How Do I solve $f(x)^2=f(2x)$ analytically?

Question

If $f$ is continuous and differentiable, how do I determine the solution for the functional equation $f(x)^2=f(2x)$. I am aware that $f(x)=e^x$ but I just don't see how I can reach that answer analytically. Any help?

Answer 1

If $f(x)=0$ for any $x$ then $f(0)=\lim_{n\to\infty} f(x2^{-n})=0$. But if $f(y)\neq 0$ for some $y$ then $f(y2^{-n})=f(y)^{1/2^n}\to 1$ as $n\to \infty$.

So continuity at $0$ means that if $f(x)=0$ for some $x$, we must have $f(x)=0$ for all $x$.

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Now assume $f(x)\neq 0$ for all $x$. Then $f(x)=f^2(x/2)>0.$

If you let $g(x)=2^{-x}\log(f(2^x))$, you get $g(x+1)=g(x)$.

If you let $h(x)=2^{-x}\log(f(-2^x))$, you get $h(x+1)=h(x)$.

So, given $h,g$ of period $1$ on $\mathbb R$, we can define:

$$f(z)=\begin{cases} e^{zg(\log_2 z)}&z>0\\ e^{-zh(\log_2(-z))}&z<0\\ 1&z=0 \end{cases}$$

If $g$ and $h$ are differentiable, then this is everywhere differentiable except for possibly at $0$.

Now, for any $a>0$ we can take $a_n=2^{-n}a$ and if there is a derivative, it would have to be equal to:

$$\lim_{n\to\infty} \frac{f(a_n)-1}{a_n} =\lim_{n\to\infty}\frac{e^{a2^{-n}g(\log_2 a)}-1}{a2^{-n}}=g(\log_2 a)$$

So $g$ must be constant, and so must $h$, and they must be additive inverses.

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This proof works even if the condition is only that $f$ is continuous and differentiable at $x=0.$