Determine $sup {xy−x^2/2mid xin [-1,1]}$
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I am trying to find the supremum of the following set ${xy−x^2/2mid xin [-1,1]}$, where $y$ is a real number. I am not sure if this is correct, but I managed to find that
$$
sup { xy-x^2/2mid xin [-1,1]}=begin{cases}
y^2/2 & text{ if } yin [-1,1] \
y-1/2 & text{ if } y>1 \
-y-1/2 & text{ if } y<-1
end{cases}
$$
convex-analysis supremum-and-infimum
asked Jan 15 at 2:31
UnknownWUnknownW
1,023922
$endgroup$
add a comment |
$begingroup$
I am trying to find the supremum of the following set ${xy−x^2/2mid xin [-1,1]}$, where $y$ is a real number. I am not sure if this is correct, but I managed to find that
$$
sup { xy-x^2/2mid xin [-1,1]}=begin{cases}
y^2/2 & text{ if } yin [-1,1] \
y-1/2 & text{ if } y>1 \
-y-1/2 & text{ if } y<-1
end{cases}
$$
convex-analysis supremum-and-infimum
asked Jan 15 at 2:31
UnknownWUnknownW
1,023922
$endgroup$
$begingroup$
Looks good to me
$endgroup$
– angryavian
Jan 15 at 2:46
$begingroup$
Yes, it's correct.
$endgroup$
– Ryan Greyling
Jan 15 at 2:58
add a comment |
$begingroup$
I am trying to find the supremum of the following set ${xy−x^2/2mid xin [-1,1]}$, where $y$ is a real number. I am not sure if this is correct, but I managed to find that
$$
sup { xy-x^2/2mid xin [-1,1]}=begin{cases}
y^2/2 & text{ if } yin [-1,1] \
y-1/2 & text{ if } y>1 \
-y-1/2 & text{ if } y<-1
end{cases}
$$
convex-analysis supremum-and-infimum
asked Jan 15 at 2:31
UnknownWUnknownW
1,023922
$endgroup$
I am trying to find the supremum of the following set ${xy−x^2/2mid xin [-1,1]}$, where $y$ is a real number. I am not sure if this is correct, but I managed to find that
$$
sup { xy-x^2/2mid xin [-1,1]}=begin{cases}
y^2/2 & text{ if } yin [-1,1] \
y-1/2 & text{ if } y>1 \
-y-1/2 & text{ if } y<-1
end{cases}
$$
convex-analysis supremum-and-infimum
convex-analysis supremum-and-infimum
asked Jan 15 at 2:31
UnknownWUnknownW
1,023922
asked Jan 15 at 2:31
UnknownWUnknownW
1,023922
asked Jan 15 at 2:31
UnknownWUnknownW
1,023922
asked Jan 15 at 2:31
UnknownWUnknownW
1,023922
asked Jan 15 at 2:31
UnknownWUnknownW
1,023922
1,023922
$begingroup$
Looks good to me
$endgroup$
– angryavian
Jan 15 at 2:46
$begingroup$
Yes, it's correct.
$endgroup$
– Ryan Greyling
Jan 15 at 2:58
add a comment |
$begingroup$
Looks good to me
$endgroup$
– angryavian
Jan 15 at 2:46
$begingroup$
Yes, it's correct.
$endgroup$
– Ryan Greyling
Jan 15 at 2:58
$begingroup$
Looks good to me
$endgroup$
– angryavian
Jan 15 at 2:46
$begingroup$
Looks good to me
$endgroup$
– angryavian
Jan 15 at 2:46
$begingroup$
Yes, it's correct.
$endgroup$
– Ryan Greyling
Jan 15 at 2:58
$begingroup$
Yes, it's correct.
$endgroup$
– Ryan Greyling
Jan 15 at 2:58
add a comment |
1 Answer
1
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$begingroup$
You are computing the convex conjugate of $(f+g)(x)$ where $f(x) = 0.5 x^2$ and $g(x) = delta(x, [-1,1])$ (taking the value 0 if $x$ is in the interval and $infty$ otherwise. Note that $f^*(y) = 0.5 y^2$ and $g^*(y) = |y|$.
$$sup_x { xy - (f+g)(x)} = (f+g)^*(y)= inf_z { f^*(z)+g^*(y-z)} $$
$$ = inf_z { 0.5 z^2+ |y-z|} $$
This is an unconstrained convex optimization problem. The derivative is $z-1$ if $y > z$, and $z+1$ if $y < z$, and the gradient is $[z-1,z+1]$ if $y=z$. We have that 0 is in the gradient if ($z=1$ and $y>z$) or ($z in [-1,1]$ and $y=z$) or ($z=-1$ and $y<z$). This gives the three cases in your question.
