(Copied and pasted from my post here Let $P(x,y)$ be the assertion that $f(xy)=\text{max}\{f(x+y),f(x)f(y) \}$. Then if we let $f(0)=a\in \mathbb{R}$, by $P(0,0)$ we have $a=\text{max}\{a,a^2\}$ so that $a \ge a^2$; thus $a\in [0,1]$. Now for $x\ge 0$, we have $P(\sqrt{x},\sqrt{x})\implies f(x)=\text{max}\{f(2\sqrt{x}) ,f(\sqrt{x})^2\}\ge f(\sqrt{x})^2\ge 0$ so if $x$ is non-negative so is $f(x)$. $P(0,x) \implies f(0)=\text{max}\{f(x),af(x) \}$. However we have $0\ge 0 \implies f(0)\ge 0 \implies \text{max}\{f(x),af(x) \}\ge 0$. Case 1 $a\not= 0$ Suppose there exists $k$ in $f$'s domain such that $f(k)<0$. Then indeed $\text{max}\{f(k),af(k) \}=af(k)$. By $P(0,k)$ we have $af(k)=f(0)\ge 0$. So $af(k)\ge 0$, which can only happen when $a=0$; a contradiction. Thus there is no $k$ such that $f(k)$ is negative. As a result $\text{max}\{f(x),af(x)\})=f(x)$ so that $f(0)=f(x)$ for all $x$ by considering $P(0,x)$. Then $f(x)=a$ for any fixed $a \in (0,1]$ ($f(x)$ is constant). Substituting this back into the original condition we see that this is indeed a solution since $a=\text{max}\{a, a^2\}$. Case 2 $a=0$ Again assume there exists $k$ such that $f(k)<0$. This can only happen if $k$ is negative, and thus $-k$ is positive, implying $f(-k)$ is non-negative. $P(k,-k)\implies f(-k^2)=\text{max}\{f(0),f(k)f(-k)\}=\text{max}\{0,f(k)f(-k)\}=0$ since $f(k)f(-k)$ is non-positive. Set $k=-\sqrt{n}$ where ($n>0$) and then by the above relation $f(-n)=0$ so that for all negative $x$, $f(x)=0$. Now use $P(0,x)$ to show that $0=f(0)=\max\{f(x),af(x)\}=\max\{f(x),0\}$. If $f(x)$ is negative, then $f(x)=0$ and this is indeed true. If $f(x)$ is positive, then $f(x)=\text{max}\{f(x),0 \}=0$. And if $x$ is neither positive nor negative, $x=0$ so that $f(x)=0$. Hence the solution $f(x)=0$ for all $x$, which indeed works. Overall all solutions are $f(x)=a$ where $a\in (0,1] \cup \{0\}= [0,1]$.

let $x=0,f(0)=max(f(y),f(0)f(y))$(0) let $y=0,f(0)\ge f^2(0)$so $0\le f(0)\le 1$ if there exists $y$,$f(y)<0$ then for this y,$f(0)=f(0)f(y)$so$f(0)=0$ then$0=max(f(y),0)$ then $f(y)\le 0$for all y.but $f(xy)\ge f(x)f(y)\ge 0$contradiction! then for all x,$f(x)\ge 0$then by (0),$f(0)=f(y)$hence f is constant.