How to solve this functional equation: $2f(x) = f(x-1)+f(x+1)$?

Question

After some calculations, I came up with this functional equation: $f(x-1)+f(x+1)=2f(x)$. I found linear function is one possible answer, but don't know how to derive it.

I don't know much about the techniques to solve this kind of equation. Can anyone help?

Edited: $f$ is a probability function, that is $0 \leq f \leq 1$, and $f$ is continuous.

Could you give the context of this question? Is anything known about the function $f$? — Mr.Topology, Jun 15 '16 at 11:14
With all these constraints, the only solution is a constant. — , Jun 15 '16 at 12:03
Bounded continuous solutions are not necessarily constant. $f(x)=\cos^2(k \pi x)$ is a solution for every integer $k$. — Batominovski, Jun 15 '16 at 12:29
@YvesDaoust It is not clear what a "probability function" means. The OP did not specify this. And a constant function is not a CDF. — Batominovski, Jun 15 '16 at 13:18
@YvesDaoust I don't know. It's still not clear hat a "probability function" means. — Batominovski, Jun 15 '16 at 13:25
@YvesDaoust I'm not going to play this stupid game of guessing. If you like to do it, then so be it. — Batominovski, Jun 15 '16 at 13:31
It cannot be not a probability function (cumulative, or density, or mass) — Henry, Jun 15 '16 at 15:41

score 3 · Accepted Answer · answered Jun 15 '16 at 11:28

3

You have $f(x+1)-f(x)=f(x)-f(x-1)$ and so $f(x+2)-f(x+1)$ is equal to the same value, and similarly $$f(x+n) - f(x+n-1)=f(x)-f(x-1)$$ for all integer $n$. This gives $$f(x+n) = (n+1)f(x)-nf(x-1)=n(f(x)-f(x-1)) + f(x)$$ which is linear in integer $n$, though not necessarily linear in real $x$.

If you fix the values of $f(y)$ on $[-1,1)$ then you can give a general solution using rounding and $$f(y) = \lfloor y \rfloor\left(f(y-\lfloor y \rfloor)- f(y-\lfloor y \rfloor-1)\right) + f(y-\lfloor y \rfloor)$$

answered Jun 15 '16 at 11:28

Henry

157,058

This isn't bounded, is it ? – Jun 15 '16 at 12:00
@YvesDaoust: It can be bounded, in which case it will be periodic – Henry Jun 15 '16 at 15:39
If $f(x)$ is continuous then $f(x+1)-f(x)$ is constant for all $x$ – Henry Jun 15 '16 at 15:43
Your expression is too general, it allows unbounded solutions. – Jun 15 '16 at 15:46
@YvesDaoust: it was not bounded or continuous when I answered the question. It cannot be a "probability function" over the real line – Henry Jun 15 '16 at 16:00

score 1 · Answer 2 · answered Jun 15 '16 at 11:17

1

Hint: $f$ is completely determined by its values on $[0,2)$. Use recursion to solve for $f$ given its values on $[0,2)$.

answered Jun 15 '16 at 11:17

Batominovski

49,629

Thanks, but I don't really get your hint. Can you elaborate more? – mathstud Jun 15 '16 at 11:23
With the given constraints, $f$ must be identical on $[0,1)$ and $[1,2)$. – Jun 15 '16 at 12:40
That is an obvious conclusion from the extra conditions the OP has included. My answer is a hint, not a solution. The OP should take his own time to arrive at this result on his own. – Batominovski Jun 15 '16 at 13:24

How to solve this functional equation: $2f(x) = f(x-1)+f(x+1)$?

2 Answers2