Prove a min function defined in the domain of another continuous function is continuous












0












$begingroup$


Assume a min real function is defined below:



$$
f(x)=minlimits_{xleq yleq h(x)}g(y)
$$



$g(x)$ is continuous. $xgeq 0$, $h(x)=ax+b$, $ageq 1, bgeq 0$. Prove $f(x)$ is continuous.





This is a problem I recently encountered when reading a paper. There may be several scenarios regarding the location of $y^ast$ ($g(y^ast)=minlimits_{xleq yleq 2x}g(y)$) in the domain $[x-delta, 2x+2delta]$. And we may prove the continuity by discussing the different scenarios. My proof is below:





$textit{Proof}.$



We must prove that for any $epsilon>0$, there exists $delta>0$, such that when $|x-x_0|<delta$,



$$|f(x)-f(x_0)|<epsilon$$



Because $g(x)$ is continous, for any $epsilon_1>0$, there exits $delta_1$, when $|x-x_0|<delta_1$,



$$|g(x)-g(x_0)|<epsilon_1tag{1}$$





$g(x)$ is also continous on $x=h(x_0)$. There exists $delta_2$, when $|h(x)-h(x_0)|<delta_2$,



$$|g(h(x))-g(h(x_0))|<epsilon_2tag{2}$$



For any two number $x_1$, $x_2$ in the domain $[x_0-delta, x_0+delta]$, let $deltaleq delta_1$, we can get



$$
begin{aligned}
|g(x_1)-g(x_2)|&=|g(x_1)-g(x_0)+g(x_0)-g(x_2)|\
&leq |g(x_1)-g(x_0)|+|g(x_0)-g(x_2)|<2epsilon_1
end{aligned}tag{3}
$$





For any two number $x_1$, $x_2$ in the domain $[x_0-delta, x_0+delta]$, because $h(x)=ax+b$,



$$
begin{aligned}
a(x_0-delta) +bleq h(x_1)leq a(x_0+delta)+b\
a(x_0-delta) +bleq h(x_2)leq a(x_0+delta)+b
end{aligned}
$$



When $deltaleq delta_2$, then $h(x_1), h(x_2)in[h(x_0)-delta_2, h(x_0)+delta_2]$, by Eq. (2), we can get



$$
begin{aligned}
|g(h(x_1))-g(h(x_2))|&=|g(h(x_1))-g(h(x_0))+g(h(x_0))-g(h(x_2))|\
&leq |g(h(x_1))-g(h(x_0))|+|g(h(x_0))-g(h(x_2))|<2epsilon_2
end{aligned}tag{4}
$$





Assume
$$g(y^ast_{0})=minlimits_{x_0leq yleq h(x_0)}g(y)$$



$$g(y^ast_{Delta})=minlimits_{x_0-deltaleq yleq h(x_0+delta)}g(y)$$



For any $xin [x_0-delta, x_0+delta]$,



$$g(y^ast_{x})=minlimits_{xleq yleq h(x)}g(y)$$



Apparently, $g(y^ast_{Delta})leq g(y^ast_{x})$, $g(y^ast_{Delta})leq g(y^ast_0)$.



And $f(x)=g(y^ast_x)$, $f(x_0)=g(y^ast_0)$.



So,



$$
begin{aligned}
|f(x)-f(x_0)|&=|g(y^ast_x)-g(y^ast_0)| \
&= |g(y^ast_x)-g(y^ast_Delta)+g(y^ast_Delta)-g(y^ast_0)|\
&leq |g(y^ast_x)-g(y^ast_Delta)|+|g(y^ast_0)-g(y^ast_Delta)|
end{aligned}tag{5}
$$



According the location of $y^ast_Delta$, there are three cases possible for $| g(y^ast_x)-g(y^ast_Delta) |$ :



(a). $y^ast_Delta= y^ast_x$, if $y^ast_Deltain [x, h(x)]$.



In this case, $| g(y^ast_x)-g(y^ast_Delta) |=0$.



(b). $y^ast_Deltain [x-delta, x]subset [x_0-delta, x_0+delta]$.



In this case, let $deltaleq delta_1$, $| g(y^ast_x)-g(y^ast_Delta) |leq | g(x)-g(y^ast_Delta) |<2epsilon_1$. (because of Eq. (3))



(c). $y^ast_Deltain [h(x), h(x_0+delta)]subset [h(x_0-delta), h(x_0+delta)]$.





