Mixing
Proving existence of a maximum of a continuous function
$begingroup$
Let $U$ be an open bounded in $mathbb{R}^n$. For any function $w:Urightarrowmathbb{R}$ set
$$
operatorname{Argmax}(w):={xin U:w(x)=max_Uw}.
$$
I have to prove that, if $win C(U)$ (i.e. $w$ is continuous in $U$), $w<0$ near $partial U$, while $w(x)>0$ for some $xin U$, then $operatorname{Argmax}(w)$ is nonempty. Some hints?
Thank Your
real-analysis
asked Mar 7 '18 at 14:46
JejiJeji
34318
$endgroup$
add a comment |
$begingroup$
Let $U$ be an open bounded in $mathbb{R}^n$. For any function $w:Urightarrowmathbb{R}$ set
$$
operatorname{Argmax}(w):={xin U:w(x)=max_Uw}.
$$
I have to prove that, if $win C(U)$ (i.e. $w$ is continuous in $U$), $w<0$ near $partial U$, while $w(x)>0$ for some $xin U$, then $operatorname{Argmax}(w)$ is nonempty. Some hints?
Thank Your
real-analysis
asked Mar 7 '18 at 14:46
JejiJeji
34318
$endgroup$
$begingroup$
You should prove that $lbrace xin U: w(x) geq 0rbrace$ is closed in $U$ (and thus closed in $mathbb{R}^n$) and then use the existence of a maximum on a compact domain. (Also: Did you perhaps mean $w:Urightarrow mathbb{R}$?)
$endgroup$
– Max Freiburghaus
Mar 7 '18 at 14:50
$begingroup$
@MaxFreiburghaus Yes, sorry. It is $w:Urightarrowmathbb{R}$. I correct it now. But is the set ${w(x)geq0}$ automatically closed if $w$ is continuous in $U$?
$endgroup$
– Jeji
Mar 7 '18 at 14:54
1
$begingroup$
It's the pre-image of a closed set and therefore closed. But you need to be a bit careful in using the information about the boundary. Generally, "closed in open" is not necessarily closed. but here, the closed set is bounded away from the boundary of the open set within which it is contained.
$endgroup$
– Max Freiburghaus
Mar 7 '18 at 14:54
$begingroup$
@MaxFreiburghaus Ok, I understand that the set ${w(x)geq0}$ is closed. The information about the boundary says that this set is bounded away from the boundary. So this set is compact. Now should I apply the existence of a maximum on a compact domain?
$endgroup$
– Jeji
Mar 7 '18 at 15:09
add a comment |
$begingroup$
Let $U$ be an open bounded in $mathbb{R}^n$. For any function $w:Urightarrowmathbb{R}$ set
$$
operatorname{Argmax}(w):={xin U:w(x)=max_Uw}.
$$
I have to prove that, if $win C(U)$ (i.e. $w$ is continuous in $U$), $w<0$ near $partial U$, while $w(x)>0$ for some $xin U$, then $operatorname{Argmax}(w)$ is nonempty. Some hints?
Thank Your
real-analysis
asked Mar 7 '18 at 14:46
JejiJeji
34318
$endgroup$
Let $U$ be an open bounded in $mathbb{R}^n$. For any function $w:Urightarrowmathbb{R}$ set
$$
operatorname{Argmax}(w):={xin U:w(x)=max_Uw}.
$$
I have to prove that, if $win C(U)$ (i.e. $w$ is continuous in $U$), $w<0$ near $partial U$, while $w(x)>0$ for some $xin U$, then $operatorname{Argmax}(w)$ is nonempty. Some hints?
Thank Your
real-analysis
real-analysis
asked Mar 7 '18 at 14:46
JejiJeji
34318
asked Mar 7 '18 at 14:46
JejiJeji
34318
edited Jan 21 at 0:58
Jeji
asked Mar 7 '18 at 14:46
JejiJeji
34318
asked Mar 7 '18 at 14:46
JejiJeji
34318
asked Mar 7 '18 at 14:46
JejiJeji
34318
34318
$begingroup$
You should prove that $lbrace xin U: w(x) geq 0rbrace$ is closed in $U$ (and thus closed in $mathbb{R}^n$) and then use the existence of a maximum on a compact domain. (Also: Did you perhaps mean $w:Urightarrow mathbb{R}$?)
