Convex and continuous function on compact set implies Lipschitz
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Let the function $f: C rightarrow mathbb{R}$ be convex and continuous, where $C subset mathbb{R}^n$ is a compact set.
Prove or disprove that $f$ is Lipschitz continuous on $C$.
Comments: If $f$ were defined on an open set $O$, then I can show that it is Lipschitz continuous on every compact subset of $O$, see Proving that a convex function is Lipschitz for the scalar case. In this case, however, I need to exploit the assumption that $f$ is continuous also on the boundary of $C$ - continuity of $f$ in the interior of $C$ already follows from the convexity assumption.
real-analysis convex-analysis compactness lipschitz-functions
edited Apr 13 '17 at 12:21
Community♦
1
asked Mar 3 '16 at 18:12
user284439user284439
425
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add a comment |
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Let the function $f: C rightarrow mathbb{R}$ be convex and continuous, where $C subset mathbb{R}^n$ is a compact set.
Prove or disprove that $f$ is Lipschitz continuous on $C$.
Comments: If $f$ were defined on an open set $O$, then I can show that it is Lipschitz continuous on every compact subset of $O$, see Proving that a convex function is Lipschitz for the scalar case. In this case, however, I need to exploit the assumption that $f$ is continuous also on the boundary of $C$ - continuity of $f$ in the interior of $C$ already follows from the convexity assumption.
real-analysis convex-analysis compactness lipschitz-functions
edited Apr 13 '17 at 12:21
Community♦
1
asked Mar 3 '16 at 18:12
user284439user284439
425
$endgroup$
3
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Take for simplicity $Csubset mathbb R^1$, where $C=[-1,1]$ and $f$ to be the lower semicircle of the unit sphere in $mathbb R^2$, i.e $f=-sqrt{1-x^2}$. This function is convex and continuous, but not Lipschitz. The problem arises on the boundary of the set, where the derivatives can be unbounded.
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– Svetoslav
Mar 3 '16 at 18:18
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This is correct, thank you.
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– user284439
Mar 3 '16 at 19:48
add a comment |
$begingroup$
Let the function $f: C rightarrow mathbb{R}$ be convex and continuous, where $C subset mathbb{R}^n$ is a compact set.
Prove or disprove that $f$ is Lipschitz continuous on $C$.
Comments: If $f$ were defined on an open set $O$, then I can show that it is Lipschitz continuous on every compact subset of $O$, see Proving that a convex function is Lipschitz for the scalar case. In this case, however, I need to exploit the assumption that $f$ is continuous also on the boundary of $C$ - continuity of $f$ in the interior of $C$ already follows from the convexity assumption.
real-analysis convex-analysis compactness lipschitz-functions
edited Apr 13 '17 at 12:21
Community♦
1
asked Mar 3 '16 at 18:12
user284439user284439
425
$endgroup$
Let the function $f: C rightarrow mathbb{R}$ be convex and continuous, where $C subset mathbb{R}^n$ is a compact set.
Prove or disprove that $f$ is Lipschitz continuous on $C$.
Comments: If $f$ were defined on an open set $O$, then I can show that it is Lipschitz continuous on every compact subset of $O$, see Proving that a convex function is Lipschitz for the scalar case. In this case, however, I need to exploit the assumption that $f$ is continuous also on the boundary of $C$ - continuity of $f$ in the interior of $C$ already follows from the convexity assumption.
real-analysis convex-analysis compactness lipschitz-functions
real-analysis convex-analysis compactness lipschitz-functions
edited Apr 13 '17 at 12:21
Community♦
1
asked Mar 3 '16 at 18:12
user284439user284439
425
edited Apr 13 '17 at 12:21
Community♦
1
asked Mar 3 '16 at 18:12
user284439user284439
425
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
asked Mar 3 '16 at 18:12
user284439user284439
425
asked Mar 3 '16 at 18:12
user284439user284439
425
asked Mar 3 '16 at 18:12
user284439user284439
425
425
3
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Take for simplicity $Csubset mathbb R^1$, where $C=[-1,1]$ and $f$ to be the lower semicircle of the unit sphere in $mathbb R^2$, i.e $f=-sqrt{1-x^2}$. This function is convex and continuous, but not Lipschitz. The problem arises on the boundary of the set, where the derivatives can be unbounded.
$endgroup$
– Svetoslav
Mar 3 '16 at 18:18
$begingroup$
This is correct, thank you.
$endgroup$
– user284439
Mar 3 '16 at 19:48
add a comment |
3
$begingroup$
Take for simplicity $Csubset mathbb R^1$, where $C=[-1,1]$ and $f$ to be the lower semicircle of the unit sphere in $mathbb R^2$, i.e $f=-sqrt{1-x^2}$. This function is convex and continuous, but not Lipschitz. The problem arises on the boundary of the set, where the derivatives can be unbounded.
$endgroup$
– Svetoslav
Mar 3 '16 at 18:18
$begingroup$
This is correct, thank you.
$endgroup$
– user284439
Mar 3 '16 at 19:48
3
3
$begingroup$
Take for simplicity $Csubset mathbb R^1$, where $C=[-1,1]$ and $f$ to be the lower semicircle of the unit sphere in $mathbb R^2$, i.e $f=-sqrt{1-x^2}$. This function is convex and continuous, but not Lipschitz. The problem arises on the boundary of the set, where the derivatives can be unbounded.