In this case, let $deltaleq delta_2$,





$| g(y^ast_x)-g(y^ast_Delta) |leq | g(h(x))-g(y^ast_Delta) |<2epsilon_2$ (according to Eq. (4)).



Therefore, from the three cases, when $deltaleqmin{delta_1, delta_2}$, we can get



$$| g(y^ast_x)-g(y^ast_Delta) |<max{2epsilon_1, 2epsilon_2}$$



Because $g(y^ast_0)$ is a special situation of $g(y^ast_x)$ when $x=x_0$, there is also



$$|g(y^ast_0)-g(y^ast_Delta)|<max{2epsilon_1, 2epsilon_2}$$



Eq. (6) can be written to be:



$$|f(x)-f(x_0)|<2max{2epsilon_1, 2epsilon_2}$$



Let $max{epsilon_1, epsilon_2}=epsilon$, we can get $|f(x)-f(x_0)|<epsilon$. The continuity of $f(x)$ is proved.
$$hspace{300pt}Box$$










share|cite|improve this question











$endgroup$












  • $begingroup$
    Thanks for the comments. I have edited this question. Is it qualified now?
    $endgroup$
    – zhen chen
    Jan 10 at 13:48






  • 3




    $begingroup$
    Not really. Where is the “include your work and thoughts on the problem” part?
    $endgroup$
    – José Carlos Santos
    Jan 10 at 13:57










  • $begingroup$
    f(0) = g(0). What is f(-1)?
    $endgroup$
    – William Elliot
    Jan 11 at 2:58










  • $begingroup$
    I have edited the problem, now the function is defined only for the positive domain.
    $endgroup$
    – zhen chen
    Jan 11 at 6:53










  • $begingroup$
    I found my question is very similar to this one: link, the author provide a version of proof.
    $endgroup$
    – zhen chen
    Jan 11 at 12:38


















0












$begingroup$


Assume a min real function is defined below:



$$
f(x)=minlimits_{xleq yleq h(x)}g(y)
$$



$g(x)$ is continuous. $xgeq 0$, $h(x)=ax+b$, $ageq 1, bgeq 0$. Prove $f(x)$ is continuous.





This is a problem I recently encountered when reading a paper. There may be several scenarios regarding the location of $y^ast$ ($g(y^ast)=minlimits_{xleq yleq 2x}g(y)$) in the domain $[x-delta, 2x+2delta]$. And we may prove the continuity by discussing the different scenarios. My proof is below:





$textit{Proof}.$



We must prove that for any $epsilon>0$, there exists $delta>0$, such that when $|x-x_0|<delta$,



$$|f(x)-f(x_0)|<epsilon$$



Because $g(x)$ is continous, for any $epsilon_1>0$, there exits $delta_1$, when $|x-x_0|<delta_1$,



$$|g(x)-g(x_0)|<epsilon_1tag{1}$$





$g(x)$ is also continous on $x=h(x_0)$. There exists $delta_2$, when $|h(x)-h(x_0)|<delta_2$,



$$|g(h(x))-g(h(x_0))|<epsilon_2tag{2}$$



For any two number $x_1$, $x_2$ in the domain $[x_0-delta, x_0+delta]$, let $deltaleq delta_1$, we can get



$$
begin{aligned}
|g(x_1)-g(x_2)|&=|g(x_1)-g(x_0)+g(x_0)-g(x_2)|\
&leq |g(x_1)-g(x_0)|+|g(x_0)-g(x_2)|<2epsilon_1
end{aligned}tag{3}
$$





For any two number $x_1$, $x_2$ in the domain $[x_0-delta, x_0+delta]$, because $h(x)=ax+b$,



$$
begin{aligned}
a(x_0-delta) +bleq h(x_1)leq a(x_0+delta)+b\
a(x_0-delta) +bleq h(x_2)leq a(x_0+delta)+b
end{aligned}
$$



When $deltaleq delta_2$, then $h(x_1), h(x_2)in[h(x_0)-delta_2, h(x_0)+delta_2]$, by Eq. (2), we can get



$$
begin{aligned}
|g(h(x_1))-g(h(x_2))|&=|g(h(x_1))-g(h(x_0))+g(h(x_0))-g(h(x_2))|\
&leq |g(h(x_1))-g(h(x_0))|+|g(h(x_0))-g(h(x_2))|<2epsilon_2
end{aligned}tag{4}
$$





Assume
$$g(y^ast_{0})=minlimits_{x_0leq yleq h(x_0)}g(y)$$



$$g(y^ast_{Delta})=minlimits_{x_0-deltaleq yleq h(x_0+delta)}g(y)$$



For any $xin [x_0-delta, x_0+delta]$,



$$g(y^ast_{x})=minlimits_{xleq yleq h(x)}g(y)$$



Apparently, $g(y^ast_{Delta})leq g(y^ast_{x})$, $g(y^ast_{Delta})leq g(y^ast_0)$.



