Prove that f is a constant function

[haruspex]

There's an error in the first line. The f(a/2) terms are k=1 in the sum, so only get a factor 3, not 9. When you correct that your last line will be much more useful.

 

[AdityaDev]

Sorry for the delay.

$$3|f(a/2)-f(-a/2)|+9|f(a)-f(-a)|\le 1$$
$$9|f(a)-f(-a)|\le 3|f(a)-f(-a)|+1$$
And $f(t)-f(-t)\le 1/3$ from post #21
Hence $$|f(a)-f(-a)|\le 2/9$$

 

[haruspex]

Sorry for the delay.

$$3|f(a/2)-f(-a/2)|+9|f(a)-f(-a)|\le 1$$
$$9|f(a)-f(-a)|\le 3|f(a)-f(-a)|+1$$
And $f(t)-f(-t)\le 1/3$ from post #21
Hence $$|f(a)-f(-a)|\le 2/9$$

Right. Again, a can be any value. In post #20 you showed $|f(a)-f(-a)|\le \frac 13$, and using that you've improved it to $\frac 29$. Can you see how to turn this process into an induction, or maybe a reductio ad absurdum, to show it must be zero?

 

[AdityaDev]

Yes. For n=3, y=a/3, you get $|f(a)-f(-a)|\le 5/27$
So it keeps decreasing.

 

[haruspex]

Yes. For n=3, y=a/3, you get $|f(a)-f(-a)|\le 5/27$
So it keeps decreasing.

Yes, it keeps decreasing, but you need to show that the limit is zero.

 

[AdityaDev]

In the original expression, the summation is only upto n.

 

[haruspex]

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.