If, for every $a in partial U$, $lim_x to af(x) = 0$, prove that there is at least a critical point of $f$ in $U$.
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Let $f: U to mathbbR$ be a differentiable function in the open $U subset mathbbR^n$. If, for every $a in partial U$, $lim_x to af(x) = 0$, prove that there is at least a critical point of $f$ in $U$.
I don't know what to do. I cannot use $lim_x to af(x) = f(a)$ because $a in partial U$. I know that $a = lim x_n$ with $x_n in U$ and I'm trying to figure out how to use it. Before, I tried to show that $U$ has a maximum or minimum point, but I couldnot develop. Can someone help me?
real-analysis continuity
asked Aug 8 at 21:35
Lucas Corrêa
1,211319
add a comment |Â
up vote
0
down vote
favorite
Let $f: U to mathbbR$ be a differentiable function in the open $U subset mathbbR^n$. If, for every $a in partial U$, $lim_x to af(x) = 0$, prove that there is at least a critical point of $f$ in $U$.
I don't know what to do. I cannot use $lim_x to af(x) = f(a)$ because $a in partial U$. I know that $a = lim x_n$ with $x_n in U$ and I'm trying to figure out how to use it. Before, I tried to show that $U$ has a maximum or minimum point, but I couldnot develop. Can someone help me?
real-analysis continuity
asked Aug 8 at 21:35
Lucas Corrêa
1,211319
2
The claim is false unless you have some additional condition on $U.$ For example, $f:U=mathbbRto mathbbR, xto x$ has no critical points. Is $U$ bounded?
– mfl
Aug 8 at 21:38
@mfl, thank you! The question is written as I announced, but is possible establish some condition (not too restrictive) over U so that this result is true?
– Lucas Corrêa
Aug 8 at 21:48
As @JoséCarlosSantos said in a comment below the result holds if $U$ is bounded.
– mfl
Aug 8 at 21:50
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f: U to mathbbR$ be a differentiable function in the open $U subset mathbbR^n$. If, for every $a in partial U$, $lim_x to af(x) = 0$, prove that there is at least a critical point of $f$ in $U$.
I don't know what to do. I cannot use $lim_x to af(x) = f(a)$ because $a in partial U$. I know that $a = lim x_n$ with $x_n in U$ and I'm trying to figure out how to use it. Before, I tried to show that $U$ has a maximum or minimum point, but I couldnot develop. Can someone help me?
real-analysis continuity
asked Aug 8 at 21:35
Lucas Corrêa
1,211319
Let $f: U to mathbbR$ be a differentiable function in the open $U subset mathbbR^n$. If, for every $a in partial U$, $lim_x to af(x) = 0$, prove that there is at least a critical point of $f$ in $U$.
I don't know what to do. I cannot use $lim_x to af(x) = f(a)$ because $a in partial U$. I know that $a = lim x_n$ with $x_n in U$ and I'm trying to figure out how to use it. Before, I tried to show that $U$ has a maximum or minimum point, but I couldnot develop. Can someone help me?
real-analysis continuity
asked Aug 8 at 21:35
Lucas Corrêa
1,211319
asked Aug 8 at 21:35
Lucas Corrêa
1,211319
asked Aug 8 at 21:35
Lucas Corrêa
1,211319
asked Aug 8 at 21:35
Lucas Corrêa
1,211319
1,211319
2
The claim is false unless you have some additional condition on $U.$ For example, $f:U=mathbbRto mathbbR, xto x$ has no critical points. Is $U$ bounded?
– mfl
Aug 8 at 21:38
@mfl, thank you! The question is written as I announced, but is possible establish some condition (not too restrictive) over U so that this result is true?
– Lucas Corrêa
Aug 8 at 21:48
As @JoséCarlosSantos said in a comment below the result holds if $U$ is bounded.
– mfl
Aug 8 at 21:50
add a comment |Â
2
The claim is false unless you have some additional condition on $U.$ For example, $f:U=mathbbRto mathbbR, xto x$ has no critical points. Is $U$ bounded?
– mfl
Aug 8 at 21:38
@mfl, thank you! The question is written as I announced, but is possible establish some condition (not too restrictive) over U so that this result is true?
– Lucas Corrêa
Aug 8 at 21:48
As @JoséCarlosSantos said in a comment below the result holds if $U$ is bounded.
