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I couldn't find this on the Internet.

We know this property of potential function $f(x) = k \cdot a^x$:

$\displaystyle \frac{f(x+C)}{f(x)} = P$ for any real $x$, constant $C$, and constant $P$.

Can we state the reverse?

If $\displaystyle \frac{f(x+C)}{f(x)} = P$ for any real $x$, constant $C$, and constant $P$, is $f(x) = k \cdot a^x$? If not, what are the other solutions?

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  • $\begingroup$ your "const" is not constant, it is $C$-dependent. $\endgroup$
    – Carlo Beenakker
    Commented Nov 10, 2013 at 15:06
  • $\begingroup$ If $C$ and $P$ fixed constants, then there are infinitely many functions satisfying your condition. E.g., for $P=1$ and $C$ arbitrary, take $f(x)=\sin(2\pi x/C) + d$, with any $d>1$ (the $d$ is there to make sure $f$ has no zeros). In fact, take any function on the interval $[0,C]$ with $f(C)=Pf(0)$ (and no zeros), then your condition allows to extend it to all of $R$. $\endgroup$
    – UwF
    Commented Nov 10, 2013 at 20:04

1 Answer 1

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If you do not add any regularity condition (such as monotonicity, continuity etc.), then of course, there plenty of counterexamples. Take $f(x)=e^{g(x)}$ where $g(x)$ is an additive function that is not continuous. Then, $$\frac{f(x+c)}{f(x)}=e^{g(x+c)-g(x)}=e^{g(c)}.$$ Once you add some regularity one can get an effective description.

Added later: Since the OP specified that we are interested in continuous positive functions, here is the proof that all such functions are of the form $f(x)=ke^{bx}.$ Indeed, since $f>0$ we can let $f(x)=e^{g(x)}.$ Our functional equation can be rewritten as $$e^{g(x+c)-g(c)}=e^{h(c)},$$ where $e^{h(c)}$ is the constant that depends on $c.$ In other words, $$g(x+c)=g(x)+h(c).$$

The rest depends on whether your $c$ is fixed or you let it vary. If $c$ is fixed there are plenty of counter examples. Namely, take $h(c)=0$ and let $g(x)$ to be any periodic function with period $c.$

Now, if your equality holds for all $c,$ then we have $$g(x+y)=g(x)+h(y)$$ for all $x,y\in\mathbb{R}.$ Taking $x=0$ we get $g(y)=g(0)+h(y)$ and thus $$g(x+y)=g(x)+g(y)-g(0).$$

Introduce, $g_1(x)=g(x)- g(0),$ and note that the last equation implies $$g_1(x+y)=g_1(x)+g_1(y).$$

This is Cauchy functional equation and once you have continuity, it implies that $g_1(x)=kx.$

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  • $\begingroup$ thank you very much. Well, $f(x)$ should be both continuous and monotonically increasing. Besides, it denotes frequency, so it will always be non-negative. $\endgroup$
    – t5d
    Commented Nov 10, 2013 at 17:03
  • $\begingroup$ then, just reverse the the previous example. Namely, you get that $g$ is continuous and thus linear. $\endgroup$
    – Alvin
    Commented Nov 10, 2013 at 17:12
  • $\begingroup$ thanks again! I understand, that I asked that incorrectly, but: is the equation $f(x+C)/f(c) = const$ an indicator that $f(x)$ is exponential function? And if it is, then why? $\endgroup$
    – t5d
    Commented Nov 10, 2013 at 17:14
  • $\begingroup$ In ohter words, I know that exponential functions match this condition, and I need to prove, e.g. that it is the only family of functions that match, or if some other family of functions match, I need to prove that only exponential and this family of functions match, etc. $\endgroup$
    – t5d
    Commented Nov 10, 2013 at 17:26
  • $\begingroup$ Thank you! That is just what I really needed. I have found that this solution is the only one for rational numbers. Can we state the same for real and complex numbers? $\endgroup$
    – t5d
    Commented Nov 12, 2013 at 1:31

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