s*********o
发帖数: 14 | 1 f(0)=0, and |f'(x)| <= M*|f(x)| for 0<=x<=L, where M and L are some
constants. Prove: f(x)=0 for 0<=x<=L. |
n*******l
发帖数: 2911 | 2 |f(x)|=|\int_0^x f'(t)|<= \int_0^x |f'(t)| <= M \int_0^x |f(t)|
Then Gronwall inequality.
Edit: changed notation to avoid confusion.
【在 s*********o 的大作中提到】
: f(0)=0, and |f'(x)| <= M*|f(x)| for 0<=x<=L, where M and L are some
: constants. Prove: f(x)=0 for 0<=x<=L.
|
f*****g
发帖数: 9098 | |
s*********o
发帖数: 14 | 4 So knowledgable! Thank you!
【在 n*******l 的大作中提到】
: |f(x)|=|\int_0^x f'(t)|<= \int_0^x |f'(t)| <= M \int_0^x |f(t)|
: Then Gronwall inequality.
: Edit: changed notation to avoid confusion.
|
s*********o
发帖数: 14 | 5 I tried to find a counterexample too, but couldn't. If you find one, please
let me know.
【在 f*****g 的大作中提到】
: 你这个问题本身就不对吧，举出反例很容易的阿
|
f*****g
发帖数: 9098 | 6 f(x)=0 for 0<=x<1/2
=2x-1 for 1/2<=x<1/4
=-2x+1 for 1/4<=x<=1
Take your M to be 4. |
h***1
发帖数: 40 | 7 f'(1/2+delta)=2 ? M(f(1/2+delta)=2M delta, M=4?
【在 f*****g 的大作中提到】
: f(x)=0 for 0<=x<1/2
: =2x-1 for 1/2<=x<1/4
: =-2x+1 for 1/4<=x<=1
: Take your M to be 4.
|
h***1
发帖数: 40 | 8 哈哈，用用f(x) and f(0)的，微分种植定理，应该很容易证明f(x)=0.要一个小的迭代
过程，要求xM<1,就可以了。然后不断的扩张到[0,L] |
