malonechan 发表于 2011-3-17 20:56
similar proof to the quasi concave function
just prove necessity
choose z belongs to this subset {Df(x)*z=0} choose a belongs to (0,1]
by taylor expansion
f(x+az)-f(x)-Df(x)*az=a^2/x*z*D^2 f(x+bz)*z for some b belongs to [0,a]
as z belongs to the subset, the above equation is equivalent to
f(x+az)-f(x)=a^2/x*z*D^2 f(x+bz)*z for some b belongs to [0,a]
if left hand side 0 ie. f(x+az)-f(x)>0   by mc.3 Df(x)*az>0. contradiction with Df(x)*z=0
if left hand side=0  ie. f(x+az)=f(x)   then as Df(x) not equal to 0, and f(x+az)>=f(x), Df(x)*(az)>0 as well by m.c.3.  so contradiction as well
so only the first case when left hand side is smaller than 0 is possilble.

malonechan 发表于 2011-3-17 21:46
4# xinshou0011

不会阿 b=0 只说明f里面是f(x) 没有关系啊 他直接说明z*D^2 f(x) * z