– mfl
Aug 8 at 21:50
2
2
The claim is false unless you have some additional condition on $U.$ For example, $f:U=mathbbRto mathbbR, xto x$ has no critical points. Is $U$ bounded?
– mfl
Aug 8 at 21:38
The claim is false unless you have some additional condition on $U.$ For example, $f:U=mathbbRto mathbbR, xto x$ has no critical points. Is $U$ bounded?
– mfl
Aug 8 at 21:38
@mfl, thank you! The question is written as I announced, but is possible establish some condition (not too restrictive) over U so that this result is true?
– Lucas Corrêa
Aug 8 at 21:48
@mfl, thank you! The question is written as I announced, but is possible establish some condition (not too restrictive) over U so that this result is true?
– Lucas Corrêa
Aug 8 at 21:48
As @JoséCarlosSantos said in a comment below the result holds if $U$ is bounded.
– mfl
Aug 8 at 21:50
As @JoséCarlosSantos said in a comment below the result holds if $U$ is bounded.
– mfl
Aug 8 at 21:50
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
This is not true. Take $U=,x>0$ and let $f(x,y)=x$. Then $partial U=(0,y),$ and$$(forall ainpartial U):lim_Xto af(X)=0.$$However, $f$ has no critical point.
answered Aug 8 at 21:41
José Carlos Santos
115k1699177
Thank you! Is possible establish some condition (not too restrictive) over $U$ so that this result is true?
– Lucas Corrêa
Aug 8 at 21:46
2
@LucasCorrêa Sure. If $U$ is bounded, then the statement is true.
– José Carlos Santos
Aug 8 at 21:48
Assuming that $U$ is bounded, I can ensure, at least, that there is critical points in $partial U$. How can I use $lim_x to af(x) = 0$ (if $a$ is the maximum or minimum, for example) for show that $f$ has a critical points in $U$?
– Lucas Corrêa
Aug 8 at 21:56
1
@LucasCorrêa Extend $f$ to $overline U$ defining $f(p)=0$ when $pinpartial U$. Then $f$ is a continuous function with compact domain. If it is the null function, then all points of $U$ is a critical point. Otherwise, eith the maximum of f$ is positive or the minimum is negative. Take the point at which the maximum or the minimum is attained.
– José Carlos Santos
Aug 8 at 22:15
add a comment |Â
up vote
1
down vote
If $U$ is bounded, then it is possible to extend $f$ to the boundary, using the condition, so that the extension is still continuous (define it as $0$ on the boundary). The domain of the extended function is still bounded, so the extended function attains its maximum $M$ and minimum $m$. If $M=m=0$, then $f=0$, so it has critical points (every point of $U$).
So WLOG assume that $M>0$. Then the maximum is attained at a point of $U$, so it has to be a critical point of $f$.
answered Aug 8 at 22:07
A. Pongrácz
3,562623
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
This is not true. Take $U=,x>0$ and let $f(x,y)=x$. Then $partial U=(0,y),$ and$$(forall ainpartial U):lim_Xto af(X)=0.$$However, $f$ has no critical point.
answered Aug 8 at 21:41
José Carlos Santos
115k1699177
Thank you! Is possible establish some condition (not too restrictive) over $U$ so that this result is true?
– Lucas Corrêa
Aug 8 at 21:46
2
@LucasCorrêa Sure. If $U$ is bounded, then the statement is true.
– José Carlos Santos
Aug 8 at 21:48
Assuming that $U$ is bounded, I can ensure, at least, that there is critical points in $partial U$. How can I use $lim_x to af(x) = 0$ (if $a$ is the maximum or minimum, for example) for show that $f$ has a critical points in $U$?
– Lucas Corrêa
Aug 8 at 21:56
1
@LucasCorrêa Extend $f$ to $overline U$ defining $f(p)=0$ when $pinpartial U$. Then $f$ is a continuous function with compact domain. If it is the null function, then all points of $U$ is a critical point. Otherwise, eith the maximum of f$ is positive or the minimum is negative. Take the point at which the maximum or the minimum is attained.
– José Carlos Santos
Aug 8 at 22:15
add a comment |Â
up vote
3
down vote
accepted
This is not true. Take $U=,x>0$ and let $f(x,y)=x$. Then $partial U=(0,y),$ and$$(forall ainpartial U):lim_Xto af(X)=0.$$However, $f$ has no critical point.
answered Aug 8 at 21:41
José Carlos Santos
115k1699177
Thank you! Is possible establish some condition (not too restrictive) over $U$ so that this result is true?
