List of functions $f(cx) = C\cdot f(x)$

Question

I was looking for some complex functions f(x), which satisfies the condition:

$$\exists (c, C) \in \Bbb C^2 \backslash\{(1,1)\}, \forall x \in \Bbb C, f(cx) = C\cdot f(x)$$

Till now I have got

$$\begin{array}{c|c} \text{Function, $f(x)$}&\text{Remarks}\\ x^d&\text{d is some constant}\\ \log(d^x)&\text{d is some constant}\\ sin(x\pi)&-\\ cos(x\pi)&-\\ tan(x\pi)&-\\ |x|&- \end{array}$$

which satisfies this condition.

But I am looking for more complex functions which satisfies the same condition.

Answer 1

If C and c are constant: $$cf'(cx)=Cf'(x)$$ $$f'(cx)=\frac Ccf'(x)=\frac{C^2}{c^2}f'(\frac xc)=\cdots=\frac{C^n}{c^n}|_{n\to\infty}f'(0)$$

• If $c>1$: $$f(cx)=\frac{C^n}{c^n}|_{n\to\infty}f'(0)x+C$$ $$f(x)=\left[\lim_{n\to\infty}\left(\frac Cc\right)^n\right]f'(0)\frac xc+\mathcal{Constant}$$

  • If $C>c$:
  • If $f'(0)=0$ , $f(x)=\mathcal{Constant}$

  • If $f'(0)\ne0$ , $f(x)\to\infty$

  • If $C<c$, $f(x)=\mathcal{Constant}$

  • If $C=c$, $\displaystyle f(x)=\left(\frac{f'(0)}c\right)x+\mathcal{Constant}$

• If $c=\pm1$
  • If $c=1$, $$f'(x)={C^n}|_{n\to\infty}f'(0)\implies f(x)={C^n}|_{n\to\infty}f'(0)x+\mathcal{Constant}$$
  • Some similiar results to previous one if $C>1,C=1,C<1$, Think yourself.
  • If $C=\pm1$, All even and odd functions.
• If $c<1$
  • Some similiar cases, Think yourself.

If C and c are variable: Every function satisfies the required condition.