For a convex function $f$, is the following set convex: $X := {x ∶ -f(x) leq 1 }$? [on hold]
up vote
0
down vote
favorite
For a convex function $f$, is the following set convex: $X := {x ∶ -f(x) leq 1 }$?
I know that the set $X := {x ∶ f(x) leq 1}$ is convex, but I'm unsure about the $-f(x)$ in the first set.
convex-analysis convex-optimization smooth-functions
edited 17 hours ago
Brahadeesh
5,78441957
asked 17 hours ago
Vuro H
1
New contributor
Vuro H is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by GNUSupporter 8964民主女神 地下教會, user302797, Rebellos, José Carlos Santos, user10354138 10 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – GNUSupporter 8964民主女神 地下教會, user302797, Rebellos, José Carlos Santos, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
0
down vote
favorite
For a convex function $f$, is the following set convex: $X := {x ∶ -f(x) leq 1 }$?
I know that the set $X := {x ∶ f(x) leq 1}$ is convex, but I'm unsure about the $-f(x)$ in the first set.
convex-analysis convex-optimization smooth-functions
edited 17 hours ago
Brahadeesh
5,78441957
asked 17 hours ago
Vuro H
1
New contributor
Vuro H is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by GNUSupporter 8964民主女神 地下教會, user302797, Rebellos, José Carlos Santos, user10354138 10 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – GNUSupporter 8964民主女神 地下教會, user302797, Rebellos, José Carlos Santos, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
For a convex function $f$, is the following set convex: $X := {x ∶ -f(x) leq 1 }$?
I know that the set $X := {x ∶ f(x) leq 1}$ is convex, but I'm unsure about the $-f(x)$ in the first set.
convex-analysis convex-optimization smooth-functions
edited 17 hours ago
Brahadeesh
5,78441957
asked 17 hours ago
Vuro H
1
New contributor
Vuro H is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
For a convex function $f$, is the following set convex: $X := {x ∶ -f(x) leq 1 }$?
I know that the set $X := {x ∶ f(x) leq 1}$ is convex, but I'm unsure about the $-f(x)$ in the first set.
convex-analysis convex-optimization smooth-functions
convex-analysis convex-optimization smooth-functions
edited 17 hours ago
Brahadeesh
5,78441957
asked 17 hours ago
Vuro H
1
New contributor
Vuro H is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 17 hours ago
Brahadeesh
5,78441957
asked 17 hours ago
Vuro H
1
New contributor
Vuro H is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 17 hours ago
Brahadeesh
5,78441957
edited 17 hours ago
Brahadeesh
5,78441957
edited 17 hours ago
Brahadeesh
5,78441957
5,78441957
asked 17 hours ago
Vuro H
1
New contributor
Vuro H is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 17 hours ago
Vuro H
1
asked 17 hours ago
Vuro H
1
1
New contributor
Vuro H is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Vuro H is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Vuro H is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by GNUSupporter 8964民主女神 地下教會, user302797, Rebellos, José Carlos Santos, user10354138 10 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – GNUSupporter 8964民主女神 地下教會, user302797, Rebellos, José Carlos Santos, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by GNUSupporter 8964民主女神 地下教會, user302797, Rebellos, José Carlos Santos, user10354138 10 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – GNUSupporter 8964民主女神 地下教會, user302797, Rebellos, José Carlos Santos, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
Not necessarily. Consider $f(x) = x^2 - 2$. This set would be ${x : 2 - x^2 leq 1}$, i.e. ${x : x^2 geq 1 }.$ But this set is not convex; for example, $-1$ and $1$ are both in the set, yet $frac{1}{2} (-1) + frac{1}{2} (1) = 0$ is not.
answered 17 hours ago
platty
2,323214
can you please provide a proof for $f$ being convex. I am unable to prove that $f(ax+(1-a)y)le a f(x)+(1-a)f(y)$ according as Wiki.
