Mixing
Convergence of sequence of minimizers
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Suppose $mathcalX$ is compact, and consider an optimization problem: $$max_x in X_n f(x)$$ where $f$ is a continuous function, and $X_nsubseteq mathcalX$ is a sequence of compact sets converging to some set $X subseteq mathcalX$ (for example, in the Hausdorff metric). Now let $x_n$ be a sequence of minimizers of $f$; i.e. $f(x_n) = min_x in X_n f(x)$. Note that $f(x)$ may have multiple minimums.
I am interested in conditions under which the arbitrary sequence of minimizers $x_n$ converges to a point in $X subset mathcalX$. Does anybody know?
Further Background:
I often see the statement: by compactness of $mathcalX$, we can assume without loss of generality that $x_n to x in mathcalX$. My problem with this statement is that it uses the fact that $x_n$ has a convergent subsequence, and says nothing about whether my original sequence converges.
This discussion is also related to the theorem of the maximum, which I believe can be used to establish that $X^*(n) = argmin f(x) : x in X_n$ is upper hemicontinuous. However, similar to the point above, it seems to me that upper hemicontinuity is not enough to guarantee that $x_n to x$, only that it has a convergent subsequence.
real-analysis optimization
asked Jul 22 at 14:52
möbius
834718
add a comment |Â
up vote
1
down vote
favorite
Suppose $mathcalX$ is compact, and consider an optimization problem: $$max_x in X_n f(x)$$ where $f$ is a continuous function, and $X_nsubseteq mathcalX$ is a sequence of compact sets converging to some set $X subseteq mathcalX$ (for example, in the Hausdorff metric). Now let $x_n$ be a sequence of minimizers of $f$; i.e. $f(x_n) = min_x in X_n f(x)$. Note that $f(x)$ may have multiple minimums.
I am interested in conditions under which the arbitrary sequence of minimizers $x_n$ converges to a point in $X subset mathcalX$. Does anybody know?
Further Background:
I often see the statement: by compactness of $mathcalX$, we can assume without loss of generality that $x_n to x in mathcalX$. My problem with this statement is that it uses the fact that $x_n$ has a convergent subsequence, and says nothing about whether my original sequence converges.
This discussion is also related to the theorem of the maximum, which I believe can be used to establish that $X^*(n) = argmin f(x) : x in X_n$ is upper hemicontinuous. However, similar to the point above, it seems to me that upper hemicontinuity is not enough to guarantee that $x_n to x$, only that it has a convergent subsequence.
real-analysis optimization
asked Jul 22 at 14:52
möbius
834718
I may need to think about this more, but off the top of my head, you may need to restrict $f$ to have only one minimizer on the set $X$. For example, take $X_n=[0,1]$ for all $n$ (and thus $X=[0,1]$), $f(x)=x^4-x^2$, and $x_n=(-1)^n/sqrt2$. Since this sequence of minimizers is alternating, it doesn’t converge. If $f$ has two minimizers on $X$, I think this kind of construction will always be possible.
– David M.
Jul 22 at 15:01
My understanding of the statement in bold is, in fact, that it refers to the existence of a subsequence with the property stated, i.e. one can assume that one started with that subsequence right in the beginning. If you want to prove existence in $X$, this is usually sufficient. Unless you really need convergence of the original sequence (e.g because $x_n$ is related to $x_n+1$ in some way you cannot give up, or if you want to combine existence with uniqueness) for some reason I'd read it that way.
– Thomas
Jul 22 at 16:54
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose $mathcalX$ is compact, and consider an optimization problem: $$max_x in X_n f(x)$$ where $f$ is a continuous function, and $X_nsubseteq mathcalX$ is a sequence of compact sets converging to some set $X subseteq mathcalX$ (for example, in the Hausdorff metric). Now let $x_n$ be a sequence of minimizers of $f$; i.e. $f(x_n) = min_x in X_n f(x)$. Note that $f(x)$ may have multiple minimums.
I am interested in conditions under which the arbitrary sequence of minimizers $x_n$ converges to a point in $X subset mathcalX$. Does anybody know?
Further Background:
I often see the statement: by compactness of $mathcalX$, we can assume without loss of generality that $x_n to x in mathcalX$. My problem with this statement is that it uses the fact that $x_n$ has a convergent subsequence, and says nothing about whether my original sequence converges.
This discussion is also related to the theorem of the maximum, which I believe can be used to establish that $X^*(n) = argmin f(x) : x in X_n$ is upper hemicontinuous. However, similar to the point above, it seems to me that upper hemicontinuity is not enough to guarantee that $x_n to x$, only that it has a convergent subsequence.
real-analysis optimization
asked Jul 22 at 14:52
möbius
834718
Suppose $mathcalX$ is compact, and consider an optimization problem: $$max_x in X_n f(x)$$ where $f$ is a continuous function, and $X_nsubseteq mathcalX$ is a sequence of compact sets converging to some set $X subseteq mathcalX$ (for example, in the Hausdorff metric). Now let $x_n$ be a sequence of minimizers of $f$; i.e. $f(x_n) = min_x in X_n f(x)$. Note that $f(x)$ may have multiple minimums.
I am interested in conditions under which the arbitrary sequence of minimizers $x_n$ converges to a point in $X subset mathcalX$. Does anybody know?
