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Why is $ sum_k=1^n x_k^2 = 1 $ not a convex set?
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We know that a function is convex if it can be written as $$sum_k=1^n lambda_k mathbfg_k(mathbfx) $$
for every $lambda_k geq 0$ and $mathbfg_k(mathbfx)$ is a convex function.
In our set, the function (I ignore the constant 1) is $$ mathbff(mathbfx) =sum_k=1^n lambda_k mathbfg_k(mathbfx) = sum_k=1^n x_k^2 $$ which is a convex function since $x_k^2$ is convex for every $x$.
Since the function that is defining the set is convex, the set must be convex. But this set is not convex, why?
convex-analysis
edited Aug 2 at 15:39
Cornman
2,27521027
asked Aug 2 at 15:30
SwedeGustaf
828
add a comment |Â
up vote
1
down vote
favorite
We know that a function is convex if it can be written as $$sum_k=1^n lambda_k mathbfg_k(mathbfx) $$
for every $lambda_k geq 0$ and $mathbfg_k(mathbfx)$ is a convex function.
In our set, the function (I ignore the constant 1) is $$ mathbff(mathbfx) =sum_k=1^n lambda_k mathbfg_k(mathbfx) = sum_k=1^n x_k^2 $$ which is a convex function since $x_k^2$ is convex for every $x$.
Since the function that is defining the set is convex, the set must be convex. But this set is not convex, why?
convex-analysis
edited Aug 2 at 15:39
Cornman
2,27521027
asked Aug 2 at 15:30
SwedeGustaf
828
Do you mean $sum_k=1^n x^2_k=1$?
– Cornman
Aug 2 at 15:37
Yes, thank you Cornman.
– SwedeGustaf
Aug 2 at 15:43
Perhaps a confusion of convex function and convex set.
– GEdgar
Aug 2 at 18:35
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
We know that a function is convex if it can be written as $$sum_k=1^n lambda_k mathbfg_k(mathbfx) $$
for every $lambda_k geq 0$ and $mathbfg_k(mathbfx)$ is a convex function.
In our set, the function (I ignore the constant 1) is $$ mathbff(mathbfx) =sum_k=1^n lambda_k mathbfg_k(mathbfx) = sum_k=1^n x_k^2 $$ which is a convex function since $x_k^2$ is convex for every $x$.
Since the function that is defining the set is convex, the set must be convex. But this set is not convex, why?
convex-analysis
edited Aug 2 at 15:39
Cornman
2,27521027
asked Aug 2 at 15:30
SwedeGustaf
828
We know that a function is convex if it can be written as $$sum_k=1^n lambda_k mathbfg_k(mathbfx) $$
for every $lambda_k geq 0$ and $mathbfg_k(mathbfx)$ is a convex function.
In our set, the function (I ignore the constant 1) is $$ mathbff(mathbfx) =sum_k=1^n lambda_k mathbfg_k(mathbfx) = sum_k=1^n x_k^2 $$ which is a convex function since $x_k^2$ is convex for every $x$.
Since the function that is defining the set is convex, the set must be convex. But this set is not convex, why?
convex-analysis
edited Aug 2 at 15:39
Cornman
2,27521027
asked Aug 2 at 15:30
SwedeGustaf
828
edited Aug 2 at 15:39
Cornman
2,27521027
edited Aug 2 at 15:39
Cornman
2,27521027
edited Aug 2 at 15:39
Cornman
2,27521027
2,27521027
asked Aug 2 at 15:30
SwedeGustaf
828
asked Aug 2 at 15:30
SwedeGustaf
828
asked Aug 2 at 15:30
SwedeGustaf
828
828
Do you mean $sum_k=1^n x^2_k=1$?
– Cornman
Aug 2 at 15:37
Yes, thank you Cornman.
– SwedeGustaf
Aug 2 at 15:43
Perhaps a confusion of convex function and convex set.
– GEdgar
Aug 2 at 18:35
add a comment |Â
Do you mean $sum_k=1^n x^2_k=1$?
