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Suppose $fin C^2[mathbbR^2] $ , it holds $fracpartial^2fpartial x^2(x,y)+fracpartial^2fpartial y^2(x,y) gt0 $
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Suppose $fin C^2[mathbbR^2] $ , and for every $(x,y)in mathbbR^2$, it holds
$$fracpartial^2fpartial x^2(x,y)+fracpartial^2fpartial y^2(x,y) gt0 $$
Prove there is no extremum (i.e. local maximum ) for $f$.
I suppose we could done this like give an arbitrary point, we can always find another point in a circle around it to have lower or greater value. But the space is $mathbbR^2$, I failed in convert the condition $fracpartial^2fpartial x^2(x,y)+fracpartial^2fpartial y^2(x,y) gt0 $ into something useful.
calculus
asked Jul 21 at 5:16
Jaqen Chou
1476
add a comment |Â
up vote
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favorite
Suppose $fin C^2[mathbbR^2] $ , and for every $(x,y)in mathbbR^2$, it holds
$$fracpartial^2fpartial x^2(x,y)+fracpartial^2fpartial y^2(x,y) gt0 $$
Prove there is no extremum (i.e. local maximum ) for $f$.
I suppose we could done this like give an arbitrary point, we can always find another point in a circle around it to have lower or greater value. But the space is $mathbbR^2$, I failed in convert the condition $fracpartial^2fpartial x^2(x,y)+fracpartial^2fpartial y^2(x,y) gt0 $ into something useful.
calculus
asked Jul 21 at 5:16
Jaqen Chou
1476
2
if $f(x,y) = x^2+y^2$, then it has a global minimum at $(0,0).$ Perhaps, you miscopied the question?
– dezdichado
Jul 21 at 5:27
@ dezdichado Forgive me. I'm a terrible English user. I took 'extremum' to mean the 'local maximum'.
– Jaqen Chou
Jul 21 at 11:40
If you know how to do it in an open domain in $mathbbR^2$, then you are basically done. Suppose $f$ attains a local maximum at $p$. Then, you can find a circle centered at $p$ in which $f(p)$ is the maximum. But you have claimed to proven that is impossible, no?
– dezdichado
Jul 21 at 19:37
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose $fin C^2[mathbbR^2] $ , and for every $(x,y)in mathbbR^2$, it holds
$$fracpartial^2fpartial x^2(x,y)+fracpartial^2fpartial y^2(x,y) gt0 $$
Prove there is no extremum (i.e. local maximum ) for $f$.
I suppose we could done this like give an arbitrary point, we can always find another point in a circle around it to have lower or greater value. But the space is $mathbbR^2$, I failed in convert the condition $fracpartial^2fpartial x^2(x,y)+fracpartial^2fpartial y^2(x,y) gt0 $ into something useful.
calculus
asked Jul 21 at 5:16
Jaqen Chou
1476
Suppose $fin C^2[mathbbR^2] $ , and for every $(x,y)in mathbbR^2$, it holds
$$fracpartial^2fpartial x^2(x,y)+fracpartial^2fpartial y^2(x,y) gt0 $$
Prove there is no extremum (i.e. local maximum ) for $f$.
I suppose we could done this like give an arbitrary point, we can always find another point in a circle around it to have lower or greater value. But the space is $mathbbR^2$, I failed in convert the condition $fracpartial^2fpartial x^2(x,y)+fracpartial^2fpartial y^2(x,y) gt0 $ into something useful.
calculus
asked Jul 21 at 5:16
Jaqen Chou
1476
edited Jul 21 at 11:25
asked Jul 21 at 5:16
Jaqen Chou
1476
asked Jul 21 at 5:16
Jaqen Chou
1476
asked Jul 21 at 5:16
Jaqen Chou
1476
1476
2
if $f(x,y) = x^2+y^2$, then it has a global minimum at $(0,0).$ Perhaps, you miscopied the question?
– dezdichado
Jul 21 at 5:27
@ dezdichado Forgive me. I'm a terrible English user. I took 'extremum' to mean the 'local maximum'.
– Jaqen Chou
Jul 21 at 11:40
If you know how to do it in an open domain in $mathbbR^2$, then you are basically done. Suppose $f$ attains a local maximum at $p$. Then, you can find a circle centered at $p$ in which $f(p)$ is the maximum. But you have claimed to proven that is impossible, no?
– dezdichado
Jul 21 at 19:37
add a comment |Â
2
if $f(x,y) = x^2+y^2$, then it has a global minimum at $(0,0).$ Perhaps, you miscopied the question?
– dezdichado
Jul 21 at 5:27
@ dezdichado Forgive me. I'm a terrible English user. I took 'extremum' to mean the 'local maximum'.
– Jaqen Chou
Jul 21 at 11:40
If you know how to do it in an open domain in $mathbbR^2$, then you are basically done. Suppose $f$ attains a local maximum at $p$. Then, you can find a circle centered at $p$ in which $f(p)$ is the maximum. But you have claimed to proven that is impossible, no?
– dezdichado
Jul 21 at 19:37
2
2
if $f(x,y) = x^2+y^2$, then it has a global minimum at $(0,0).$ Perhaps, you miscopied the question?
– dezdichado
Jul 21 at 5:27
if $f(x,y) = x^2+y^2$, then it has a global minimum at $(0,0).$ Perhaps, you miscopied the question?
– dezdichado
Jul 21 at 5:27
@ dezdichado Forgive me. I'm a terrible English user. I took 'extremum' to mean the 'local maximum'.
– Jaqen Chou
Jul 21 at 11:40
@ dezdichado Forgive me. I'm a terrible English user. I took 'extremum' to mean the 'local maximum'.
– Jaqen Chou
Jul 21 at 11:40
If you know how to do it in an open domain in $mathbbR^2$, then you are basically done. Suppose $f$ attains a local maximum at $p$. Then, you can find a circle centered at $p$ in which $f(p)$ is the maximum. But you have claimed to proven that is impossible, no?
– dezdichado
Jul 21 at 19:37
If you know how to do it in an open domain in $mathbbR^2$, then you are basically done. Suppose $f$ attains a local maximum at $p$. Then, you can find a circle centered at $p$ in which $f(p)$ is the maximum. But you have claimed to proven that is impossible, no?
– dezdichado
Jul 21 at 19:37
add a comment |Â
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2
if $f(x,y) = x^2+y^2$, then it has a global minimum at $(0,0).$ Perhaps, you miscopied the question?
– dezdichado
Jul 21 at 5:27
@ dezdichado Forgive me. I'm a terrible English user. I took 'extremum' to mean the 'local maximum'.
– Jaqen Chou
Jul 21 at 11:40
If you know how to do it in an open domain in $mathbbR^2$, then you are basically done. Suppose $f$ attains a local maximum at $p$. Then, you can find a circle centered at $p$ in which $f(p)$ is the maximum. But you have claimed to proven that is impossible, no?
– dezdichado
Jul 21 at 19:37