L********3
发帖数: 204 | 1 f(x):[0,\infty)->R^+, f is strictly increasing and strictly convex, i.e.
f>0, f'>0, f''>=0 on (0,\infty). Moreover f(0)=0, f'(0)=0.
Show that: 2x^2f''(x)-xf'(x)+f(x)>0 on (0,\infty).
Counterexample is wellcome as well.
You can also relax the condition on the domain
(instead of on (0,\infty), any interval (0, a), a>0 is ok).
You can also neglect the condition f'(0)=0.
Thanks. | f*c
发帖数: 687 | 2 好像既有正例也有反例。
可能的反例：令f渐近趋于y=x-1,f'渐近趋于1，让f" 渐近 O(x^{-3})...
【在 L********3 的大作中提到】
: f(x):[0,\infty)->R^+, f is strictly increasing and strictly convex, i.e.
: f>0, f'>0, f''>=0 on (0,\infty). Moreover f(0)=0, f'(0)=0.
: Show that: 2x^2f''(x)-xf'(x)+f(x)>0 on (0,\infty).
: Counterexample is wellcome as well.
: You can also relax the condition on the domain
: (instead of on (0,\infty), any interval (0, a), a>0 is ok).
: You can also neglect the condition f'(0)=0.
: Thanks.
| f***r
发帖数: 1126 | 3 举个反例吧：f(x)=e^{.1x}+e^{-.1x}-2
2x^2f"(x)-xf'(x)+f(x)=-0.15x^2+O(x^3) when x\to 0 | L********3
发帖数: 204 | 4 Thanks, but your calculation seems to be incorrect
the last line ...=0.03x^2+O(x^3)?
【在 f***r 的大作中提到】
: 举个反例吧：f(x)=e^{.1x}+e^{-.1x}-2
: 2x^2f"(x)-xf'(x)+f(x)=-0.15x^2+O(x^3) when x\to 0
| L********3
发帖数: 204 | 5 the problem is: can you do it with the restriction f(0)=0, etc.
Thanks.
【在 f*c 的大作中提到】
: 好像既有正例也有反例。
: 可能的反例：令f渐近趋于y=x-1,f'渐近趋于1，让f" 渐近 O(x^{-3})...
| f***r
发帖数: 1126 | 6 不好意思，真算错了。不过这个问题的答案应该是很符合直觉的：只要取一个x使得f"(
x)=0，那么原来的不等式就等价于
f(x)=int_0^x f'(s)ds>xf'(x),
Impossible!(因为f'(s)
【在 L********3 的大作中提到】
: Thanks, but your calculation seems to be incorrect
: the last line ...=0.03x^2+O(x^3)?
| L********3
发帖数: 204 | 7 多谢用心哈
在x趋于零的时候，貌似答案是肯定的：
f(x)=f(0)+f'(0)x+f''(0)x^2/2+O(x^3)
f'(x)=f'(0)+f''(0)x+O(x^2)
now: f(x) ->f''(0)x^2
f'(x)->f''(0)x
f''(x)->f''(0), (suppose f is smooth enough)
then 2x^2f''-xf'+f->x^2(2f''(0)-f''(0)+f''(0)/2)=3/2*x^2*f''(0)>0, when x
small
你的思路的问题在于：当f''(x)接近0，f接近线性， f'接近常数
所以f=int_0^x f'(s)ds虽然不大于xf'(x),但是差的余量能被f''补上。
不知道你同意不?
"(
【在 f***r 的大作中提到】
: 不好意思，真算错了。不过这个问题的答案应该是很符合直觉的：只要取一个x使得f"(
: x)=0，那么原来的不等式就等价于
: f(x)=int_0^x f'(s)ds>xf'(x),
: Impossible!(因为f'(s)
| M****o
发帖数: 4860 | 8 since f'(x)<=xf''(x)
2x^2f''(x)-xf'(x)+f(x)>= x^2*f''(x)+f(x)>0 for x>0
【在 L********3 的大作中提到】
: f(x):[0,\infty)->R^+, f is strictly increasing and strictly convex, i.e.
: f>0, f'>0, f''>=0 on (0,\infty). Moreover f(0)=0, f'(0)=0.
: Show that: 2x^2f''(x)-xf'(x)+f(x)>0 on (0,\infty).
: Counterexample is wellcome as well.
: You can also relax the condition on the domain
: (instead of on (0,\infty), any interval (0, a), a>0 is ok).
: You can also neglect the condition f'(0)=0.
: Thanks.
| f***r
发帖数: 1126 | 9 你的前面那段是对的。
后面的：f"(x)可以构造得任意小，但是 xf'(x)-f(x)是个递增函数。
【在 L********3 的大作中提到】
: 多谢用心哈
: 在x趋于零的时候，貌似答案是肯定的：
: f(x)=f(0)+f'(0)x+f''(0)x^2/2+O(x^3)
: f'(x)=f'(0)+f''(0)x+O(x^2)
: now: f(x) ->f''(0)x^2
: f'(x)->f''(0)x
: f''(x)->f''(0), (suppose f is smooth enough)
: then 2x^2f''-xf'+f->x^2(2f''(0)-f''(0)+f''(0)/2)=3/2*x^2*f''(0)>0, when x
: small
: 你的思路的问题在于：当f''(x)接近0，f接近线性， f'接近常数
| L********3
发帖数: 204 | 10 en, you are talking about when x->inf?
【在 f***r 的大作中提到】
: 你的前面那段是对的。
: 后面的：f"(x)可以构造得任意小，但是 xf'(x)-f(x)是个递增函数。
| L********3
发帖数: 204 | 11 sorry, f' is not necessarily convex, right?
【在 M****o 的大作中提到】
: since f'(x)<=xf''(x)
: 2x^2f''(x)-xf'(x)+f(x)>= x^2*f''(x)+f(x)>0 for x>0
| L********3
发帖数: 204 | 12 e.g. f(x)=e^(-x)-1+x
so it seemingly only work for x near 0
【在 L********3 的大作中提到】
: sorry, f' is not necessarily convex, right?
| b*******g
发帖数: 363 | 13
at least not true for some such f(x)
【在 L********3 的大作中提到】
: e.g. f(x)=e^(-x)-1+x
: so it seemingly only work for x near 0
| z*********g
发帖数: 37 | 14 For the functions with continuous 3rd order derivatives, the statement is
true near 0. The statement is obviously true when f'''>=0. We now
try to construct some functions that will have
2x^2f''(x)-xf'(x)+f(x)=0 for some x>0.
Choose, f=bx-4x^{1/2}, for some $b>0$
f'(x)=b-2x^{-1/2},
f''(x)=x^{-3/2},
Clearly, for $x>16/b^2, all of them are positive. For x in [0, 16/b^2],
you patch it by some nice convex and positive functions with x^2 order
at x=0.
Now, 2x^2f''(x)-xf'(x)+f(x)=2x^{1/2}-bx+2x^{1/2}+bx-4x^{1/2}=0.
Therefore, your statement is not true for general functions.
Is this an acceptable counter-example?
【在 L********3 的大作中提到】
: e.g. f(x)=e^(-x)-1+x
: so it seemingly only work for x near 0
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