Strong convexity
I have to prove to things about strong convexity and I want to know if my demonstrations are corrects:
1: All of it in $R^n$.I have $f(x)$ strongly convex in $R^n$ and $g(x)=x-w$ with $w$ different from $0$. Then $f circ g$ is strongly convex.
2:I have $f(x)$ strongly convex in $R^n$ and $g(x)=Ax-w$ where $g(x)$ belong to $R^m$, the matrix $A$ is $ntimes m$ and different from $0$, and $w$ belongs to $R^n$.Then $fcirc g$ is strong convexity.
Prove:
1- I thought that when I do $f circ g$ what I´m doing is a translation, and geometric properties of $f$ will not change with it. So $f circ g$ is strongly convex like $f$.
2- I use the idea that $f circ g$ can be a rotation if I take $w=0$ and $A$ like the matrix rotation. So if I take $A$ like the matrix rotation that rotates $f$ through $180$ degrees I can said it would not be a convex function and then will not be strongly convex.
If some explanation is not so clear just tell me and I will try to solve it.
Thanks in advance for your help.
geometry convex-optimization
edited Dec 26 at 10:40
postmortes
1,78511016
asked Dec 26 at 10:20
Meliodas
194
New contributor
Meliodas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I have to prove to things about strong convexity and I want to know if my demonstrations are corrects:
1: All of it in $R^n$.I have $f(x)$ strongly convex in $R^n$ and $g(x)=x-w$ with $w$ different from $0$. Then $f circ g$ is strongly convex.
2:I have $f(x)$ strongly convex in $R^n$ and $g(x)=Ax-w$ where $g(x)$ belong to $R^m$, the matrix $A$ is $ntimes m$ and different from $0$, and $w$ belongs to $R^n$.Then $fcirc g$ is strong convexity.
Prove:
1- I thought that when I do $f circ g$ what I´m doing is a translation, and geometric properties of $f$ will not change with it. So $f circ g$ is strongly convex like $f$.
2- I use the idea that $f circ g$ can be a rotation if I take $w=0$ and $A$ like the matrix rotation. So if I take $A$ like the matrix rotation that rotates $f$ through $180$ degrees I can said it would not be a convex function and then will not be strongly convex.
If some explanation is not so clear just tell me and I will try to solve it.
Thanks in advance for your help.
geometry convex-optimization
edited Dec 26 at 10:40
postmortes
1,78511016
asked Dec 26 at 10:20
Meliodas
194
New contributor
Meliodas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I have to prove to things about strong convexity and I want to know if my demonstrations are corrects:
1: All of it in $R^n$.I have $f(x)$ strongly convex in $R^n$ and $g(x)=x-w$ with $w$ different from $0$. Then $f circ g$ is strongly convex.
2:I have $f(x)$ strongly convex in $R^n$ and $g(x)=Ax-w$ where $g(x)$ belong to $R^m$, the matrix $A$ is $ntimes m$ and different from $0$, and $w$ belongs to $R^n$.Then $fcirc g$ is strong convexity.
Prove:
1- I thought that when I do $f circ g$ what I´m doing is a translation, and geometric properties of $f$ will not change with it. So $f circ g$ is strongly convex like $f$.
2- I use the idea that $f circ g$ can be a rotation if I take $w=0$ and $A$ like the matrix rotation. So if I take $A$ like the matrix rotation that rotates $f$ through $180$ degrees I can said it would not be a convex function and then will not be strongly convex.
If some explanation is not so clear just tell me and I will try to solve it.
Thanks in advance for your help.
geometry convex-optimization
edited Dec 26 at 10:40
postmortes
1,78511016
asked Dec 26 at 10:20
Meliodas
194
New contributor
Meliodas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have to prove to things about strong convexity and I want to know if my demonstrations are corrects:
1: All of it in $R^n$.I have $f(x)$ strongly convex in $R^n$ and $g(x)=x-w$ with $w$ different from $0$. Then $f circ g$ is strongly convex.
2:I have $f(x)$ strongly convex in $R^n$ and $g(x)=Ax-w$ where $g(x)$ belong to $R^m$, the matrix $A$ is $ntimes m$ and different from $0$, and $w$ belongs to $R^n$.Then $fcirc g$ is strong convexity.
