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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?

Charles Hudgins
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    This is not true for $f(x)=0$ – Arctic Char Oct 18 '20 at 19:01
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    To obtain from multiplicativity exponent we need additionally $f(0)=1, f(1)=a>0$ and continuity in $0$. Can you add this also to your question? – zkutch Oct 18 '20 at 19:07
  • @ArcticChar Fair point. I'll specify that $f$ is nonzero. – Charles Hudgins Oct 18 '20 at 20:11
  • @zkutch The point of the question is to see if we can show the limit exists without first constructing exponential functions. Also, $f(0) = 1$ follows immediately from $f$ being nonzero and the multiplicative property. $f(x) = f(x + 0) = f(x)f(0)$ for all $x$, so $f(0) = 1$. We don't need $f(1) = a > 0$. A simple argument with continuity shows that $f$ must be positive everywhere. – Charles Hudgins Oct 18 '20 at 20:12
  • Maybe this is more clear (if it is, I'll delete this post and make a new one): Is it possible to construct $\log a$ as the value of the above limit with $f(1) = a$ without first constructing the exponential function, defining e, using any differential equations, etc.? I'm looking for an elementary argument that the above limit exists. – Charles Hudgins Oct 18 '20 at 20:24
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    Is exploiting convexity fair game? [$f$ is convex, hence differentiable a.e., and since it's a homomorphism it follows that it's differentiable everywhere.] – Daniel Fischer Oct 18 '20 at 20:36
  • @DanielFischer Hmm. That seems fairly elementary to me. The first part seems like a standard result. Could you sketch the second for me? Also, If you post an answer, could you sketch a proof of both statements please? – Charles Hudgins Oct 18 '20 at 20:38
  • If $f$ is differentiable at $x_0$, then $$\frac{f(x_1 + h) - f(x_1)}{h} = f(x_1 - x_0)\cdot \frac{f(x_0+h) - f(x_0)}{h}$$ and the limit on the right hand side exists by assumption. – Daniel Fischer Oct 18 '20 at 20:41
  • @DanielFischer Oh duh. The whole point (which I knew, but forgot) is that we only need differentiability at a point for the function to be differentiable everywhere. – Charles Hudgins Oct 18 '20 at 20:44
  • @Charles Hudgins. You have not $f$ nonzero from scratch, when I wrote my comment, so your comment about not needing $f(0)=1$ is not fair. My comment is about without requirement $f$ nonzero and having brought by me 3 conditions can be proved that there is only one such function. My way is, that differentiability is equivalence to OP limit. – zkutch Oct 18 '20 at 23:48
  • @zkutch Fair enough. I might have come off more dismissive than I intended. I just meant to say that $f(0) = 1$ and $f(1) > 0$ only fail to be true in the edge case where $f \equiv 0$, which, you're right, I had not added at the time of your post. – Charles Hudgins Oct 19 '20 at 03:09

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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

  • $D_-f(s) \leqslant D_+f(s)$ for all $s$,
  • $D_-f$ and $D_+f$ are both monotnically increasing (nonstrictly in general),
  • $s < t$ implies $D_+f(s) \leqslant D_-f(t)$.

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 $x_1$.

Daniel Fischer
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  • I find it interesting that this approach to the differentiability of such functions ultimately rests on a pigeonhole principle argument. I wonder if there is a more constructive way to see that this limit should exist, but, obviously, this post is a great answer to my question as written. – Charles Hudgins Oct 19 '20 at 03:13