# For fixed $n \in \mathbb{R}$, what are all the continuous functions that satisfy $nf(x) = f(nx)$?

For fixed $n \in \mathbb{R}$, what are all the continuous functions that satisfy $nf(x) = f(nx)$?

I thought it would just be functions of the form $f(x) = kx$ but, for example, in the $n=2$ case we have that $f(x) = x \cos (\frac{2\pi \ln x }{ \ln 2})$ also works.

• math.stackexchange.com/questions/1033898/… – Tsemo Aristide May 1 '18 at 0:47
• If $n=0$, iff $f$ is continuous s.t. $f(0)$=0,it satisfies your condition. I think you can exclude the case. Also, you can exclude $n=1$ case. Well... I haven't get idea for other case. Also, how clear do you want the 'all continuous function that satisfy $nf(x)=f(nx)$' be? For $n=0$, is 'all continuous function s.t. $f(0)=0$ clear enough? By the way, It is weird to Let $n\in\mathbb R$ (Confusing) and I recommend you to mention the $dom(f)=\mathbb R$. – Tony Ma May 1 '18 at 0:48
• It seems to be a linear function times a $n$-periodic one. – A.Γ. May 1 '18 at 1:00
• $n$ can be any real number? Please choose a better variable! – Connor Harris May 15 '18 at 13:25

Let's start with $n$ positive but not $1$. We can separate positive and negative $x$, so take any periodic continuous functions $g_1(x)$ and $g_2(x)$ with $g_1(x\pm 1)=g_1(x)$ and $g_2(x\pm 1)=g_2(x)$ - a possibility is they are both constant, while another is that they each have period $1$, and they can be the fame function

• for $x \gt 0$ let $f(x)=x g_1({\log_n (x)})$,
• for $x \lt 0$ let $f(x)=x g_2({\log_n (-x)})$,
• and $f(0)=0$

while for $n=1$ you simply have

• $f(x)$ is any continuous function

and with $n=0$, almost as simply,

• $f(x)$ is any continuous function with $f(0)=0$

It gets slightly more complicated for negative $n \not = -1$. Take a continuous function $h(x)$ on $[0,1]$ with $h(1)=-h(0)$ and extend it to $\mathbb R$ with $h(x\pm 1)=-h(x)$, so if $h$ is ever non-zero then it has period $2$ and $|h|$ has period $1$

• for $x \gt 0$ let $f(x)=x h({\log_{-n} (x)})$,
• for $x \lt 0$ let $f(x)=x h(1+{\log_{-n} (-x)})$,
• and $f(0)=0$

while for $n=-1$ you simply have

• $f(x)$ is any odd continuous function so $f(-x)=-f(x)$ and $f(0)=0$

In particular, $f(x)=0$ is a possible solution for every value of $n$