If $V:mathbb{R}^2rightarrowmathbb{R}$ is $C^1$ and positive definite does its level curves form a continuum...
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Let $V:mathbb{R}^2rightarrowmathbb{R}$ be continuously differentiable and positive definite. I'm currently reading a book where the authors claim that $V$ has a continuum of closed level curves around the origin. While I certainly find this to be plausible it is not immediately evident to me. How is this proven? If the proof is extensive a simple sketch will suffice and I will do my best to fill in the gaps.
Note that in this context $V$ being positive definite means that $V(textbf{0})=0$ and $textbf{x}inmathbb{R}^2setminus{textbf{0}}Rightarrow V(textbf{x}) > 0$.
real-analysis calculus analysis multivariable-calculus functions
asked Dec 14 '18 at 19:15
DavidDavid
679416
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add a comment |
$begingroup$
Let $V:mathbb{R}^2rightarrowmathbb{R}$ be continuously differentiable and positive definite. I'm currently reading a book where the authors claim that $V$ has a continuum of closed level curves around the origin. While I certainly find this to be plausible it is not immediately evident to me. How is this proven? If the proof is extensive a simple sketch will suffice and I will do my best to fill in the gaps.
Note that in this context $V$ being positive definite means that $V(textbf{0})=0$ and $textbf{x}inmathbb{R}^2setminus{textbf{0}}Rightarrow V(textbf{x}) > 0$.
real-analysis calculus analysis multivariable-calculus functions
asked Dec 14 '18 at 19:15
DavidDavid
679416
$endgroup$
$begingroup$
Mind defining "continuum of curves"?
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– user25959
Dec 14 '18 at 20:04
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Is it something like, $exists$ neighborhood $B ni mathbf{0}$ such that if $V^{-1}(c) subset B$ then all $V^{-1}(d)$ with $d<c$ are contained in $B$, and each is a closed curve?
$endgroup$
– user25959
Dec 14 '18 at 20:19
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Something along those lines I suppose. The authors gave no strict definition.
$endgroup$
– David
Dec 14 '18 at 20:23
$begingroup$
Closed curve (i.e. topological $S^1$) is probably true close enough to the origin. Do you already know how to do something simpler: Show that there is some $c>0$ such that $V^{-1}(c)$ is compact.
$endgroup$
– user25959
Dec 14 '18 at 20:37
add a comment |
$begingroup$
Let $V:mathbb{R}^2rightarrowmathbb{R}$ be continuously differentiable and positive definite. I'm currently reading a book where the authors claim that $V$ has a continuum of closed level curves around the origin. While I certainly find this to be plausible it is not immediately evident to me. How is this proven? If the proof is extensive a simple sketch will suffice and I will do my best to fill in the gaps.
Note that in this context $V$ being positive definite means that $V(textbf{0})=0$ and $textbf{x}inmathbb{R}^2setminus{textbf{0}}Rightarrow V(textbf{x}) > 0$.
real-analysis calculus analysis multivariable-calculus functions
asked Dec 14 '18 at 19:15
DavidDavid
679416
$endgroup$
Let $V:mathbb{R}^2rightarrowmathbb{R}$ be continuously differentiable and positive definite. I'm currently reading a book where the authors claim that $V$ has a continuum of closed level curves around the origin. While I certainly find this to be plausible it is not immediately evident to me. How is this proven? If the proof is extensive a simple sketch will suffice and I will do my best to fill in the gaps.
Note that in this context $V$ being positive definite means that $V(textbf{0})=0$ and $textbf{x}inmathbb{R}^2setminus{textbf{0}}Rightarrow V(textbf{x}) > 0$.
real-analysis calculus analysis multivariable-calculus functions
real-analysis calculus analysis multivariable-calculus functions
asked Dec 14 '18 at 19:15
DavidDavid
679416
asked Dec 14 '18 at 19:15
DavidDavid
679416
asked Dec 14 '18 at 19:15
DavidDavid
679416
asked Dec 14 '18 at 19:15
DavidDavid
679416
asked Dec 14 '18 at 19:15
DavidDavid
679416
679416
$begingroup$
Mind defining "continuum of curves"?
