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Positive directional derivatives on sphere
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Let $f:mathbbR^mrightarrowmathbbR$ be continuous such that all the directional derivatives exist in $mathbbR^m$. If $fracpartial fpartial u(u)>0$ for every $uin S^m-1$, then there exists $ain mathbbR^m$ such that $fracpartial fpartial v(a)=0,,forall v.$
Tried using MVT for the straight lines through zero, but we don't have that the partial derivatives are continuous... Any tips?
multivariable-calculus partial-derivative
asked Jul 22 at 18:55
MathNewbie
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up vote
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Let $f:mathbbR^mrightarrowmathbbR$ be continuous such that all the directional derivatives exist in $mathbbR^m$. If $fracpartial fpartial u(u)>0$ for every $uin S^m-1$, then there exists $ain mathbbR^m$ such that $fracpartial fpartial v(a)=0,,forall v.$
Tried using MVT for the straight lines through zero, but we don't have that the partial derivatives are continuous... Any tips?
multivariable-calculus partial-derivative
asked Jul 22 at 18:55
MathNewbie
232211
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f:mathbbR^mrightarrowmathbbR$ be continuous such that all the directional derivatives exist in $mathbbR^m$. If $fracpartial fpartial u(u)>0$ for every $uin S^m-1$, then there exists $ain mathbbR^m$ such that $fracpartial fpartial v(a)=0,,forall v.$
Tried using MVT for the straight lines through zero, but we don't have that the partial derivatives are continuous... Any tips?
multivariable-calculus partial-derivative
asked Jul 22 at 18:55
MathNewbie
232211
Let $f:mathbbR^mrightarrowmathbbR$ be continuous such that all the directional derivatives exist in $mathbbR^m$. If $fracpartial fpartial u(u)>0$ for every $uin S^m-1$, then there exists $ain mathbbR^m$ such that $fracpartial fpartial v(a)=0,,forall v.$
Tried using MVT for the straight lines through zero, but we don't have that the partial derivatives are continuous... Any tips?
multivariable-calculus partial-derivative
asked Jul 22 at 18:55
MathNewbie
232211
asked Jul 22 at 18:55
MathNewbie
232211
asked Jul 22 at 18:55
MathNewbie
232211
asked Jul 22 at 18:55
MathNewbie
232211
232211
add a comment |Â
add a comment |Â
1 Answer
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1
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Denote by $m$ the least value of (a continuous) $f$ on the (compact) unit ball $B^m$. $m$ is attained somewhere on $B^m$.
Suppose to the contrary that there is $u in S^m - 1$ such that $f(u) = m$. From the assumption it follows that $f(u - h u) < m$ for $h > 0$ sufficiently small. But $u - h u in B^m$, which contradicts the choice of $m$.
Therefore, there is $a in B^m setminus S^m - 1$ such that $f(a) = m$. So $f$ has a (local) minimum at $a$. All directional derivatives of $f$ at $a$ must thus be zero.
answered Jul 22 at 21:03
user539887
1,4591313
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Denote by $m$ the least value of (a continuous) $f$ on the (compact) unit ball $B^m$. $m$ is attained somewhere on $B^m$.
Suppose to the contrary that there is $u in S^m - 1$ such that $f(u) = m$. From the assumption it follows that $f(u - h u) < m$ for $h > 0$ sufficiently small. But $u - h u in B^m$, which contradicts the choice of $m$.
Therefore, there is $a in B^m setminus S^m - 1$ such that $f(a) = m$. So $f$ has a (local) minimum at $a$. All directional derivatives of $f$ at $a$ must thus be zero.
answered Jul 22 at 21:03
user539887
1,4591313
add a comment |Â
up vote
1
down vote
accepted
Denote by $m$ the least value of (a continuous) $f$ on the (compact) unit ball $B^m$. $m$ is attained somewhere on $B^m$.
Suppose to the contrary that there is $u in S^m - 1$ such that $f(u) = m$. From the assumption it follows that $f(u - h u) < m$ for $h > 0$ sufficiently small. But $u - h u in B^m$, which contradicts the choice of $m$.
Therefore, there is $a in B^m setminus S^m - 1$ such that $f(a) = m$. So $f$ has a (local) minimum at $a$. All directional derivatives of $f$ at $a$ must thus be zero.
answered Jul 22 at 21:03
user539887
1,4591313
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Denote by $m$ the least value of (a continuous) $f$ on the (compact) unit ball $B^m$. $m$ is attained somewhere on $B^m$.
Suppose to the contrary that there is $u in S^m - 1$ such that $f(u) = m$. From the assumption it follows that $f(u - h u) < m$ for $h > 0$ sufficiently small. But $u - h u in B^m$, which contradicts the choice of $m$.
Therefore, there is $a in B^m setminus S^m - 1$ such that $f(a) = m$. So $f$ has a (local) minimum at $a$. All directional derivatives of $f$ at $a$ must thus be zero.
answered Jul 22 at 21:03
user539887
1,4591313
Denote by $m$ the least value of (a continuous) $f$ on the (compact) unit ball $B^m$. $m$ is attained somewhere on $B^m$.
Suppose to the contrary that there is $u in S^m - 1$ such that $f(u) = m$. From the assumption it follows that $f(u - h u) < m$ for $h > 0$ sufficiently small. But $u - h u in B^m$, which contradicts the choice of $m$.
Therefore, there is $a in B^m setminus S^m - 1$ such that $f(a) = m$. So $f$ has a (local) minimum at $a$. All directional derivatives of $f$ at $a$ must thus be zero.
answered Jul 22 at 21:03
user539887
1,4591313
answered Jul 22 at 21:03
user539887
1,4591313
answered Jul 22 at 21:03
user539887
1,4591313
answered Jul 22 at 21:03
user539887
1,4591313
1,4591313
add a comment |Â
add a comment |Â
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