– Cornman
Aug 2 at 15:37
Yes, thank you Cornman.
– SwedeGustaf
Aug 2 at 15:43
Perhaps a confusion of convex function and convex set.
– GEdgar
Aug 2 at 18:35
Do you mean $sum_k=1^n x^2_k=1$?
– Cornman
Aug 2 at 15:37
Do you mean $sum_k=1^n x^2_k=1$?
– Cornman
Aug 2 at 15:37
Yes, thank you Cornman.
– SwedeGustaf
Aug 2 at 15:43
Yes, thank you Cornman.
– SwedeGustaf
Aug 2 at 15:43
Perhaps a confusion of convex function and convex set.
– GEdgar
Aug 2 at 18:35
Perhaps a confusion of convex function and convex set.
– GEdgar
Aug 2 at 18:35
add a comment |Â
5 Answers
5
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oldest
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2
down vote
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When we say that a function $f: mathbbR^n rightarrow mathbbR$ is convex, what we mean is that the set of everything above its graph - i.e. $ f(mathbfx) leq y $ - is convex.
Your set is defined by a function, but not as the space above the graph, so there's no reason to expect it to be convex.
answered Aug 2 at 15:40
Chessanator
1,592210
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Your set is the surface of the unit ball, which clearly isn't convex because it misses the origin (and it also misses the entire open unit ball).
answered Aug 2 at 15:36
Math_QED
6,30831344
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You are looking at the boundary of the set $ x:f(x) le 1$, which is convex (the set, not it's boundary).
answered Aug 2 at 15:36
Thomas
15.7k21429
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0
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I believe you compare apples with pears, or here: convex functions with convex sets.
A subset of $mathbbR$ is convex if it is path connected. Thus for every two arbitrary points of the set the points on the line in between must also belong to the set.
This is certainly not the case for a $n$-dimensional unit sphere $S_n$:
$$
e_1, -e_1 in S_n quad 0 notin S
$$
answered Aug 2 at 15:37
mvw
30.2k22250
add a comment |Â
up vote
0
down vote
To add to the other answers, work from the definition. A set $S$ is convex if for any two points $(x,y)$ in the set $S$, $theta x + (1-theta) y in S ;forall theta in [0,1]$. So the line from $x$ to $y$ has to lie completely in $S$. Check this for $x=(1,0,dots,0),y=(0,1,0,dots,0)$ both in $mathbbR^n$ and your set $S=xin mathbbR^n$. Then $theta^2 1 + (1-theta)^2 1 = 1 - 2theta $ needs to equal $1$, which implies it cannot hold for all $thetain [0,1]$. This already shows the claim of $S$ being convex is false.
answered Aug 2 at 16:00
WalterJ
704610
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
When we say that a function $f: mathbbR^n rightarrow mathbbR$ is convex, what we mean is that the set of everything above its graph - i.e. $ f(mathbfx) leq y $ - is convex.
Your set is defined by a function, but not as the space above the graph, so there's no reason to expect it to be convex.
answered Aug 2 at 15:40
Chessanator
1,592210
add a comment |Â
up vote
2
down vote
accepted
When we say that a function $f: mathbbR^n rightarrow mathbbR$ is convex, what we mean is that the set of everything above its graph - i.e. $ f(mathbfx) leq y $ - is convex.
Your set is defined by a function, but not as the space above the graph, so there's no reason to expect it to be convex.
answered Aug 2 at 15:40
Chessanator
1,592210
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
When we say that a function $f: mathbbR^n rightarrow mathbbR$ is convex, what we mean is that the set of everything above its graph - i.e. $ f(mathbfx) leq y $ - is convex.
Your set is defined by a function, but not as the space above the graph, so there's no reason to expect it to be convex.
answered Aug 2 at 15:40
Chessanator
1,592210
When we say that a function $f: mathbbR^n rightarrow mathbbR$ is convex, what we mean is that the set of everything above its graph - i.e. $ f(mathbfx) leq y $ - is convex.
