f********t
发帖数: 14 | 1 一个貌似简单的问题，觉得对但又不敢确定，请高手指教。
U（x,y） is a continuously differentiable function of (x,y).
For three fixed values y1, y2 and x0,
a*U(x0,y2)+(1-a)*U(x0,y1)>U(x0,a*y2+(1-a)*y1),
where 0=
Add a small term k*a*(1-a) to x0, for k small enough, I claim the inequality
is still true, as follows,
a*U(x0+k*a*(1-a),y2)+(1-a)*U(x0+k*a*(1-a),y1)>U(x0,a*y2+(1-a)*y1).
Let f(a)=a*U(x0+k*a*(1-a),y2)+(1-a)*U(x0+k*a*(1-a),y1)-U(x0,a*y2+(1-a)*y1),
the diff | G******i
发帖数: 163 | 2 This does not necessarily hold, if we only assume the conditions you gave.
> Add a small term k*a*(1-a) to x0, for k small enough, I claim the
inequality
is still true, as follows,
a*U(x0+k*a*(1-a),y2)+(1-a)*U(x0+k*a*(1-a),y1)>U(x0,a*y2+(1-a)*y1). | f********t
发帖数: 14 | 3 Thanks, could you please say a little bit more?
【在 G******i 的大作中提到】
: This does not necessarily hold, if we only assume the conditions you gave.
: > Add a small term k*a*(1-a) to x0, for k small enough, I claim the
: inequality
: is still true, as follows,
: a*U(x0+k*a*(1-a),y2)+(1-a)*U(x0+k*a*(1-a),y1)>U(x0,a*y2+(1-a)*y1).
| G******i
发帖数: 163 | 4 Let x0=0, y1=0,y2=1.
Consider U(x,y)=(1-x)f(y),
where f(0)=0, f(1)=1.
Your condition
a*U(x0,y2)+(1-a)*U(x0,y1)>U(x0,a*y2+(1-a)*y1) for 0
<=> f(a)
Your claim
a*U(x0+k*a*(1-a),y2)+(1-a)*U(x0+k*a*(1-a),y1)
>U(x0,a*y2+(1-a)*y1) for 0
<=> f(a)< a -k*a^2*(1-a) for 0
So, essentially, you are saying:
if f(0)=0, f(1)=1, f(a)
then there exists a small k0>0 such that
f(a)< a -k*a^2*(1-a) for 0
But this is incorrect, considering the case f(a)=a-a^3( | f********t
发帖数: 14 | 5 Thank you, that's really helpful.
But in your case, does it mean that
if lim(a->0){(a-f(a))/(a^2(1-a))}>0, there exists k small enough 0
))/(a^2(1-a))), so that a>f(a) can imply f(a)< a -k*a^2*(1-a) for 0
【在 G******i 的大作中提到】
: Let x0=0, y1=0,y2=1.
: Consider U(x,y)=(1-x)f(y),
: where f(0)=0, f(1)=1.
: Your condition
: a*U(x0,y2)+(1-a)*U(x0,y1)>U(x0,a*y2+(1-a)*y1) for 0: <=> f(a): Your claim
: a*U(x0+k*a*(1-a),y2)+(1-a)*U(x0+k*a*(1-a),y1)
: >U(x0,a*y2+(1-a)*y1) for 0: <=> f(a)< a -k*a^2*(1-a) for 0
| f********t
发帖数: 14 | | G******i
发帖数: 163 | 7 Yes. That's enough for us to see your original claim was wrong.
Consider this one:
U(x,y)=[1-g(x)]*[y-y(1-y)h(y)], with
g(x)=exp(-1/x) *sin(1/x),
h(y)=exp(-1/y^2).
In this example, whatever nonzero k you take,
the desired inequality does not hold for a near 0.
【在 f********t 的大作中提到】
| f********t
发帖数: 14 | 8 Thank you so much!
I see your points.
If I modify the small term, and let it be any positive number, so the desired inequality
is
a*U(x0+emisilon,y2)+(1-a)*U(x0+emisilon,y1)>U(x0,a*y2+(1-a)y1). Do you think
if we assume U(x,y) is absolutely continuous, is the inequality true?
【在 G******i 的大作中提到】
: Yes. That's enough for us to see your original claim was wrong.
: Consider this one:
: U(x,y)=[1-g(x)]*[y-y(1-y)h(y)], with
: g(x)=exp(-1/x) *sin(1/x),
: h(y)=exp(-1/y^2).
: In this example, whatever nonzero k you take,
: the desired inequality does not hold for a near 0.
| G******i
发帖数: 163 | 9 更加不对了
desired inequality
think
【在 f********t 的大作中提到】
: Thank you so much!
: I see your points.
: If I modify the small term, and let it be any positive number, so the desired inequality
: is
: a*U(x0+emisilon,y2)+(1-a)*U(x0+emisilon,y1)>U(x0,a*y2+(1-a)y1). Do you think
: if we assume U(x,y) is absolutely continuous, is the inequality true?
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