# 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)?

• @MichaelMorrow: What exactly do you mean by 'delete bits of the graph'? If you mean 'restrict the domain of $f$', then notice that $f$ is defined as a function from the reals to the reals.
– Joe
Commented Jun 7, 2021 at 15:54
• @MichaelMorrow The function is stated to be defined on the whole of $\mathbb R$ Commented Jun 7, 2021 at 15:55
• You are right. Thanks Commented Jun 7, 2021 at 15:55

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 $$b = e.$$

• nice. In fact, because you've already established $f(x)=b^x$ for some $b>0$, you know $f$ is differentiable at the origin (even analytic on $\Bbb{R}$)so in your final paragraph, it suffices to only consider the limit as $x\to 0^+$. You're guaranteed that the limit $x\to 0^-$ will agree with this, so in fact no case-work is necessary. Commented Jun 7, 2021 at 19:57
• This is a very elegant answer. I especially like the argument that $f$ is continuous, which settles my question definitively. Thanks @ stochasticboy321.
– Joe
Commented Jun 7, 2021 at 21:10

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 $$-y\lt -x$$