# Prove or disprove: if $f$ and $fg$ are continuous then $g$ is continuous.

Prove of provide a counterexample:

Suppose that $f$ and $g$ are defined and finite valued on an open interval $I$ which contains $a$, that $f$ is continuous at $a$, and that $f(a)\neq 0$. Then $g$ is continuous at $a$ if and only if $fg$ is continuous at $a$.

I don't suppose it's true, based on the fact that the common theorem '$f, g$ continuous implies $fg$ continuous' is not stated as true both ways; obviously, this implies exceptions. The only ones I can think of, however, are one's that don't fit the "open interval" or "$f(a)\neq 0$" parts, or ones where both f and g are discontinuous.

I've also tried proving it, but with no luck.

Help? :-S

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I suppose you actually meant: $\,f\,$ is a continuous function, then for a function $\,g\,$ we have that $\,fg\,$ is continuous iff $\,g\,$ is continuous. – DonAntonio Jun 2 '13 at 19:33
@DonAntonio, I did not, but, maybe I'm mistaken, does that not mean the same thing? "$fg$ is continuous iff $g$ is continuous" is logically equivalent to "$g$ is continuous iff $fg$ is continuous". – user80696 Jun 2 '13 at 20:34
you title reads "$\,f\,$ and $\,fg\,$ are cont. iff $\,g\,$ is continuous"...this, of course, is far from being true, but I think this is not what you really meant. – DonAntonio Jun 2 '13 at 22:37
@DonAntonio, Ah, now I see what you meant. I was not looking at the title when I read your last comment - I thought you meant the actual proposition. I agree, the title is incorrect; it should read "Prove or disprove: if $f$ and $fg$ are continuous then $g$ is continuous" (the other direction being trivial). I'll fix it. – user80696 Jun 2 '13 at 23:20
What does "finite valued" mean? It seems to reveal a misunderstanding of what "defined" means. – TonyK Jun 2 '13 at 23:38

$f$ is continuous and nonzero at $a$, hence $1/f$ is continuous at $a$. Since $fg$ is continuous at $a$ as well, so is their product $(fg)(1/f)=g$.
How can you deduce that $f$ must be continuous on an interval around $a$? – Javier Jun 2 '13 at 21:29
@JavierBadia, I suppose I can't; I need $f$ to be continuous on an interval around $a$, in addition to the other assumptions stated. Otherwise, there might be some sequence $a_1, a_2, \ldots$ with $a_i\rightarrow a$ and $f(a_i)=0$, which kills my proof. – vadim123 Jun 3 '13 at 2:54
@vadim123: No, wait. If there was such a sequence, $f$ would not be continuous at $a$. You don't need $f$ to be continuous on an interval; I believe algebra of limits implies that if $f$ is continuous and non-zero at $a$, then so is $1/f$. – Javier Jun 3 '13 at 10:34
@JavierBadia Badia Yes, that's actually how I reworked the proof once you pointed that out. Simply removing the open interval around $a$ and discussing the point $a$ itself works. Thanks to you both! – user80696 Jun 3 '13 at 17:09