In summary, the given function f:R->R satisfies the inequality ##|\sum\limits_{k=1}^n3^k[f(x+ky)-f(x-ky)]|<=1## for every positive integer n and all real x, y. To prove that f is a constant function, we can use the mean value theorem and consider the difference quotient f(a)-f(-a) for different values of n and y. By making suitable choices for these variables, we can bound the difference and show that it approaches zero as n and y increase. This implies that f'(c)=0
#36
AdityaDev
527
33
In the original expression, the summation is only upto n.
In the original expression, the summation is only upto n.
I have two responses to that.
First, having shown |f(a)−f(−a)|≤2/9, you could plug that back into the same process, still with n=2, and obtain a tighter bound still. I'm not sure whether that sequence will get you to a limit of zero, though.
Secondly, it's for all n (see posts #6, #7). You can make n as large as you like.