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?

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    $\begingroup$ Nicely done!${}{}{}$ $\endgroup$ – André Nicolas May 29 '12 at 4:26
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    $\begingroup$ The additive version of this question is also on the site. Do we consider them abstract duplicates? $\endgroup$ – Zev Chonoles May 29 '12 at 4:47
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    $\begingroup$ We also have a few question about this functional equation, see here and the linked question. $\endgroup$ – Martin Sleziak May 29 '12 at 4:55
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    $\begingroup$ @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. $\endgroup$ – user17762 May 29 '12 at 4:57
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    $\begingroup$ 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. $\endgroup$ – 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 $0$.

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    $\begingroup$ "Easy to see"? But no easier than the problem we are doing here... $\endgroup$ – GEdgar May 29 '12 at 17:43

If this is for a first year real analysis class, I think you might get dinged for not at least saying about the zero case that it's a constant function, and constant functions are continuous. Also, you probably should learn your favorite form of LaTeX; graders tend to like this sort of thing, and happy graders tends to mean better scores, in my experience.

Other than that, excellent work.

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    $\begingroup$ I mean not to intrude, but, thus far as I can judge, is this not a comment? $\endgroup$ – awllower May 29 '12 at 5:34
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    $\begingroup$ @awllower: If the question is "Is my work correct" then saying "Yes" is an answer, as is "yes, but..." $\endgroup$ – Henry May 29 '12 at 6:52
  • $\begingroup$ @awllower: I was wondering about that, but I ended up deciding what Henry said. $\endgroup$ – Eric Stucky May 29 '12 at 23:55

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