I am searching for a non-constant function $f:\mathbb{R}\rightarrow \mathbb{R}$ with the following properties:

1) $f(a+b)=f(a)f(b)$

2) $\lim\limits_{x\rightarrow -\infty} f(x) = 1$.

Is it possible to find such a function or is there a reason why such a function can not exist?

  • 5
    $\begingroup$ How about $f \equiv 1$? $\endgroup$ – user384138 Dec 31 '16 at 9:12
  • 1
    $\begingroup$ Ok. I meant not constant $\endgroup$ – Thorsten Dec 31 '16 at 9:13

By induction, you get $f (-n )=f(-1)^{n} $. The limit condition now forces $f (-1)=1$. But then, for any $x $, $$f (-n+x)=f (-1)^nf (x)=f (x), $$ and now the limit condition gives $f (x)=1$.


The only function $f : \Bbb{R} \to \Bbb{R}$ satisfying both 1) and 2) is the constant function $f \equiv 1$.

  1. $f$ is positive. First, if $f(a) = 0$, then $f(x) = f(x-a)f(a) = 0$ for all $x$ and thus $f(x) \to 0$ as $x \to -\infty$, which contradicts 2). So $f$ never vanishes. Then $f(x) = f(x/2)^2 > 0$ and hence $f$ is always positive.

  2. Now let $g(x) = \log f(x)$. This functions satisfies the Cauchy functional equation $$g(x+y) = g(x) + g(y).$$ Since $g(x) \to 0$ as $x \to -\infty$, the graph of $g$ cannot be dense in $\Bbb{R}^2$ and thus $g$ is linear: $g(x) = cx$ for some constant $c$.

  3. The only possibility for $f(x) = \mathrm{e}^{cx}$ to satisfy 2) is that $c = 0$. This corresponds to $f \equiv 1$.

(In fact, the above argument classifies all functions that satisfy 1): either $f \equiv 0$ or $\log f$ solves the Cauchy functional equation.)

  • 1
    $\begingroup$ @DanRust, No problem! $\endgroup$ – Sangchul Lee Dec 31 '16 at 9:29

$f(x)=1$ works but not at all satisfying.

  • $\begingroup$ Thanks. I edited the question to exclude this constant function $\endgroup$ – Thorsten Dec 31 '16 at 9:15

Rough proof.

Assuming $f$ continuous. I invert the condition so that $f(x) \to 1$ as $x \to +\infty$. Also $f \neq 1$ anywhere.

Since the function has a limit of $1$, so does the sequence $a_k = f(2^kx), k \in \mathbb{N}, x > 0$.

Pick a term from this sequence $a_p$ such that $|a_p - 1| < 1$.

Since $a_{k+1} = {a_k}^2$, if $a_p = 1 + s > 1$, then $a_{p+k+1} > a_{p+k}$, so $|a_j - 1|$ is increasing. (not limiting). Similar for $a_p = 1-t < 1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.