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Prove that, for $f:\Bbb{R} \rightarrow \Bbb{R}$, if $f(x+y)=f(x)f(y)$ and f is continuous at $x=0,$ then it is continuous in all $\Bbb{R}$.

I haven't figured out how to prove this. What would you suggest?

I've already tried finding and expression for $f(0)$ which is either $0$ or $1$.

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    $\begingroup$ Note that if $f(0)=0$, then $f(x)=0,\forall x\in\Bbb{R}$, hence the result is done. $\endgroup$ – DiegoMath Sep 16 '14 at 4:27
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Actually, we do not need to know the value of $f(0)$. For any $a \in \mathbb{R}$, \begin{align*} \lim_{x \rightarrow a} f(x) & = \lim_{x \rightarrow a} f\big((x-a)+a\big) = \lim_{x \rightarrow a} f(x-a)f(a) \\ & = f\Big(\lim_{x \rightarrow a} (x-a)\Big)f(a) \quad \text{because } \lim_{x \rightarrow a} (x-a) = 0 \text{ and } f \text{ is continuous at } 0 \\ & =f(0)f(a) = f(0+a) = f(a). \end{align*}

Hence $f$ is continuous at $a$.

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Case $1$: If $f(0) = 1$, then we have

$$\lim_{y \to 0} \;[f(x + y) - f(x) ] = \lim_{y \to 0} \; [f(x)f(y) - f(x)] = \lim_{y \to 0} \;f(x)[f(y) - 1]= f(x)(f(0) - 1) = 0.$$

Case $2$: If $f(0) = 0$, then we are done because the problem told us $f$ was continuous.

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  • $\begingroup$ Can you clarify your remark for case 2? I would have said $f(x) = f(x+0) = f(x)f(0) = 0$ for all $x$, hence $f$ is continuous. $\endgroup$ – Bungo Sep 16 '14 at 4:42
  • $\begingroup$ @Bungo, it's exactly what you wrote. I didn't write anything because I thought OP found those values through your calculations. $\endgroup$ – IAmNoOne Sep 16 '14 at 4:50
  • $\begingroup$ Oh, I interpreted his remark to mean that he noted that $f(0) = f(0+0) = f(0)f(0)$, so $f(0)$ has to be either $0$ or $1$. But I don't see that he concluded that if $f(0) = 0$ then $f$ identically zero. I might just be misinterpreting the OP, though. $\endgroup$ – Bungo Sep 16 '14 at 4:55

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