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












1












$begingroup$


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.






continuity







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why should these sequences exist? What if $g(x)=frac1x$ for $x>0$ and $g(x)=0$ for $xleq0$?
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 13:50










  • $begingroup$
    @SmileyCraft Notice that $f(x) neq 0$.
    $endgroup$
    – mengdie1982
    Dec 13 '18 at 13:52
















1












$begingroup$


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.






continuity







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why should these sequences exist? What if $g(x)=frac1x$ for $x>0$ and $g(x)=0$ for $xleq0$?
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 13:50










  • $begingroup$
    @SmileyCraft Notice that $f(x) neq 0$.
    $endgroup$
    – mengdie1982
    Dec 13 '18 at 13:52














1












1








1





$begingroup$


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.






continuity







share|cite|improve this question











$endgroup$




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.






continuity




continuity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 13:45







mengdie1982

















asked Dec 13 '18 at 13:40









mengdie1982mengdie1982

4,932618




4,932618












  • $begingroup$
    Why should these sequences exist? What if $g(x)=frac1x$ for $x>0$ and $g(x)=0$ for $xleq0$?
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 13:50










  • $begingroup$
    @SmileyCraft Notice that $f(x) neq 0$.
    $endgroup$
    – mengdie1982
    Dec 13 '18 at 13:52


















  • $begingroup$
    Why should these sequences exist? What if $g(x)=frac1x$ for $x>0$ and $g(x)=0$ for $xleq0$?
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 13:50










  • $begingroup$
    @SmileyCraft Notice that $f(x) neq 0$.
    $endgroup$
    – mengdie1982
    Dec 13 '18 at 13:52
















$begingroup$
Why should these sequences exist? What if $g(x)=frac1x$ for $x>0$ and $g(x)=0$ for $xleq0$?
$endgroup$
– SmileyCraft
Dec 13 '18 at 13:50




$begingroup$
Why should these sequences exist? What if $g(x)=frac1x$ for $x>0$ and $g(x)=0$ for $xleq0$?
$endgroup$
– SmileyCraft
Dec 13 '18 at 13:50












$begingroup$
@SmileyCraft Notice that $f(x) neq 0$.
$endgroup$
– mengdie1982
Dec 13 '18 at 13:52




$begingroup$
@SmileyCraft Notice that $f(x) neq 0$.
$endgroup$
– mengdie1982
Dec 13 '18 at 13:52










2 Answers
2






active

oldest

votes


















3












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Good reply ,thanks!
    $endgroup$
    – mengdie1982
    Dec 13 '18 at 13:58



















0












$begingroup$

This illustrate my first answer:



Take $f(x) = 1 + x^2$ and $g(x)= 1$ if $xgeq 0$ 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 $-1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    If you want to prove " $f(x)g(x)$ has necessarily a discontinuity point", one example is not enough...
    $endgroup$
    – mengdie1982
    Dec 13 '18 at 13:49










  • $begingroup$
    My point is not to give a proof...
    $endgroup$
    – dallonsi
    Dec 13 '18 at 14:01











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Good reply ,thanks!
    $endgroup$
    – mengdie1982
    Dec 13 '18 at 13:58
















3












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Good reply ,thanks!
    $endgroup$
    – mengdie1982
    Dec 13 '18 at 13:58














3












3








3





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 13 '18 at 13:53









CrostulCrostul

27.9k22352




27.9k22352












  • $begingroup$
    Good reply ,thanks!
    $endgroup$
    – mengdie1982
    Dec 13 '18 at 13:58


















  • $begingroup$
    Good reply ,thanks!
    $endgroup$
    – mengdie1982
    Dec 13 '18 at 13:58
















$begingroup$
Good reply ,thanks!
$endgroup$
– mengdie1982
Dec 13 '18 at 13:58




$begingroup$
Good reply ,thanks!
$endgroup$
– mengdie1982
Dec 13 '18 at 13:58











0












$begingroup$

This illustrate my first answer:



Take $f(x) = 1 + x^2$ and $g(x)= 1$ if $xgeq 0$ 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 $-1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    If you want to prove " $f(x)g(x)$ has necessarily a discontinuity point", one example is not enough...
    $endgroup$
    – mengdie1982
    Dec 13 '18 at 13:49










  • $begingroup$
    My point is not to give a proof...
    $endgroup$
    – dallonsi
    Dec 13 '18 at 14:01
















0












$begingroup$

This illustrate my first answer:



Take $f(x) = 1 + x^2$ and $g(x)= 1$ if $xgeq 0$ 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 $-1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    If you want to prove " $f(x)g(x)$ has necessarily a discontinuity point", one example is not enough...
    $endgroup$
    – mengdie1982
    Dec 13 '18 at 13:49










  • $begingroup$
    My point is not to give a proof...
    $endgroup$
    – dallonsi
    Dec 13 '18 at 14:01














0












0








0





$begingroup$

This illustrate my first answer:



Take $f(x) = 1 + x^2$ and $g(x)= 1$ if $xgeq 0$ 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 $-1$.






share|cite|improve this answer











$endgroup$



This illustrate my first answer:



Take $f(x) = 1 + x^2$ and $g(x)= 1$ if $xgeq 0$ 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 $-1$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 13 '18 at 14:02

























answered Dec 13 '18 at 13:46









dallonsidallonsi

1187




1187












  • $begingroup$
    If you want to prove " $f(x)g(x)$ has necessarily a discontinuity point", one example is not enough...
    $endgroup$
    – mengdie1982
    Dec 13 '18 at 13:49










  • $begingroup$
    My point is not to give a proof...
    $endgroup$
    – dallonsi
    Dec 13 '18 at 14:01


















  • $begingroup$
    If you want to prove " $f(x)g(x)$ has necessarily a discontinuity point", one example is not enough...
    $endgroup$
    – mengdie1982
    Dec 13 '18 at 13:49










  • $begingroup$
    My point is not to give a proof...
    $endgroup$
    – dallonsi
    Dec 13 '18 at 14:01
















$begingroup$
If you want to prove " $f(x)g(x)$ has necessarily a discontinuity point", one example is not enough...
$endgroup$
– mengdie1982
Dec 13 '18 at 13:49




$begingroup$
If you want to prove " $f(x)g(x)$ has necessarily a discontinuity point", one example is not enough...
$endgroup$
– mengdie1982
Dec 13 '18 at 13:49












$begingroup$
My point is not to give a proof...
$endgroup$
– dallonsi
Dec 13 '18 at 14:01




$begingroup$
My point is not to give a proof...
$endgroup$
– dallonsi
Dec 13 '18 at 14:01


















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