Question

f为(a,b)的凸函数证明f的两个单侧导数f'+(x)和f'-(x)在（a,b)的任意子闭区间[c,d]上黎曼可积且 f

Answer 1

定理:f为(a,b)的凸函数,则其左右导数f'{-},f'{+}

存在,且

1.f'{-},f'{+}递减.

2.f'{-}(c)≥f'{+}(c)

3.c,d∈(a,b),则

f'{+}(c)≥[f(d)-f(c)]/[d-c]≥f'{-}(d)

1.

由于f'{-},f'{+}递减.

只需研究

In=(d-c)/n∑{0≤k≤n-1}f'{+}(c+k(d-c)/n)和

Jn=(d-c)/n∑{0≤k≤n-1}f'{-}(c+k(d-c)/n)的极限.

2.In≤Jn.(定理的2.)

3.由定理的3.得

In≥[(d-c)/n]*

∑{0≤k≤n-1}[f(c+(k+1)(d-c)/n)-f(c+k(d-c)/n)]/[(d-c)/n]=

=f(d)-f(c)

4.由定理的3.得

Jn≤[(d-c)/n]f'{-}(c)+[(d-c)/n]*

∑{1≤k≤n-1}[f(c+k(d-c)/n)-f(c+(k-1)(d-c)/n)]/[(d-c)/n]=

=[(d-c)/n]f'{-}(c)+f(d-(d-c)/n)-f(c)

5.由f的连续性得,

Lim{n→∞}{[(d-c)/n]f'{-}(c)+f(d-(d-c)/n)-f(c)}=

=f(d)-f(c)

所以

Lim{n→∞}In=Lim{n→∞}Jn=f(d)-f(c)

==>f'{-},f'{+}黎曼可积,且

∫{c→d}f'{-}(x)dx=∫{c→d}f'{+}(x)dx=f(d)-f(c).