Given a function $f: [0,\infty) \rightarrow (0,\infty)$ which admits primitives, and let $F$ be a primitive such that: $F(x) = f([x]) \cdot f(\{x\})$ for all x, and there exists $a>0$ s.t. $f(a)=F([a]) \cdot F(\{a\})$, where $[x]$ refers to the integer part, and the other to the fractional part. Find every function $f$ that verifies these properties.
I have only been able to find that small relations between $F$ and $f$ for intervals of the form $[n,n+1)$ and that $f$ seems to have a recursive formula for natural numbers, yet nothing that would point out a certain clear path..
Let $(c_n)$ be the sequence in $(0,+\infty)$ defined by $c_n = f(n)$.
Let $x\in[0,1)$. We have that $F$ is continuous on $[0,1)$ and differentiable on $(0,1)$ by the property of being antiderivative of $f$. The condition $F(x) = f([x])f(\{x\})$ simplifies to $F(x) = c_0 f(x)$, i.e. $f(x) = \frac 1{c_0} F(x)$, so $f$ is continuous on $[0,1)$ and differentiable on $(0,1)$ and we have differential equation $f'(x) = \frac 1{c_0}f(x)$, $f(0) = c_0,$ which solves to $f(x) = c_0e^{x/c_0}$ and $F(x) = c_0^2e^{x/c_0}$.
Now let $x\in[n,n+1)$. The condition $F(x) = f([x])f(\{x\})$ simplifies to $F(x) = c_nf(x-n) = c_nc_0e^{(x-n)/c_0}$, so by differentiating we have $f(x) = c_ne^{(x-n)/c_0}$ on $[n,n+1)$.
Since $F$ is a continuous function, we get $F(n+1) = \lim_{x\to(n+1)^-}F(x)$, i.e. $c_{n+1}c_0e^{(x-(n+1))/c_0} = \lim_{x\to(n+1)^-}c_nc_0e^{(x-n)/c_0}$ or $c_{n+1} = c_ne^{1/c_0}$.
Finally, let $n$ be such that $a\in[n,n+1)$. From the condition $f(a) = F([a])F(\{a\})$ we get $c_ne^{(a-n)/c_0} = c_nc_0c_0^2e^{(a-n)/c_0}$ which simplifies to $c_0^3 = 1$, or $c_0 = 1$.
This now gives $c_n = e^n$ and $f(x) = c_ne^{(x-n)/c_0} = e^x.$