Finding functions such that $f(g(x)) = f(x)g(x)$

Question

Let's assume the functions in question are continuously differentiable. If $f(g(x)) = f(x)g(x)$, then: $$f'(g(x))g'(x) = f'(x)g(x)+f(x)g'(x)$$ I attempted to find a particular solution by letting $g(x) = x$. Then, $$f'(x) = xf'(x) + f(x)$$ $$df = xdf + fdx$$ $$\frac{df}{f} = \frac{dx}{1-x}$$ $$\ln(f(x)) = -\ln(1-x)+C$$ $$f(x) = \frac {A}{1-x}$$ Where $A$ is a constant. This solution was fairly unsatisfying to me, and upon checking it, I see it doesn't actually work, as letting $g(x) = x$ leads to $f(x) = xf(x)$. I was wondering if there were better answers, or at least a less trivial way to go about the problem.

Answer 1

The problem with your approach is, that you look for solutions to $$ \tag{1} f(x) = xf(x) + c $$ for some $c \in \mathbf R$, as all those functions fulfill $$ f'(x) = xf'(x) + f(x) $$ your function fulfills (1) with $c = A$.

(1) Of course $f = 0$ fulfills your equation with any $g$.

(2) Let's look for polynomials satisfying your equation, as for general functions I doubt there is a wholesome answer, you have to restrict yourself to some function class. If $f$ and $g$ are polynomials of degree $n$ and $m$ say, then $f \circ g$ has degree $nm$ and $fg$ has degree $n +m$. Hence, we must have (let's ignote the zero polynomial here, we discussed the 0-function above) $$ nm = n +m \iff n(m-1) = m \iff n = \frac{m}{m-1}$$ As $\frac m{m-1}$ is only integer for $m \in\{0,2\}$, we must have $n=m=2$ or $n=m=0$. If both $f$ and $g$ are of degree $0$, i. e. constants, say $f(x) = a_0$ and $g(x) = b_0$, our eqation reads $$ (f \circ g)(x) = f\bigl(g(x)\bigr) = a_0 \stackrel!= a_0b_0 = f(x)g(x) $$ which is correct for $b_0 = 1$.
For $m = n = 2$, say $f(x) = a_2x^2 + a_1x + a_0$ and $g(x) = b_2x^2 + b_1x + b_0$ with $a_2b_2 \ne 0$, we have $$ f\bigl(g(x)\bigr) = a_2b_2^2x^4 + r_1(x),\quad f(x)g(x) = a_2b_2x^4 + r_2(x) $$ for some polynomials $r_1$, $r_2$ of degree $3$. This gives $b_2 = 1$. Now let's have a look at the degree $3$ terms, we have \begin{align*} f\bigl(g(x)\bigr) &= a_2x^4 + 2a_2b_1x^3 + r_3(x)\\ f(x)g(x) &= a_2x^4 + (a_2b_1 + a_1)x^3 + r_4(x) \end{align*} for some polynomials of degree at most $2$. This gives $$ a_1 = a_2b_1 $$

\begin{align*} f\bigl(g(x)\bigr) &= a_2(b_2x^2 + b_1 x + b_0) + a_1(b_2x^2+ b_1x + b_0) + a_0\\ &= a_2b_2^2x^4 + 2a_2b_1b_2x^3 + (2a_2b_2b_0 + a_2b_1^2 + a_1b_2)x^2\\&\qquad{} + (2a_2b_1b_0 + a_1b_1)x + (a_2b_0^2 + a_1b_0 + a_0)\\ f(x)g(x) &= (a_2x^2 + a_1x + a_0)(b_2x^2 + b_1x + b_0)\\ &= a_2b_2x^4 + (a_2b_1 + a_1b_2)x^3 + (a_2b_0 + a_1b_1 + a_0b_2)x^2 + (a_1b_0 + a_0b_1)x + a_0b_0 \end{align*} This gives \begin{align*} a_2b_2 &= a_2b_2^2\\ 2a_2b_1b_2 &= a_2b_1 + a_1b_2\\ 2a_2b_2b_0 + a_2b_1^2 + a_1b_2 &= a_2b_0 + a_1b_1 + a_0b_2\\ 2a_2b_1b_0 + a_1b_1 &= a_1b_0 + a_0b_1\\ a_2b_0^2 + a_1b_0 + a_0 &= a_0b_0 \end{align*} The first equation gives (due to $a_2b_2 \ne 0$ that $b_2 = 1$), we are left with \begin{align*} a_2b_1 &= a_1\\ a_2b_0 + a_2b_1^2 + a_1 &= a_1b_1 + a_0\\ 2a_2b_1b_0 + a_1b_1 &= a_1b_0 + a_0b_1\\ a_2b_0^2 + a_1b_0 + a_0 &= a_0b_0 \end{align*} If we use the first equation in the second and the third to simplify further, we get \begin{align*} a_2b_1 &= a_1\\ a_2b_0 + a_1 &= a_0\\ a_1b_0 + a_1b_1 &= a_0b_1\\ a_2b_0^2 + a_1b_0 + a_0 &= a_0b_0 \end{align*} Now note that the third equation follows from the first two (multiply the second with $b_1$ and use the first), so we can drop it, giving \begin{align*} a_2b_1 &= a_1\\ a_2b_0 + a_1 &= a_0\\ a_2b_0^2 + a_1b_0 + a_0 &= a_0b_0 \end{align*} If we multiply the second equation with $b_0$ and subtract it from the third, we get $$ a_0= 0 $$ Hence, we have \begin{align*} a_2b_1 &= a_1\\ a_2b_0 + a_1 &= 0\\ \end{align*} Due to $a_2 \ne 0$, this gives $b_0 = -b_1$ and $a_1 = a_2b_1$. Hence we have \begin{align*} f(x) &= a_2x^2 + a_2b_1x\\ g(x) &= x^2 + b_1x - b_1 \end{align*} as a soultion for any $a_2 \in \mathbf R \setminus\{0\}$, $b_1 \in \mathbf R$.