Mixing
Inner product identity for cones
$begingroup$
Let $emptyset neq Csubseteq mathbb R^n$ be a convex, open cone with the property that $operatorname{int }C^* neq emptyset$, where $C^*$ denotes the dual cone defined by $$C^* = {x in mathbb R^n: langle x,y rangle geq 0 quad forall y in C}.$$
(always a closed and convex cone). Then we have for each $yin C$
$$inf_{xin C^* cap S^{n-1}} langle x,y rangle geq c_y lVert y rVert$$ for some constant $c_y >0$.
I was unable to show this. I know that $C^* cap S^{n-1}$ is compact and the inner product is continuous, so this attains a maximum. But I failed to see why it is impossible for this minimum to be $0$. Any help appreciated!
real-analysis analysis convex-optimization convex-cone
asked Jan 9 at 19:48
Staki42Staki42
1,154618
$endgroup$
|
show 5 more comments
$begingroup$
Let $emptyset neq Csubseteq mathbb R^n$ be a convex, open cone with the property that $operatorname{int }C^* neq emptyset$, where $C^*$ denotes the dual cone defined by $$C^* = {x in mathbb R^n: langle x,y rangle geq 0 quad forall y in C}.$$
(always a closed and convex cone). Then we have for each $yin C$
$$inf_{xin C^* cap S^{n-1}} langle x,y rangle geq c_y lVert y rVert$$ for some constant $c_y >0$.
I was unable to show this. I know that $C^* cap S^{n-1}$ is compact and the inner product is continuous, so this attains a maximum. But I failed to see why it is impossible for this minimum to be $0$. Any help appreciated!
real-analysis analysis convex-optimization convex-cone
asked Jan 9 at 19:48
Staki42Staki42
1,154618
$endgroup$
$begingroup$
If you take $x=0$, then the inf is clearly 0. What's to stop you from doing this?
$endgroup$
– Brian Borchers
Jan 9 at 20:54
$begingroup$
$x = 0$ does not lie in the unit sphere.
$endgroup$
– Staki42
Jan 9 at 20:56
$begingroup$
Sorry- I was taking that to be a space of symmetric matrices...
$endgroup$
– Brian Borchers
Jan 9 at 21:18
$begingroup$
$c_{y}$ is allowed to depend on the particular value of $y$, right?
$endgroup$
– Brian Borchers
Jan 9 at 21:21
$begingroup$
What is $c_y$? ${}{}$
$endgroup$
– copper.hat
Jan 9 at 21:39
|
show 5 more comments
$begingroup$
Let $emptyset neq Csubseteq mathbb R^n$ be a convex, open cone with the property that $operatorname{int }C^* neq emptyset$, where $C^*$ denotes the dual cone defined by $$C^* = {x in mathbb R^n: langle x,y rangle geq 0 quad forall y in C}.$$
(always a closed and convex cone). Then we have for each $yin C$
$$inf_{xin C^* cap S^{n-1}} langle x,y rangle geq c_y lVert y rVert$$ for some constant $c_y >0$.
I was unable to show this. I know that $C^* cap S^{n-1}$ is compact and the inner product is continuous, so this attains a maximum. But I failed to see why it is impossible for this minimum to be $0$. Any help appreciated!
real-analysis analysis convex-optimization convex-cone
asked Jan 9 at 19:48
Staki42Staki42
1,154618
$endgroup$
Let $emptyset neq Csubseteq mathbb R^n$ be a convex, open cone with the property that $operatorname{int }C^* neq emptyset$, where $C^*$ denotes the dual cone defined by $$C^* = {x in mathbb R^n: langle x,y rangle geq 0 quad forall y in C}.$$
(always a closed and convex cone). Then we have for each $yin C$
$$inf_{xin C^* cap S^{n-1}} langle x,y rangle geq c_y lVert y rVert$$ for some constant $c_y >0$.