$endgroup$
– Max Freiburghaus
Mar 7 '18 at 14:50
$begingroup$
@MaxFreiburghaus Yes, sorry. It is $w:Urightarrowmathbb{R}$. I correct it now. But is the set ${w(x)geq0}$ automatically closed if $w$ is continuous in $U$?
$endgroup$
– Jeji
Mar 7 '18 at 14:54
1
$begingroup$
It's the pre-image of a closed set and therefore closed. But you need to be a bit careful in using the information about the boundary. Generally, "closed in open" is not necessarily closed. but here, the closed set is bounded away from the boundary of the open set within which it is contained.
$endgroup$
– Max Freiburghaus
Mar 7 '18 at 14:54
$begingroup$
@MaxFreiburghaus Ok, I understand that the set ${w(x)geq0}$ is closed. The information about the boundary says that this set is bounded away from the boundary. So this set is compact. Now should I apply the existence of a maximum on a compact domain?
$endgroup$
– Jeji
Mar 7 '18 at 15:09
add a comment |
$begingroup$
You should prove that $lbrace xin U: w(x) geq 0rbrace$ is closed in $U$ (and thus closed in $mathbb{R}^n$) and then use the existence of a maximum on a compact domain. (Also: Did you perhaps mean $w:Urightarrow mathbb{R}$?)
$endgroup$
– Max Freiburghaus
Mar 7 '18 at 14:50
$begingroup$
@MaxFreiburghaus Yes, sorry. It is $w:Urightarrowmathbb{R}$. I correct it now. But is the set ${w(x)geq0}$ automatically closed if $w$ is continuous in $U$?
$endgroup$
– Jeji
Mar 7 '18 at 14:54
1
$begingroup$
It's the pre-image of a closed set and therefore closed. But you need to be a bit careful in using the information about the boundary. Generally, "closed in open" is not necessarily closed. but here, the closed set is bounded away from the boundary of the open set within which it is contained.
$endgroup$
– Max Freiburghaus
Mar 7 '18 at 14:54
$begingroup$
@MaxFreiburghaus Ok, I understand that the set ${w(x)geq0}$ is closed. The information about the boundary says that this set is bounded away from the boundary. So this set is compact. Now should I apply the existence of a maximum on a compact domain?
$endgroup$
– Jeji
Mar 7 '18 at 15:09
$begingroup$
You should prove that $lbrace xin U: w(x) geq 0rbrace$ is closed in $U$ (and thus closed in $mathbb{R}^n$) and then use the existence of a maximum on a compact domain. (Also: Did you perhaps mean $w:Urightarrow mathbb{R}$?)
$endgroup$
– Max Freiburghaus
Mar 7 '18 at 14:50
$begingroup$
You should prove that $lbrace xin U: w(x) geq 0rbrace$ is closed in $U$ (and thus closed in $mathbb{R}^n$) and then use the existence of a maximum on a compact domain. (Also: Did you perhaps mean $w:Urightarrow mathbb{R}$?)
$endgroup$
– Max Freiburghaus
Mar 7 '18 at 14:50
$begingroup$
@MaxFreiburghaus Yes, sorry. It is $w:Urightarrowmathbb{R}$. I correct it now. But is the set ${w(x)geq0}$ automatically closed if $w$ is continuous in $U$?
$endgroup$
– Jeji
Mar 7 '18 at 14:54
$begingroup$
@MaxFreiburghaus Yes, sorry. It is $w:Urightarrowmathbb{R}$. I correct it now. But is the set ${w(x)geq0}$ automatically closed if $w$ is continuous in $U$?
$endgroup$
– Jeji
Mar 7 '18 at 14:54
1
1
$begingroup$
It's the pre-image of a closed set and therefore closed. But you need to be a bit careful in using the information about the boundary. Generally, "closed in open" is not necessarily closed. but here, the closed set is bounded away from the boundary of the open set within which it is contained.
$endgroup$
– Max Freiburghaus
Mar 7 '18 at 14:54
$begingroup$
It's the pre-image of a closed set and therefore closed. But you need to be a bit careful in using the information about the boundary. Generally, "closed in open" is not necessarily closed. but here, the closed set is bounded away from the boundary of the open set within which it is contained.
$endgroup$
– Max Freiburghaus
Mar 7 '18 at 14:54
$begingroup$
@MaxFreiburghaus Ok, I understand that the set ${w(x)geq0}$ is closed. The information about the boundary says that this set is bounded away from the boundary. So this set is compact. Now should I apply the existence of a maximum on a compact domain?
