Mixing
How to prove that this set is convex.
$begingroup$
Let $mathbb{C}$ and $mathbb{D}$ be two convex sets. Prove that the set $E:=bigcuplimits_{lambda in[0,1]}((1-lambda)Ccaplambda D)$ is also convex.
I have two problems here: I don't know how to manage this expresion to prove that it is convex and I can't see this type of set in $mathbb{R}^2$
convex-analysis
asked Jan 29 at 16:11
LecterLecter
11910
$endgroup$
add a comment |
$begingroup$
Let $mathbb{C}$ and $mathbb{D}$ be two convex sets. Prove that the set $E:=bigcuplimits_{lambda in[0,1]}((1-lambda)Ccaplambda D)$ is also convex.
I have two problems here: I don't know how to manage this expresion to prove that it is convex and I can't see this type of set in $mathbb{R}^2$
convex-analysis
asked Jan 29 at 16:11
LecterLecter
11910
$endgroup$
add a comment |
$begingroup$
Let $mathbb{C}$ and $mathbb{D}$ be two convex sets. Prove that the set $E:=bigcuplimits_{lambda in[0,1]}((1-lambda)Ccaplambda D)$ is also convex.
I have two problems here: I don't know how to manage this expresion to prove that it is convex and I can't see this type of set in $mathbb{R}^2$
convex-analysis
asked Jan 29 at 16:11
LecterLecter
11910
$endgroup$
Let $mathbb{C}$ and $mathbb{D}$ be two convex sets. Prove that the set $E:=bigcuplimits_{lambda in[0,1]}((1-lambda)Ccaplambda D)$ is also convex.
I have two problems here: I don't know how to manage this expresion to prove that it is convex and I can't see this type of set in $mathbb{R}^2$
convex-analysis
convex-analysis
asked Jan 29 at 16:11
LecterLecter
11910
asked Jan 29 at 16:11
LecterLecter
11910
asked Jan 29 at 16:11
LecterLecter
11910
asked Jan 29 at 16:11
LecterLecter
11910
asked Jan 29 at 16:11
LecterLecter
11910
11910
add a comment |
add a comment |
2 Answers
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oldest
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$begingroup$
Suppose $e_1,e_2in E$ and let $0le tle 1.$ There are $0le lambdale 1; c_1,c_2in C; d_1,d_2in D$ such that $e_1=(1-lambda) c_1=lambda d_1$ and similarly for $e_2.$ We need to show that $x:=(1-t)e_1+te_2in E.$
Now, it is easy to see that $(1-lambda)C$ and $lambda D$ are convex sets.
Then,
$x=(1-t)e_1+te_2=(1-t)(1-lambda) c_1+t(1-lambda) c_2in (1-lambda)C.$
and
$x=(1-t)e_1+te_2=(1-t)lambda d_1+tlambda d_2in lambda D.$
Thus, $xin (1-lambda)Ccap lambda D$, so $E$ is convex.
As for the geometric interpretation of these sets, take two intersecting disks and see what happens when you multiply one by $lambda$ and the other by $lambda$. Then, try strips. Etc. In short, play with shapes!
answered Jan 29 at 17:17
MatematletaMatematleta
12k21020
$endgroup$
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$begingroup$
Let $x,yin E$, so that there exists $lambda, muin [0,1]$ such that
$$
x = (1-lambda) c_x + lambda d_x,
qquad
y = (1-mu) c_y + mu d_y,
$$
with $c_x, c_yin C$ and $d_x, d_yin D$.
Let $tin (0,1)$ and let us prove that $(1-t)x + ty in E$.
Setting $s := (1-t)lambda + t mu$, it is easy to check that
$$
begin{gather*}
frac{(1-t)(1-lambda)}{1-s} c_x + frac{t(1-mu)}{1-s} c_y =: c_z in C,\
frac{(1-t)lambda}{s} d_x + frac{tmu}{s} d_y =: d_z in D
end{gather*}
$$
(by the convexity of $C$ and $D$), and
$$
(1-t)x + t y = (1-s) c_z + s d_z in E.
$$
(Here we have implicitly assumed that $sin (0,1)$; the cases $s=0$ and $s=1$ are trivial.)
answered Jan 29 at 17:34
RigelRigel
11.4k11320
$endgroup$
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2 Answers
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2 Answers
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$begingroup$
Suppose $e_1,e_2in E$ and let $0le tle 1.$ There are $0le lambdale 1; c_1,c_2in C; d_1,d_2in D$ such that $e_1=(1-lambda) c_1=lambda d_1$ and similarly for $e_2.$ We need to show that $x:=(1-t)e_1+te_2in E.$
Now, it is easy to see that $(1-lambda)C$ and $lambda D$ are convex sets.
