# Prove that f is a constant function

Tags:
1. Apr 28, 2015

1. The problem statement, all variables and given/known data
Suppose that f:R->R satisfies the inequality $|\sum\limits_{k=1}^n3^k[f(x+ky)-f(x-ky)]|<=1$ for every positive integer k, for all real x, y. Prove that f is a constant function.

2. Relevant equations

None
3. The attempt at a solution
I tried taking f(x)=sinx and then using sinC-sinD=2cos(C/2+D/2)sin(C/2-D/2). Then you will get 2cosx.sinky. |cosx| is always less than or equal to one, but greater than zero. So I ignored cosx from the sum since it is just like a positive constant. and you will be left with
$$2\sum3^ksinky=2.Im\sum3^ke^{iky}$$

2. Apr 28, 2015

### Dick

To prove a function is constant you often prove that the derivative is zero.

3. Apr 29, 2015

There are 2 variables x and y. Should I differentiate the expression with respect to x or with respect to y?

4. Apr 29, 2015

### haruspex

As far as f is concerned, it is a function of a single variable.

5. Apr 29, 2015

If I differentiate w.r.t t, then you will have $$\sum\limits_{k=1}^n3^k[f'(x+ky). (\frac{dx}{dt}+k\frac{dy}{dt})-f'(x-ky). (\frac{dx}{dt}-k\frac{dy}{dt})]$$

6. Apr 29, 2015

### haruspex

I didn't mean to imply endorsement of differentiating. Since Dick suggested it, I'm prepared to believe that helps, but I don't see it yet.
I think there's something not quite right in the OP. Should it say "for every positive integer n, for all real x, y"?

7. Apr 29, 2015

It does.

8. Apr 29, 2015

I thinks its just <sum expression> for all real x, y.

9. Apr 29, 2015

### haruspex

But n is also a variable for that sum, and the terms inside the sum can change sign as k changes, so I think it's important to include "for all n".

10. Apr 29, 2015

### Dick

Don't differentiate with respect to either one. Remember that the derivative is the limit of a difference quotient. Let y approach zero and take the limit.

11. Apr 29, 2015

### Dick

Actually letting y actually approach zero doesn't cut it. I think you need some sort of mean value theorem type of thing to get the difference to actually equal $f'(c) 2ky$ for some value of $c$ and nonzero $y$. Or you need to make some sort of approximation argument about the difference in terms a derivative. THEN you can let $k$ approach infinity to conclude $f'(c)=0$. Maybe I haven't thought this through enough, so you try and fill in the details for me.

Last edited: Apr 29, 2015
12. Apr 29, 2015

### haruspex

Fix on some x to begin with. For simplicity I'll take x=0.
Consider f(a)-f(-a). By suitable choices of n and y, what bounds can you put on it?
Next, consider n=2, but changing y such that the extreme ends of the sum are still at -a and +a, you should now be able to put tighter bounds on that difference. And so on.

13. Apr 29, 2015

For x=0, then you get f(a) for n=1, y=a.
For n=2, the expression has two pairs: f(y)-f(-y) + f(2y)-f(-2y)
then there is no value of y for which y=a and -2y=-a.

14. Apr 29, 2015

According to MVT, $$f'(c)=\frac{f(x+ky)-f(x-ky)}{2ky}$$
Hence $f'(c)2ky=g(k)$ where g(k)=f(x+ky)-f(x-ky)

15. Apr 29, 2015

If I take a point x, f(x) on the curve Y=f(X), then the two points x-ky and x+ky on the X-axis are aymmetrically placed about X=x. So can I say that the arbitary point c will remain same for any value of k? That is, can I say that the lines formed by connecting (x-2y,f(x-2y)) and (x+2y, f(x+2y)) will be parallel to that formed by (x-y, f(x-y)) and (x+y,f(x+y))?

16. Apr 29, 2015

### Dick

Good point, and no you can't say that. So this won't work in the way I was thinking.

17. Apr 29, 2015

### haruspex

Yes, but what inequality do you get for |f(a)-f(-a)|?
Sorry, a small error in what I posted may have misled you:
I meant the extreme top end of the sum, singular. I.e. a = ny. What inequality do you get now? How can you combine this with the n=1 inequality to get even tighter bounds on f(a)-f(-a)?

18. Apr 30, 2015

For n=1, y=a, the inequality will be |f(a)-f(-a)|<=1. So f(a)-f(-a) is greater than -1 and less than 1.
for n=2,y=a, you get f(a)-f(-a)+f(2a)-f(-2a) and this should again lie between -1 and 1. Let f(a)-f(-a)=A. So---> -1-A<=f(2a)-f(-2a)<=1-A.

19. Apr 30, 2015

### haruspex

No, the given inequality has 3k, not 3k-1, and this is the k=n term.
I meant you to pick y such that the k=n (=2) term corresponds to f(a), but I don't think it matters.
What does matter is that you've left out the power-of-three factors again.

20. Apr 30, 2015

And $-1/3\le f(a)-f(-a) \le 1/3$