– Lucas Corrêa
Aug 8 at 21:46
2
@LucasCorrêa Sure. If $U$ is bounded, then the statement is true.
– José Carlos Santos
Aug 8 at 21:48
Assuming that $U$ is bounded, I can ensure, at least, that there is critical points in $partial U$. How can I use $lim_x to af(x) = 0$ (if $a$ is the maximum or minimum, for example) for show that $f$ has a critical points in $U$?
– Lucas Corrêa
Aug 8 at 21:56
1
@LucasCorrêa Extend $f$ to $overline U$ defining $f(p)=0$ when $pinpartial U$. Then $f$ is a continuous function with compact domain. If it is the null function, then all points of $U$ is a critical point. Otherwise, eith the maximum of f$ is positive or the minimum is negative. Take the point at which the maximum or the minimum is attained.
– José Carlos Santos
Aug 8 at 22:15
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
This is not true. Take $U=,x>0$ and let $f(x,y)=x$. Then $partial U=(0,y),$ and$$(forall ainpartial U):lim_Xto af(X)=0.$$However, $f$ has no critical point.
answered Aug 8 at 21:41
José Carlos Santos
115k1699177
This is not true. Take $U=,x>0$ and let $f(x,y)=x$. Then $partial U=(0,y),$ and$$(forall ainpartial U):lim_Xto af(X)=0.$$However, $f$ has no critical point.
answered Aug 8 at 21:41
José Carlos Santos
115k1699177
answered Aug 8 at 21:41
José Carlos Santos
115k1699177
answered Aug 8 at 21:41
José Carlos Santos
115k1699177
answered Aug 8 at 21:41
José Carlos Santos
115k1699177
115k1699177
Thank you! Is possible establish some condition (not too restrictive) over $U$ so that this result is true?
– Lucas Corrêa
Aug 8 at 21:46
2
@LucasCorrêa Sure. If $U$ is bounded, then the statement is true.
– José Carlos Santos
Aug 8 at 21:48
Assuming that $U$ is bounded, I can ensure, at least, that there is critical points in $partial U$. How can I use $lim_x to af(x) = 0$ (if $a$ is the maximum or minimum, for example) for show that $f$ has a critical points in $U$?
– Lucas Corrêa
Aug 8 at 21:56
1
@LucasCorrêa Extend $f$ to $overline U$ defining $f(p)=0$ when $pinpartial U$. Then $f$ is a continuous function with compact domain. If it is the null function, then all points of $U$ is a critical point. Otherwise, eith the maximum of f$ is positive or the minimum is negative. Take the point at which the maximum or the minimum is attained.
– José Carlos Santos
Aug 8 at 22:15
add a comment |Â
Thank you! Is possible establish some condition (not too restrictive) over $U$ so that this result is true?
– Lucas Corrêa
Aug 8 at 21:46
2
@LucasCorrêa Sure. If $U$ is bounded, then the statement is true.
– José Carlos Santos
Aug 8 at 21:48
Assuming that $U$ is bounded, I can ensure, at least, that there is critical points in $partial U$. How can I use $lim_x to af(x) = 0$ (if $a$ is the maximum or minimum, for example) for show that $f$ has a critical points in $U$?
– Lucas Corrêa
Aug 8 at 21:56
1
@LucasCorrêa Extend $f$ to $overline U$ defining $f(p)=0$ when $pinpartial U$. Then $f$ is a continuous function with compact domain. If it is the null function, then all points of $U$ is a critical point. Otherwise, eith the maximum of f$ is positive or the minimum is negative. Take the point at which the maximum or the minimum is attained.
– José Carlos Santos
Aug 8 at 22:15
Thank you! Is possible establish some condition (not too restrictive) over $U$ so that this result is true?
– Lucas Corrêa
Aug 8 at 21:46
Thank you! Is possible establish some condition (not too restrictive) over $U$ so that this result is true?
– Lucas Corrêa
Aug 8 at 21:46
2
2
@LucasCorrêa Sure. If $U$ is bounded, then the statement is true.
– José Carlos Santos
Aug 8 at 21:48
@LucasCorrêa Sure. If $U$ is bounded, then the statement is true.