Further Background:
I often see the statement: by compactness of $mathcalX$, we can assume without loss of generality that $x_n to x in mathcalX$. My problem with this statement is that it uses the fact that $x_n$ has a convergent subsequence, and says nothing about whether my original sequence converges.
This discussion is also related to the theorem of the maximum, which I believe can be used to establish that $X^*(n) = argmin f(x) : x in X_n$ is upper hemicontinuous. However, similar to the point above, it seems to me that upper hemicontinuity is not enough to guarantee that $x_n to x$, only that it has a convergent subsequence.
real-analysis optimization
asked Jul 22 at 14:52
möbius
834718
asked Jul 22 at 14:52
möbius
834718
asked Jul 22 at 14:52
möbius
834718
asked Jul 22 at 14:52
möbius
834718
834718
I may need to think about this more, but off the top of my head, you may need to restrict $f$ to have only one minimizer on the set $X$. For example, take $X_n=[0,1]$ for all $n$ (and thus $X=[0,1]$), $f(x)=x^4-x^2$, and $x_n=(-1)^n/sqrt2$. Since this sequence of minimizers is alternating, it doesn’t converge. If $f$ has two minimizers on $X$, I think this kind of construction will always be possible.
– David M.
Jul 22 at 15:01
My understanding of the statement in bold is, in fact, that it refers to the existence of a subsequence with the property stated, i.e. one can assume that one started with that subsequence right in the beginning. If you want to prove existence in $X$, this is usually sufficient. Unless you really need convergence of the original sequence (e.g because $x_n$ is related to $x_n+1$ in some way you cannot give up, or if you want to combine existence with uniqueness) for some reason I'd read it that way.
– Thomas
Jul 22 at 16:54
add a comment |Â
I may need to think about this more, but off the top of my head, you may need to restrict $f$ to have only one minimizer on the set $X$. For example, take $X_n=[0,1]$ for all $n$ (and thus $X=[0,1]$), $f(x)=x^4-x^2$, and $x_n=(-1)^n/sqrt2$. Since this sequence of minimizers is alternating, it doesn’t converge. If $f$ has two minimizers on $X$, I think this kind of construction will always be possible.
– David M.
Jul 22 at 15:01
My understanding of the statement in bold is, in fact, that it refers to the existence of a subsequence with the property stated, i.e. one can assume that one started with that subsequence right in the beginning. If you want to prove existence in $X$, this is usually sufficient. Unless you really need convergence of the original sequence (e.g because $x_n$ is related to $x_n+1$ in some way you cannot give up, or if you want to combine existence with uniqueness) for some reason I'd read it that way.
– Thomas
Jul 22 at 16:54
I may need to think about this more, but off the top of my head, you may need to restrict $f$ to have only one minimizer on the set $X$. For example, take $X_n=[0,1]$ for all $n$ (and thus $X=[0,1]$), $f(x)=x^4-x^2$, and $x_n=(-1)^n/sqrt2$. Since this sequence of minimizers is alternating, it doesn’t converge. If $f$ has two minimizers on $X$, I think this kind of construction will always be possible.
– David M.
Jul 22 at 15:01
I may need to think about this more, but off the top of my head, you may need to restrict $f$ to have only one minimizer on the set $X$. For example, take $X_n=[0,1]$ for all $n$ (and thus $X=[0,1]$), $f(x)=x^4-x^2$, and $x_n=(-1)^n/sqrt2$. Since this sequence of minimizers is alternating, it doesn’t converge. If $f$ has two minimizers on $X$, I think this kind of construction will always be possible.
– David M.
Jul 22 at 15:01
My understanding of the statement in bold is, in fact, that it refers to the existence of a subsequence with the property stated, i.e. one can assume that one started with that subsequence right in the beginning. If you want to prove existence in $X$, this is usually sufficient. Unless you really need convergence of the original sequence (e.g because $x_n$ is related to $x_n+1$ in some way you cannot give up, or if you want to combine existence with uniqueness) for some reason I'd read it that way.
– Thomas
Jul 22 at 16:54
My understanding of the statement in bold is, in fact, that it refers to the existence of a subsequence with the property stated, i.e. one can assume that one started with that subsequence right in the beginning. If you want to prove existence in $X$, this is usually sufficient. Unless you really need convergence of the original sequence (e.g because $x_n$ is related to $x_n+1$ in some way you cannot give up, or if you want to combine existence with uniqueness) for some reason I'd read it that way.
– Thomas
Jul 22 at 16:54
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2859473%2fconvergence-of-sequence-of-minimizers%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
I may need to think about this more, but off the top of my head, you may need to restrict $f$ to have only one minimizer on the set $X$. For example, take $X_n=[0,1]$ for all $n$ (and thus $X=[0,1]$), $f(x)=x^4-x^2$, and $x_n=(-1)^n/sqrt2$. Since this sequence of minimizers is alternating, it doesn’t converge. If $f$ has two minimizers on $X$, I think this kind of construction will always be possible.
– David M.
Jul 22 at 15:01
My understanding of the statement in bold is, in fact, that it refers to the existence of a subsequence with the property stated, i.e. one can assume that one started with that subsequence right in the beginning. If you want to prove existence in $X$, this is usually sufficient. Unless you really need convergence of the original sequence (e.g because $x_n$ is related to $x_n+1$ in some way you cannot give up, or if you want to combine existence with uniqueness) for some reason I'd read it that way.
– Thomas
Jul 22 at 16:54