Pseudo convex quadratic forms are quasi convex
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I have been trying to show: A Pseudoconvex quadratic form is Quasiconvex in the same set X, by using the following definitions. These definitions are equivalent to the standard definitions of Pseudo and Quasiconvexity of the quadratic forms. In the proof lemma 8, it is written that this follows from continuity. I am not able to understand how to use continuity to prove this.
Thanks in advance!
matrices optimization convex-analysis quadratic-forms
asked Dec 27 '18 at 9:56
Shivani GoelShivani Goel
226115
$endgroup$
add a comment |
$begingroup$
I have been trying to show: A Pseudoconvex quadratic form is Quasiconvex in the same set X, by using the following definitions. These definitions are equivalent to the standard definitions of Pseudo and Quasiconvexity of the quadratic forms. In the proof lemma 8, it is written that this follows from continuity. I am not able to understand how to use continuity to prove this.
Thanks in advance!
matrices optimization convex-analysis quadratic-forms
asked Dec 27 '18 at 9:56
Shivani GoelShivani Goel
226115
$endgroup$
$begingroup$
I understood by going through some texts on convexity that the continuity the author is talking about is the continuity of the Quadratic form due to which all the different type of strict and non-strict convexities coincides in a way. And that makes the above lemma to be true trivially.
$endgroup$
– Shivani Goel
Dec 27 '18 at 17:48
add a comment |
$begingroup$
I have been trying to show: A Pseudoconvex quadratic form is Quasiconvex in the same set X, by using the following definitions. These definitions are equivalent to the standard definitions of Pseudo and Quasiconvexity of the quadratic forms. In the proof lemma 8, it is written that this follows from continuity. I am not able to understand how to use continuity to prove this.
Thanks in advance!
matrices optimization convex-analysis quadratic-forms
asked Dec 27 '18 at 9:56
Shivani GoelShivani Goel
226115
$endgroup$
I have been trying to show: A Pseudoconvex quadratic form is Quasiconvex in the same set X, by using the following definitions. These definitions are equivalent to the standard definitions of Pseudo and Quasiconvexity of the quadratic forms. In the proof lemma 8, it is written that this follows from continuity. I am not able to understand how to use continuity to prove this.
Thanks in advance!
matrices optimization convex-analysis quadratic-forms
matrices optimization convex-analysis quadratic-forms
asked Dec 27 '18 at 9:56
Shivani GoelShivani Goel
226115
asked Dec 27 '18 at 9:56
Shivani GoelShivani Goel
226115
asked Dec 27 '18 at 9:56
Shivani GoelShivani Goel
226115
asked Dec 27 '18 at 9:56
Shivani GoelShivani Goel
226115
asked Dec 27 '18 at 9:56
Shivani GoelShivani Goel
226115
226115
$begingroup$
I understood by going through some texts on convexity that the continuity the author is talking about is the continuity of the Quadratic form due to which all the different type of strict and non-strict convexities coincides in a way. And that makes the above lemma to be true trivially.
$endgroup$
– Shivani Goel
Dec 27 '18 at 17:48
add a comment |
$begingroup$
I understood by going through some texts on convexity that the continuity the author is talking about is the continuity of the Quadratic form due to which all the different type of strict and non-strict convexities coincides in a way. And that makes the above lemma to be true trivially.
$endgroup$
– Shivani Goel
Dec 27 '18 at 17:48
$begingroup$
I understood by going through some texts on convexity that the continuity the author is talking about is the continuity of the Quadratic form due to which all the different type of strict and non-strict convexities coincides in a way. And that makes the above lemma to be true trivially.
$endgroup$
– Shivani Goel
Dec 27 '18 at 17:48
$begingroup$
I understood by going through some texts on convexity that the continuity the author is talking about is the continuity of the Quadratic form due to which all the different type of strict and non-strict convexities coincides in a way. And that makes the above lemma to be true trivially.
$endgroup$
– Shivani Goel
Dec 27 '18 at 17:48
add a comment |
1 Answer
1
active
oldest
votes
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Suppose $Q$ is Pseudoconvex and let $x$ and $y$ be in $X$.
If $x'Cx-y'Cy >0$, then by pseudoconvexity of $Q$, it follows that $(x-y)'Cx > 0$.
Suppose $x'Cx-y'Cy =0$. Assume on the contrary that $(x-y)'Cx < 0$, i.e $x'Cx-y'Cx<0$. So
$$(x-y)'C(x-y) = x'Cx+y'Cy-2y'Cx = 2x'Cx-2y'Cx <0$$
Therefore
$$0'C0-(x-y)'C(x-y)>0$$
By the pseudoconvexity of $Q$, it follows that
$$(0-x+y)'C0>0$$
i.e $$0>0$$
Which is absurd.
