Do the properties $f(x)f(y)=f(x+y)$ and $f(x) \geq 1+x$ uniquely characterise the function $f(x)=e^x$?

Question

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, Jun 07 '21 at 15:54
@MichaelMorrow The function is stated to be defined on the whole of $\mathbb R$ — Mark Bennet, Jun 07 '21 at 15:55

stochasticboy321 · Accepted Answer · 2021-06-07T18:09:27.217

7

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 $b = e.$

edited Jun 07 '21 at 18:09

answered Jun 07 '21 at 16:10

stochasticboy321

9,003

4

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. – peek-a-boo Jun 07 '21 at 19:57
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This is a very elegant answer. I especially like the argument that $f$ is continuous, which settles my question definitively. Thanks @ stochasticboy321. – Joe Jun 07 '21 at 21:10

score 3 · Answer 2 · answered Jun 07 '21 at 16:05

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 $-y\lt -x$

Do the properties $f(x)f(y)=f(x+y)$ and $f(x) \geq 1+x$ uniquely characterise the function $f(x)=e^x$?

2 Answers2