$endgroup$
– Jeji
Mar 7 '18 at 15:09
$begingroup$
@MaxFreiburghaus Ok, I understand that the set ${w(x)geq0}$ is closed. The information about the boundary says that this set is bounded away from the boundary. So this set is compact. Now should I apply the existence of a maximum on a compact domain?
$endgroup$
– Jeji
Mar 7 '18 at 15:09
add a comment |
1 Answer
1
active
oldest
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$begingroup$
Since $ w < 0 $ near $ partial U $ and $ w(x) > 0 $ for some $ x epsilon U $, we know by the intermediate value theorem that $ 0 epsilon w (U) $. Then, $ w^{-1} [0, infty) $ is nonempty. But $ [0, infty) $ is a closed set, then, because $ w $ is continuous, $ w^{-1} [0, infty) $ is closed. Let's call this set A. Since $ U $ is bounded and $ A subset U $, $ A $ is closed and bounded, and therefore compact. Then, $ w | A $ (w restricted to $ A $) being a continuous function in a compact set, it reaches it's maximum by Weierstrass' theorem, that is,
$$ exists : u : epsilon : U : st. w(u) = max_{x epsilon A} w(x) $$
Since $ w < 0 $ in $ U-A $, this maximum in $ A $ should be a maximum in $ U $ too, that is,
$$ w(u) = max_{x epsilon U} w(x) $$
Therefore
$$ u : epsilon : argmax_{x epsilon U} w(x) $$
showing this set is nonempty.
Edit: I think the nontrivial part is that the only thing I proved about $ A $ is that it is closed relative to $ U $. But to prove that $ A $ is compact I need to show that it is bounded and closed with respect to $ mathbb R $. There you use that $ w < 0 $ close to the boundary of $ U $, because this means that the set $ w >= 0 $ is contained in a closed set inside $ U $. That is, $ A $ is a closed set with respect to a closed set and hence closed in $ mathbb R $.
answered Mar 7 '18 at 15:12
David MoselerDavid Moseler
1314
$endgroup$
1
$begingroup$
We know that $w^{-1}[0,infty)$ is closed in the space $U$ because $w:Uto Bbb R$ is continuous. But you must show that it is closed in $Bbb R^n$ to conclude that it is compact.
$endgroup$
– DanielWainfleet
Mar 8 '18 at 1:53
1
$begingroup$
@DanielWainfleet probably he shows that $A$ is closed in $mathbb{R}^n$ at the end of the post (see the edit). Anyway thank you.
$endgroup$
– Jeji
Mar 8 '18 at 11:02
$begingroup$
To the proposer: The edit appeared after my comment.... I interpret "$; w<0$ near $partial U;$" to mean $(partial U) cap overline {w^{-1}[0,infty )}=emptyset.$.... So if $v$ is a limit point of $S=w^{-1}[0,infty)$ then $vin (overline U)setminus partial U=U$ (as $U$ is open), so $vin dom (w) cap overline S,$ so $vin S$ by continuity of $w$... (Closure bars denoting closure in $Bbb R^n.)$
$endgroup$
– DanielWainfleet
Mar 8 '18 at 21:42
$begingroup$
This is a nice way to put it, @DanielWainfleet. To exploit the argument of relative closedness, however, my idea would be to use the fact mentioned by you that, denoting $ S = w^{-1} [0, infty) $, $ (partial U) cap bar{S} = emptyset $ to take $ a = inf S $, $ b = sup S $, and note that there is some $ epsilon > 0 $ for which $ S subset [a - epsilon, a + epsilon] subset U $. S being closed with respect to U means S is closed with respect with this closed interval, which in turn is closed with respect to $ mathbb R $, making S closed in $ mathbb R $.
$endgroup$
– David Moseler
Mar 10 '18 at 4:55
add a comment |
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$begingroup$
Since $ w < 0 $ near $ partial U $ and $ w(x) > 0 $ for some $ x epsilon U $, we know by the intermediate value theorem that $ 0 epsilon w (U) $. Then, $ w^{-1} [0, infty) $ is nonempty. But $ [0, infty) $ is a closed set, then, because $ w $ is continuous, $ w^{-1} [0, infty) $ is closed. Let's call this set A. Since $ U $ is bounded and $ A subset U $, $ A $ is closed and bounded, and therefore compact. Then, $ w | A $ (w restricted to $ A $) being a continuous function in a compact set, it reaches it's maximum by Weierstrass' theorem, that is,
$$ exists : u : epsilon : U : st. w(u) = max_{x epsilon A} w(x) $$
Since $ w < 0 $ in $ U-A $, this maximum in $ A $ should be a maximum in $ U $ too, that is,
$$ w(u) = max_{x epsilon U} w(x) $$
Therefore
$$ u : epsilon : argmax_{x epsilon U} w(x) $$
showing this set is nonempty.