And $f(x)=g(y^ast_x)$, $f(x_0)=g(y^ast_0)$.



So,



$$
begin{aligned}
|f(x)-f(x_0)|&=|g(y^ast_x)-g(y^ast_0)| \
&= |g(y^ast_x)-g(y^ast_Delta)+g(y^ast_Delta)-g(y^ast_0)|\
&leq |g(y^ast_x)-g(y^ast_Delta)|+|g(y^ast_0)-g(y^ast_Delta)|
end{aligned}tag{5}
$$



According the location of $y^ast_Delta$, there are three cases possible for $| g(y^ast_x)-g(y^ast_Delta) |$ :



(a). $y^ast_Delta= y^ast_x$, if $y^ast_Deltain [x, h(x)]$.



In this case, $| g(y^ast_x)-g(y^ast_Delta) |=0$.



(b). $y^ast_Deltain [x-delta, x]subset [x_0-delta, x_0+delta]$.



In this case, let $deltaleq delta_1$, $| g(y^ast_x)-g(y^ast_Delta) |leq | g(x)-g(y^ast_Delta) |<2epsilon_1$. (because of Eq. (3))



(c). $y^ast_Deltain [h(x), h(x_0+delta)]subset [h(x_0-delta), h(x_0+delta)]$.





In this case, let $deltaleq delta_2$,





$| g(y^ast_x)-g(y^ast_Delta) |leq | g(h(x))-g(y^ast_Delta) |<2epsilon_2$ (according to Eq. (4)).



Therefore, from the three cases, when $deltaleqmin{delta_1, delta_2}$, we can get



$$| g(y^ast_x)-g(y^ast_Delta) |<max{2epsilon_1, 2epsilon_2}$$



Because $g(y^ast_0)$ is a special situation of $g(y^ast_x)$ when $x=x_0$, there is also



$$|g(y^ast_0)-g(y^ast_Delta)|<max{2epsilon_1, 2epsilon_2}$$



Eq. (6) can be written to be:



$$|f(x)-f(x_0)|<2max{2epsilon_1, 2epsilon_2}$$



Let $max{epsilon_1, epsilon_2}=epsilon$, we can get $|f(x)-f(x_0)|<epsilon$. The continuity of $f(x)$ is proved.
$$hspace{300pt}Box$$










share|cite|improve this question











$endgroup$












  • $begingroup$
    Thanks for the comments. I have edited this question. Is it qualified now?
    $endgroup$
    – zhen chen
    Jan 10 at 13:48






  • 3




    $begingroup$
    Not really. Where is the “include your work and thoughts on the problem” part?
    $endgroup$
    – José Carlos Santos
    Jan 10 at 13:57










  • $begingroup$
    f(0) = g(0). What is f(-1)?
    $endgroup$
    – William Elliot
    Jan 11 at 2:58










  • $begingroup$
    I have edited the problem, now the function is defined only for the positive domain.
    $endgroup$
    – zhen chen
    Jan 11 at 6:53










  • $begingroup$
    I found my question is very similar to this one: link, the author provide a version of proof.
    $endgroup$
    – zhen chen
    Jan 11 at 12:38
















0












0








0


1



$begingroup$


Assume a min real function is defined below:



$$
f(x)=minlimits_{xleq yleq h(x)}g(y)
$$



$g(x)$ is continuous. $xgeq 0$, $h(x)=ax+b$, $ageq 1, bgeq 0$. Prove $f(x)$ is continuous.





