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So far I have shown that the equality only holds when $x=0$ or $y=0$. I also have found out that $f(nx) \leq f(x)^n$ for natural $n$ (otherwise the summation doesn't make sense).

But I do not know how to deduce the continuity. My idea is to show it with the sequence criterion for continuity, but the only thing I've shown with it is that for $n \rightarrow 0$ the function $f(nx) \rightarrow f(0)=1$, which doesn't help a lot.

I also could use that $f$ is continuous in $x=0$ to make any $f(x)$ a product of $f(1)$, but it only holds for natural numbers, thus not leading to a solution.

I would appreciate hints, not solutions. Thank you for help.

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Hint:

Take $x=x_0$ and $y=\frac{1}{n}$

$$f\left(x_0+\frac{1}{n}\right) \leq f(x_0)f\left(\frac{1}{n}\right)$$

Now take $x=x_0+\frac{1}{n}$ and $y=-\frac{1}{n}$

$$f\left(x_0+\frac{1}{n}\right) \geq \frac{f(x_0)}{f\left(-\frac{1}{n}\right)}$$

Now make $n \rightarrow \infty$

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  • $\begingroup$ Hello, @Arnaldo Nascimento, thank you for your hint. Would it not be better to take $a_n$ with $a_n \rightarrow 0$ as $n \rightarrow \infty$ instead of $1/n$ and then use the claim that $f$ is continous at $x=0$, thus $f(a_n)$ and $f(-a_n)$ both converge to $f(0)=1$? Would it not be the solution? $\endgroup$ – 77and33is100 Dec 12 '16 at 23:34
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    $\begingroup$ Hi @B.Schnebbler, that is the same idea. I just wanted to take a easy sequence like $\frac{1}{n}$ to be more clear. $\endgroup$ – Arnaldo Dec 12 '16 at 23:38
  • $\begingroup$ You have assumed that $f(-1/n)$ is not equal to 0. Is there a way to prove this? I am struggling with it. $\endgroup$ – 77and33is100 Dec 13 '16 at 0:02
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    $\begingroup$ If it is you will have $f(x_0) \leq 0$, for any $x_0$, which is not true if you take $x_0=0$. $\endgroup$ – Arnaldo Dec 13 '16 at 0:09
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    $\begingroup$ No need excuse! It is a pleasure! $\endgroup$ – Arnaldo Dec 13 '16 at 0:14
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Hint

$f$ is continuous at $x_0$ if $$\forall \epsilon>0\exists\delta>0:|h|<\delta\implies |f(x_0+h)-f(x_0)|<\epsilon.$$

Now, $$f(x_0+h)\le f(x_0)f(h)$$ and since $f$ is continuous at $0$ we have that $f(h)\to 1$ as $h\to 0.$

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  • $\begingroup$ Thank you for your hint @mfl. Does it suffice to say that $f(x_0 +h) \leq f(x_0)f(h)$ implies that $f(x_0 +h)-f(x_0)f(h) \leq 0$ where the left term approaches $0$ as $h$ approaches $0$ and thus for every $\epsilon>0$ there exists a $\delta > 0$ such that the continuity criterion is satisfied? $\endgroup$ – 77and33is100 Dec 12 '16 at 23:12

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