Do the properties $f(x)f(y)=f(x+y)$ and $f(x) \geq 1+x$ uniquely characterise the function $f(x)=e^x$? In this post about possible definitions of the exponential function, it is mentioned that $e^x$ is the unique function $f:\Bbb{R}\mapsto\Bbb{R}$ satisfying

*

*$f(x)f(y)=f(x+y)$

*$f(x)\geq 1+x$
Do these properties alone really uniquely characterise the exponential function, or do we have to impose further requirements (e.g. that $f$ be continuous)?
 A: Notice that $f(0) = 1$ (since $0 < 1 + 0$). Further, $ f(-x) = \frac{1}{f(x)}$ for all $x > -1$.
Thus, for all $1 > x > -1,$ $$ \frac{1}{1+x} \ge \frac{1}{f(x)} =  f(-x) \ge 1-x. $$ By squeezing, we conclude that $\lim_{x \to 0} f(x) =  1 = f(0).$
But then $$ \lim_{x \to y} f(x) = \lim_{\delta \to 0} f(y + \delta) = f(y) \lim_{\delta \to 0} f(\delta) = f(y),$$ and the function is consequently continuous everywhere. Now we can use the usual argument with continuity to conclude that the function is of the form $b^x$ for some $b > 0$.
Further, by using the bounds on $f(x)$ in the above, we can show that $\lim_{x \to 0} \frac{f(x) - 1}{x} = 1$ (this is easiest with some casework - if $1 > x > 0$ then $1 \le (f(x) - 1)/x  \le 1/({1-x})$. If $-1 < x \le 0,$ the inequalities are reversed. In either case squeezing leads to the limit $1$). But the derivative of $b^x$ at $0$ is $\ln b$, which is equal to $1$ iff $b = e.$
A: It is easy to see that $f(nx)=f(x)^n$ and $f\left(\frac xn\right)=f(x)^{\frac 1n}$. Also $f(0)=1$ and $f(-x)=\frac 1{f(x)}$
Now suppose there is a value of $x$ with $f(x)\lt e^x$ so that $f(x)=e^y$ for some $y$ with $y\lt x$ (don't assume $x$ positive)
Then $f\left(\frac xn\right)=e^{\frac yn}=1+\frac yn + \dots$ and with $y\lt x$ I think you can choose $n$ large enough that this is $\lt 1+\frac xn$
If $f(x)=e^y$ with $y\gt x$ then we use $f(-x)=e^{-y}$ and $-y\lt -x$
