I am trying to solve the following problem which stipulates certain conditions on the target function $f$. In other words, I want to find a non-constant function $f: \mathbb{R} \to [0,1]$ such that
$$ 1) \ \text{ The derivative } f':\mathbb{R} \to \mathbb{R} \text{ is an even function } \\ 2) \ f(z_1) \cdot f(z_2) \ldots f(z_n)= b_0+ \sum_{k} b_k \cdot f(a_{k1}z_1+\ldots+a_{kn}z_n), $$ where $a_{k,i}, b_0$ are some real numbers. The second condition states that the product of the functions can be written as linear combination of the function whose arguments are again linear combinations of the original arguments. Does there exist any such function satisfying these conditions?
My attempt: The closest I could think of is that $f$ is a shifted $\sin$ function, i.e, $f(z)=\frac{1+\sin z}{2}$. Clearly the derivative is an even function. But the product $f(z_1) \cdot f(z_2) \ldots f(z_n)$ can be written as linear combination of $\sin$ functions only if $n$ is odd. It fails for $n$ even.
Another function I have in mind is $f(z)=\frac{1}{1+e^{-z}}$ but I an unable to verify if this works.
I suppose that such a function exists (I suppose that you want (2) to be true for all $n$). Take $n=2$, and $z_1=z_2=z$. Then you have $$f(z)^2=\sum b_k f(c_kz)$$ with $c_k=a_{k,1}+a_{k,2}$. One compute the derivative
$$2f^{\prime}(z)f(z)=\sum b_k c_k f^{\prime}(c_k z)$$ Now replace $z$ by $-z$. As $f^{\prime}$ is even, you get that $2f^{\prime}(z)f(-z)=2f^{\prime}(z)f(z)$.
Hence we get that $2f^{\prime}(z)(f(z)-f(-z))=0$. Now if $g(z)=f(z)-f(-z)$, we have that $g^{\prime}(z)=f^{\prime}(z)+f^{\prime}(-z)=2f^{\prime}(z)$. Hence $g^{\prime}(z)g(z)=0$, $g(z)^2$ is a constant, and as $g(0)=0$, $g=0$. This means that $f$ is even, and then $f^ {\prime}$ is odd. But as $f^{\prime}$ is also even, we have $f^{\prime}=0$, and $f$ is constant.