- #36
Lazerlike42
- 20
- 0
lurflurf said:without loss of generality assume x<=y to avoid considering the cases y<x
The integral was a wrong turn.
I noticed this is Baby Rudin exercise 5.26.
Show f=0 on [0,1] and f=0 on R will follow.
0<x<1
f(x)=f(x)-f(0)=(x-0)f'(t)=x f'(t) (mean value theorem with 0<t<x<1)
|f(x)|=x|f'(t)|<=x|f(t)| (by given inequality)
M:=sup(t|0<t<x||f(x)|)
|f(x)|<=x M
M<=x M
(1-x)M<=0
M=0
|f|=0 on [0,1]
(f(1)=f(1-)=0
Looks like or similar to a more concise version of what I did, but what are you saying when you say M:=sup(t|0<t<x||f(x)|) ? I'm unsure what t|0<t<x||f(x)| means.