Your set is defined by a function, but not as the space above the graph, so there's no reason to expect it to be convex.
answered Aug 2 at 15:40
Chessanator
1,592210
answered Aug 2 at 15:40
Chessanator
1,592210
answered Aug 2 at 15:40
Chessanator
1,592210
answered Aug 2 at 15:40
Chessanator
1,592210
1,592210
add a comment |Â
add a comment |Â
up vote
2
down vote
Your set is the surface of the unit ball, which clearly isn't convex because it misses the origin (and it also misses the entire open unit ball).
answered Aug 2 at 15:36
Math_QED
6,30831344
add a comment |Â
up vote
2
down vote
Your set is the surface of the unit ball, which clearly isn't convex because it misses the origin (and it also misses the entire open unit ball).
answered Aug 2 at 15:36
Math_QED
6,30831344
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Your set is the surface of the unit ball, which clearly isn't convex because it misses the origin (and it also misses the entire open unit ball).
answered Aug 2 at 15:36
Math_QED
6,30831344
Your set is the surface of the unit ball, which clearly isn't convex because it misses the origin (and it also misses the entire open unit ball).
answered Aug 2 at 15:36
Math_QED
6,30831344
answered Aug 2 at 15:36
Math_QED
6,30831344
answered Aug 2 at 15:36
Math_QED
6,30831344
answered Aug 2 at 15:36
Math_QED
6,30831344
6,30831344
add a comment |Â
add a comment |Â
up vote
0
down vote
You are looking at the boundary of the set $ x:f(x) le 1$, which is convex (the set, not it's boundary).
answered Aug 2 at 15:36
Thomas
15.7k21429
add a comment |Â
up vote
0
down vote
You are looking at the boundary of the set $ x:f(x) le 1$, which is convex (the set, not it's boundary).
answered Aug 2 at 15:36
Thomas
15.7k21429
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You are looking at the boundary of the set $ x:f(x) le 1$, which is convex (the set, not it's boundary).
answered Aug 2 at 15:36
Thomas
15.7k21429
You are looking at the boundary of the set $ x:f(x) le 1$, which is convex (the set, not it's boundary).
answered Aug 2 at 15:36
Thomas
15.7k21429
answered Aug 2 at 15:36
Thomas
15.7k21429
answered Aug 2 at 15:36
Thomas
15.7k21429
answered Aug 2 at 15:36
Thomas
15.7k21429
15.7k21429
add a comment |Â
add a comment |Â
up vote
0
down vote
I believe you compare apples with pears, or here: convex functions with convex sets.
A subset of $mathbbR$ is convex if it is path connected. Thus for every two arbitrary points of the set the points on the line in between must also belong to the set.
This is certainly not the case for a $n$-dimensional unit sphere $S_n$:
$$
e_1, -e_1 in S_n quad 0 notin S
$$
answered Aug 2 at 15:37
mvw
30.2k22250
add a comment |Â
up vote
0
down vote
I believe you compare apples with pears, or here: convex functions with convex sets.
A subset of $mathbbR$ is convex if it is path connected. Thus for every two arbitrary points of the set the points on the line in between must also belong to the set.
This is certainly not the case for a $n$-dimensional unit sphere $S_n$:
$$
e_1, -e_1 in S_n quad 0 notin S
$$
answered Aug 2 at 15:37
mvw
30.2k22250
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I believe you compare apples with pears, or here: convex functions with convex sets.
A subset of $mathbbR$ is convex if it is path connected. Thus for every two arbitrary points of the set the points on the line in between must also belong to the set.
This is certainly not the case for a $n$-dimensional unit sphere $S_n$:
$$
e_1, -e_1 in S_n quad 0 notin S
$$
answered Aug 2 at 15:37
mvw
30.2k22250
I believe you compare apples with pears, or here: convex functions with convex sets.
