Mixing
convex function - global minimum
$begingroup$
Suppose that $f(x):R^p rightarrow R$ is a convex function with global minimum, say 0.
Let $C=(x: f(x)=0)$, i.e. the set of the global minimum. Suppose that there exist at least one point $y$ such that $f(y) = 0$,
It is easy to see that $C$ is convex subset.
Let $a_{lambda}$ such that $f(a_{lambda})$ approach 0 as $lambda$ approach 0 and let $a_{lambda}^1$ be the closest point to $a_{lambda}$ in $C$.
Prove that $|a_{lambda}^1-a_{lambda}|$ converge to 0, i.e. $a_{lambda}$ approaches $C$ as $lambda$ approach 0.
real-analysis convex-analysis convex-optimization
asked Feb 7 '15 at 20:50
nirnir
13610
$endgroup$
add a comment |
$begingroup$
Suppose that $f(x):R^p rightarrow R$ is a convex function with global minimum, say 0.
Let $C=(x: f(x)=0)$, i.e. the set of the global minimum. Suppose that there exist at least one point $y$ such that $f(y) = 0$,
It is easy to see that $C$ is convex subset.
Let $a_{lambda}$ such that $f(a_{lambda})$ approach 0 as $lambda$ approach 0 and let $a_{lambda}^1$ be the closest point to $a_{lambda}$ in $C$.
Prove that $|a_{lambda}^1-a_{lambda}|$ converge to 0, i.e. $a_{lambda}$ approaches $C$ as $lambda$ approach 0.
real-analysis convex-analysis convex-optimization
asked Feb 7 '15 at 20:50
nirnir
13610
$endgroup$
$begingroup$
What about $f(x)=e^x$?
$endgroup$
– daw
Feb 7 '15 at 20:53
$begingroup$
the global minimum is attinable, for $e^x$ the global minimum is 0 but for $x=-infty$,
$endgroup$
– nir
Feb 7 '15 at 20:56
$begingroup$
I saw the following: math.stackexchange.com/questions/345865/…
$endgroup$
– nir
Feb 7 '15 at 22:10
$begingroup$
Maybe, I need to add another restriction. If I assume that C is compact then it is probably true by argument like the link above
$endgroup$
– nir
Feb 7 '15 at 22:18
add a comment |
$begingroup$
Suppose that $f(x):R^p rightarrow R$ is a convex function with global minimum, say 0.
Let $C=(x: f(x)=0)$, i.e. the set of the global minimum. Suppose that there exist at least one point $y$ such that $f(y) = 0$,
It is easy to see that $C$ is convex subset.
Let $a_{lambda}$ such that $f(a_{lambda})$ approach 0 as $lambda$ approach 0 and let $a_{lambda}^1$ be the closest point to $a_{lambda}$ in $C$.
Prove that $|a_{lambda}^1-a_{lambda}|$ converge to 0, i.e. $a_{lambda}$ approaches $C$ as $lambda$ approach 0.
real-analysis convex-analysis convex-optimization
asked Feb 7 '15 at 20:50
nirnir
13610
$endgroup$
Suppose that $f(x):R^p rightarrow R$ is a convex function with global minimum, say 0.
Let $C=(x: f(x)=0)$, i.e. the set of the global minimum. Suppose that there exist at least one point $y$ such that $f(y) = 0$,
It is easy to see that $C$ is convex subset.
Let $a_{lambda}$ such that $f(a_{lambda})$ approach 0 as $lambda$ approach 0 and let $a_{lambda}^1$ be the closest point to $a_{lambda}$ in $C$.
Prove that $|a_{lambda}^1-a_{lambda}|$ converge to 0, i.e. $a_{lambda}$ approaches $C$ as $lambda$ approach 0.
real-analysis convex-analysis convex-optimization
real-analysis convex-analysis convex-optimization
asked Feb 7 '15 at 20:50
nirnir
13610
asked Feb 7 '15 at 20:50
nirnir
13610
edited Feb 7 '15 at 20:57
nir
asked Feb 7 '15 at 20:50
nirnir
13610
asked Feb 7 '15 at 20:50
nirnir
13610
asked Feb 7 '15 at 20:50
nirnir
13610
13610
$begingroup$
What about $f(x)=e^x$?
$endgroup$
– daw
Feb 7 '15 at 20:53
$begingroup$
the global minimum is attinable, for $e^x$ the global minimum is 0 but for $x=-infty$,
$endgroup$
– nir
Feb 7 '15 at 20:56
$begingroup$
I saw the following: math.stackexchange.com/questions/345865/…
$endgroup$
– nir
Feb 7 '15 at 22:10
$begingroup$
Maybe, I need to add another restriction. If I assume that C is compact then it is probably true by argument like the link above
$endgroup$
– nir
Feb 7 '15 at 22:18
add a comment |
$begingroup$
What about $f(x)=e^x$?
$endgroup$
– daw
Feb 7 '15 at 20:53
$begingroup$
the global minimum is attinable, for $e^x$ the global minimum is 0 but for $x=-infty$,
$endgroup$
– nir
Feb 7 '15 at 20:56
$begingroup$
I saw the following: math.stackexchange.com/questions/345865/…
$endgroup$
– nir
Feb 7 '15 at 22:10
$begingroup$
Maybe, I need to add another restriction. If I assume that C is compact then it is probably true by argument like the link above
$endgroup$
– nir
Feb 7 '15 at 22:18
$begingroup$
What about $f(x)=e^x$?
