Functional equation $f(x+1)+f(x-1)=\sqrt{2}\cdot f(x)$

Question

Can $ f: \mathbb{R}\rightarrow \mathbb{R}$ which satisfies functional equation $$f(x+1)+f(x-1)=\sqrt{2}\cdot f(x)$$ be periodic?

No idea how to prove this - $f(x+T)=f(x-T)=f(x)...$

Answer 1

$f(x-2)+f(x)+f(x)+f(x+2)=\sqrt2f(x-1)+\sqrt2f(x+1)=\sqrt2\sqrt2f(x)$

We conclude $f(x-2)+f(x+2)=0$

Hence $f(x+4)=-f(x)$ and so $f(x+8)=f(x)$.

Answer 2

Let $a=\sqrt{2}$, then $$f(x+1)= af(x)-f(x-1)$$

so \begin{align}f(x+2) &= af(x+1)-f(x) \\ &= \underbrace{a^2f(x) -af(x-1)}_{af(x+1)}-f(x)\\ &= f(x)-af(x-1) \\ &= \underbrace{af(x-1)-f(x-2)}_{f(x)}-af(x-1)\\ & =- f(x-2) \end{align}

So $$f(x+8) = -f(x+4) = f(x)$$

Answer 3

This is a tricky solution not based on functional equations.

Consider the recurrence relation $$f_{x+1}+f_{x-1}=\sqrt{2}\, f_x$$ Let $f_x=2^{-x/2} g(x)$ to face $$g_{x+1}=2(g_x-g_{x-1}) \qquad\implies\qquad g_x= c_1 (1-i)^x+c_2 (1+i)^x$$ Going from complex numbers to trigonometric functions $$g_x=2^{x/2}\left(\left(c_1+c_2\right) \cos \left(\frac{\pi }{4}x\right)+i \left(c_2-c_1\right) \sin \left(\frac{\pi }{4}x\right)\right)$$ $$f_x=\left(c_1+c_2\right) \cos \left(\frac{\pi }{4}x\right)+i \left(c_2-c_1\right) \sin \left(\frac{\pi }{4}x\right)$$ Now, consider $x$ as a continuous variable and you see the period.

Answer 4

This solution employs the operator technique. Define $Sg(x)=g(x+1)$ for any $g:\Bbb R\to \Bbb R$. We have by the problem statement $$(S^2-\sqrt{2}S+1)f=0.$$ However $$s^8-1=(s^4-1)(s^4+1)=(s^4-1)\big((s^2+1)^2-2s^2\big)=(s^4-1)(s^2+\sqrt2s+1)(s^2-\sqrt2s+1).$$ This means $$(S^8-1)f= (S^4-1)(S^2+\sqrt2S+1)(S^2-\sqrt2S+1)f=(S^4-1)(S^2+\sqrt2S+1)0=0.$$ Thus $S^8f=f$. Therefore $f$ is periodic with period $8$.