I was unable to show this. I know that $C^* cap S^{n-1}$ is compact and the inner product is continuous, so this attains a maximum. But I failed to see why it is impossible for this minimum to be $0$. Any help appreciated!
real-analysis analysis convex-optimization convex-cone
real-analysis analysis convex-optimization convex-cone
asked Jan 9 at 19:48
Staki42Staki42
1,154618
asked Jan 9 at 19:48
Staki42Staki42
1,154618
edited Jan 9 at 22:19
Staki42
asked Jan 9 at 19:48
Staki42Staki42
1,154618
asked Jan 9 at 19:48
Staki42Staki42
1,154618
asked Jan 9 at 19:48
Staki42Staki42
1,154618
1,154618
$begingroup$
If you take $x=0$, then the inf is clearly 0. What's to stop you from doing this?
$endgroup$
– Brian Borchers
Jan 9 at 20:54
$begingroup$
$x = 0$ does not lie in the unit sphere.
$endgroup$
– Staki42
Jan 9 at 20:56
$begingroup$
Sorry- I was taking that to be a space of symmetric matrices...
$endgroup$
– Brian Borchers
Jan 9 at 21:18
$begingroup$
$c_{y}$ is allowed to depend on the particular value of $y$, right?
$endgroup$
– Brian Borchers
Jan 9 at 21:21
$begingroup$
What is $c_y$? ${}{}$
$endgroup$
– copper.hat
Jan 9 at 21:39
|
show 5 more comments
$begingroup$
If you take $x=0$, then the inf is clearly 0. What's to stop you from doing this?
$endgroup$
– Brian Borchers
Jan 9 at 20:54
$begingroup$
$x = 0$ does not lie in the unit sphere.
$endgroup$
– Staki42
Jan 9 at 20:56
$begingroup$
Sorry- I was taking that to be a space of symmetric matrices...
$endgroup$
– Brian Borchers
Jan 9 at 21:18
$begingroup$
$c_{y}$ is allowed to depend on the particular value of $y$, right?
$endgroup$
– Brian Borchers
Jan 9 at 21:21
$begingroup$
What is $c_y$? ${}{}$
$endgroup$
– copper.hat
Jan 9 at 21:39
$begingroup$
If you take $x=0$, then the inf is clearly 0. What's to stop you from doing this?
$endgroup$
– Brian Borchers
Jan 9 at 20:54
$begingroup$
If you take $x=0$, then the inf is clearly 0. What's to stop you from doing this?
$endgroup$
– Brian Borchers
Jan 9 at 20:54
$begingroup$
$x = 0$ does not lie in the unit sphere.
$endgroup$
– Staki42
Jan 9 at 20:56
$begingroup$
$x = 0$ does not lie in the unit sphere.
$endgroup$
– Staki42
Jan 9 at 20:56
$begingroup$
Sorry- I was taking that to be a space of symmetric matrices...
$endgroup$
– Brian Borchers
Jan 9 at 21:18
$begingroup$
Sorry- I was taking that to be a space of symmetric matrices...
$endgroup$
– Brian Borchers
Jan 9 at 21:18
$begingroup$
$c_{y}$ is allowed to depend on the particular value of $y$, right?
$endgroup$
– Brian Borchers
Jan 9 at 21:21
$begingroup$
$c_{y}$ is allowed to depend on the particular value of $y$, right?
$endgroup$
– Brian Borchers
Jan 9 at 21:21
$begingroup$
What is $c_y$? ${}{}$
$endgroup$
– copper.hat
Jan 9 at 21:39
$begingroup$
What is $c_y$? ${}{}$
$endgroup$
– copper.hat
Jan 9 at 21:39
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Big Hint:
I think the key is that $C$ is a non-empty open cone.
If $C$ were a close cone, then $c_y$ would be 0 for any non-zero $y$ on the boundary of $C$. For example, suppose $Csubset R^2$ and $C={(x,0) | xgeq 0}.$ Then $C^*={(x,y)| xgeq 0, yinmathbb R}$. Notice that
$C^*cap S^{n-1} = {(sin t, cos t) | 0leq tleq pi}$, so for any fixed $yin C$, $y=(y_0, 0)$ and
$$
inf_{xin C^*cap S^{n-1}} langle y,xrangle = langle(y_0, 0),(0,1)rangle=0.
$$
Hence $c_y$ would be 0.
