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Boundedness of a sublevel set of a convex function
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Let $f : mathbf R^n to mathbf R$ be a convex function with $f(0)=0$ and $displaystylelim_ttoinftyf(tx)=infty$ for any $xinmathbf R-0$. Is $$A:=x mid f(x)le 1 $$ bounded?
I was thinking about projecting $A$ onto the unit spherical surface, but did not know how to proceed.
real-analysis convex-analysis
asked Aug 19 at 16:58
Hans
4,18221028
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up vote
1
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Let $f : mathbf R^n to mathbf R$ be a convex function with $f(0)=0$ and $displaystylelim_ttoinftyf(tx)=infty$ for any $xinmathbf R-0$. Is $$A:=x mid f(x)le 1 $$ bounded?
I was thinking about projecting $A$ onto the unit spherical surface, but did not know how to proceed.
real-analysis convex-analysis
asked Aug 19 at 16:58
Hans
4,18221028
2
From Rockafellar's "Convex Analysis", Theorem 8.4: "A non-empty closed convex set $C$ in $mathbbR^n$ is bounded if and only if its recession cone $0^+C$ consists of the zero vector alone".
– copper.hat
Aug 19 at 20:17
1
@copper.hat: Thank you for pointing out this is a classical result.
– Hans
Aug 19 at 20:40
add a comment |Â
up vote
1
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favorite
up vote
1
down vote
favorite
Let $f : mathbf R^n to mathbf R$ be a convex function with $f(0)=0$ and $displaystylelim_ttoinftyf(tx)=infty$ for any $xinmathbf R-0$. Is $$A:=x mid f(x)le 1 $$ bounded?
I was thinking about projecting $A$ onto the unit spherical surface, but did not know how to proceed.
real-analysis convex-analysis
asked Aug 19 at 16:58
Hans
4,18221028
Let $f : mathbf R^n to mathbf R$ be a convex function with $f(0)=0$ and $displaystylelim_ttoinftyf(tx)=infty$ for any $xinmathbf R-0$. Is $$A:=x mid f(x)le 1 $$ bounded?
I was thinking about projecting $A$ onto the unit spherical surface, but did not know how to proceed.
real-analysis convex-analysis
asked Aug 19 at 16:58
Hans
4,18221028
edited Aug 19 at 20:49
asked Aug 19 at 16:58
Hans
4,18221028
asked Aug 19 at 16:58
Hans
4,18221028
asked Aug 19 at 16:58
Hans
4,18221028
4,18221028
2
From Rockafellar's "Convex Analysis", Theorem 8.4: "A non-empty closed convex set $C$ in $mathbbR^n$ is bounded if and only if its recession cone $0^+C$ consists of the zero vector alone".
– copper.hat
Aug 19 at 20:17
1
@copper.hat: Thank you for pointing out this is a classical result.
– Hans
Aug 19 at 20:40
add a comment |Â
2
From Rockafellar's "Convex Analysis", Theorem 8.4: "A non-empty closed convex set $C$ in $mathbbR^n$ is bounded if and only if its recession cone $0^+C$ consists of the zero vector alone".
– copper.hat
Aug 19 at 20:17
1
@copper.hat: Thank you for pointing out this is a classical result.
– Hans
Aug 19 at 20:40
2
2
From Rockafellar's "Convex Analysis", Theorem 8.4: "A non-empty closed convex set $C$ in $mathbbR^n$ is bounded if and only if its recession cone $0^+C$ consists of the zero vector alone".
– copper.hat
Aug 19 at 20:17
From Rockafellar's "Convex Analysis", Theorem 8.4: "A non-empty closed convex set $C$ in $mathbbR^n$ is bounded if and only if its recession cone $0^+C$ consists of the zero vector alone".
– copper.hat
Aug 19 at 20:17
1
1
@copper.hat: Thank you for pointing out this is a classical result.
– Hans
Aug 19 at 20:40
@copper.hat: Thank you for pointing out this is a classical result.
– Hans
Aug 19 at 20:40
add a comment |Â
2 Answers
2
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oldest
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up vote
2
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accepted
$A$ is indeed bounded
We can prove it by contradiction, using the result that for finite dimensional real vector spaces an unbounded convex subset contains a ray. See my website for the proof.