– José Carlos Santos
Aug 8 at 21:48
Assuming that $U$ is bounded, I can ensure, at least, that there is critical points in $partial U$. How can I use $lim_x to af(x) = 0$ (if $a$ is the maximum or minimum, for example) for show that $f$ has a critical points in $U$?
– Lucas Corrêa
Aug 8 at 21:56
Assuming that $U$ is bounded, I can ensure, at least, that there is critical points in $partial U$. How can I use $lim_x to af(x) = 0$ (if $a$ is the maximum or minimum, for example) for show that $f$ has a critical points in $U$?
– Lucas Corrêa
Aug 8 at 21:56
1
1
@LucasCorrêa Extend $f$ to $overline U$ defining $f(p)=0$ when $pinpartial U$. Then $f$ is a continuous function with compact domain. If it is the null function, then all points of $U$ is a critical point. Otherwise, eith the maximum of f$ is positive or the minimum is negative. Take the point at which the maximum or the minimum is attained.
– José Carlos Santos
Aug 8 at 22:15
@LucasCorrêa Extend $f$ to $overline U$ defining $f(p)=0$ when $pinpartial U$. Then $f$ is a continuous function with compact domain. If it is the null function, then all points of $U$ is a critical point. Otherwise, eith the maximum of f$ is positive or the minimum is negative. Take the point at which the maximum or the minimum is attained.
– José Carlos Santos
Aug 8 at 22:15
add a comment |Â
up vote
1
down vote
If $U$ is bounded, then it is possible to extend $f$ to the boundary, using the condition, so that the extension is still continuous (define it as $0$ on the boundary). The domain of the extended function is still bounded, so the extended function attains its maximum $M$ and minimum $m$. If $M=m=0$, then $f=0$, so it has critical points (every point of $U$).
So WLOG assume that $M>0$. Then the maximum is attained at a point of $U$, so it has to be a critical point of $f$.
answered Aug 8 at 22:07
A. Pongrácz
3,562623
add a comment |Â
up vote
1
down vote
If $U$ is bounded, then it is possible to extend $f$ to the boundary, using the condition, so that the extension is still continuous (define it as $0$ on the boundary). The domain of the extended function is still bounded, so the extended function attains its maximum $M$ and minimum $m$. If $M=m=0$, then $f=0$, so it has critical points (every point of $U$).
So WLOG assume that $M>0$. Then the maximum is attained at a point of $U$, so it has to be a critical point of $f$.
answered Aug 8 at 22:07
A. Pongrácz
3,562623
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If $U$ is bounded, then it is possible to extend $f$ to the boundary, using the condition, so that the extension is still continuous (define it as $0$ on the boundary). The domain of the extended function is still bounded, so the extended function attains its maximum $M$ and minimum $m$. If $M=m=0$, then $f=0$, so it has critical points (every point of $U$).
So WLOG assume that $M>0$. Then the maximum is attained at a point of $U$, so it has to be a critical point of $f$.
answered Aug 8 at 22:07
A. Pongrácz
3,562623
If $U$ is bounded, then it is possible to extend $f$ to the boundary, using the condition, so that the extension is still continuous (define it as $0$ on the boundary). The domain of the extended function is still bounded, so the extended function attains its maximum $M$ and minimum $m$. If $M=m=0$, then $f=0$, so it has critical points (every point of $U$).
So WLOG assume that $M>0$. Then the maximum is attained at a point of $U$, so it has to be a critical point of $f$.
answered Aug 8 at 22:07
A. Pongrácz
3,562623
answered Aug 8 at 22:07
A. Pongrácz
3,562623
answered Aug 8 at 22:07
A. Pongrácz
3,562623
answered Aug 8 at 22:07
A. Pongrácz
3,562623
3,562623
add a comment |Â
add a comment |Â
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2
The claim is false unless you have some additional condition on $U.$ For example, $f:U=mathbbRto mathbbR, xto x$ has no critical points. Is $U$ bounded?
– mfl
Aug 8 at 21:38
@mfl, thank you! The question is written as I announced, but is possible establish some condition (not too restrictive) over U so that this result is true?
– Lucas Corrêa
Aug 8 at 21:48
As @JoséCarlosSantos said in a comment below the result holds if $U$ is bounded.
– mfl
Aug 8 at 21:50