Functional equation. Can a function be periodic?

Question

For any real number, a function (not a constant) is specified:

$$f(x) = \frac{f(x+1) + f(x-1)}{\sqrt{2}}$$

Can this function be periodic, if so, find it?

Answer 1

Let $x=t-1$, then $f(t-1) =\frac{f(t) + f(t-2)}{\sqrt{2}}$ so $f(t)=\sqrt{2} f(t-1)-f(t-2)$. On the other hand $f(t) = \frac{f(t+1) + f(t-1)}{\sqrt{2}}=\sqrt{2} f(t-1)-f(t-2)$. Hence $f(t-2) = \frac{f(t-1) - f(t+1)}{\sqrt{2}}$. Also $f(t-2) = \frac{f(t-1) + f(t-3)}{\sqrt{2}}$. Now $ \frac{f(t-1) - f(t+1)}{\sqrt{2}}=\frac{f(t-1) + f(t-3)}{\sqrt{2}}$. That $f(t-3)=-f(t+1)----> f(u-4)=-f(u)$.

Answer 2

In terms of a general solution, the given functional equation is essentially a second order linear recurrence relation. You can rewrite the equation as follows: $$f(x) = \sqrt{2}f(x - 1) - f(x - 2).$$ The value at each point $x \in \mathbb{R}$ depends only on the points of the form $x + n$, where $n \in \mathbb{Z}$, and so a (doubly infinite) sequence can be chosen independently for each equivalence class.

First, let's solve the recurrence relation $a_n = \sqrt{2} a_{n-1} - a_{n-2}$. Consider the characteristic equation: $$r^2 = \sqrt{2} r - 1.$$ The solutions to this are $$r = \frac{1 \pm i}{\sqrt{2}} = e^{\pm i\frac{\pi}{4}},$$ which mean the solutions to the recurrence relation are of the form $$a_n = Ae^{in\frac{\pi}{4}} + Be^{-in\frac{\pi}{4}},$$ where $A, B \in \mathbb{C}$, or in terms of real coefficents, $$a_n = A \cos\left(n\frac{\pi}{4}\right) + B \sin\left(n\frac{\pi}{4}\right),$$ where $A, B \in \mathbb{R}$.

If $f$ is a solution and $x \in \mathbb{R}$, then $f(x + n)$ must be a sequence of the above form. Conversely, if we choose $A = A(x)$ and $B = B(x)$ as functions of $x \in [0, 1)$, then we determine a solution over all of $\mathbb{R}$ uniquely. Hence, our set of solutions is parameterised by two completely arbitrary functions $A, B : [0, 1) \to \mathbb{R}$.

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For example, if we take $A(x) = \cos\left(x\frac{\pi}{4}\right)$ and $B(x) = -\sin\left(x\frac{\pi}{4}\right)$, we get the solution, \begin{align*} f(x + n) &= \cos\left(x\frac{\pi}{4}\right)\cos\left(n\frac{\pi}{4}\right) - \sin\left(x\frac{\pi}{4}\right)\sin\left(n\frac{\pi}{4}\right) \\ &= \cos\left((x + n)\frac{\pi}{4}\right), \end{align*} for $x \in [0, 1)$ and $n \in \mathbb{Z}$. That is to say, $$f(x) = \cos\left(x\frac{\pi}{4}\right)$$ for all $x \in \mathbb{R}$. You can also form a similar expression with $\sin$, or indeed any linear combination of these functions.

Answer 3

The linear homogeneous difference equation

$$ \sqrt 2 f(x) - f(x + 1) - f(x-1) = 0 $$

after the substitution $f(x) = c \alpha^x$ has the general solution

$$ f(x) = c_1 (1-i)^x 2^{-x/2}+c_2 (1+i)^x 2^{-x/2} = \left(c_1 (1-i)^x+c_2 (1+i)^x \right) 2^{-x/2} $$

NOTE

Due to

$$ (1\pm i)^x = 2^{\frac x2}e^{\pm i\frac{\pi}{4}x} $$

this solution can be read also as

$$ f(x) = C_1\sin\left(\frac{\pi}{4}x\right)+C_2\cos\left(\frac{\pi}{4}x\right) $$