$A$ being a level set of $f$ is convex and not empty as $0 in A$. $0$ is even an interior point of $A$ as $f$ is continuous. According to the result above, $A$ contains a ray $[0, infty v )$ with $v in A$. By definition of $A$, the value of $f$ on this ray is less or equal to $1$. In contradiction with the hypothesis $displaystylelim_ttoinftyf(tv)=infty$.
answered Aug 19 at 19:22
mathcounterexamples.net
25.4k21755
How does this differ from the answer below?
– copper.hat
Aug 19 at 19:59
@copper.hat: I suppose the difference is that the proof here of "an unbounded closed convex subset of a finite dimensional real vector space contains a ray" is simpler than the accepted answer in the linked question in the answer of Wraith1995.
– Hans
Aug 19 at 20:44
add a comment |Â
up vote
2
down vote
We already know that $A$ is convex as the sub-level sets of a convex function are convex. If $A$ is unbounded, then for every $xin A$, there is some nonzero vector $h_x$ such that $x+th_xin A$ for all $tgeq 0$. Since $f(0)=0$, we can find $v:=h_0$ so $th_0in A$ for all $tgeq 0$. Thus, $f(th_0)leq 1$ for all $tgeq 0$. This contradicts that $lim_ttoinfty f(th_0) = infty$.
answered Aug 19 at 17:45
Wraith1995
588315
Just a comment, as shown in the link, it's important that $A$ is closed.
– user512723
Aug 19 at 18:18
@Balgani: It is sufficient in this case to have $0 in operatornameri A$, and since $0 in A^circ$ that is satisfied.
– copper.hat
Aug 19 at 20:13
+1. Nice answer. I hope you do not mind me accepting mathcounterexample.net's answer as the proof of his for the existence of ray is simpler than the accepted answer of the linked question.
– Hans
Aug 19 at 20:47
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
$A$ is indeed bounded
We can prove it by contradiction, using the result that for finite dimensional real vector spaces an unbounded convex subset contains a ray. See my website for the proof.
$A$ being a level set of $f$ is convex and not empty as $0 in A$. $0$ is even an interior point of $A$ as $f$ is continuous. According to the result above, $A$ contains a ray $[0, infty v )$ with $v in A$. By definition of $A$, the value of $f$ on this ray is less or equal to $1$. In contradiction with the hypothesis $displaystylelim_ttoinftyf(tv)=infty$.
answered Aug 19 at 19:22
mathcounterexamples.net
25.4k21755
How does this differ from the answer below?
– copper.hat
Aug 19 at 19:59
@copper.hat: I suppose the difference is that the proof here of "an unbounded closed convex subset of a finite dimensional real vector space contains a ray" is simpler than the accepted answer in the linked question in the answer of Wraith1995.
– Hans
Aug 19 at 20:44
add a comment |Â
up vote
2
down vote
accepted
$A$ is indeed bounded
We can prove it by contradiction, using the result that for finite dimensional real vector spaces an unbounded convex subset contains a ray. See my website for the proof.
$A$ being a level set of $f$ is convex and not empty as $0 in A$. $0$ is even an interior point of $A$ as $f$ is continuous. According to the result above, $A$ contains a ray $[0, infty v )$ with $v in A$. By definition of $A$, the value of $f$ on this ray is less or equal to $1$. In contradiction with the hypothesis $displaystylelim_ttoinftyf(tv)=infty$.
answered Aug 19 at 19:22
mathcounterexamples.net
25.4k21755
How does this differ from the answer below?
– copper.hat
Aug 19 at 19:59
@copper.hat: I suppose the difference is that the proof here of "an unbounded closed convex subset of a finite dimensional real vector space contains a ray" is simpler than the accepted answer in the linked question in the answer of Wraith1995.
– Hans
Aug 19 at 20:44
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
$A$ is indeed bounded
We can prove it by contradiction, using the result that for finite dimensional real vector spaces an unbounded convex subset contains a ray. See my website for the proof.
