find period

[man111]

If [tex]f(x)[/tex] be a real value function which satisfies [tex]f(x+\frac{3}{2})+f(x)=f(x+1)+f(x+\frac{1}{2})[/tex] and [tex]|f(x)|\leq 2, \forall x \in R,[/tex] .then period of [tex]f (x)[/tex] is

[martin123456]

(1) [tex]f(x+\frac{3}{2})-f(x+\frac{1}{2})=f(x+1)-f(x)[/tex]
Giving [tex]x[/tex] the value [tex]x+\frac{1}{2}[/tex] and substituting in (1) we get [tex]f(x+2)-f(x+1)=f(x+\frac{3}{2})-f(x+\frac{1}{2})[/tex] and so [tex]f(x+2)-f(x+1)=f(x+1)-f(x)[/tex]. Now giving [tex]x[/tex] the value [tex]x+1[/tex] and substituting in the previous result we get [tex]f(x+3)-f(x+2)=f(x+2)-f(x+1)[/tex] so [tex]f(x+3)-f(x+2)=f(x+1)-f(x)[/tex]. Summing this equalities we get [tex]f(x+n)=nf(x+1)-(n-1)f(x)[/tex] for [tex]n \in \mathbb{N}[/tex] (2).

Let [tex]g(x)=f(x+1)-f(x)[/tex]. From (1) we get [tex]g(x+\frac{1}{2})=g(x)[/tex]. So [tex]g[/tex] is periodic with a period [tex]T_g=\frac{1}{2k}[/tex], [tex]k \in \mathbb{N}[/tex]. Let [tex]T_f[/tex] be the period of [tex]f[/tex], then [tex]g(x+T_f)=f(x+1+T_f)-f(x+T_f)=f(x+1)-f(x)=g(x)[/tex] and so [tex]T_f=\frac{n}{2k}[/tex], where [tex]n \in \mathbb{N}[/tex] (3). So [tex]f(x+n)=f(x)[/tex]. From (2) we get [tex]f(x)=f(x+1)[/tex] and by an induction [tex]f(x)=f(x+1)=f(x+2)=\cdots[/tex] (4). So [tex]\exist s \in \mathbb{N}[/tex]: [tex]ns=2k[/tex](5).

so i dont get why there is [tex]\|f(x)\| \le 2[/tex].
?

[man111]

thanks martin123456

trypo error

[allier]

How do you know that f is periodic at all? You do need that f is bounded ...

[martin123456]

yeah but allier can u say what is the period of f

[mkmarinov]

I'll start from [tex]f(x+2)-f(x+1)=f(x+1)-f(x)[/tex].
Let us look at how f(x) behaves over the field of integers. Let [tex]f(x)=a_x[/tex], we get the recurrent relation [tex]a_{x+2}-2a_{x+1}+a_x=0[/tex], which means [tex]a_x[/tex] is an arithmetic progression. If it is non-constant, it is also unbounded, which contradicts [tex]|f(x)| \le 2[/tex]. So f(x) is constant over the integers. Which means its period is not greater than 1.
But this is as far as one can go with the current conditions. E.g. any constant function, bounded by -2 and 2 satisfies these conditions, and we all know what the period of constant functions is .

[allier]

Well, mkmarinov, you have proved that f is constant over the integers which is correct. However, is there a function which is non-constant globally which satisties the conditions of the problem.

@martin: The question is obviously not stated correctly. Your solution works (but for each period value, one should still construct a proper function which has this value as its minimal period) assuming we know in advance f is periodic, which is to some extent suggested by the question "What is the period of f?". I do believe, though, that one should go the extra step to prove f is indeed periodic ...

[int]

It is obvious that the question is not formulated correctly. In fact there are many functions of different periods that satisfy the given equation. For example for each fixed natural [tex]m[/tex], each function [tex]f[/tex] of period [tex]\frac{1}{2m}[/tex] satisfies the given equation. With the condition [tex]|f(x)|\le 2[/tex] you just restrict a bit the class of functions which satisfy the given equation.

[martin123456]

Personally, I dont see how it can be proven that the function is periodic.

[int]

You have not to prove anything. Take a natural [tex]m[/tex] and a function [tex]f[/tex] of period [tex]\frac{1}{2m}[/tex] that is [tex]f(x+\frac{1}{2m})=f(x)[/tex] for all real [tex]x.[/tex]
Then on one hand, for the left side of the given equation you have [tex]f(x+\frac{3}{2})+f(x)=f(x+3m\frac{1}{2m})+f(x)=2f(x).[/tex]
On the other hand, for the right side you have [tex]f(x+1)+f(x+\frac{1}{2})=f(x+2m\frac{1}{2m})+f(x+m\frac{1}{2m})=2f(x)[/tex] so the equation is satisfied for all such functions.

[mkmarinov]

Well, mkmarinov, you have proved that f is constant over the integers which is correct. However, is there a function which is non-constant globally which satisties the conditions of the problem.

Well, I may have not expressed myself correctly.
First, yes, f is constant over the integers. But it can also be proven in the same way that it is constant for arguments with integral differences, i.e. f(x+1)=f(x+2)=...
Second, I did not say that since f is constant over the integers, it is constant in its whole domain. I just gave this as an example. If f is constant it satisfies all the given conditions, yet can be considered periodic with any period.
If you are looking for a non-constant function, [tex]2sin(2\pi x)[/tex] could be considered an example.

[int]

I don't understand why you continue to argue. Didn't you get the answer here? Let's analyse the question once again.
You have given function that satisfies certain functional equation and one asks, what is the period of that function? Indeed if the function is not periodic the question formulated in that way has no sense. If we know that f is periodic, then as I wrote before the answer of the question stated above is "We can't say anything". That's because it is possible nonconstant functions with different period to satisfy the equation.