讨论班的题整理

函数的凸性

1.设(f)是一个下凸函数，且满足(lim_limits{x o-infty} f(x) = -infty)

证明必有 (lim_limits{x o+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 o+infty}f(0)) (le) (lim_limits{x o+infty} frac{1}{2} f(x)) (+) (lim_limits{x o+infty} frac{1}{2} f(-x)) (<) (M).

与条件矛盾，故(lim_limits{x o+infty}f(x) = +infty).

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

证明：f下凸 ( ightarrow) 对(forall x_1, x_2 in I tin(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_3in 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}egin{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(egin{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, eta+h] subset (a, b), forall x_1, x_2 in [alpha, eta], 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{3pi}{4}) = 0, f(frac{7pi}{4}) = 0)

则存在 (xi in (frac{3pi}{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 o 0}(frac{1}{x^2}-frac{cotx}{x}) \ =lim_limits{x o 0} frac{simx - xcosx}{x^2sinx} \ = lim_limits{x o 0}frac{xsinx}{2xsinx+x^2cosx} \ = lim_limits{x o 0}frac{cosx}{3cosx-xsinx} \ =lim_limits{x o 0}frac{1}{3-xtanx} = frac{1}{3})

(lim_limits{x o +infty}e^{2x}(frac{e^x+e^{-x}}{e^x-e^{-x}}) \ =lim_limits{x o +infty}(frac{t^2+1}{t^2-1})^{t^2} \ = lim_limits{x o +infty}(1+frac{2}{t^2-1})^{t^2 cdot frac{t^2-1}{2} cdot frac{2}{t^2-1}} \ =e^2)