For all real numbers $x$, a function $f(x)$ satisfies the functional equation:
$$ f(x+1) + f(x-1) =\sqrt{2} f(x). $$
Show that it is periodic.
Since the function satisfies the above equation for all real values, we can shift the variable $x$ by $-1$. This gives:
$$ f(x) + f(x-2) =\sqrt{2} f(x-1). $$
If we substitute $f(x)$ from the above equation into the first equation, we obtain:
$$ f(x+1) - f(x-1) = -\sqrt{2} f(x-2). $$
Now by shifting $x$ by $2$ in the last equation we get:
$$ f(x+3) - f(x+1) = -\sqrt{2} f(x). $$
Adding the original equation to the above equation we find
$$ f(x+3) + f(x-1) = 0. $$
From this we find $f(x) = - f(x\pm 4)$. We conclude that the function $f(x)$ is periodic with period $8$.
Huge thanks to Rainer Nase (Germany), David Galarza (Bogotá), Fabio Skilnik (Brazil), Aritra (Maryland), Kayra BOLAT (Istanbul), Oguzhan Kasikci (Istanbul), Olivier Massicot (Champaign, IL), Subhadeep Pal, (Bangalore India), Sayan Sinha Ray, jc (New York), B Sashi Kanth (Hyderabad, India), Monbikash Gayan (Pune), Nicolas Markossian (Lebanon), and Fabio Skilnik (Brazil) for submitting solutions to this week’s challenge problem.