A subset of $mathbbR$ is convex if it is path connected. Thus for every two arbitrary points of the set the points on the line in between must also belong to the set.
This is certainly not the case for a $n$-dimensional unit sphere $S_n$:
$$
e_1, -e_1 in S_n quad 0 notin S
$$
answered Aug 2 at 15:37
mvw
30.2k22250
edited Aug 2 at 16:02
answered Aug 2 at 15:37
mvw
30.2k22250
answered Aug 2 at 15:37
mvw
30.2k22250
answered Aug 2 at 15:37
mvw
30.2k22250
30.2k22250
add a comment |Â
add a comment |Â
up vote
0
down vote
To add to the other answers, work from the definition. A set $S$ is convex if for any two points $(x,y)$ in the set $S$, $theta x + (1-theta) y in S ;forall theta in [0,1]$. So the line from $x$ to $y$ has to lie completely in $S$. Check this for $x=(1,0,dots,0),y=(0,1,0,dots,0)$ both in $mathbbR^n$ and your set $S=xin mathbbR^n$. Then $theta^2 1 + (1-theta)^2 1 = 1 - 2theta $ needs to equal $1$, which implies it cannot hold for all $thetain [0,1]$. This already shows the claim of $S$ being convex is false.
answered Aug 2 at 16:00
WalterJ
704610
add a comment |Â
up vote
0
down vote
To add to the other answers, work from the definition. A set $S$ is convex if for any two points $(x,y)$ in the set $S$, $theta x + (1-theta) y in S ;forall theta in [0,1]$. So the line from $x$ to $y$ has to lie completely in $S$. Check this for $x=(1,0,dots,0),y=(0,1,0,dots,0)$ both in $mathbbR^n$ and your set $S=xin mathbbR^n$. Then $theta^2 1 + (1-theta)^2 1 = 1 - 2theta $ needs to equal $1$, which implies it cannot hold for all $thetain [0,1]$. This already shows the claim of $S$ being convex is false.
answered Aug 2 at 16:00
WalterJ
704610
add a comment |Â
up vote
0
down vote
up vote
0
down vote
To add to the other answers, work from the definition. A set $S$ is convex if for any two points $(x,y)$ in the set $S$, $theta x + (1-theta) y in S ;forall theta in [0,1]$. So the line from $x$ to $y$ has to lie completely in $S$. Check this for $x=(1,0,dots,0),y=(0,1,0,dots,0)$ both in $mathbbR^n$ and your set $S=xin mathbbR^n$. Then $theta^2 1 + (1-theta)^2 1 = 1 - 2theta $ needs to equal $1$, which implies it cannot hold for all $thetain [0,1]$. This already shows the claim of $S$ being convex is false.
answered Aug 2 at 16:00
WalterJ
704610
To add to the other answers, work from the definition. A set $S$ is convex if for any two points $(x,y)$ in the set $S$, $theta x + (1-theta) y in S ;forall theta in [0,1]$. So the line from $x$ to $y$ has to lie completely in $S$. Check this for $x=(1,0,dots,0),y=(0,1,0,dots,0)$ both in $mathbbR^n$ and your set $S=xin mathbbR^n$. Then $theta^2 1 + (1-theta)^2 1 = 1 - 2theta $ needs to equal $1$, which implies it cannot hold for all $thetain [0,1]$. This already shows the claim of $S$ being convex is false.
answered Aug 2 at 16:00
WalterJ
704610
edited Aug 2 at 16:08
answered Aug 2 at 16:00
WalterJ
704610
answered Aug 2 at 16:00
WalterJ
704610
answered Aug 2 at 16:00
WalterJ
704610
704610
add a comment |Â
add a comment |Â
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Do you mean $sum_k=1^n x^2_k=1$?
– Cornman
Aug 2 at 15:37
Yes, thank you Cornman.
– SwedeGustaf
Aug 2 at 15:43
Perhaps a confusion of convex function and convex set.
– GEdgar
Aug 2 at 18:35