– Sujit Bhattacharyya
17 hours ago
You should be able to prove that $x^2$ is convex. Translating a function will not change this. Alternatively, you can use the alternative definition: $f(x)$ is twice-differentiable and its second derivative is always positive, so $f(x)$ is convex.
– platty
17 hours ago
Oh! yes, thanks for the tip. I forgot the second definition.
– Sujit Bhattacharyya
17 hours ago
Twice differentiability is a criterion, not a definition. There are non differentiable convex functions.
– GNUSupporter 8964民主女神 地下教會
17 hours ago
Right; if I'm being precise, I should specify that, if $f$ is twice-differentiable, then the second derivative being positive is equivalent to being convex. Since this holds here, this suffices as proof.
– platty
17 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Not necessarily. Consider $f(x) = x^2 - 2$. This set would be ${x : 2 - x^2 leq 1}$, i.e. ${x : x^2 geq 1 }.$ But this set is not convex; for example, $-1$ and $1$ are both in the set, yet $frac{1}{2} (-1) + frac{1}{2} (1) = 0$ is not.
answered 17 hours ago
platty
2,323214
can you please provide a proof for $f$ being convex. I am unable to prove that $f(ax+(1-a)y)le a f(x)+(1-a)f(y)$ according as Wiki.
– Sujit Bhattacharyya
17 hours ago
You should be able to prove that $x^2$ is convex. Translating a function will not change this. Alternatively, you can use the alternative definition: $f(x)$ is twice-differentiable and its second derivative is always positive, so $f(x)$ is convex.
– platty
17 hours ago
Oh! yes, thanks for the tip. I forgot the second definition.
– Sujit Bhattacharyya
17 hours ago
Twice differentiability is a criterion, not a definition. There are non differentiable convex functions.
– GNUSupporter 8964民主女神 地下教會
17 hours ago
Right; if I'm being precise, I should specify that, if $f$ is twice-differentiable, then the second derivative being positive is equivalent to being convex. Since this holds here, this suffices as proof.
– platty
17 hours ago
add a comment |
up vote
3
down vote
Not necessarily. Consider $f(x) = x^2 - 2$. This set would be ${x : 2 - x^2 leq 1}$, i.e. ${x : x^2 geq 1 }.$ But this set is not convex; for example, $-1$ and $1$ are both in the set, yet $frac{1}{2} (-1) + frac{1}{2} (1) = 0$ is not.
answered 17 hours ago
platty
2,323214
can you please provide a proof for $f$ being convex. I am unable to prove that $f(ax+(1-a)y)le a f(x)+(1-a)f(y)$ according as Wiki.
– Sujit Bhattacharyya
17 hours ago
You should be able to prove that $x^2$ is convex. Translating a function will not change this. Alternatively, you can use the alternative definition: $f(x)$ is twice-differentiable and its second derivative is always positive, so $f(x)$ is convex.
– platty
17 hours ago
Oh! yes, thanks for the tip. I forgot the second definition.
– Sujit Bhattacharyya
17 hours ago
Twice differentiability is a criterion, not a definition. There are non differentiable convex functions.
– GNUSupporter 8964民主女神 地下教會
17 hours ago
Right; if I'm being precise, I should specify that, if $f$ is twice-differentiable, then the second derivative being positive is equivalent to being convex. Since this holds here, this suffices as proof.
– platty
17 hours ago
add a comment |
up vote
3
down vote
up vote
3
down vote
Not necessarily. Consider $f(x) = x^2 - 2$. This set would be ${x : 2 - x^2 leq 1}$, i.e. ${x : x^2 geq 1 }.$ But this set is not convex; for example, $-1$ and $1$ are both in the set, yet $frac{1}{2} (-1) + frac{1}{2} (1) = 0$ is not.
answered 17 hours ago
platty
2,323214
Not necessarily. Consider $f(x) = x^2 - 2$. This set would be ${x : 2 - x^2 leq 1}$, i.e. ${x : x^2 geq 1 }.$ But this set is not convex; for example, $-1$ and $1$ are both in the set, yet $frac{1}{2} (-1) + frac{1}{2} (1) = 0$ is not.
answered 17 hours ago
platty
2,323214
answered 17 hours ago
platty
2,323214
answered 17 hours ago
platty
2,323214
answered 17 hours ago
platty
2,323214
2,323214
can you please provide a proof for $f$ being convex. I am unable to prove that $f(ax+(1-a)y)le a f(x)+(1-a)f(y)$ according as Wiki.