$A$ being a level set of $f$ is convex and not empty as $0 in A$. $0$ is even an interior point of $A$ as $f$ is continuous. According to the result above, $A$ contains a ray $[0, infty v )$ with $v in A$. By definition of $A$, the value of $f$ on this ray is less or equal to $1$. In contradiction with the hypothesis $displaystylelim_ttoinftyf(tv)=infty$.
answered Aug 19 at 19:22
mathcounterexamples.net
25.4k21755
$A$ is indeed bounded
We can prove it by contradiction, using the result that for finite dimensional real vector spaces an unbounded convex subset contains a ray. See my website for the proof.
$A$ being a level set of $f$ is convex and not empty as $0 in A$. $0$ is even an interior point of $A$ as $f$ is continuous. According to the result above, $A$ contains a ray $[0, infty v )$ with $v in A$. By definition of $A$, the value of $f$ on this ray is less or equal to $1$. In contradiction with the hypothesis $displaystylelim_ttoinftyf(tv)=infty$.
answered Aug 19 at 19:22
mathcounterexamples.net
25.4k21755
answered Aug 19 at 19:22
mathcounterexamples.net
25.4k21755
answered Aug 19 at 19:22
mathcounterexamples.net
25.4k21755
answered Aug 19 at 19:22
mathcounterexamples.net
25.4k21755
25.4k21755
How does this differ from the answer below?
– copper.hat
Aug 19 at 19:59
@copper.hat: I suppose the difference is that the proof here of "an unbounded closed convex subset of a finite dimensional real vector space contains a ray" is simpler than the accepted answer in the linked question in the answer of Wraith1995.
– Hans
Aug 19 at 20:44
add a comment |Â
How does this differ from the answer below?
– copper.hat
Aug 19 at 19:59
@copper.hat: I suppose the difference is that the proof here of "an unbounded closed convex subset of a finite dimensional real vector space contains a ray" is simpler than the accepted answer in the linked question in the answer of Wraith1995.
– Hans
Aug 19 at 20:44
How does this differ from the answer below?
– copper.hat
Aug 19 at 19:59
How does this differ from the answer below?
– copper.hat
Aug 19 at 19:59
@copper.hat: I suppose the difference is that the proof here of "an unbounded closed convex subset of a finite dimensional real vector space contains a ray" is simpler than the accepted answer in the linked question in the answer of Wraith1995.
– Hans
Aug 19 at 20:44
@copper.hat: I suppose the difference is that the proof here of "an unbounded closed convex subset of a finite dimensional real vector space contains a ray" is simpler than the accepted answer in the linked question in the answer of Wraith1995.
– Hans
Aug 19 at 20:44
add a comment |Â
up vote
2
down vote
We already know that $A$ is convex as the sub-level sets of a convex function are convex. If $A$ is unbounded, then for every $xin A$, there is some nonzero vector $h_x$ such that $x+th_xin A$ for all $tgeq 0$. Since $f(0)=0$, we can find $v:=h_0$ so $th_0in A$ for all $tgeq 0$. Thus, $f(th_0)leq 1$ for all $tgeq 0$. This contradicts that $lim_ttoinfty f(th_0) = infty$.
answered Aug 19 at 17:45
Wraith1995
588315
Just a comment, as shown in the link, it's important that $A$ is closed.
– user512723
Aug 19 at 18:18
@Balgani: It is sufficient in this case to have $0 in operatornameri A$, and since $0 in A^circ$ that is satisfied.
– copper.hat
Aug 19 at 20:13
+1. Nice answer. I hope you do not mind me accepting mathcounterexample.net's answer as the proof of his for the existence of ray is simpler than the accepted answer of the linked question.
– Hans
Aug 19 at 20:47
add a comment |Â
up vote
2
down vote
We already know that $A$ is convex as the sub-level sets of a convex function are convex. If $A$ is unbounded, then for every $xin A$, there is some nonzero vector $h_x$ such that $x+th_xin A$ for all $tgeq 0$. Since $f(0)=0$, we can find $v:=h_0$ so $th_0in A$ for all $tgeq 0$. Thus, $f(th_0)leq 1$ for all $tgeq 0$. This contradicts that $lim_ttoinfty f(th_0) = infty$.
answered Aug 19 at 17:45
Wraith1995
588315
Just a comment, as shown in the link, it's important that $A$ is closed.
