Is it possible that?

I will try to find all functions satisfying the three first properties and show that there are exactly the exponential functions. Therefore we will conclude that no function satisfies the four properties.

Let: $\\ (1) \ f \ \text{is} \ C^1\\ (2) \ \forall(x,y)\in\mathbb{R}^2,\ f(x+y)\leq f(x)f(y)\\ (3) \ f(0)=1$

Let $f:\mathbb{R}\longrightarrow\mathbb{R}$ satisfying $(1),\ (2), \ (3)$ .

Let $x\in\mathbb{R}$ . Then by $(2)$ , for any $h\neq 0$ we have:

$\begin{aligned} \dfrac{f(x+h)-f(x)}{h} & \leq \dfrac{f(x)f(h)-f(x)}{h}\\ & =f(x)\dfrac{f(h)-1}{h}\\ & =f(x)\dfrac{f(h)-f(0)}{h} \end{aligned}$

By letting $h\to 0$ , because $f$ is $C^1$ , we get:

$f'(x)\leq f(x)f'(0)$

Similarly, for any $h\in\mathbb{R}^*$ ,

$\begin{aligned} \dfrac{f(x)-f(x-h)}{h} & \geq \dfrac{f(x)-f(x)f(-h)}{h}\\ & = f(x)\dfrac{f(0)-f(-h)}{h} \end{aligned}$

Letting $h\to 0$ :

$f'(x)\geq f(x)f'(0)$

Therefore, $\forall x\in\mathbb{R}, \ f'(x)=f'(0)f(x)$ .

So (by solving the ODE and by $(3)$ ) there exists $\alpha\in\mathbb{R}, \ \text{such that} \ f=\exp(\alpha \ id_\mathbb{R})$ . Note: $\alpha=f'(0)$

Verification: for any $\alpha\in\mathbb{R}, \ x\in\mathbb{R}\mapsto e^{\alpha x}$ satisfies the tree properties. $\square$