This is a problem I recently encountered when reading a paper. There may be several scenarios regarding the location of $y^ast$ ($g(y^ast)=minlimits_{xleq yleq 2x}g(y)$) in the domain $[x-delta, 2x+2delta]$. And we may prove the continuity by discussing the different scenarios. My proof is below:





$textit{Proof}.$



We must prove that for any $epsilon>0$, there exists $delta>0$, such that when $|x-x_0|<delta$,



$$|f(x)-f(x_0)|<epsilon$$



Because $g(x)$ is continous, for any $epsilon_1>0$, there exits $delta_1$, when $|x-x_0|<delta_1$,



$$|g(x)-g(x_0)|<epsilon_1tag{1}$$





$g(x)$ is also continous on $x=h(x_0)$. There exists $delta_2$, when $|h(x)-h(x_0)|<delta_2$,



$$|g(h(x))-g(h(x_0))|<epsilon_2tag{2}$$



For any two number $x_1$, $x_2$ in the domain $[x_0-delta, x_0+delta]$, let $deltaleq delta_1$, we can get



$$
begin{aligned}
|g(x_1)-g(x_2)|&=|g(x_1)-g(x_0)+g(x_0)-g(x_2)|\
&leq |g(x_1)-g(x_0)|+|g(x_0)-g(x_2)|<2epsilon_1
end{aligned}tag{3}
$$





For any two number $x_1$, $x_2$ in the domain $[x_0-delta, x_0+delta]$, because $h(x)=ax+b$,



$$
begin{aligned}
a(x_0-delta) +bleq h(x_1)leq a(x_0+delta)+b\
a(x_0-delta) +bleq h(x_2)leq a(x_0+delta)+b
end{aligned}
$$



When $deltaleq delta_2$, then $h(x_1), h(x_2)in[h(x_0)-delta_2, h(x_0)+delta_2]$, by Eq. (2), we can get



$$
begin{aligned}
|g(h(x_1))-g(h(x_2))|&=|g(h(x_1))-g(h(x_0))+g(h(x_0))-g(h(x_2))|\
&leq |g(h(x_1))-g(h(x_0))|+|g(h(x_0))-g(h(x_2))|<2epsilon_2
end{aligned}tag{4}
$$





Assume
$$g(y^ast_{0})=minlimits_{x_0leq yleq h(x_0)}g(y)$$



$$g(y^ast_{Delta})=minlimits_{x_0-deltaleq yleq h(x_0+delta)}g(y)$$



For any $xin [x_0-delta, x_0+delta]$,



$$g(y^ast_{x})=minlimits_{xleq yleq h(x)}g(y)$$



Apparently, $g(y^ast_{Delta})leq g(y^ast_{x})$, $g(y^ast_{Delta})leq g(y^ast_0)$.



And $f(x)=g(y^ast_x)$, $f(x_0)=g(y^ast_0)$.



So,



$$
begin{aligned}
|f(x)-f(x_0)|&=|g(y^ast_x)-g(y^ast_0)| \
&= |g(y^ast_x)-g(y^ast_Delta)+g(y^ast_Delta)-g(y^ast_0)|\
&leq |g(y^ast_x)-g(y^ast_Delta)|+|g(y^ast_0)-g(y^ast_Delta)|
end{aligned}tag{5}
$$



According the location of $y^ast_Delta$, there are three cases possible for $| g(y^ast_x)-g(y^ast_Delta) |$ :



(a). $y^ast_Delta= y^ast_x$, if $y^ast_Deltain [x, h(x)]$.



In this case, $| g(y^ast_x)-g(y^ast_Delta) |=0$.



(b). $y^ast_Deltain [x-delta, x]subset [x_0-delta, x_0+delta]$.



In this case, let $deltaleq delta_1$, $| g(y^ast_x)-g(y^ast_Delta) |leq | g(x)-g(y^ast_Delta) |<2epsilon_1$. (because of Eq. (3))



(c). $y^ast_Deltain [h(x), h(x_0+delta)]subset [h(x_0-delta), h(x_0+delta)]$.





In this case, let $deltaleq delta_2$,





$| g(y^ast_x)-g(y^ast_Delta) |leq | g(h(x))-g(y^ast_Delta) |<2epsilon_2$ (according to Eq. (4)).