$endgroup$
– Svetoslav
Mar 3 '16 at 18:18
$begingroup$
Take for simplicity $Csubset mathbb R^1$, where $C=[-1,1]$ and $f$ to be the lower semicircle of the unit sphere in $mathbb R^2$, i.e $f=-sqrt{1-x^2}$. This function is convex and continuous, but not Lipschitz. The problem arises on the boundary of the set, where the derivatives can be unbounded.
$endgroup$
– Svetoslav
Mar 3 '16 at 18:18
$begingroup$
This is correct, thank you.
$endgroup$
– user284439
Mar 3 '16 at 19:48
$begingroup$
This is correct, thank you.
$endgroup$
– user284439
Mar 3 '16 at 19:48
add a comment |
2 Answers
2
active
oldest
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As noted by Svetoslav in a comment, the statement is false: a counterexample is given by $f(x) = -sqrt{1-x^2}$ on $[-1,1]$.
A slightly simpler counterexample is $f(x) = -sqrt{x}$ on $[0,1]$.
In both cases, the issue is that $f'$ is not bounded. A function with unbounded derivative cannot be Lipschitz.
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add a comment |
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This statement is true if instead of dealing with arbitrary compact subsets, you deal with compact subsets of the interior of the domain. This is Theorem 3.3.1 in these notes.
answered Dec 22 '17 at 22:29
Alfred YergerAlfred Yerger
10.6k2249
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2 Answers
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$begingroup$
As noted by Svetoslav in a comment, the statement is false: a counterexample is given by $f(x) = -sqrt{1-x^2}$ on $[-1,1]$.
A slightly simpler counterexample is $f(x) = -sqrt{x}$ on $[0,1]$.
In both cases, the issue is that $f'$ is not bounded. A function with unbounded derivative cannot be Lipschitz.
$endgroup$
add a comment |
$begingroup$
As noted by Svetoslav in a comment, the statement is false: a counterexample is given by $f(x) = -sqrt{1-x^2}$ on $[-1,1]$.
A slightly simpler counterexample is $f(x) = -sqrt{x}$ on $[0,1]$.
In both cases, the issue is that $f'$ is not bounded. A function with unbounded derivative cannot be Lipschitz.
$endgroup$
add a comment |
$begingroup$
As noted by Svetoslav in a comment, the statement is false: a counterexample is given by $f(x) = -sqrt{1-x^2}$ on $[-1,1]$.
A slightly simpler counterexample is $f(x) = -sqrt{x}$ on $[0,1]$.
In both cases, the issue is that $f'$ is not bounded. A function with unbounded derivative cannot be Lipschitz.
$endgroup$
As noted by Svetoslav in a comment, the statement is false: a counterexample is given by $f(x) = -sqrt{1-x^2}$ on $[-1,1]$.
A slightly simpler counterexample is $f(x) = -sqrt{x}$ on $[0,1]$.
In both cases, the issue is that $f'$ is not bounded. A function with unbounded derivative cannot be Lipschitz.
answered Dec 5 '17 at 22:14
user357151
add a comment |
add a comment |
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This statement is true if instead of dealing with arbitrary compact subsets, you deal with compact subsets of the interior of the domain. This is Theorem 3.3.1 in these notes.
answered Dec 22 '17 at 22:29
Alfred YergerAlfred Yerger
10.6k2249
$endgroup$
add a comment |
$begingroup$
This statement is true if instead of dealing with arbitrary compact subsets, you deal with compact subsets of the interior of the domain. This is Theorem 3.3.1 in these notes.
answered Dec 22 '17 at 22:29
Alfred YergerAlfred Yerger
10.6k2249
$endgroup$
add a comment |
$begingroup$
This statement is true if instead of dealing with arbitrary compact subsets, you deal with compact subsets of the interior of the domain. This is Theorem 3.3.1 in these notes.
answered Dec 22 '17 at 22:29
Alfred YergerAlfred Yerger
10.6k2249
$endgroup$
This statement is true if instead of dealing with arbitrary compact subsets, you deal with compact subsets of the interior of the domain. This is Theorem 3.3.1 in these notes.
answered Dec 22 '17 at 22:29
Alfred YergerAlfred Yerger
10.6k2249
answered Dec 22 '17 at 22:29
Alfred YergerAlfred Yerger
10.6k2249
answered Dec 22 '17 at 22:29
Alfred YergerAlfred Yerger
10.6k2249
answered Dec 22 '17 at 22:29
Alfred YergerAlfred Yerger
10.6k2249
10.6k2249
add a comment |
add a comment |
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$begingroup$
Take for simplicity $Csubset mathbb R^1$, where $C=[-1,1]$ and $f$ to be the lower semicircle of the unit sphere in $mathbb R^2$, i.e $f=-sqrt{1-x^2}$. This function is convex and continuous, but not Lipschitz. The problem arises on the boundary of the set, where the derivatives can be unbounded.
$endgroup$
– Svetoslav
Mar 3 '16 at 18:18
$begingroup$
This is correct, thank you.
$endgroup$
– user284439
Mar 3 '16 at 19:48