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Consider the exponential functional equation on the diagonal \begin{align} f(x+x) = f(x)f(x)\,, \quad f(0) = 1\,. \end{align}

What are the solutions for this problem? Either for all $x $ or for non-negative $x $.

EDIT: Apparently the following is wrong: I guess the exponential functions $x \mapsto e^{cx}$ are the only solutions if we assume measurability, continuity or some other regularity property.

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    $\begingroup$ And do you mean $f(0)=1$? $\endgroup$ – Hagen von Eitzen Sep 27 '16 at 13:04
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    $\begingroup$ Your proposed solution does not satisfy $f(0)=0$. $\endgroup$ – Thomas Rot Sep 27 '16 at 13:06
  • $\begingroup$ I meant f (0)=1. $\endgroup$ – desos Sep 27 '16 at 13:08
  • $\begingroup$ Note that : $f(nx)=f(x)^n $ and $ f(x)f(-x)=1 $ $\endgroup$ – user369582 Sep 27 '16 at 13:11
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    $\begingroup$ Let $f(x) = 2^{x2^{g(\log_2 x)}}$ where $g$ is any continuous function with period $1$. $\endgroup$ – Rahul Sep 27 '16 at 13:43
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Summary: There is an uncountably infinite family of $\mathcal{C}^\infty$ functions on $(0, +\infty)$ satisfying this equation, and these functions are also continuous at $x = 0$ if we demand $f(0) = 1$. If we further demand that the function be differentiable at $x = 0$, $f(x)$ must be an exponential: $f(x) = e^{cx}$ for some constant $c$.


Let $g(x): [1,2] \to \mathbb{R}$ be any function satisfying $g(2) = 2g(1)$. Define a function $h(x): [0, +\infty) \to \mathbb{R}$ recursively: $$ h(x) = \begin{cases} g(x) & 1 < x \leq 2 \\ \frac{1}{2} h(2x) & 0 < x \leq 1 \\ 2 h(x/2) & x > 2 \\ 0 & x = 0 \end{cases} $$ For example, by this definition $$ h(3) = 2 h(3/2) = 2 g(3/2) $$ and $$ h(3/16) = \frac{1}{2} h(3/8) = \frac{1}{4} h(3/4) = \frac{1}{8} h(3/2) = \frac{1}{8} g(3/2). $$ The graph of this function would be self-similar: enter image description here

By construction, $h(2x) = 2 h(x)$ for all $x$. But we can then define $f(x) = e^{h(x)}$; and therefore $f(2x) = e^{h(2x)} = e^{2 h(x)} = (e^{h(x)})^2 = f(x)^2$. Moreover, we have $$ \lim_{x \to 0_+} h(x) = 0 \quad \Rightarrow \quad \lim_{x \to 0_+} f(x) = 1 $$ and so the function is continuous at $x = 0$.

If you want the function to satisfy any particular derivative conditions, you can also demand that $g^{(n)}(1) = 2^{n-1} g^{(n)}(2)$ for any and all values of $n$ you desire. In particular, you can choose $g$ to be a $\mathcal{C}^{\infty}$ function for which all of the derivatives match at $x = 1$ and $x = 2$, and $f(x)$ will be a $\mathcal{C}^{\infty}$ function as well.

The exponential family of functions you found correspond to choosing $g(x)$ to be a linear function on $[1,2]$. A similar construction can be made to extend $h(x)$ (and thereby $f(x)$) to the negative real numbers as well; the $g(x)$ we choose to construct $f(x)$ on this domain need not be the same function as the one we used for the positive reals. @mfl's answer corresponds to taking $g(x) = x$ for the positive reals and $g(x) = -x$ for the negative reals.


EDIT: After further thought, I'm convinced that even if $g(x)$ is a $\mathcal{C}^n$ function on its domain, the resulting function $h(x)$ will not be $\mathcal{C}^n$ at $x = 0$ unless $n = 0$. This can be seen by noting that we have $$ h^{(n)}(x) = \begin{cases} g^{(n)}(x) & 1 < x \leq 2 \\ 2^{n-1} h^{(n)}(2x) & 0 < x \leq 1 \\ 2^{1-n} h^{(n)}(x/2) & x > 2 \\ 0 & x = 0 \end{cases} $$ For $n > 1$, consider the sequence of values $x_m = a/2^m$, with $1<a<2$. It can be seen that $h^{(n)}(x_m) = 2^{m(n-1)} g^{(n)}(a)$. This sequence of values will diverge as $x_m \to 0$, and so the $n$th derivative of $h$ cannot approach a finite value as $x \to 0$. The only exception here is if $g^{(n)}(a) = 0$ for all $a$, i.e., $g(x)$ is a linear function. In this case, the resulting $f(x)$ is an exponential.

For $n = 1$, meanwhile, $h'(x)$ will take on all possible values of $g'(x)$ in every interval containing the origin; and so $h'(x)$ cannot have a well-behaved limit as $x \to 0$ either. The only exception here is if $g'(x)$ is a constant over its domain, in which case we get $f(x)$ as an exponential function as well.

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  • $\begingroup$ I believe that the function proposed by @Rahul in the comments on the original question is equivalent to this method, though I'm not sure. $\endgroup$ – Michael Seifert Sep 27 '16 at 14:16
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Your claim is false. For example

$$f(x)= \begin{cases} e^x & x\ge 0 \\ e^{-x} & x<0 \end{cases}$$

satisfies the functional equation, is continuous and is not an exponential function.

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Be $f(x):=\sum\limits_{k=0}^\infty a_k x^k$ and $a_0:=1$.

With $f(2x)=f(x)^2$ follows $\sum\limits_{v=0}^k a_v a_{k-v}=2^k a_k$.

This means, that $a_1$ can be any value.

We get the exponential function $e^{a_1 x}$. $\enspace$(as $\pi r8$ and Michael Seifert have mentioned below)

Short hint: $\enspace\sum\limits_{v=0}^k \frac{(a_1)^v}{v!} \frac{(a_1)^{k-v}}{(k-v)!}=a_1^k\frac{(1+1)^k}{k!}=2^k \frac{(a_1)^k}{k!}$

This shows that functions which can be represented by Taylor series, can only be exponentially, if they solve the claim above.

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    $\begingroup$ Right, but that wouldn't necessarily be incompatible with OP's statement - $e^{a_1x}$ is in the family of solutions proposed. $\endgroup$ – πr8 Sep 27 '16 at 15:33
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    $\begingroup$ @πr8: Moreover, if you go through the first few values of $k$ above, you find that $a_2 = a_1^2/2$, $a_3 = a_1^3/6$, and $a_4 = a_1^4/24$. I suspect that continuing this will just yield the power series for $e^{a_1 x}$. This might then serve as a proof that requiring $f(x)$ to be analytic at $x = 0$ (as opposed to just $\mathcal{C}^{\infty}$) implies that it is an exponential function. $\endgroup$ – Michael Seifert Sep 27 '16 at 16:15
  • $\begingroup$ @Michael Seifert : Yes thanks, right. I will delete this post soon. $\endgroup$ – user90369 Sep 27 '16 at 17:06
  • $\begingroup$ I wouldn't necessarily delete it, actually. I suspect it shows that if a power series about $x = 0$ exists (i.e., the function is analytic there), then it must be an exponential. Ideally, one would want to prove this rigorously by showing that $a_n = a_1^n/n!$ for all $n$ (not just the few that I calculated.) $\endgroup$ – Michael Seifert Sep 27 '16 at 17:32
  • $\begingroup$ @Michael Seifert : Thank you, I have corrected. :-) $\endgroup$ – user90369 Sep 27 '16 at 19:30

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