Therefore, from the three cases, when $deltaleqmin{delta_1, delta_2}$, we can get



$$| g(y^ast_x)-g(y^ast_Delta) |<max{2epsilon_1, 2epsilon_2}$$



Because $g(y^ast_0)$ is a special situation of $g(y^ast_x)$ when $x=x_0$, there is also



$$|g(y^ast_0)-g(y^ast_Delta)|<max{2epsilon_1, 2epsilon_2}$$



Eq. (6) can be written to be:



$$|f(x)-f(x_0)|<2max{2epsilon_1, 2epsilon_2}$$



Let $max{epsilon_1, epsilon_2}=epsilon$, we can get $|f(x)-f(x_0)|<epsilon$. The continuity of $f(x)$ is proved.
$$hspace{300pt}Box$$










share|cite|improve this question











$endgroup$




Assume a min real function is defined below:



$$
f(x)=minlimits_{xleq yleq h(x)}g(y)
$$



$g(x)$ is continuous. $xgeq 0$, $h(x)=ax+b$, $ageq 1, bgeq 0$. Prove $f(x)$ is continuous.





This is a problem I recently encountered when reading a paper. There may be several scenarios regarding the location of $y^ast$ ($g(y^ast)=minlimits_{xleq yleq 2x}g(y)$) in the domain $[x-delta, 2x+2delta]$. And we may prove the continuity by discussing the different scenarios. My proof is below:





$textit{Proof}.$



We must prove that for any $epsilon>0$, there exists $delta>0$, such that when $|x-x_0|<delta$,



$$|f(x)-f(x_0)|<epsilon$$



Because $g(x)$ is continous, for any $epsilon_1>0$, there exits $delta_1$, when $|x-x_0|<delta_1$,



$$|g(x)-g(x_0)|<epsilon_1tag{1}$$





$g(x)$ is also continous on $x=h(x_0)$. There exists $delta_2$, when $|h(x)-h(x_0)|<delta_2$,



$$|g(h(x))-g(h(x_0))|<epsilon_2tag{2}$$



For any two number $x_1$, $x_2$ in the domain $[x_0-delta, x_0+delta]$, let $deltaleq delta_1$, we can get



$$
begin{aligned}
|g(x_1)-g(x_2)|&=|g(x_1)-g(x_0)+g(x_0)-g(x_2)|\
&leq |g(x_1)-g(x_0)|+|g(x_0)-g(x_2)|<2epsilon_1
end{aligned}tag{3}
$$





For any two number $x_1$, $x_2$ in the domain $[x_0-delta, x_0+delta]$, because $h(x)=ax+b$,



$$
begin{aligned}
a(x_0-delta) +bleq h(x_1)leq a(x_0+delta)+b\
a(x_0-delta) +bleq h(x_2)leq a(x_0+delta)+b
end{aligned}
$$



When $deltaleq delta_2$, then $h(x_1), h(x_2)in[h(x_0)-delta_2, h(x_0)+delta_2]$, by Eq. (2), we can get



$$
begin{aligned}
|g(h(x_1))-g(h(x_2))|&=|g(h(x_1))-g(h(x_0))+g(h(x_0))-g(h(x_2))|\
&leq |g(h(x_1))-g(h(x_0))|+|g(h(x_0))-g(h(x_2))|<2epsilon_2
end{aligned}tag{4}
$$





Assume
$$g(y^ast_{0})=minlimits_{x_0leq yleq h(x_0)}g(y)$$



$$g(y^ast_{Delta})=minlimits_{x_0-deltaleq yleq h(x_0+delta)}g(y)$$



For any $xin [x_0-delta, x_0+delta]$,



$$g(y^ast_{x})=minlimits_{xleq yleq h(x)}g(y)$$



Apparently, $g(y^ast_{Delta})leq g(y^ast_{x})$, $g(y^ast_{Delta})leq g(y^ast_0)$.



And $f(x)=g(y^ast_x)$, $f(x_0)=g(y^ast_0)$.



So,



$$
begin{aligned}
|f(x)-f(x_0)|&=|g(y^ast_x)-g(y^ast_0)| \
&= |g(y^ast_x)-g(y^ast_Delta)+g(y^ast_Delta)-g(y^ast_0)|\
&leq |g(y^ast_x)-g(y^ast_Delta)|+|g(y^ast_0)-g(y^ast_Delta)|
end{aligned}tag{5}
$$



According the location of $y^ast_Delta$, there are three cases possible for $| g(y^ast_x)-g(y^ast_Delta) |$ :



(a). $y^ast_Delta= y^ast_x$, if $y^ast_Deltain [x, h(x)]$.



In this case, $| g(y^ast_x)-g(y^ast_Delta) |=0$.



(b). $y^ast_Deltain [x-delta, x]subset [x_0-delta, x_0+delta]$.