Then,
$x=(1-t)e_1+te_2=(1-t)(1-lambda) c_1+t(1-lambda) c_2in (1-lambda)C.$
and
$x=(1-t)e_1+te_2=(1-t)lambda d_1+tlambda d_2in lambda D.$
Thus, $xin (1-lambda)Ccap lambda D$, so $E$ is convex.
As for the geometric interpretation of these sets, take two intersecting disks and see what happens when you multiply one by $lambda$ and the other by $lambda$. Then, try strips. Etc. In short, play with shapes!
answered Jan 29 at 17:17
MatematletaMatematleta
12k21020
$endgroup$
add a comment |
$begingroup$
Suppose $e_1,e_2in E$ and let $0le tle 1.$ There are $0le lambdale 1; c_1,c_2in C; d_1,d_2in D$ such that $e_1=(1-lambda) c_1=lambda d_1$ and similarly for $e_2.$ We need to show that $x:=(1-t)e_1+te_2in E.$
Now, it is easy to see that $(1-lambda)C$ and $lambda D$ are convex sets.
Then,
$x=(1-t)e_1+te_2=(1-t)(1-lambda) c_1+t(1-lambda) c_2in (1-lambda)C.$
and
$x=(1-t)e_1+te_2=(1-t)lambda d_1+tlambda d_2in lambda D.$
Thus, $xin (1-lambda)Ccap lambda D$, so $E$ is convex.
As for the geometric interpretation of these sets, take two intersecting disks and see what happens when you multiply one by $lambda$ and the other by $lambda$. Then, try strips. Etc. In short, play with shapes!
answered Jan 29 at 17:17
MatematletaMatematleta
12k21020
$endgroup$
add a comment |
$begingroup$
Suppose $e_1,e_2in E$ and let $0le tle 1.$ There are $0le lambdale 1; c_1,c_2in C; d_1,d_2in D$ such that $e_1=(1-lambda) c_1=lambda d_1$ and similarly for $e_2.$ We need to show that $x:=(1-t)e_1+te_2in E.$
Now, it is easy to see that $(1-lambda)C$ and $lambda D$ are convex sets.
Then,
$x=(1-t)e_1+te_2=(1-t)(1-lambda) c_1+t(1-lambda) c_2in (1-lambda)C.$
and
$x=(1-t)e_1+te_2=(1-t)lambda d_1+tlambda d_2in lambda D.$
Thus, $xin (1-lambda)Ccap lambda D$, so $E$ is convex.
As for the geometric interpretation of these sets, take two intersecting disks and see what happens when you multiply one by $lambda$ and the other by $lambda$. Then, try strips. Etc. In short, play with shapes!
answered Jan 29 at 17:17
MatematletaMatematleta
12k21020
$endgroup$
Suppose $e_1,e_2in E$ and let $0le tle 1.$ There are $0le lambdale 1; c_1,c_2in C; d_1,d_2in D$ such that $e_1=(1-lambda) c_1=lambda d_1$ and similarly for $e_2.$ We need to show that $x:=(1-t)e_1+te_2in E.$
Now, it is easy to see that $(1-lambda)C$ and $lambda D$ are convex sets.
Then,
$x=(1-t)e_1+te_2=(1-t)(1-lambda) c_1+t(1-lambda) c_2in (1-lambda)C.$
and
$x=(1-t)e_1+te_2=(1-t)lambda d_1+tlambda d_2in lambda D.$
Thus, $xin (1-lambda)Ccap lambda D$, so $E$ is convex.
As for the geometric interpretation of these sets, take two intersecting disks and see what happens when you multiply one by $lambda$ and the other by $lambda$. Then, try strips. Etc. In short, play with shapes!
answered Jan 29 at 17:17
MatematletaMatematleta
12k21020
edited Jan 29 at 17:23
answered Jan 29 at 17:17
MatematletaMatematleta
12k21020
answered Jan 29 at 17:17
MatematletaMatematleta
12k21020
answered Jan 29 at 17:17
MatematletaMatematleta
12k21020
12k21020
add a comment |
add a comment |
$begingroup$
Let $x,yin E$, so that there exists $lambda, muin [0,1]$ such that
$$
x = (1-lambda) c_x + lambda d_x,
qquad
y = (1-mu) c_y + mu d_y,
$$
with $c_x, c_yin C$ and $d_x, d_yin D$.