answered Jan 15 at 14:55
LinAlgLinAlg
10.1k1521
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
You are computing the convex conjugate of $(f+g)(x)$ where $f(x) = 0.5 x^2$ and $g(x) = delta(x, [-1,1])$ (taking the value 0 if $x$ is in the interval and $infty$ otherwise. Note that $f^*(y) = 0.5 y^2$ and $g^*(y) = |y|$.
$$sup_x { xy - (f+g)(x)} = (f+g)^*(y)= inf_z { f^*(z)+g^*(y-z)} $$
$$ = inf_z { 0.5 z^2+ |y-z|} $$
This is an unconstrained convex optimization problem. The derivative is $z-1$ if $y > z$, and $z+1$ if $y < z$, and the gradient is $[z-1,z+1]$ if $y=z$. We have that 0 is in the gradient if ($z=1$ and $y>z$) or ($z in [-1,1]$ and $y=z$) or ($z=-1$ and $y<z$). This gives the three cases in your question.
answered Jan 15 at 14:55
LinAlgLinAlg
10.1k1521
$endgroup$
add a comment |
$begingroup$
You are computing the convex conjugate of $(f+g)(x)$ where $f(x) = 0.5 x^2$ and $g(x) = delta(x, [-1,1])$ (taking the value 0 if $x$ is in the interval and $infty$ otherwise. Note that $f^*(y) = 0.5 y^2$ and $g^*(y) = |y|$.
$$sup_x { xy - (f+g)(x)} = (f+g)^*(y)= inf_z { f^*(z)+g^*(y-z)} $$
$$ = inf_z { 0.5 z^2+ |y-z|} $$
This is an unconstrained convex optimization problem. The derivative is $z-1$ if $y > z$, and $z+1$ if $y < z$, and the gradient is $[z-1,z+1]$ if $y=z$. We have that 0 is in the gradient if ($z=1$ and $y>z$) or ($z in [-1,1]$ and $y=z$) or ($z=-1$ and $y<z$). This gives the three cases in your question.
answered Jan 15 at 14:55
LinAlgLinAlg
10.1k1521
$endgroup$
add a comment |
$begingroup$
You are computing the convex conjugate of $(f+g)(x)$ where $f(x) = 0.5 x^2$ and $g(x) = delta(x, [-1,1])$ (taking the value 0 if $x$ is in the interval and $infty$ otherwise. Note that $f^*(y) = 0.5 y^2$ and $g^*(y) = |y|$.
$$sup_x { xy - (f+g)(x)} = (f+g)^*(y)= inf_z { f^*(z)+g^*(y-z)} $$
$$ = inf_z { 0.5 z^2+ |y-z|} $$
This is an unconstrained convex optimization problem. The derivative is $z-1$ if $y > z$, and $z+1$ if $y < z$, and the gradient is $[z-1,z+1]$ if $y=z$. We have that 0 is in the gradient if ($z=1$ and $y>z$) or ($z in [-1,1]$ and $y=z$) or ($z=-1$ and $y<z$). This gives the three cases in your question.
answered Jan 15 at 14:55
LinAlgLinAlg
10.1k1521
$endgroup$
You are computing the convex conjugate of $(f+g)(x)$ where $f(x) = 0.5 x^2$ and $g(x) = delta(x, [-1,1])$ (taking the value 0 if $x$ is in the interval and $infty$ otherwise. Note that $f^*(y) = 0.5 y^2$ and $g^*(y) = |y|$.
$$sup_x { xy - (f+g)(x)} = (f+g)^*(y)= inf_z { f^*(z)+g^*(y-z)} $$
$$ = inf_z { 0.5 z^2+ |y-z|} $$
This is an unconstrained convex optimization problem. The derivative is $z-1$ if $y > z$, and $z+1$ if $y < z$, and the gradient is $[z-1,z+1]$ if $y=z$. We have that 0 is in the gradient if ($z=1$ and $y>z$) or ($z in [-1,1]$ and $y=z$) or ($z=-1$ and $y<z$). This gives the three cases in your question.
answered Jan 15 at 14:55
LinAlgLinAlg
10.1k1521
answered Jan 15 at 14:55
LinAlgLinAlg
10.1k1521
answered Jan 15 at 14:55
LinAlgLinAlg
10.1k1521
answered Jan 15 at 14:55
LinAlgLinAlg
10.1k1521
10.1k1521
add a comment |
add a comment |
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$begingroup$
Looks good to me
$endgroup$
– angryavian
Jan 15 at 2:46
$begingroup$
Yes, it's correct.
$endgroup$
– Ryan Greyling
Jan 15 at 2:58