In this case, let $deltaleq delta_1$, $| g(y^ast_x)-g(y^ast_Delta) |leq | g(x)-g(y^ast_Delta) |<2epsilon_1$. (because of Eq. (3))



(c). $y^ast_Deltain [h(x), h(x_0+delta)]subset [h(x_0-delta), h(x_0+delta)]$.





In this case, let $deltaleq delta_2$,





$| g(y^ast_x)-g(y^ast_Delta) |leq | g(h(x))-g(y^ast_Delta) |<2epsilon_2$ (according to Eq. (4)).



Therefore, from the three cases, when $deltaleqmin{delta_1, delta_2}$, we can get



$$| g(y^ast_x)-g(y^ast_Delta) |<max{2epsilon_1, 2epsilon_2}$$



Because $g(y^ast_0)$ is a special situation of $g(y^ast_x)$ when $x=x_0$, there is also



$$|g(y^ast_0)-g(y^ast_Delta)|<max{2epsilon_1, 2epsilon_2}$$



Eq. (6) can be written to be:



$$|f(x)-f(x_0)|<2max{2epsilon_1, 2epsilon_2}$$



Let $max{epsilon_1, epsilon_2}=epsilon$, we can get $|f(x)-f(x_0)|<epsilon$. The continuity of $f(x)$ is proved.
$$hspace{300pt}Box$$







real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 21 at 8:21







zhen chen

















asked Jan 10 at 13:33









zhen chenzhen chen

61




61












  • $begingroup$
    Thanks for the comments. I have edited this question. Is it qualified now?
    $endgroup$
    – zhen chen
    Jan 10 at 13:48






  • 3




    $begingroup$
    Not really. Where is the “include your work and thoughts on the problem” part?
    $endgroup$
    – José Carlos Santos
    Jan 10 at 13:57










  • $begingroup$
    f(0) = g(0). What is f(-1)?
    $endgroup$
    – William Elliot
    Jan 11 at 2:58










  • $begingroup$
    I have edited the problem, now the function is defined only for the positive domain.
    $endgroup$
    – zhen chen
    Jan 11 at 6:53










  • $begingroup$
    I found my question is very similar to this one: link, the author provide a version of proof.
    $endgroup$
    – zhen chen
    Jan 11 at 12:38




















  • $begingroup$
    Thanks for the comments. I have edited this question. Is it qualified now?
    $endgroup$
    – zhen chen
    Jan 10 at 13:48






  • 3




    $begingroup$
    Not really. Where is the “include your work and thoughts on the problem” part?
    $endgroup$
    – José Carlos Santos
    Jan 10 at 13:57










  • $begingroup$
    f(0) = g(0). What is f(-1)?
    $endgroup$
    – William Elliot
    Jan 11 at 2:58










  • $begingroup$
    I have edited the problem, now the function is defined only for the positive domain.
    $endgroup$
    – zhen chen
    Jan 11 at 6:53










  • $begingroup$
    I found my question is very similar to this one: link, the author provide a version of proof.
    $endgroup$
    – zhen chen
    Jan 11 at 12:38


















$begingroup$
Thanks for the comments. I have edited this question. Is it qualified now?
$endgroup$
– zhen chen
Jan 10 at 13:48




$begingroup$
Thanks for the comments. I have edited this question. Is it qualified now?
$endgroup$
– zhen chen
Jan 10 at 13:48




3




3




$begingroup$
Not really. Where is the “include your work and thoughts on the problem” part?
$endgroup$
– José Carlos Santos
Jan 10 at 13:57




$begingroup$
Not really. Where is the “include your work and thoughts on the problem” part?
$endgroup$
– José Carlos Santos
Jan 10 at 13:57












$begingroup$
f(0) = g(0). What is f(-1)?
$endgroup$
– William Elliot
Jan 11 at 2:58




$begingroup$
f(0) = g(0). What is f(-1)?
$endgroup$
– William Elliot
Jan 11 at 2:58












$begingroup$
I have edited the problem, now the function is defined only for the positive domain.
$endgroup$
– zhen chen
Jan 11 at 6:53




$begingroup$
I have edited the problem, now the function is defined only for the positive domain.
$endgroup$
– zhen chen
Jan 11 at 6:53












$begingroup$
I found my question is very similar to this one: link, the author provide a version of proof.
$endgroup$
– zhen chen
Jan 11 at 12:38






$begingroup$
I found my question is very similar to this one: link, the author provide a version of proof.
$endgroup$
– zhen chen
Jan 11 at 12:38












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