Hence $(x-y)'Cx ge 0$ i.e $Q$ is Quasiconvex.
answered Dec 27 '18 at 17:40
Shivani GoelShivani Goel
226115
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
Suppose $Q$ is Pseudoconvex and let $x$ and $y$ be in $X$.
If $x'Cx-y'Cy >0$, then by pseudoconvexity of $Q$, it follows that $(x-y)'Cx > 0$.
Suppose $x'Cx-y'Cy =0$. Assume on the contrary that $(x-y)'Cx < 0$, i.e $x'Cx-y'Cx<0$. So
$$(x-y)'C(x-y) = x'Cx+y'Cy-2y'Cx = 2x'Cx-2y'Cx <0$$
Therefore
$$0'C0-(x-y)'C(x-y)>0$$
By the pseudoconvexity of $Q$, it follows that
$$(0-x+y)'C0>0$$
i.e $$0>0$$
Which is absurd.
Hence $(x-y)'Cx ge 0$ i.e $Q$ is Quasiconvex.
answered Dec 27 '18 at 17:40
Shivani GoelShivani Goel
226115
$endgroup$
add a comment |
$begingroup$
Suppose $Q$ is Pseudoconvex and let $x$ and $y$ be in $X$.
If $x'Cx-y'Cy >0$, then by pseudoconvexity of $Q$, it follows that $(x-y)'Cx > 0$.
Suppose $x'Cx-y'Cy =0$. Assume on the contrary that $(x-y)'Cx < 0$, i.e $x'Cx-y'Cx<0$. So
$$(x-y)'C(x-y) = x'Cx+y'Cy-2y'Cx = 2x'Cx-2y'Cx <0$$
Therefore
$$0'C0-(x-y)'C(x-y)>0$$
By the pseudoconvexity of $Q$, it follows that
$$(0-x+y)'C0>0$$
i.e $$0>0$$
Which is absurd.
Hence $(x-y)'Cx ge 0$ i.e $Q$ is Quasiconvex.
answered Dec 27 '18 at 17:40
Shivani GoelShivani Goel
226115
$endgroup$
add a comment |
$begingroup$
Suppose $Q$ is Pseudoconvex and let $x$ and $y$ be in $X$.
If $x'Cx-y'Cy >0$, then by pseudoconvexity of $Q$, it follows that $(x-y)'Cx > 0$.
Suppose $x'Cx-y'Cy =0$. Assume on the contrary that $(x-y)'Cx < 0$, i.e $x'Cx-y'Cx<0$. So
$$(x-y)'C(x-y) = x'Cx+y'Cy-2y'Cx = 2x'Cx-2y'Cx <0$$
Therefore
$$0'C0-(x-y)'C(x-y)>0$$
By the pseudoconvexity of $Q$, it follows that
$$(0-x+y)'C0>0$$
i.e $$0>0$$
Which is absurd.
Hence $(x-y)'Cx ge 0$ i.e $Q$ is Quasiconvex.
answered Dec 27 '18 at 17:40
Shivani GoelShivani Goel
226115
$endgroup$
Suppose $Q$ is Pseudoconvex and let $x$ and $y$ be in $X$.
If $x'Cx-y'Cy >0$, then by pseudoconvexity of $Q$, it follows that $(x-y)'Cx > 0$.
Suppose $x'Cx-y'Cy =0$. Assume on the contrary that $(x-y)'Cx < 0$, i.e $x'Cx-y'Cx<0$. So
$$(x-y)'C(x-y) = x'Cx+y'Cy-2y'Cx = 2x'Cx-2y'Cx <0$$
Therefore
$$0'C0-(x-y)'C(x-y)>0$$
By the pseudoconvexity of $Q$, it follows that
$$(0-x+y)'C0>0$$
i.e $$0>0$$
Which is absurd.
Hence $(x-y)'Cx ge 0$ i.e $Q$ is Quasiconvex.
answered Dec 27 '18 at 17:40
Shivani GoelShivani Goel
226115
answered Dec 27 '18 at 17:40
Shivani GoelShivani Goel
226115
answered Dec 27 '18 at 17:40
Shivani GoelShivani Goel
226115
answered Dec 27 '18 at 17:40
Shivani GoelShivani Goel
226115
226115
add a comment |
add a comment |
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$begingroup$
I understood by going through some texts on convexity that the continuity the author is talking about is the continuity of the Quadratic form due to which all the different type of strict and non-strict convexities coincides in a way. And that makes the above lemma to be true trivially.
$endgroup$
– Shivani Goel
Dec 27 '18 at 17:48