Prove:
1- I thought that when I do $f circ g$ what I´m doing is a translation, and geometric properties of $f$ will not change with it. So $f circ g$ is strongly convex like $f$.
2- I use the idea that $f circ g$ can be a rotation if I take $w=0$ and $A$ like the matrix rotation. So if I take $A$ like the matrix rotation that rotates $f$ through $180$ degrees I can said it would not be a convex function and then will not be strongly convex.
If some explanation is not so clear just tell me and I will try to solve it.
Thanks in advance for your help.
geometry convex-optimization
geometry convex-optimization
edited Dec 26 at 10:40
postmortes
1,78511016
asked Dec 26 at 10:20
Meliodas
194
New contributor
Meliodas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 26 at 10:40
postmortes
1,78511016
asked Dec 26 at 10:20
Meliodas
194
New contributor
Meliodas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 26 at 10:40
postmortes
1,78511016
edited Dec 26 at 10:40
postmortes
1,78511016
edited Dec 26 at 10:40
postmortes
1,78511016
1,78511016
asked Dec 26 at 10:20
Meliodas
194
New contributor
Meliodas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 26 at 10:20
Meliodas
194
asked Dec 26 at 10:20
Meliodas
194
194
New contributor
Meliodas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Meliodas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Meliodas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
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votes
Your argument for (2) is not correct. If $g$ is a rotation and $f$ (strongly) convex then $f circ g$ is (strongly) convex as well.
Generally, $g(x) = Ax + w$ with $A in Bbb R^{ntimes m}$ and $w in Bbb R^n$ is an affine transformation from $ Bbb R^m$ to $ Bbb R^n$. In particular,
$$
g(lambda x + (1-lambda) y) = lambda g(x)+ (1-lambda) g(y)
$$
for $x, y in Bbb R^m$ and $0 < lambda < 1$. Therefore, if $f : Bbb R^n to Bbb R$ is convex then
$$
f(g(lambda x + (1-lambda) y)) = f( lambda g(x)+ (1-lambda) g(y)) \
le lambda f(g(x))+ (1-lambda) f(g(y))
$$
which shows that $f circ g$ is convex as well.
If $f$ is strongly convex and $A$ has full column rank then $f circ g$ is strongly convex as well: $g$ is injective, so that $x ne y$ implies
$g(x) ne g(y)$, and therefore
$$
f(g(lambda x + (1-lambda) y)) = f( lambda g(x)+ (1-lambda) g(y)) \
< lambda f(g(x))+ (1-lambda) f(g(y))
$$
for $x ne y in Bbb R^m$ and $0 < lambda < 1$.
answered Dec 26 at 11:48
Martin R
27k33252
I understand, thanks and I will try to follow all the steps of your prove. And about my first argument?
– Meliodas
Dec 26 at 14:01
@Meliodas: Your conclusion in (1) is current, but I would not call "what I'm doing is a translation, and geometric properties of f will not change" a strict proof. Note also that (1) is a special case of (2).
– Martin R
Dec 26 at 14:25
thanks, yes is not a very strict prove. I wanted to prove it without the calcul but I understand is better if I do all the calculus work.
– Meliodas
Dec 26 at 14:35
add a comment |
Your Answer
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1 Answer
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1 Answer
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Your argument for (2) is not correct. If $g$ is a rotation and $f$ (strongly) convex then $f circ g$ is (strongly) convex as well.
Generally, $g(x) = Ax + w$ with $A in Bbb R^{ntimes m}$ and $w in Bbb R^n$ is an affine transformation from $ Bbb R^m$ to $ Bbb R^n$. In particular,
$$
g(lambda x + (1-lambda) y) = lambda g(x)+ (1-lambda) g(y)
$$
for $x, y in Bbb R^m$ and $0 < lambda < 1$. Therefore, if $f : Bbb R^n to Bbb R$ is convex then
$$
f(g(lambda x + (1-lambda) y)) = f( lambda g(x)+ (1-lambda) g(y)) \
le lambda f(g(x))+ (1-lambda) f(g(y))
$$
which shows that $f circ g$ is convex as well.
