Subadditivity of Linear Functional defined via Inf
$begingroup$
Let $B$ be a the set of all real valued functions bounded by $M$. Let $Y$ be a linear subspace of non-negative functions. Define $l$ to be a positive linear functional on $Y$ and let $$p(x) = inf_{x leq y} l(y) quad forall xin B, y in Y$$ We'd like to show $p$ is subadditive.
The proof I'm looking at looks proceeds as follows:
Consider $x_1$ and $x_2$ in $B$ and $y_1$ and $y_2$ in $Y$ such that $x_1 leq y_1$ and $x_2 leq y_2$. Adding the two we obtain $x_1 + x_2 leq y_1 + y_2$ so:
$$p(x_1 + x_2) = inf_{x_1 + x_2 leq y}l(y) \ leqinf_{x_1 leq y_1 \x_2 leq y_2} l(y) quad (*)\ = inf_{x_1 leq y_1} l(y_1) + inf_{x_2 leq y_2} l(y_2) quad (*) \ p(x_1) + p(x_2)$$
I didn't understand how properties of infs were used in the two starred steps. In particular:
- Why inequality between first and second line? Its not obvious to me that the set ${x_1+x_2 leq y} subset {x_1 leq y_1 cup x_2 leq y_2}$
- What's going on with the equality from the second to third line?
Thanks!
functional-analysis convex-analysis proof-explanation
asked Dec 25 '18 at 15:04
yoshiyoshi
1,196817
$endgroup$
add a comment |
$begingroup$
Let $B$ be a the set of all real valued functions bounded by $M$. Let $Y$ be a linear subspace of non-negative functions. Define $l$ to be a positive linear functional on $Y$ and let $$p(x) = inf_{x leq y} l(y) quad forall xin B, y in Y$$ We'd like to show $p$ is subadditive.
The proof I'm looking at looks proceeds as follows:
Consider $x_1$ and $x_2$ in $B$ and $y_1$ and $y_2$ in $Y$ such that $x_1 leq y_1$ and $x_2 leq y_2$. Adding the two we obtain $x_1 + x_2 leq y_1 + y_2$ so:
$$p(x_1 + x_2) = inf_{x_1 + x_2 leq y}l(y) \ leqinf_{x_1 leq y_1 \x_2 leq y_2} l(y) quad (*)\ = inf_{x_1 leq y_1} l(y_1) + inf_{x_2 leq y_2} l(y_2) quad (*) \ p(x_1) + p(x_2)$$
I didn't understand how properties of infs were used in the two starred steps. In particular:
- Why inequality between first and second line? Its not obvious to me that the set ${x_1+x_2 leq y} subset {x_1 leq y_1 cup x_2 leq y_2}$
- What's going on with the equality from the second to third line?
Thanks!
functional-analysis convex-analysis proof-explanation
asked Dec 25 '18 at 15:04
yoshiyoshi
1,196817
$endgroup$
1
$begingroup$
What do you mean by "linear subspace of non-negative functions"? I don't see how that is a linear space. What is the ordering for functions?
$endgroup$
– LinAlg
Dec 25 '18 at 17:01
$begingroup$
Suppose $S$ is the domain of functions in $B$. there's a partial ordering on functions. $x < y$ implies $x(s) < y(s) ,, forall s in S$
$endgroup$
– yoshi
Dec 25 '18 at 17:45
1
$begingroup$
what do you mean by $inf (y)$ and how does $y$ relate to $y_1$ and $y_2$?
$endgroup$
– LinAlg
Dec 25 '18 at 18:46
$begingroup$
oh no! those are typos, they are fixed. I meant $l(y)$, $l(y_1)$, and $l(y_2)$
$endgroup$
– yoshi
Dec 25 '18 at 22:35
add a comment |
$begingroup$
Let $B$ be a the set of all real valued functions bounded by $M$. Let $Y$ be a linear subspace of non-negative functions. Define $l$ to be a positive linear functional on $Y$ and let $$p(x) = inf_{x leq y} l(y) quad forall xin B, y in Y$$ We'd like to show $p$ is subadditive.
The proof I'm looking at looks proceeds as follows:
Consider $x_1$ and $x_2$ in $B$ and $y_1$ and $y_2$ in $Y$ such that $x_1 leq y_1$ and $x_2 leq y_2$. Adding the two we obtain $x_1 + x_2 leq y_1 + y_2$ so:
$$p(x_1 + x_2) = inf_{x_1 + x_2 leq y}l(y) \ leqinf_{x_1 leq y_1 \x_2 leq y_2} l(y) quad (*)\ = inf_{x_1 leq y_1} l(y_1) + inf_{x_2 leq y_2} l(y_2) quad (*) \ p(x_1) + p(x_2)$$
I didn't understand how properties of infs were used in the two starred steps. In particular:
- Why inequality between first and second line? Its not obvious to me that the set ${x_1+x_2 leq y} subset {x_1 leq y_1 cup x_2 leq y_2}$
- What's going on with the equality from the second to third line?
Thanks!
functional-analysis convex-analysis proof-explanation
asked Dec 25 '18 at 15:04
yoshiyoshi
1,196817
$endgroup$
Let $B$ be a the set of all real valued functions bounded by $M$. Let $Y$ be a linear subspace of non-negative functions. Define $l$ to be a positive linear functional on $Y$ and let $$p(x) = inf_{x leq y} l(y) quad forall xin B, y in Y$$ We'd like to show $p$ is subadditive.
