Suppose $f : \mathbb{R} \to \mathbb{R}$ is a continuous, nonzero function such that $f(x + y) = f(x) f(y)$. Prove that
$$
\lim_{h \to 0} \frac{f(h) – 1}{h}
$$
exists.
Is there a way to do this without using the fact that $f(x) = e^{kx}$ for some $k \in \mathbb{R}$? That is, is there a way to show that this limit exists just using continuity and the above multiplicative property?
Best Answer
Such a function is convex. First we note that $f(x) > 0$ for all $x$. Then $$f\biggl(\frac{x+y}{2}\biggr) = f\biggl(\frac{x}{2}\biggr)f\biggl(\frac{y}{2}\biggr) = \sqrt{f(x)f(y)} \leqslant \frac{f(x) + f(y)}{2}$$ by the homomorphism property and the AM-GM inequality. Hence $f$ is midpoint-convex, and since it's also continuous it follows that $f$ is convex. [Standard argument using the fact that rationals of the form $k/2^n$ are dense.]
Then we use that for a convex function $f$ we have $$\frac{f(t) - f(s)}{t - s} \leqslant \frac{f(v) - f(u)}{v-u} \tag{1}$$ whenever $s < t$, $u < v$, $s \leqslant u$, $t \leqslant v$. Apply $(1)$ with $s = u < t \leqslant v$ to deduce that $$D_+f(s) = \lim_{\substack{t \to s \\ t > s}} \frac{f(t) - f(s)}{t-s}$$ exists for all $s$. Apply it with $s \leqslant u < t = v$ to deduce that $$D_- f(t) = \lim_{\substack{s \to t \\ s < t}} \frac{f(t) - f(s)}{t-s}$$ also exists for every $t$. Furthermore, from $(1)$ we can deduce that
It follows that we have $D_-f(s) = D_+f(s)$ at all points where $D_-f$ (or $D_+f$) is continuous, hence at all but at most countably many points. Thus a convex $f$ is differentiable except at at most countably many points.
The homomorphism property then implies that $f$ is differentiable everywhere, since $$\frac{f(x_1+h) - f(x_1)}{h} = f(x_1 - x_0)\cdot \frac{f(x_0+h) - f(x_0)}{h}\,.$$ If $f$ is differentiable at $x_0$ it follows that it is also differentiable at $x_1$.