On the other hand, if $C$ is a open cone and $yin Csetminus{0}$, then there exist a positive number $epsilon>0$ such that $B(y,epsilon)subset C$. We know that $langle c^*, y_brangle geq 0$ for all $y_bin B(y,epsilon)$ and $c^* in C^*$ by the definition of $C^*$ and the fact that $B(y,epsilon)subset C$. That implies that $inf_{y_b in B(y,epsilon) } langle c^*, y_brangle geq 0$ for all $ c^*in C^*$. So now we can write that for any $c^*in S^{n-1}cap C^*$,
$$langle B(y, epsilon), c^* rangle geq 0,$$
$$langle y - epsilon c^*, c^* rangle geq 0,$$
$$langle y , c^* rangle - epsilon||c^*||^2geq 0,....$$
That line of reasoning should lead you to the answer.
answered Jan 10 at 19:47
irchansirchans
1,07739
$endgroup$
add a comment |
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$begingroup$
Big Hint:
I think the key is that $C$ is a non-empty open cone.
If $C$ were a close cone, then $c_y$ would be 0 for any non-zero $y$ on the boundary of $C$. For example, suppose $Csubset R^2$ and $C={(x,0) | xgeq 0}.$ Then $C^*={(x,y)| xgeq 0, yinmathbb R}$. Notice that
$C^*cap S^{n-1} = {(sin t, cos t) | 0leq tleq pi}$, so for any fixed $yin C$, $y=(y_0, 0)$ and
$$
inf_{xin C^*cap S^{n-1}} langle y,xrangle = langle(y_0, 0),(0,1)rangle=0.
$$
Hence $c_y$ would be 0.
On the other hand, if $C$ is a open cone and $yin Csetminus{0}$, then there exist a positive number $epsilon>0$ such that $B(y,epsilon)subset C$. We know that $langle c^*, y_brangle geq 0$ for all $y_bin B(y,epsilon)$ and $c^* in C^*$ by the definition of $C^*$ and the fact that $B(y,epsilon)subset C$. That implies that $inf_{y_b in B(y,epsilon) } langle c^*, y_brangle geq 0$ for all $ c^*in C^*$. So now we can write that for any $c^*in S^{n-1}cap C^*$,
$$langle B(y, epsilon), c^* rangle geq 0,$$
$$langle y - epsilon c^*, c^* rangle geq 0,$$
$$langle y , c^* rangle - epsilon||c^*||^2geq 0,....$$
That line of reasoning should lead you to the answer.
answered Jan 10 at 19:47
irchansirchans
1,07739
$endgroup$
add a comment |
$begingroup$
Big Hint:
I think the key is that $C$ is a non-empty open cone.
If $C$ were a close cone, then $c_y$ would be 0 for any non-zero $y$ on the boundary of $C$. For example, suppose $Csubset R^2$ and $C={(x,0) | xgeq 0}.$ Then $C^*={(x,y)| xgeq 0, yinmathbb R}$. Notice that
$C^*cap S^{n-1} = {(sin t, cos t) | 0leq tleq pi}$, so for any fixed $yin C$, $y=(y_0, 0)$ and
$$
inf_{xin C^*cap S^{n-1}} langle y,xrangle = langle(y_0, 0),(0,1)rangle=0.
$$
Hence $c_y$ would be 0.
On the other hand, if $C$ is a open cone and $yin Csetminus{0}$, then there exist a positive number $epsilon>0$ such that $B(y,epsilon)subset C$. We know that $langle c^*, y_brangle geq 0$ for all $y_bin B(y,epsilon)$ and $c^* in C^*$ by the definition of $C^*$ and the fact that $B(y,epsilon)subset C$. That implies that $inf_{y_b in B(y,epsilon) } langle c^*, y_brangle geq 0$ for all $ c^*in C^*$. So now we can write that for any $c^*in S^{n-1}cap C^*$,
$$langle B(y, epsilon), c^* rangle geq 0,$$
$$langle y - epsilon c^*, c^* rangle geq 0,$$
$$langle y , c^* rangle - epsilon||c^*||^2geq 0,....$$
That line of reasoning should lead you to the answer.
answered Jan 10 at 19:47
irchansirchans
1,07739
$endgroup$
add a comment |
$begingroup$
Big Hint:
I think the key is that $C$ is a non-empty open cone.