The proof I'm looking at looks proceeds as follows:
Consider $x_1$ and $x_2$ in $B$ and $y_1$ and $y_2$ in $Y$ such that $x_1 leq y_1$ and $x_2 leq y_2$. Adding the two we obtain $x_1 + x_2 leq y_1 + y_2$ so:
$$p(x_1 + x_2) = inf_{x_1 + x_2 leq y}l(y) \ leqinf_{x_1 leq y_1 \x_2 leq y_2} l(y) quad (*)\ = inf_{x_1 leq y_1} l(y_1) + inf_{x_2 leq y_2} l(y_2) quad (*) \ p(x_1) + p(x_2)$$
I didn't understand how properties of infs were used in the two starred steps. In particular:
- Why inequality between first and second line? Its not obvious to me that the set ${x_1+x_2 leq y} subset {x_1 leq y_1 cup x_2 leq y_2}$
- What's going on with the equality from the second to third line?
Thanks!
functional-analysis convex-analysis proof-explanation
functional-analysis convex-analysis proof-explanation
asked Dec 25 '18 at 15:04
yoshiyoshi
1,196817
asked Dec 25 '18 at 15:04
yoshiyoshi
1,196817
edited Dec 25 '18 at 22:34
yoshi
asked Dec 25 '18 at 15:04
yoshiyoshi
1,196817
asked Dec 25 '18 at 15:04
yoshiyoshi
1,196817
asked Dec 25 '18 at 15:04
yoshiyoshi
1,196817
1,196817
1
$begingroup$
What do you mean by "linear subspace of non-negative functions"? I don't see how that is a linear space. What is the ordering for functions?
$endgroup$
– LinAlg
Dec 25 '18 at 17:01
$begingroup$
Suppose $S$ is the domain of functions in $B$. there's a partial ordering on functions. $x < y$ implies $x(s) < y(s) ,, forall s in S$
$endgroup$
– yoshi
Dec 25 '18 at 17:45
1
$begingroup$
what do you mean by $inf (y)$ and how does $y$ relate to $y_1$ and $y_2$?
$endgroup$
– LinAlg
Dec 25 '18 at 18:46
$begingroup$
oh no! those are typos, they are fixed. I meant $l(y)$, $l(y_1)$, and $l(y_2)$
$endgroup$
– yoshi
Dec 25 '18 at 22:35
add a comment |
1
$begingroup$
What do you mean by "linear subspace of non-negative functions"? I don't see how that is a linear space. What is the ordering for functions?
$endgroup$
– LinAlg
Dec 25 '18 at 17:01
$begingroup$
Suppose $S$ is the domain of functions in $B$. there's a partial ordering on functions. $x < y$ implies $x(s) < y(s) ,, forall s in S$
$endgroup$
– yoshi
Dec 25 '18 at 17:45
1
$begingroup$
what do you mean by $inf (y)$ and how does $y$ relate to $y_1$ and $y_2$?
$endgroup$
– LinAlg
Dec 25 '18 at 18:46
$begingroup$
oh no! those are typos, they are fixed. I meant $l(y)$, $l(y_1)$, and $l(y_2)$
$endgroup$
– yoshi
Dec 25 '18 at 22:35
1
1
$begingroup$
What do you mean by "linear subspace of non-negative functions"? I don't see how that is a linear space. What is the ordering for functions?
$endgroup$
– LinAlg
Dec 25 '18 at 17:01
$begingroup$
What do you mean by "linear subspace of non-negative functions"? I don't see how that is a linear space. What is the ordering for functions?
$endgroup$
– LinAlg
Dec 25 '18 at 17:01
$begingroup$
Suppose $S$ is the domain of functions in $B$. there's a partial ordering on functions. $x < y$ implies $x(s) < y(s) ,, forall s in S$
$endgroup$
– yoshi
Dec 25 '18 at 17:45
$begingroup$
Suppose $S$ is the domain of functions in $B$. there's a partial ordering on functions. $x < y$ implies $x(s) < y(s) ,, forall s in S$
$endgroup$
– yoshi
Dec 25 '18 at 17:45
1
1
$begingroup$
what do you mean by $inf (y)$ and how does $y$ relate to $y_1$ and $y_2$?
$endgroup$
– LinAlg
Dec 25 '18 at 18:46
$begingroup$
what do you mean by $inf (y)$ and how does $y$ relate to $y_1$ and $y_2$?
$endgroup$
– LinAlg
Dec 25 '18 at 18:46
$begingroup$
oh no! those are typos, they are fixed. I meant $l(y)$, $l(y_1)$, and $l(y_2)$
$endgroup$
– yoshi
Dec 25 '18 at 22:35
$begingroup$
oh no! those are typos, they are fixed. I meant $l(y)$, $l(y_1)$, and $l(y_2)$
$endgroup$
– yoshi
Dec 25 '18 at 22:35
add a comment |
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$begingroup$
What do you mean by "linear subspace of non-negative functions"? I don't see how that is a linear space. What is the ordering for functions?
$endgroup$
– LinAlg
Dec 25 '18 at 17:01
$begingroup$
Suppose $S$ is the domain of functions in $B$. there's a partial ordering on functions. $x < y$ implies $x(s) < y(s) ,, forall s in S$
$endgroup$
– yoshi
Dec 25 '18 at 17:45
1
$begingroup$
what do you mean by $inf (y)$ and how does $y$ relate to $y_1$ and $y_2$?
$endgroup$
– LinAlg
Dec 25 '18 at 18:46
$begingroup$
oh no! those are typos, they are fixed. I meant $l(y)$, $l(y_1)$, and $l(y_2)$
$endgroup$
– yoshi
Dec 25 '18 at 22:35