$endgroup$
– daw
Feb 7 '15 at 20:53
$begingroup$
What about $f(x)=e^x$?
$endgroup$
– daw
Feb 7 '15 at 20:53
$begingroup$
the global minimum is attinable, for $e^x$ the global minimum is 0 but for $x=-infty$,
$endgroup$
– nir
Feb 7 '15 at 20:56
$begingroup$
the global minimum is attinable, for $e^x$ the global minimum is 0 but for $x=-infty$,
$endgroup$
– nir
Feb 7 '15 at 20:56
$begingroup$
I saw the following: math.stackexchange.com/questions/345865/…
$endgroup$
– nir
Feb 7 '15 at 22:10
$begingroup$
I saw the following: math.stackexchange.com/questions/345865/…
$endgroup$
– nir
Feb 7 '15 at 22:10
$begingroup$
Maybe, I need to add another restriction. If I assume that C is compact then it is probably true by argument like the link above
$endgroup$
– nir
Feb 7 '15 at 22:18
$begingroup$
Maybe, I need to add another restriction. If I assume that C is compact then it is probably true by argument like the link above
$endgroup$
– nir
Feb 7 '15 at 22:18
add a comment |
1 Answer
1
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oldest
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The answer is NO:
Consider in $X = mathbb{R}^2$ the function $f(x,y) = x+sqrt{x^2+y^2}$.
Then the set of minimizers $C=mathbb{R}_-times{0}$.
Set $mathbf{x}_n=(-n,1)$.
Then $f(mathbf{x}_n)to 0$ but the distance from $mathbf{x}_n$ to $C$ is $1$.
This is Example 11.24 in Bauschke-Combettes Convex Analysis and Monotone Operator Theory, second edition. That section contains other (some even crazier) examples.
answered Jan 31 at 22:14
max_zornmax_zorn
3,44061429
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The answer is NO:
Consider in $X = mathbb{R}^2$ the function $f(x,y) = x+sqrt{x^2+y^2}$.
Then the set of minimizers $C=mathbb{R}_-times{0}$.
Set $mathbf{x}_n=(-n,1)$.
Then $f(mathbf{x}_n)to 0$ but the distance from $mathbf{x}_n$ to $C$ is $1$.
This is Example 11.24 in Bauschke-Combettes Convex Analysis and Monotone Operator Theory, second edition. That section contains other (some even crazier) examples.
answered Jan 31 at 22:14
max_zornmax_zorn
3,44061429
$endgroup$
add a comment |
$begingroup$
The answer is NO:
Consider in $X = mathbb{R}^2$ the function $f(x,y) = x+sqrt{x^2+y^2}$.
Then the set of minimizers $C=mathbb{R}_-times{0}$.
Set $mathbf{x}_n=(-n,1)$.
Then $f(mathbf{x}_n)to 0$ but the distance from $mathbf{x}_n$ to $C$ is $1$.
This is Example 11.24 in Bauschke-Combettes Convex Analysis and Monotone Operator Theory, second edition. That section contains other (some even crazier) examples.
answered Jan 31 at 22:14
max_zornmax_zorn
3,44061429
$endgroup$
add a comment |
$begingroup$
The answer is NO:
Consider in $X = mathbb{R}^2$ the function $f(x,y) = x+sqrt{x^2+y^2}$.
Then the set of minimizers $C=mathbb{R}_-times{0}$.
Set $mathbf{x}_n=(-n,1)$.
Then $f(mathbf{x}_n)to 0$ but the distance from $mathbf{x}_n$ to $C$ is $1$.
This is Example 11.24 in Bauschke-Combettes Convex Analysis and Monotone Operator Theory, second edition. That section contains other (some even crazier) examples.
answered Jan 31 at 22:14
max_zornmax_zorn
3,44061429
$endgroup$
The answer is NO:
Consider in $X = mathbb{R}^2$ the function $f(x,y) = x+sqrt{x^2+y^2}$.
Then the set of minimizers $C=mathbb{R}_-times{0}$.
Set $mathbf{x}_n=(-n,1)$.
Then $f(mathbf{x}_n)to 0$ but the distance from $mathbf{x}_n$ to $C$ is $1$.
This is Example 11.24 in Bauschke-Combettes Convex Analysis and Monotone Operator Theory, second edition. That section contains other (some even crazier) examples.
answered Jan 31 at 22:14
max_zornmax_zorn
3,44061429
answered Jan 31 at 22:14
max_zornmax_zorn
3,44061429
answered Jan 31 at 22:14
max_zornmax_zorn
3,44061429
answered Jan 31 at 22:14
max_zornmax_zorn
3,44061429
3,44061429
add a comment |
add a comment |
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$begingroup$
What about $f(x)=e^x$?
$endgroup$
– daw
Feb 7 '15 at 20:53
$begingroup$
the global minimum is attinable, for $e^x$ the global minimum is 0 but for $x=-infty$,
$endgroup$
– nir
Feb 7 '15 at 20:56
$begingroup$
I saw the following: math.stackexchange.com/questions/345865/…
$endgroup$
– nir
Feb 7 '15 at 22:10
$begingroup$
Maybe, I need to add another restriction. If I assume that C is compact then it is probably true by argument like the link above
$endgroup$
– nir
Feb 7 '15 at 22:18