Edit: I think the nontrivial part is that the only thing I proved about $ A $ is that it is closed relative to $ U $. But to prove that $ A $ is compact I need to show that it is bounded and closed with respect to $ mathbb R $. There you use that $ w < 0 $ close to the boundary of $ U $, because this means that the set $ w >= 0 $ is contained in a closed set inside $ U $. That is, $ A $ is a closed set with respect to a closed set and hence closed in $ mathbb R $.
answered Mar 7 '18 at 15:12
David MoselerDavid Moseler
1314
$endgroup$
1
$begingroup$
We know that $w^{-1}[0,infty)$ is closed in the space $U$ because $w:Uto Bbb R$ is continuous. But you must show that it is closed in $Bbb R^n$ to conclude that it is compact.
$endgroup$
– DanielWainfleet
Mar 8 '18 at 1:53
1
$begingroup$
@DanielWainfleet probably he shows that $A$ is closed in $mathbb{R}^n$ at the end of the post (see the edit). Anyway thank you.
$endgroup$
– Jeji
Mar 8 '18 at 11:02
$begingroup$
To the proposer: The edit appeared after my comment.... I interpret "$; w<0$ near $partial U;$" to mean $(partial U) cap overline {w^{-1}[0,infty )}=emptyset.$.... So if $v$ is a limit point of $S=w^{-1}[0,infty)$ then $vin (overline U)setminus partial U=U$ (as $U$ is open), so $vin dom (w) cap overline S,$ so $vin S$ by continuity of $w$... (Closure bars denoting closure in $Bbb R^n.)$
$endgroup$
– DanielWainfleet
Mar 8 '18 at 21:42
$begingroup$
This is a nice way to put it, @DanielWainfleet. To exploit the argument of relative closedness, however, my idea would be to use the fact mentioned by you that, denoting $ S = w^{-1} [0, infty) $, $ (partial U) cap bar{S} = emptyset $ to take $ a = inf S $, $ b = sup S $, and note that there is some $ epsilon > 0 $ for which $ S subset [a - epsilon, a + epsilon] subset U $. S being closed with respect to U means S is closed with respect with this closed interval, which in turn is closed with respect to $ mathbb R $, making S closed in $ mathbb R $.
$endgroup$
– David Moseler
Mar 10 '18 at 4:55
add a comment |
$begingroup$
Since $ w < 0 $ near $ partial U $ and $ w(x) > 0 $ for some $ x epsilon U $, we know by the intermediate value theorem that $ 0 epsilon w (U) $. Then, $ w^{-1} [0, infty) $ is nonempty. But $ [0, infty) $ is a closed set, then, because $ w $ is continuous, $ w^{-1} [0, infty) $ is closed. Let's call this set A. Since $ U $ is bounded and $ A subset U $, $ A $ is closed and bounded, and therefore compact. Then, $ w | A $ (w restricted to $ A $) being a continuous function in a compact set, it reaches it's maximum by Weierstrass' theorem, that is,
$$ exists : u : epsilon : U : st. w(u) = max_{x epsilon A} w(x) $$
Since $ w < 0 $ in $ U-A $, this maximum in $ A $ should be a maximum in $ U $ too, that is,
$$ w(u) = max_{x epsilon U} w(x) $$
Therefore
$$ u : epsilon : argmax_{x epsilon U} w(x) $$
showing this set is nonempty.
Edit: I think the nontrivial part is that the only thing I proved about $ A $ is that it is closed relative to $ U $. But to prove that $ A $ is compact I need to show that it is bounded and closed with respect to $ mathbb R $. There you use that $ w < 0 $ close to the boundary of $ U $, because this means that the set $ w >= 0 $ is contained in a closed set inside $ U $. That is, $ A $ is a closed set with respect to a closed set and hence closed in $ mathbb R $.
answered Mar 7 '18 at 15:12
David MoselerDavid Moseler
1314
$endgroup$
1
$begingroup$
We know that $w^{-1}[0,infty)$ is closed in the space $U$ because $w:Uto Bbb R$ is continuous. But you must show that it is closed in $Bbb R^n$ to conclude that it is compact.
