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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?

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

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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.)

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    $\begingroup$ @DanRust, No problem! $\endgroup$ – Sangchul Lee Dec 31 '16 at 9:29
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$f(x)=1$ works but not at all satisfying.

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  • $\begingroup$ Thanks. I edited the question to exclude this constant function $\endgroup$ – Thorsten Dec 31 '16 at 9:15
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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 $a_p = 1-t < 1$.

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