Functional equation $f\left( x^{2}\right) =2f\left( x\right)$and the Cauchy problem $f\left( xy\right) =f\left( x\right) +f\left( y\right)$

Question

Is it possible to show that if a continuous function $f:\mathbb{R} _{+}\rightarrow \mathbb{R}$ verifies the equation $$f\left( x^{2}\right) =2f\left( x\right)$$ then $f$ also satisfies the Cauchy equation $$f\left( xy\right) =f\left( x\right) +f\left( y\right)$$ WITHOUT SOLVING THEM? ( Otherwise, given that $f\left( x\right) =k\cdot \log _{a}x$, the question becomes trivial....)
Thanks for some suggestions...

Answer 1

It is not possible. Consider:

$$f : \mathbb{R}^+ \to \mathbb{R}, \;\; x \mapsto |\ln(x)|$$

Then, it is easy to verify (by casework) that this satisfies $f(x^2) = 2f(x)$, but $f$ is not a solution to the logarithmic Cauchy functional equation. Therefore, your conjecture is not true.

Note that the key difference between the two conditions is that $|x| < 1 \iff |x^2| < 1$, whereas $x+y$ is arbitrary for $x$ fixed.

Edit: if you require that $f(\mathbb{R}^+) = \mathbb{R}$, then note that the above implies that you can stitch together any two logarithms at $x = 1$ to find a solution to the functional equation; that is to say:

$$ f(x) = \begin{cases} a \ln(x) & x \le 1 \\ b \ln(x) & x \ge 1 \\ \end{cases}$$

gives an infinite family of solutions, which are further counterexamples when $a \neq b$ and $ab > 0$.

(It can be noted that the original counterexample was $(a,b) = (-1,1)$.)

Answer 2

legionwhale has already provided a simple counterexample to your conjecture, let me now give a wide source of examples that are surjective.

Notice letting $g(t)=f(e^t)$ gives an equivalent formulation
$$g(2t)=2g(t)\tag{1}$$

To construct counterexamples, let $h\colon [1,2]\to [1,2]$ be continuous, with $h(1)=1$, $h(2)=2$, but $h$ not the identity. Define $g(2^nt)=2^nh(t)$ for all $n\in \mathbb Z$ and $t\in[1,2]$, and likewise $g(-2^nt)=-2^nh(t)$, and observe this is well-defined, continuous, and surjective from $\mathbb R\backslash \{0\}\to \mathbb R\backslash \{0\}$, and satisfies $\lim_{t\to 0}g(t)=0$, so we may extend $g$ to a surjection $\mathbb R\to\mathbb R$.

Observe that from the construction $g$ satisfies (1), so $f(x)=g(\log(x))$ satisfies the initial criterion. On the other hand, if $f$ satisfied the Cauchy equation, then $f(x)=k\log x$, whereby $g(t)=f(e^t)=kt$, and evaluating at points $t=2^n$ forces $k=1$, which would imply $g$ is the identity, a contradiction.

Remark

Let me just add that the condition you may have wanted to assume was likely $$f(x^n)=nf(x)\tag{2}$$ for arbitrary $n\in\mathbb Z$. Then it is indeed true that a continuous function satisfying (2) must satisfy $$f(xy)=f(x)+f(y)\tag{3}$$

To prove this, first observe that $$f(x)=f(x^{\frac{q}{q}})=qf(x^{\frac{1}{q}})\text{,}$$ so $$f(x^{\frac{1}{q}})=\frac{1}{q}f(x)\text{,}$$ giving us $$f(x^{\frac{p}{q}})=\frac{p}{q}f(x)\text{,}$$ for all interger $p,q\neq0$. By the density of $\mathbb Q$ in $\mathbb R$ and continuity we have $$f(x^\alpha)=\alpha f(x)$$ for all $\alpha\in \mathbb R$.

But now let $x=e^\alpha,y=e^\beta\in \mathbb R^+$, then we have \begin{align*} f(xy) &=f(e^\alpha e^\beta)=f(e^{\alpha+\beta})=(\alpha+\beta)f(e)\\ &=\alpha f(e)+\beta f(e) = f(e^\alpha)+f(e^\beta) = f(x)+f(y)\text{.} \end{align*}

So we never explicitly calculated the general form of the solution for (2), though we did arguably come very close, since the equation $f(e^\alpha)=\alpha f(e)$ is really all you need to get to an explicit solution. Your mileage may vary on whether this counts as proving (3) without solving.