(-x)\) \(<\) \(M\).

与条件矛盾，故\(\lim_\limits{x \to+\infty}f(x) = +\infty\).

2.设\(f, g\)是下凸函数， 且\(g\)单调递增，证明\(h = g\circ f\)是下凸函数

证明：f下凸 \(\rightarrow\) 对\(\forall x_1, x_2 \in I t\in(0,1)\)有

\(f[tx_1+(1-t)x_2] \le tf(x_1)+(1-t)f(x_2)\) 又因为\(g(x)\)单调递增，则

\(g(f[tx_1+(1-t)x_2]) \le g(f(x_1)+(1-t)f(x_2)) \le tg(f(x_1)) + (1-t) g(f(x_2))\)

故\(h(x) = g(f(x))\)下凸

3.(下凸函数的几种等价叙述) \(x_1 <x_2 < x_3\), \(x_1, x_2, x_3\in I\)

i) \(f(x)\)在\(I\)是下凸函数

ii) \(\frac{f(x_2)-f(x_1)}{x_2-x_1} \le \frac{f(x_3)-f(x_1)}{x_3-x_1}\)

iii) \(\frac{f(x_3)-f(x_1)}{x_3-x_1} \le \frac{f(x_3)-f(x_2)}{x_3-x_2}\)

iv) \(\frac{f(x_2)-f(x_1)}{x_2-x_1} \le \frac{f(x_3)-f(x_2)}{x_3-x_2}\)

v) 曲线y = f(x)上三点 \(A(x_1, f(x_1)), B(x_2, f(x_2)), C(x_3,f(x_3))\)所围的有向面积

\(\frac{1}{2}\begin{vmatrix}1&x_1&f(x_1)\\1&x_2&f(x_2)\\1&x_3&f(x_3)\end{vmatrix} \ge 0\)

vi) \(\forall x_0 \in I, \exists \alpha \in R, s.t. \forall x \in I\)有

\(f(x) \ge \alpha(x-x_0) + f(x_0)\)

证明以上定义等价

证明：

由i)得： \(f[tx_1+(1-t)x_2] \le tf(x_1)+(1-t)f(x_2)\)

由ii)得：\(f(x_2) \le \frac{x_2-x_1}{x_3-x_1}f(x_3)+\frac{x_3-x_2}{x_3-x_1}f(x_1)\)

令\(\lambda = \frac{x_2-x_1}{x_3-x_1}\) 则 \(i \Rightarrow ii\) 反之令\(x_2 - \lambda x_3 + (1-\lambda)x_1\) =则\(i \Leftarrow ii\)

iii) iv)同理

由ii)得：\((x_3-x_2)f(x_1)+(x_1-x_3)f(x_2)+(x_2-x_1)f(x_3) \ge 0\)

即\(\begin{vmatrix}1&x_1&f(x_1)\\1&x_2&f(x_2)\\1&x_3&f(x_3)\end{vmatrix} \ge 0\)\(ii) \Rightarrow iii)\)

因\(f(X)\)为下凸函数，则\(F(x) = \frac{f(x)-f(x_0)}{x-x_0}，(x_0 \in I)\)单调递增

故\(\forall \alpha \ge f‘(x_0)\),当\(x < x_0\)时，\(f(x) \le \alpha(x-x_0)+f(x_0)\)

同理\(\forall \alpha \le f‘(x_0)\),当\(x > x_0\)时，\(f(x) \le \alpha(x-x_0)+f(x_0)\)

故i)\(\Rightarrow\)vi)

设\(x_1<x_2<x_3 \in I, \exists \alpha s.t. f(x) \ge \alpha(x-x_2)+f(x_2), \forall x \in I\)

分别将\(x_1, x_3\)代入得\(\frac{f(x_3)-f(x_2)}{x_3-x_2} \ge \alpha \ge \frac{f(x_1)-f(x_2)}{x_1-x_2}\)

vi)\(\Rightarrow\)i)

