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In this answer I established the following characterization of $\exp(x)$:

If $f:\mathbb{R} \to \mathbb{R}$ is a function such that

  • $f(x) \geq 1 + x$ for all $x \in \mathbb{R}$
  • $f(x + y) = f(x)f(y)$ for all $x, y \in \mathbb{R}$

then $f(x) = \exp(x)$.

Actually I established in the linked answer that $f(0) = 1, f'(0) = 1$ and this implies $f'(x) = f(x)$ so that the characterization of $f$ as the exponential function is complete.

Similarly it can be proved that the following characterization for $\log x$ holds:

If $g: \mathbb{R}^{+} \to \mathbb{R}$ is a function such that

  • $g(x) \leq x - 1$ for all $x \in \mathbb{R}^{+}$
  • $g(xy) = g(x) + g(y)$ for all $x, y \in \mathbb{R}^{+}$

then $g(x) = \log x$.

These characterizations don't mention anything about continuity or differentiability and instead rely on inequalities and the functional equation.

Do we have any other characterizations of exponential and logarithmic functions which don't rely on analytic properties (continuity, derivatives etc) and instead rely on properties which are purely algebraic in nature?

I am thinking of monotone nature. If we augment the functional relation with the requirement that the function is strictly increasing on its domain, will it be sufficient to determine the functions $f, g$ uniquely? I guess it does not work. The function $f(x) = a^{x}$ with $a > 1$ is strictly monotone and satisfies the functional equation, so we still don't get uniqueness. My bad!

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Paramanand Singh
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  • How does $f(0)=f'(0)=1$ implies $f(x)\equiv f'(x)$? – Kenny Lau May 08 '16 at 15:17
  • Well, all of the characterizations will ultimately be equivalent, so it'd say there are infinitely many. – orion May 08 '16 at 15:17
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    @KennyLau: we have $$f'(x) = \lim \frac{f(x + h) - f(x)}{h} = \lim f(x)\frac{f(h) - f(0)}{h} = f(x)f'(0) = f(x)$$ here we have used functional equation to get $f(x + h) = f(x)f(h)$ and $f(x) = f(x)f(0)$. – Paramanand Singh May 08 '16 at 15:18
  • @orion: agree all the characterizations have to be equivalent because they characterize the same thing, but still it makes sense to know many alternative formulations (focus being to find simpler and simpler ones) so as to get to the essence of the simplest properties which characterize these important elementary functions. – Paramanand Singh May 08 '16 at 15:22
  • That way every property of a function is its valid characterisation ! – Narasimham May 08 '16 at 15:36
  • @Narasimham: Well, not every property alone can be a characterization, but yes a set of properties do characterize the function uniquely. Like $f'(x) = f(x)$ does not lead to unique function $\exp(x)$, but combined with $f(0) = 1$ it does lead to $\exp(x)$. Similarly $f(x + y) = f(x)f(y)$ alone is not sufficient, it needs at least continuity and the fact that $f(0) = 1$ (or the inequality which I have mentioned). By "characterize" I intended to mean "determine uniquely/completely". – Paramanand Singh May 08 '16 at 15:39
  • The elegance of your characterization is that, with $f(x) \geq 1 + x$, not only do you eliminate $a^x$ for $a \neq e$ but also eliminate pathological solutions with a single condition (with no reference to derivatives or calculus). That will be difficult to match. – MathematicsStudent1122 May 30 '16 at 02:07
  • @MathematicsStudent1122: I am also of the same view and this question is posed here with the hope that perhaps there is some other characterization of these functions which is not widely known. – Paramanand Singh May 30 '16 at 04:27

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