$endgroup$
– user25959
Dec 14 '18 at 20:04
$begingroup$
Is it something like, $exists$ neighborhood $B ni mathbf{0}$ such that if $V^{-1}(c) subset B$ then all $V^{-1}(d)$ with $d<c$ are contained in $B$, and each is a closed curve?
$endgroup$
– user25959
Dec 14 '18 at 20:19
$begingroup$
Something along those lines I suppose. The authors gave no strict definition.
$endgroup$
– David
Dec 14 '18 at 20:23
$begingroup$
Closed curve (i.e. topological $S^1$) is probably true close enough to the origin. Do you already know how to do something simpler: Show that there is some $c>0$ such that $V^{-1}(c)$ is compact.
$endgroup$
– user25959
Dec 14 '18 at 20:37
add a comment |
$begingroup$
Mind defining "continuum of curves"?
$endgroup$
– user25959
Dec 14 '18 at 20:04
$begingroup$
Is it something like, $exists$ neighborhood $B ni mathbf{0}$ such that if $V^{-1}(c) subset B$ then all $V^{-1}(d)$ with $d<c$ are contained in $B$, and each is a closed curve?
$endgroup$
– user25959
Dec 14 '18 at 20:19
$begingroup$
Something along those lines I suppose. The authors gave no strict definition.
$endgroup$
– David
Dec 14 '18 at 20:23
$begingroup$
Closed curve (i.e. topological $S^1$) is probably true close enough to the origin. Do you already know how to do something simpler: Show that there is some $c>0$ such that $V^{-1}(c)$ is compact.
$endgroup$
– user25959
Dec 14 '18 at 20:37
$begingroup$
Mind defining "continuum of curves"?
$endgroup$
– user25959
Dec 14 '18 at 20:04
$begingroup$
Mind defining "continuum of curves"?
$endgroup$
– user25959
Dec 14 '18 at 20:04
$begingroup$
Is it something like, $exists$ neighborhood $B ni mathbf{0}$ such that if $V^{-1}(c) subset B$ then all $V^{-1}(d)$ with $d<c$ are contained in $B$, and each is a closed curve?
$endgroup$
– user25959
Dec 14 '18 at 20:19
$begingroup$
Is it something like, $exists$ neighborhood $B ni mathbf{0}$ such that if $V^{-1}(c) subset B$ then all $V^{-1}(d)$ with $d<c$ are contained in $B$, and each is a closed curve?
$endgroup$
– user25959
Dec 14 '18 at 20:19
$begingroup$
Something along those lines I suppose. The authors gave no strict definition.
$endgroup$
– David
Dec 14 '18 at 20:23
$begingroup$
Something along those lines I suppose. The authors gave no strict definition.
$endgroup$
– David
Dec 14 '18 at 20:23
$begingroup$
Closed curve (i.e. topological $S^1$) is probably true close enough to the origin. Do you already know how to do something simpler: Show that there is some $c>0$ such that $V^{-1}(c)$ is compact.
$endgroup$
– user25959
Dec 14 '18 at 20:37
$begingroup$
Closed curve (i.e. topological $S^1$) is probably true close enough to the origin. Do you already know how to do something simpler: Show that there is some $c>0$ such that $V^{-1}(c)$ is compact.
$endgroup$
– user25959
Dec 14 '18 at 20:37
add a comment |
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$begingroup$
Mind defining "continuum of curves"?
$endgroup$
– user25959
Dec 14 '18 at 20:04
$begingroup$
Is it something like, $exists$ neighborhood $B ni mathbf{0}$ such that if $V^{-1}(c) subset B$ then all $V^{-1}(d)$ with $d<c$ are contained in $B$, and each is a closed curve?
$endgroup$
– user25959
Dec 14 '18 at 20:19
$begingroup$
Something along those lines I suppose. The authors gave no strict definition.
$endgroup$
– David
Dec 14 '18 at 20:23
$begingroup$
Closed curve (i.e. topological $S^1$) is probably true close enough to the origin. Do you already know how to do something simpler: Show that there is some $c>0$ such that $V^{-1}(c)$ is compact.
$endgroup$
– user25959
Dec 14 '18 at 20:37