If $f$ is strongly convex and $A$ has full column rank then $f circ g$ is strongly convex as well: $g$ is injective, so that $x ne y$ implies
$g(x) ne g(y)$, and therefore
$$
f(g(lambda x + (1-lambda) y)) = f( lambda g(x)+ (1-lambda) g(y)) \
< lambda f(g(x))+ (1-lambda) f(g(y))
$$
for $x ne y in Bbb R^m$ and $0 < lambda < 1$.
answered Dec 26 at 11:48
Martin R
27k33252
I understand, thanks and I will try to follow all the steps of your prove. And about my first argument?
– Meliodas
Dec 26 at 14:01
@Meliodas: Your conclusion in (1) is current, but I would not call "what I'm doing is a translation, and geometric properties of f will not change" a strict proof. Note also that (1) is a special case of (2).
– Martin R
Dec 26 at 14:25
thanks, yes is not a very strict prove. I wanted to prove it without the calcul but I understand is better if I do all the calculus work.
– Meliodas
Dec 26 at 14:35
add a comment |
Your argument for (2) is not correct. If $g$ is a rotation and $f$ (strongly) convex then $f circ g$ is (strongly) convex as well.
Generally, $g(x) = Ax + w$ with $A in Bbb R^{ntimes m}$ and $w in Bbb R^n$ is an affine transformation from $ Bbb R^m$ to $ Bbb R^n$. In particular,
$$
g(lambda x + (1-lambda) y) = lambda g(x)+ (1-lambda) g(y)
$$
for $x, y in Bbb R^m$ and $0 < lambda < 1$. Therefore, if $f : Bbb R^n to Bbb R$ is convex then
$$
f(g(lambda x + (1-lambda) y)) = f( lambda g(x)+ (1-lambda) g(y)) \
le lambda f(g(x))+ (1-lambda) f(g(y))
$$
which shows that $f circ g$ is convex as well.
If $f$ is strongly convex and $A$ has full column rank then $f circ g$ is strongly convex as well: $g$ is injective, so that $x ne y$ implies
$g(x) ne g(y)$, and therefore
$$
f(g(lambda x + (1-lambda) y)) = f( lambda g(x)+ (1-lambda) g(y)) \
< lambda f(g(x))+ (1-lambda) f(g(y))
$$
for $x ne y in Bbb R^m$ and $0 < lambda < 1$.
answered Dec 26 at 11:48
Martin R
27k33252
I understand, thanks and I will try to follow all the steps of your prove. And about my first argument?
– Meliodas
Dec 26 at 14:01
@Meliodas: Your conclusion in (1) is current, but I would not call "what I'm doing is a translation, and geometric properties of f will not change" a strict proof. Note also that (1) is a special case of (2).
– Martin R
Dec 26 at 14:25
thanks, yes is not a very strict prove. I wanted to prove it without the calcul but I understand is better if I do all the calculus work.
– Meliodas
Dec 26 at 14:35
add a comment |
Your argument for (2) is not correct. If $g$ is a rotation and $f$ (strongly) convex then $f circ g$ is (strongly) convex as well.
Generally, $g(x) = Ax + w$ with $A in Bbb R^{ntimes m}$ and $w in Bbb R^n$ is an affine transformation from $ Bbb R^m$ to $ Bbb R^n$. In particular,
$$
g(lambda x + (1-lambda) y) = lambda g(x)+ (1-lambda) g(y)
$$
for $x, y in Bbb R^m$ and $0 < lambda < 1$. Therefore, if $f : Bbb R^n to Bbb R$ is convex then
$$
f(g(lambda x + (1-lambda) y)) = f( lambda g(x)+ (1-lambda) g(y)) \
le lambda f(g(x))+ (1-lambda) f(g(y))
$$
which shows that $f circ g$ is convex as well.