– user512723
Aug 19 at 18:18
@Balgani: It is sufficient in this case to have $0 in operatornameri A$, and since $0 in A^circ$ that is satisfied.
– copper.hat
Aug 19 at 20:13
+1. Nice answer. I hope you do not mind me accepting mathcounterexample.net's answer as the proof of his for the existence of ray is simpler than the accepted answer of the linked question.
– Hans
Aug 19 at 20:47
add a comment |Â
up vote
2
down vote
up vote
2
down vote
We already know that $A$ is convex as the sub-level sets of a convex function are convex. If $A$ is unbounded, then for every $xin A$, there is some nonzero vector $h_x$ such that $x+th_xin A$ for all $tgeq 0$. Since $f(0)=0$, we can find $v:=h_0$ so $th_0in A$ for all $tgeq 0$. Thus, $f(th_0)leq 1$ for all $tgeq 0$. This contradicts that $lim_ttoinfty f(th_0) = infty$.
answered Aug 19 at 17:45
Wraith1995
588315
We already know that $A$ is convex as the sub-level sets of a convex function are convex. If $A$ is unbounded, then for every $xin A$, there is some nonzero vector $h_x$ such that $x+th_xin A$ for all $tgeq 0$. Since $f(0)=0$, we can find $v:=h_0$ so $th_0in A$ for all $tgeq 0$. Thus, $f(th_0)leq 1$ for all $tgeq 0$. This contradicts that $lim_ttoinfty f(th_0) = infty$.
answered Aug 19 at 17:45
Wraith1995
588315
answered Aug 19 at 17:45
Wraith1995
588315
answered Aug 19 at 17:45
Wraith1995
588315
answered Aug 19 at 17:45
Wraith1995
588315
588315
Just a comment, as shown in the link, it's important that $A$ is closed.
– user512723
Aug 19 at 18:18
@Balgani: It is sufficient in this case to have $0 in operatornameri A$, and since $0 in A^circ$ that is satisfied.
– copper.hat
Aug 19 at 20:13
+1. Nice answer. I hope you do not mind me accepting mathcounterexample.net's answer as the proof of his for the existence of ray is simpler than the accepted answer of the linked question.
– Hans
Aug 19 at 20:47
add a comment |Â
Just a comment, as shown in the link, it's important that $A$ is closed.
– user512723
Aug 19 at 18:18
@Balgani: It is sufficient in this case to have $0 in operatornameri A$, and since $0 in A^circ$ that is satisfied.
– copper.hat
Aug 19 at 20:13
+1. Nice answer. I hope you do not mind me accepting mathcounterexample.net's answer as the proof of his for the existence of ray is simpler than the accepted answer of the linked question.
– Hans
Aug 19 at 20:47
Just a comment, as shown in the link, it's important that $A$ is closed.
– user512723
Aug 19 at 18:18
Just a comment, as shown in the link, it's important that $A$ is closed.
– user512723
Aug 19 at 18:18
@Balgani: It is sufficient in this case to have $0 in operatornameri A$, and since $0 in A^circ$ that is satisfied.
– copper.hat
Aug 19 at 20:13
@Balgani: It is sufficient in this case to have $0 in operatornameri A$, and since $0 in A^circ$ that is satisfied.
– copper.hat
Aug 19 at 20:13
+1. Nice answer. I hope you do not mind me accepting mathcounterexample.net's answer as the proof of his for the existence of ray is simpler than the accepted answer of the linked question.
– Hans
Aug 19 at 20:47
+1. Nice answer. I hope you do not mind me accepting mathcounterexample.net's answer as the proof of his for the existence of ray is simpler than the accepted answer of the linked question.
– Hans
Aug 19 at 20:47
add a comment |Â
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2
From Rockafellar's "Convex Analysis", Theorem 8.4: "A non-empty closed convex set $C$ in $mathbbR^n$ is bounded if and only if its recession cone $0^+C$ consists of the zero vector alone".
– copper.hat
Aug 19 at 20:17
1
@copper.hat: Thank you for pointing out this is a classical result.
– Hans
Aug 19 at 20:40