4.证明：取\(\lambda = \frac{x-a}{b-a}, \lambda \in (0,1)\),则 \(x = \lambda b + (1-\lambda)a\)

由于\(\exists M s.t. f(a) \le M, f(b) \le M\)

即\(f(x) \le \lambda f(b) + (1 - \lambda)f(a) \le \lambda M + (1 - \lambda)M = M\)

即\(\exists M s.t. f(x) \le M\) \(f(x)\) 有上界

设\(c = \frac{a + b}{2} ,f(c) \le \frac{f(x_1)+f(x_2)}{2} \le \frac{1}{2}f(x_1)+\frac{1}{2}M (\frac{x_1+x_2}{2} = c)\)

\(\forall x \in [a, b], f(x) \ge 2f(x)-M\)

故\(f(x)\)有下界

（2）设有\(h\)充分小\(s.t.\)\([\alpha-h, \beta+h] \subset (a, b), \forall x_1, x_2 \in [\alpha, \beta], x_3 = x_2 + b\), 令\(x_1 < x_2\)

则\(\frac{f(x_2)-f(x_1)}{x_2-x_1} \le \frac{f(x_3)-f(x_2)}{x_3-x_2} \le \frac{M-m}{b}\)

\(|f(x_2)-f(x_1)| \le \frac{M-m}{b}|x_1-x_2|\)

令\(L = \frac{M-m}{b}\), 则\(|f(x_1)-f(x_2)| \le L|x_1-x_2|\)

中值定理的应用

1.

证明:(1)设\(f(x)\in C[0, +\infty)\),且在\([0, +\infty)\)上可导,则

\(\forall x \in [0, +\infty), \exists \xi \in (0, x), s.t. f(x) = (1+\xi)ln(1+x)f‘(\xi)\).

\(\frac{f(x)}{ln(1+x)} = (1+\xi)f‘(\xi)\).

由柯西中值定理：\(\exists \xi \in (0, x) s.t. \frac{f(x)-f(0)}{g(x)-g(0)} = \frac{f(x)}{ln(1+x)} = \frac{f‘(\xi)}{g‘(\xi)} = (1+\xi)f‘(\xi)\)

(2) 设\(f(x)\)在\([\frac{3 \pi}{4}, \frac{7 \pi}{4}]\)上可导，且\(f(\frac{3\pi}{4}) = 0, f(\frac{7\pi}{4}) = 0\)

则存在 \(\xi \in (\frac{3\pi}{4}, \frac{7 \pi}{4}) s.t. f(\xi) + f‘(\xi) = cos \xi\)

令\(F(x) = e^x(f(x) - \frac{cosx+sinx}{2})\)

\(\exists \xi \in(\frac{3 \pi}{4}, \frac{7 \pi}{4}), s.t. F‘(\xi) = \frac{F(\frac{7 \pi}{4}) - F(\frac{3 \pi}{4})}{\pi} = 0 \Rightarrow f(\xi) + f‘(\xi) = cos \xi\)

2.

解：

\(\lim_\limits{x \to 0}(\frac{1}{x^2}-\frac{cotx}{x}) \\ =\lim_\limits{x \to 0} \frac{simx - xcosx}{x^2sinx} \\ = \lim_\limits{x \to 0}\frac{xsinx}{2xsinx+x^2cosx} \\ = \lim_\limits{x \to 0}\frac{cosx}{3cosx-xsinx} \\ =\lim_\limits{x \to 0}\frac{1}{3-xtanx} = \frac{1}{3}\)

\(\lim_\limits{x \to +\infty}e^{2x}(\frac{e^x+e^{-x}}{e^x-e^{-x}}) \\ =\lim_\limits{x \to +\infty}(\frac{t^2+1}{t^2-1})^{t^2} \\ = \lim_\limits{x \to +\infty}(1+\frac{2}{t^2-1})^{t^2 \cdot \frac{t^2-1}{2} \cdot \frac{2}{t^2-1}} \\ =e^2\)

讨论班的题整理

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原文地址：https://www.cnblogs.com/fseject-2002/p/14643620.html