How about $f(x)g(x)$?

Question

Question

Define $f(x)$ and $g(x)$ over $(-\infty,+\infty)$. $f(x)$ is continuous, and $f(x) \neq 0$ for any $x \in \mathbb{R}$. $g(x)$ is not continuous, in another word, it has at least one discontinuity point. How about $f(x)g(x)$? Can it be continuous？ Or it necessariy has a discontinuity point？

I guess $f(x)g(x)$ may be continuous, but I failed to find an example. Anyone can help? Thanks in advance.

Answer 1

I prove the contrapositive statement: if $f(x)g(x)$ is continuous, then $g(x)$ is continuous.

Since $f(x) \neq 0$, $1/f(x)$ is continous and defined over $\Bbb R$. If $f(x) g(x)$ is continuous, then $$g(x)= f(x) g(x) \cdot 1/f(x)$$ is continuous. This concludes the proof.

Answer 2

This illustrate my first answer:

Take $f(x) = 1 + x^2$ and $g(x)= 1$ if $x\geq 0$ and $g(x) = -1$ if $x<0$. Take the sequences $(u_n) = (\frac{1}{n})$ and $(v_n) = (-u_n)$. Then $f(u_n) g(u_n)$ converges to 1 but $f(v_n) g(v_n)$ converges to $-1$.