If $f$ is strongly convex and $A$ has full column rank then $f circ g$ is strongly convex as well: $g$ is injective, so that $x ne y$ implies
$g(x) ne g(y)$, and therefore
$$
f(g(lambda x + (1-lambda) y)) = f( lambda g(x)+ (1-lambda) g(y)) \
< lambda f(g(x))+ (1-lambda) f(g(y))
$$
for $x ne y in Bbb R^m$ and $0 < lambda < 1$.
answered Dec 26 at 11:48
Martin R
27k33252
Your argument for (2) is not correct. If $g$ is a rotation and $f$ (strongly) convex then $f circ g$ is (strongly) convex as well.
Generally, $g(x) = Ax + w$ with $A in Bbb R^{ntimes m}$ and $w in Bbb R^n$ is an affine transformation from $ Bbb R^m$ to $ Bbb R^n$. In particular,
$$
g(lambda x + (1-lambda) y) = lambda g(x)+ (1-lambda) g(y)
$$
for $x, y in Bbb R^m$ and $0 < lambda < 1$. Therefore, if $f : Bbb R^n to Bbb R$ is convex then
$$
f(g(lambda x + (1-lambda) y)) = f( lambda g(x)+ (1-lambda) g(y)) \
le lambda f(g(x))+ (1-lambda) f(g(y))
$$
which shows that $f circ g$ is convex as well.
If $f$ is strongly convex and $A$ has full column rank then $f circ g$ is strongly convex as well: $g$ is injective, so that $x ne y$ implies
$g(x) ne g(y)$, and therefore
$$
f(g(lambda x + (1-lambda) y)) = f( lambda g(x)+ (1-lambda) g(y)) \
< lambda f(g(x))+ (1-lambda) f(g(y))
$$
for $x ne y in Bbb R^m$ and $0 < lambda < 1$.
answered Dec 26 at 11:48
Martin R
27k33252
answered Dec 26 at 11:48
Martin R
27k33252
answered Dec 26 at 11:48
Martin R
27k33252
answered Dec 26 at 11:48
Martin R
27k33252
27k33252
I understand, thanks and I will try to follow all the steps of your prove. And about my first argument?
– Meliodas
Dec 26 at 14:01
@Meliodas: Your conclusion in (1) is current, but I would not call "what I'm doing is a translation, and geometric properties of f will not change" a strict proof. Note also that (1) is a special case of (2).
– Martin R
Dec 26 at 14:25
thanks, yes is not a very strict prove. I wanted to prove it without the calcul but I understand is better if I do all the calculus work.
– Meliodas
Dec 26 at 14:35
add a comment |
I understand, thanks and I will try to follow all the steps of your prove. And about my first argument?
– Meliodas
Dec 26 at 14:01
@Meliodas: Your conclusion in (1) is current, but I would not call "what I'm doing is a translation, and geometric properties of f will not change" a strict proof. Note also that (1) is a special case of (2).
– Martin R
Dec 26 at 14:25
thanks, yes is not a very strict prove. I wanted to prove it without the calcul but I understand is better if I do all the calculus work.
– Meliodas
Dec 26 at 14:35
I understand, thanks and I will try to follow all the steps of your prove. And about my first argument?
– Meliodas
Dec 26 at 14:01
I understand, thanks and I will try to follow all the steps of your prove. And about my first argument?
– Meliodas
Dec 26 at 14:01
@Meliodas: Your conclusion in (1) is current, but I would not call "what I'm doing is a translation, and geometric properties of f will not change" a strict proof. Note also that (1) is a special case of (2).
– Martin R
Dec 26 at 14:25
@Meliodas: Your conclusion in (1) is current, but I would not call "what I'm doing is a translation, and geometric properties of f will not change" a strict proof. Note also that (1) is a special case of (2).
– Martin R
Dec 26 at 14:25
thanks, yes is not a very strict prove. I wanted to prove it without the calcul but I understand is better if I do all the calculus work.
– Meliodas
Dec 26 at 14:35
thanks, yes is not a very strict prove. I wanted to prove it without the calcul but I understand is better if I do all the calculus work.
– Meliodas
Dec 26 at 14:35
add a comment |
Meliodas is a new contributor. Be nice, and check out our Code of Conduct.
Meliodas is a new contributor. Be nice, and check out our Code of Conduct.
Meliodas is a new contributor. Be nice, and check out our Code of Conduct.
Meliodas is a new contributor. Be nice, and check out our Code of Conduct.
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