Prove that f is a constant function

[AdityaDev]

Homework Statement

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.

Homework Equations

None

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
$$2\sum3^ksinky=2.Im\sum3^ke^{iky}$$
I am not sure about this procedure. Is there a general method?

 

[Dick]

Homework Statement

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.

Homework Equations

None

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
$$2\sum3^ksinky=2.Im\sum3^ke^{iky}$$
I am not sure about this procedure. Is there a general method?

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

 

[AdityaDev]

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

 

[haruspex]

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

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

 

[AdityaDev]

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

 

[haruspex]

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

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"?

 

[AdityaDev]

It does.

 

[AdityaDev]

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

 

[haruspex]

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

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".

 

[Dick]

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

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.

 

[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.

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.

 

[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.

 

[AdityaDev]

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.

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.

 

[AdityaDev]

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.

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)

 

[AdityaDev]

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))?

 

[Dick]

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))?

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

 

[haruspex]

For x=0, then you get f(a) for n=1, y=a.

Yes, but what inequality do you get for |f(a)-f(-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.

Sorry, a small error in what I posted may have misled you:

changing y such that the extreme ends of the sum are still at -a and +a

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)?

 

[AdityaDev]

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.

 

[haruspex]

For n=1, y=a, the inequality will be |f(a)-f(-a)|<=1

No, the given inequality has 3^k, not 3^k-1, and this is the k=n term.

for n=2,y=a, you get f(a)-f(-a)+f(2a)-f(-2a)

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.

 

[AdityaDev]

No, the given inequality has 3^k, not 3^k-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.

Sorry.
For n=2, y can be taken as a/2
So 3[f(a/2)-f(-a/2)] + 9[f(a)-f(-a)]
And $-1/3\le f(a)-f(-a) \le 1/3$

 

[haruspex]

Sorry.
For n=2, y can be taken as a/2
So 3[f(a/2)-f(-a/2)] + 9[f(a)-f(-a)]
And $-1/3\le f(a)-f(-a) \le 1/3$

Right, but properly: |3[f(a/2)-f(-a/2)] + 9[f(a)-f(-a)]|≤1
Now, in $-1/3\le f(a)-f(-a) \le 1/3$, a was arbitrary, so it applies equally to a/2, right?
How can you use that to eliminate the f(a/2) terms from the n=2 inequality?

 

[AdityaDev]

Yes. It applies to a/2.
$$-1/3\le f(a)-f(-a)\le 1/3$$
$$-3\le 9f(a)-9f(-a)\le 3$$
And,
$$-1/3\le f(a/2)-f(-a/2)\le 1/3$$
$$-1\le 3f(a/2)-3f(-a/2)\le 1$$
adding them, $-4\le A+B\le 4$

 

[haruspex]

Yes. It applies to a/2.
$$-1/3\le f(a)-f(-a)\le 1/3$$
$$-3\le 9f(a)-9f(-a)\le 3$$
And,
$$-1/3\le f(a/2)-f(-a/2)\le 1/3$$
$$-1\le 3f(a/2)-3f(-a/2)\le 1$$
adding them, $-4\le A+B\le 4$

You have to be careful with the modulus signs here.
If |x+y|<|z|, what bound can you put on |x| in terms of y and z?

 

[AdityaDev]

-z<x+y<z
-z-y<x<z-y
So |-z-y|<|x|<|z-y|

 

[haruspex]

-z<x+y<z
-z-y<x<z-y
So |-z-y|<|x|<|z-y|

No, the last step is not valid. E.g. x=1, y=2, z=4 satisfies the original inequality, as does x=-2, y=3, z=4.

 

[AdityaDev]

For x+y>0 and z>0, x<z-y
for x+y> 0,z<0 then x<-z-y
For x+y<0, z<0 then x>z-y
for x+y<0,z> 0 then x>-z-y

 

[haruspex]

For x+y>0 and z>0, x<z-y
for x+y> 0,z<0 then x<-z-y
For x+y<0, z<0 then x>z-y
for x+y<0,z> 0 then x>-z-y

You need to get it into the form |x|< some function of y and z. I'll give you a lead, it will be a function of |y| and |z|. A quite simple one.

 

[AdityaDev]

For example, |x-2|>|4|
Then x can be -1,0,1,2,3,4,5
So x<|y|+|z| and greater than |y|-|z|

 

[haruspex]

For example, |x-2|>|4|
Then x can be -1,0,1,2,3,4,5
So x<|y|+|z| and greater than |y|-|z|

I'll settle for |x|<|y|+|z|.
Now to resume the main discusion. You found |f(a/2)-f(-a/2)| <1/3. Now write out the inequality for the case n=2, ny=a. Use the |x+y|<|z| implies |x|<|y|+|z| result to rearrange it into the form |f(a)-f(-a)|< ... form.

 

[AdityaDev]

so for n=2,y=a/2,
$$|f(a/2)-f(a/2)+f(a)-f(-a)|\le 1/9$$
So $$|f(a)-f(-a)|\le|f(a/2)-f(-a/2)|+1/9$$
and $$|f(a/2)-f(-a/2)|\le 1/3$$
so $$|f(a)-f(-a)|\le 1/3+1/9$$

 

[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.