For a convex function $f$, is the following set convex: $X := {x ∶ -f(x) leq 1 }$? [closed]
0
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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 Nov 30 '18 at 9:52
Brahadeesh
6,15742361
asked Nov 30 '18 at 9:45
Vuro HVuro H
1
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closed as off-topic by GNUSupporter 8964民主女神 地下教會, Saad, Rebellos, José Carlos Santos, user10354138 Nov 30 '18 at 16:38
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, Saad, 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 |
0
$begingroup$
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 Nov 30 '18 at 9:52
Brahadeesh
6,15742361
asked Nov 30 '18 at 9:45
Vuro HVuro H
1
$endgroup$
closed as off-topic by GNUSupporter 8964民主女神 地下教會, Saad, Rebellos, José Carlos Santos, user10354138 Nov 30 '18 at 16:38
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, Saad, 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 |
0
0
0
$begingroup$
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 Nov 30 '18 at 9:52
Brahadeesh
6,15742361
asked Nov 30 '18 at 9:45
Vuro HVuro H
1
$endgroup$
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 Nov 30 '18 at 9:52
Brahadeesh
6,15742361
asked Nov 30 '18 at 9:45
Vuro HVuro H
1
edited Nov 30 '18 at 9:52
Brahadeesh
6,15742361
asked Nov 30 '18 at 9:45
Vuro HVuro H
1
edited Nov 30 '18 at 9:52
Brahadeesh
6,15742361
edited Nov 30 '18 at 9:52
Brahadeesh
6,15742361
edited Nov 30 '18 at 9:52
Brahadeesh
6,15742361
6,15742361
asked Nov 30 '18 at 9:45
Vuro HVuro H
1
asked Nov 30 '18 at 9:45
Vuro HVuro H
1
asked Nov 30 '18 at 9:45
Vuro HVuro H
1
1
closed as off-topic by GNUSupporter 8964民主女神 地下教會, Saad, Rebellos, José Carlos Santos, user10354138 Nov 30 '18 at 16:38
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, Saad, Rebellos, José Carlos Santos, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by GNUSupporter 8964民主女神 地下教會, Saad, Rebellos, José Carlos Santos, user10354138 Nov 30 '18 at 16:38
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, Saad, 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
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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 Nov 30 '18 at 9:53
plattyplatty
3,370320
$endgroup$
$begingroup$
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.
$endgroup$
– Sujit Bhattacharyya
Nov 30 '18 at 9:57
$begingroup$
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.
$endgroup$
– platty
Nov 30 '18 at 9:59
$begingroup$
Oh! yes, thanks for the tip. I forgot the second definition.
$endgroup$
– Sujit Bhattacharyya
Nov 30 '18 at 10:02
$begingroup$
Twice differentiability is a criterion, not a definition. There are non differentiable convex functions.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Nov 30 '18 at 10:06
$begingroup$
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.
$endgroup$
– platty
Nov 30 '18 at 10:09
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
3
$begingroup$
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 Nov 30 '18 at 9:53
plattyplatty
3,370320
$endgroup$
$begingroup$
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.
$endgroup$
– Sujit Bhattacharyya
Nov 30 '18 at 9:57
$begingroup$
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.
$endgroup$
– platty
Nov 30 '18 at 9:59
$begingroup$
Oh! yes, thanks for the tip. I forgot the second definition.
$endgroup$
– Sujit Bhattacharyya
Nov 30 '18 at 10:02
$begingroup$
Twice differentiability is a criterion, not a definition. There are non differentiable convex functions.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Nov 30 '18 at 10:06
$begingroup$
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.
$endgroup$
– platty
Nov 30 '18 at 10:09
add a comment |
3
$begingroup$
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 Nov 30 '18 at 9:53
plattyplatty
3,370320
$endgroup$
$begingroup$
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.
$endgroup$
– Sujit Bhattacharyya
Nov 30 '18 at 9:57
$begingroup$
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.
$endgroup$
– platty
Nov 30 '18 at 9:59
$begingroup$
Oh! yes, thanks for the tip. I forgot the second definition.
$endgroup$
– Sujit Bhattacharyya
Nov 30 '18 at 10:02
$begingroup$
Twice differentiability is a criterion, not a definition. There are non differentiable convex functions.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Nov 30 '18 at 10:06
$begingroup$
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.
$endgroup$
– platty
Nov 30 '18 at 10:09
add a comment |
3
3
3
$begingroup$
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 Nov 30 '18 at 9:53
plattyplatty
3,370320
$endgroup$
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 Nov 30 '18 at 9:53
plattyplatty
3,370320
answered Nov 30 '18 at 9:53
plattyplatty
3,370320
answered Nov 30 '18 at 9:53
plattyplatty
3,370320
answered Nov 30 '18 at 9:53
plattyplatty
3,370320
3,370320
$begingroup$
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.
$endgroup$
– Sujit Bhattacharyya
Nov 30 '18 at 9:57
$begingroup$
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.
$endgroup$
– platty
Nov 30 '18 at 9:59
$begingroup$
Oh! yes, thanks for the tip. I forgot the second definition.
$endgroup$
– Sujit Bhattacharyya
Nov 30 '18 at 10:02
$begingroup$
Twice differentiability is a criterion, not a definition. There are non differentiable convex functions.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Nov 30 '18 at 10:06
$begingroup$
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.
$endgroup$
– platty
Nov 30 '18 at 10:09
add a comment |
$begingroup$
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.
$endgroup$
– Sujit Bhattacharyya
Nov 30 '18 at 9:57
$begingroup$
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.
$endgroup$
– platty
Nov 30 '18 at 9:59
$begingroup$
Oh! yes, thanks for the tip. I forgot the second definition.
$endgroup$
– Sujit Bhattacharyya
Nov 30 '18 at 10:02
$begingroup$
Twice differentiability is a criterion, not a definition. There are non differentiable convex functions.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Nov 30 '18 at 10:06
$begingroup$
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.
$endgroup$
– platty
Nov 30 '18 at 10:09
$begingroup$
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.
$endgroup$
– Sujit Bhattacharyya
Nov 30 '18 at 9:57
$begingroup$
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.
$endgroup$
– Sujit Bhattacharyya
Nov 30 '18 at 9:57
$begingroup$
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.
$endgroup$
– platty
Nov 30 '18 at 9:59
$begingroup$
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.
$endgroup$
– platty
Nov 30 '18 at 9:59
$begingroup$
Oh! yes, thanks for the tip. I forgot the second definition.
$endgroup$
– Sujit Bhattacharyya
Nov 30 '18 at 10:02
$begingroup$
Oh! yes, thanks for the tip. I forgot the second definition.
$endgroup$
– Sujit Bhattacharyya
Nov 30 '18 at 10:02
$begingroup$
Twice differentiability is a criterion, not a definition. There are non differentiable convex functions.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Nov 30 '18 at 10:06
$begingroup$
Twice differentiability is a criterion, not a definition. There are non differentiable convex functions.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Nov 30 '18 at 10:06
$begingroup$
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.
$endgroup$
– platty
Nov 30 '18 at 10:09
$begingroup$
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.
$endgroup$
– platty
Nov 30 '18 at 10:09
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