Let $tin (0,1)$ and let us prove that $(1-t)x + ty in E$.
Setting $s := (1-t)lambda + t mu$, it is easy to check that
$$
begin{gather*}
frac{(1-t)(1-lambda)}{1-s} c_x + frac{t(1-mu)}{1-s} c_y =: c_z in C,\
frac{(1-t)lambda}{s} d_x + frac{tmu}{s} d_y =: d_z in D
end{gather*}
$$
(by the convexity of $C$ and $D$), and
$$
(1-t)x + t y = (1-s) c_z + s d_z in E.
$$
(Here we have implicitly assumed that $sin (0,1)$; the cases $s=0$ and $s=1$ are trivial.)
answered Jan 29 at 17:34
RigelRigel
11.4k11320
$endgroup$
add a comment |
$begingroup$
Let $x,yin E$, so that there exists $lambda, muin [0,1]$ such that
$$
x = (1-lambda) c_x + lambda d_x,
qquad
y = (1-mu) c_y + mu d_y,
$$
with $c_x, c_yin C$ and $d_x, d_yin D$.
Let $tin (0,1)$ and let us prove that $(1-t)x + ty in E$.
Setting $s := (1-t)lambda + t mu$, it is easy to check that
$$
begin{gather*}
frac{(1-t)(1-lambda)}{1-s} c_x + frac{t(1-mu)}{1-s} c_y =: c_z in C,\
frac{(1-t)lambda}{s} d_x + frac{tmu}{s} d_y =: d_z in D
end{gather*}
$$
(by the convexity of $C$ and $D$), and
$$
(1-t)x + t y = (1-s) c_z + s d_z in E.
$$
(Here we have implicitly assumed that $sin (0,1)$; the cases $s=0$ and $s=1$ are trivial.)
answered Jan 29 at 17:34
RigelRigel
11.4k11320
$endgroup$
add a comment |
$begingroup$
Let $x,yin E$, so that there exists $lambda, muin [0,1]$ such that
$$
x = (1-lambda) c_x + lambda d_x,
qquad
y = (1-mu) c_y + mu d_y,
$$
with $c_x, c_yin C$ and $d_x, d_yin D$.
Let $tin (0,1)$ and let us prove that $(1-t)x + ty in E$.
Setting $s := (1-t)lambda + t mu$, it is easy to check that
$$
begin{gather*}
frac{(1-t)(1-lambda)}{1-s} c_x + frac{t(1-mu)}{1-s} c_y =: c_z in C,\
frac{(1-t)lambda}{s} d_x + frac{tmu}{s} d_y =: d_z in D
end{gather*}
$$
(by the convexity of $C$ and $D$), and
$$
(1-t)x + t y = (1-s) c_z + s d_z in E.
$$
(Here we have implicitly assumed that $sin (0,1)$; the cases $s=0$ and $s=1$ are trivial.)
answered Jan 29 at 17:34
RigelRigel
11.4k11320
$endgroup$
Let $x,yin E$, so that there exists $lambda, muin [0,1]$ such that
$$
x = (1-lambda) c_x + lambda d_x,
qquad
y = (1-mu) c_y + mu d_y,
$$
with $c_x, c_yin C$ and $d_x, d_yin D$.
Let $tin (0,1)$ and let us prove that $(1-t)x + ty in E$.
Setting $s := (1-t)lambda + t mu$, it is easy to check that
$$
begin{gather*}
frac{(1-t)(1-lambda)}{1-s} c_x + frac{t(1-mu)}{1-s} c_y =: c_z in C,\
frac{(1-t)lambda}{s} d_x + frac{tmu}{s} d_y =: d_z in D
end{gather*}
$$
(by the convexity of $C$ and $D$), and
$$
(1-t)x + t y = (1-s) c_z + s d_z in E.
$$
(Here we have implicitly assumed that $sin (0,1)$; the cases $s=0$ and $s=1$ are trivial.)
answered Jan 29 at 17:34
RigelRigel
11.4k11320
answered Jan 29 at 17:34
RigelRigel
11.4k11320
answered Jan 29 at 17:34
RigelRigel
11.4k11320
answered Jan 29 at 17:34
RigelRigel
11.4k11320
11.4k11320
add a comment |
add a comment |
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