If $C$ were a close cone, then $c_y$ would be 0 for any non-zero $y$ on the boundary of $C$. For example, suppose $Csubset R^2$ and $C={(x,0) | xgeq 0}.$ Then $C^*={(x,y)| xgeq 0, yinmathbb R}$. Notice that
$C^*cap S^{n-1} = {(sin t, cos t) | 0leq tleq pi}$, so for any fixed $yin C$, $y=(y_0, 0)$ and
$$
inf_{xin C^*cap S^{n-1}} langle y,xrangle = langle(y_0, 0),(0,1)rangle=0.
$$
Hence $c_y$ would be 0.
On the other hand, if $C$ is a open cone and $yin Csetminus{0}$, then there exist a positive number $epsilon>0$ such that $B(y,epsilon)subset C$. We know that $langle c^*, y_brangle geq 0$ for all $y_bin B(y,epsilon)$ and $c^* in C^*$ by the definition of $C^*$ and the fact that $B(y,epsilon)subset C$. That implies that $inf_{y_b in B(y,epsilon) } langle c^*, y_brangle geq 0$ for all $ c^*in C^*$. So now we can write that for any $c^*in S^{n-1}cap C^*$,
$$langle B(y, epsilon), c^* rangle geq 0,$$
$$langle y - epsilon c^*, c^* rangle geq 0,$$
$$langle y , c^* rangle - epsilon||c^*||^2geq 0,....$$
That line of reasoning should lead you to the answer.
answered Jan 10 at 19:47
irchansirchans
1,07739
$endgroup$
Big Hint:
I think the key is that $C$ is a non-empty open cone.
If $C$ were a close cone, then $c_y$ would be 0 for any non-zero $y$ on the boundary of $C$. For example, suppose $Csubset R^2$ and $C={(x,0) | xgeq 0}.$ Then $C^*={(x,y)| xgeq 0, yinmathbb R}$. Notice that
$C^*cap S^{n-1} = {(sin t, cos t) | 0leq tleq pi}$, so for any fixed $yin C$, $y=(y_0, 0)$ and
$$
inf_{xin C^*cap S^{n-1}} langle y,xrangle = langle(y_0, 0),(0,1)rangle=0.
$$
Hence $c_y$ would be 0.
On the other hand, if $C$ is a open cone and $yin Csetminus{0}$, then there exist a positive number $epsilon>0$ such that $B(y,epsilon)subset C$. We know that $langle c^*, y_brangle geq 0$ for all $y_bin B(y,epsilon)$ and $c^* in C^*$ by the definition of $C^*$ and the fact that $B(y,epsilon)subset C$. That implies that $inf_{y_b in B(y,epsilon) } langle c^*, y_brangle geq 0$ for all $ c^*in C^*$. So now we can write that for any $c^*in S^{n-1}cap C^*$,
$$langle B(y, epsilon), c^* rangle geq 0,$$
$$langle y - epsilon c^*, c^* rangle geq 0,$$
$$langle y , c^* rangle - epsilon||c^*||^2geq 0,....$$
That line of reasoning should lead you to the answer.
answered Jan 10 at 19:47
irchansirchans
1,07739
answered Jan 10 at 19:47
irchansirchans
1,07739
answered Jan 10 at 19:47
irchansirchans
1,07739
answered Jan 10 at 19:47
irchansirchans
1,07739
1,07739
add a comment |
add a comment |
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$begingroup$
If you take $x=0$, then the inf is clearly 0. What's to stop you from doing this?
$endgroup$
– Brian Borchers
Jan 9 at 20:54
$begingroup$
$x = 0$ does not lie in the unit sphere.
$endgroup$
– Staki42
Jan 9 at 20:56
$begingroup$
Sorry- I was taking that to be a space of symmetric matrices...
$endgroup$
– Brian Borchers
Jan 9 at 21:18
$begingroup$
$c_{y}$ is allowed to depend on the particular value of $y$, right?
$endgroup$
– Brian Borchers
Jan 9 at 21:21
$begingroup$
What is $c_y$? ${}{}$
$endgroup$
– copper.hat
Jan 9 at 21:39