# If $f\colon \mathbb{R} \to \mathbb{R}$ is such that $f (x + y) = f (x) f (y)$ and continuous at $0$, then continuous everywhere

Prove that if $f\colon\mathbb{R}\to\mathbb{R}$ is such that $f(x+y)=f(x)f(y)$ for all $x,y$, and $f$ is continuous at $0$, then it is continuous everywhere.

If there exists $c \in \mathbb{R}$ such that $f(c) = 0$, then $$f(x + c) = f(x)f(c) = 0.$$ As every real number $y$ can be written as $y = x + c$ for some real $x$, this function is either everywhere zero or nowhere zero. The latter case is the interesting one. So let's consider the case that $f$ is not the constant function $f = 0$.

To prove continuity in this case, note that for any $x \in \mathbb{R}$ $$f(x) = f(x + 0) = f(x)f(0) \implies f(0) = 1.$$

Continuity at $0$ tells us that given any $\varepsilon_0 > 0$, we can find $\delta_0 > 0$ such that $|x| < \delta_0$ implies $$|f(x) - 1| < \varepsilon_0.$$

Okay, so let $c \in \mathbb{R}$ be fixed arbitrarily (recall that $f(c)$ is nonzero). Let $\varepsilon > 0$. By continuity of $f$ at $0$, we can choose $\delta > 0$ such that $$|x - c| < \delta\implies |f(x - c) - 1| < \frac{\varepsilon}{|f(c)|}.$$

Now notice that for all $x$ such that $|x - c| < \delta$, we have \begin{align*} |f(x) - f(c)| &= |f(x - c + c) - f(c)|\\ &= |f(x - c)f(c) - f(c)|\\ &= |f(c)| |f(x - c) - 1|\\ &\lt |f(c)| \frac{\varepsilon}{|f(c)|}\\ &= \varepsilon. \end{align*} Hence $f$ is continuous at $c$. Since $c$ was arbitrary, $f$ is continuous on all of $\mathbb{R}$.

Is my procedure correct?

• Nicely done!${}{}{}$ – André Nicolas May 29 '12 at 4:26
• The additive version of this question is also on the site. Do we consider them abstract duplicates? – Zev Chonoles May 29 '12 at 4:47
• We also have a few question about this functional equation, see here and the linked question. – Martin Sleziak May 29 '12 at 4:55
• @ZevChonoles Yes. I think it would be good if someone can write a generalized version of this and call it, say Cauchy functional equation, add an answer to the generalized question. All the rest should be closed as dupe of this. – user17762 May 29 '12 at 4:57
• I am not sure if closing question of the type please check my proof should be closed as duplicate question. (Even if we had a question on this exact result, where some proof is given, in this kind of question I would expect comments on the OP's proof strategy, mistakes - if there are any, writing style etc. So it is kind of different from answer to question of the type I would like to see a proof or a reference for this. – Martin Sleziak May 29 '12 at 5:02

One easier thing to do is to notice that $f(x)=(f(x/2))^2$ so $f$ is positive, and assume that it is never zero, since then the function is identically zero. Then you can define $g(x)=\ln f(x)$ and this function $g$ will satisfy the Cauchy functional equation $$g(x+y)=g(x)+g(y)$$ and the theory for this functional equation is well known, and it is easy to see that $g$ is continuous if and only if it is continuous at $0$.