– Sujit Bhattacharyya
17 hours ago
You should be able to prove that $x^2$ is convex. Translating a function will not change this. Alternatively, you can use the alternative definition: $f(x)$ is twice-differentiable and its second derivative is always positive, so $f(x)$ is convex.
– platty
17 hours ago
Oh! yes, thanks for the tip. I forgot the second definition.
– Sujit Bhattacharyya
17 hours ago
Twice differentiability is a criterion, not a definition. There are non differentiable convex functions.
– GNUSupporter 8964民主女神 地下教會
17 hours ago
Right; if I'm being precise, I should specify that, if $f$ is twice-differentiable, then the second derivative being positive is equivalent to being convex. Since this holds here, this suffices as proof.
– platty
17 hours ago
add a comment |
can you please provide a proof for $f$ being convex. I am unable to prove that $f(ax+(1-a)y)le a f(x)+(1-a)f(y)$ according as Wiki.
– Sujit Bhattacharyya
17 hours ago
You should be able to prove that $x^2$ is convex. Translating a function will not change this. Alternatively, you can use the alternative definition: $f(x)$ is twice-differentiable and its second derivative is always positive, so $f(x)$ is convex.
– platty
17 hours ago
Oh! yes, thanks for the tip. I forgot the second definition.
– Sujit Bhattacharyya
17 hours ago
Twice differentiability is a criterion, not a definition. There are non differentiable convex functions.
– GNUSupporter 8964民主女神 地下教會
17 hours ago
Right; if I'm being precise, I should specify that, if $f$ is twice-differentiable, then the second derivative being positive is equivalent to being convex. Since this holds here, this suffices as proof.
– platty
17 hours ago
can you please provide a proof for $f$ being convex. I am unable to prove that $f(ax+(1-a)y)le a f(x)+(1-a)f(y)$ according as Wiki.
– Sujit Bhattacharyya
17 hours ago
can you please provide a proof for $f$ being convex. I am unable to prove that $f(ax+(1-a)y)le a f(x)+(1-a)f(y)$ according as Wiki.
– Sujit Bhattacharyya
17 hours ago
You should be able to prove that $x^2$ is convex. Translating a function will not change this. Alternatively, you can use the alternative definition: $f(x)$ is twice-differentiable and its second derivative is always positive, so $f(x)$ is convex.
– platty
17 hours ago
You should be able to prove that $x^2$ is convex. Translating a function will not change this. Alternatively, you can use the alternative definition: $f(x)$ is twice-differentiable and its second derivative is always positive, so $f(x)$ is convex.
– platty
17 hours ago
Oh! yes, thanks for the tip. I forgot the second definition.
– Sujit Bhattacharyya
17 hours ago
Oh! yes, thanks for the tip. I forgot the second definition.
– Sujit Bhattacharyya
17 hours ago
Twice differentiability is a criterion, not a definition. There are non differentiable convex functions.
– GNUSupporter 8964民主女神 地下教會
17 hours ago
Twice differentiability is a criterion, not a definition. There are non differentiable convex functions.
– GNUSupporter 8964民主女神 地下教會
17 hours ago
Right; if I'm being precise, I should specify that, if $f$ is twice-differentiable, then the second derivative being positive is equivalent to being convex. Since this holds here, this suffices as proof.
– platty
17 hours ago
Right; if I'm being precise, I should specify that, if $f$ is twice-differentiable, then the second derivative being positive is equivalent to being convex. Since this holds here, this suffices as proof.
– platty
17 hours ago
add a comment |