$endgroup$
– DanielWainfleet
Mar 8 '18 at 1:53
1
$begingroup$
@DanielWainfleet probably he shows that $A$ is closed in $mathbb{R}^n$ at the end of the post (see the edit). Anyway thank you.
$endgroup$
– Jeji
Mar 8 '18 at 11:02
$begingroup$
To the proposer: The edit appeared after my comment.... I interpret "$; w<0$ near $partial U;$" to mean $(partial U) cap overline {w^{-1}[0,infty )}=emptyset.$.... So if $v$ is a limit point of $S=w^{-1}[0,infty)$ then $vin (overline U)setminus partial U=U$ (as $U$ is open), so $vin dom (w) cap overline S,$ so $vin S$ by continuity of $w$... (Closure bars denoting closure in $Bbb R^n.)$
$endgroup$
– DanielWainfleet
Mar 8 '18 at 21:42
$begingroup$
This is a nice way to put it, @DanielWainfleet. To exploit the argument of relative closedness, however, my idea would be to use the fact mentioned by you that, denoting $ S = w^{-1} [0, infty) $, $ (partial U) cap bar{S} = emptyset $ to take $ a = inf S $, $ b = sup S $, and note that there is some $ epsilon > 0 $ for which $ S subset [a - epsilon, a + epsilon] subset U $. S being closed with respect to U means S is closed with respect with this closed interval, which in turn is closed with respect to $ mathbb R $, making S closed in $ mathbb R $.
$endgroup$
– David Moseler
Mar 10 '18 at 4:55
add a comment |
$begingroup$
Since $ w < 0 $ near $ partial U $ and $ w(x) > 0 $ for some $ x epsilon U $, we know by the intermediate value theorem that $ 0 epsilon w (U) $. Then, $ w^{-1} [0, infty) $ is nonempty. But $ [0, infty) $ is a closed set, then, because $ w $ is continuous, $ w^{-1} [0, infty) $ is closed. Let's call this set A. Since $ U $ is bounded and $ A subset U $, $ A $ is closed and bounded, and therefore compact. Then, $ w | A $ (w restricted to $ A $) being a continuous function in a compact set, it reaches it's maximum by Weierstrass' theorem, that is,
$$ exists : u : epsilon : U : st. w(u) = max_{x epsilon A} w(x) $$
Since $ w < 0 $ in $ U-A $, this maximum in $ A $ should be a maximum in $ U $ too, that is,
$$ w(u) = max_{x epsilon U} w(x) $$
Therefore
$$ u : epsilon : argmax_{x epsilon U} w(x) $$
showing this set is nonempty.
Edit: I think the nontrivial part is that the only thing I proved about $ A $ is that it is closed relative to $ U $. But to prove that $ A $ is compact I need to show that it is bounded and closed with respect to $ mathbb R $. There you use that $ w < 0 $ close to the boundary of $ U $, because this means that the set $ w >= 0 $ is contained in a closed set inside $ U $. That is, $ A $ is a closed set with respect to a closed set and hence closed in $ mathbb R $.
answered Mar 7 '18 at 15:12
David MoselerDavid Moseler
1314
$endgroup$
Since $ w < 0 $ near $ partial U $ and $ w(x) > 0 $ for some $ x epsilon U $, we know by the intermediate value theorem that $ 0 epsilon w (U) $. Then, $ w^{-1} [0, infty) $ is nonempty. But $ [0, infty) $ is a closed set, then, because $ w $ is continuous, $ w^{-1} [0, infty) $ is closed. Let's call this set A. Since $ U $ is bounded and $ A subset U $, $ A $ is closed and bounded, and therefore compact. Then, $ w | A $ (w restricted to $ A $) being a continuous function in a compact set, it reaches it's maximum by Weierstrass' theorem, that is,
$$ exists : u : epsilon : U : st. w(u) = max_{x epsilon A} w(x) $$
Since $ w < 0 $ in $ U-A $, this maximum in $ A $ should be a maximum in $ U $ too, that is,
$$ w(u) = max_{x epsilon U} w(x) $$
Therefore
$$ u : epsilon : argmax_{x epsilon U} w(x) $$
showing this set is nonempty.
Edit: I think the nontrivial part is that the only thing I proved about $ A $ is that it is closed relative to $ U $. But to prove that $ A $ is compact I need to show that it is bounded and closed with respect to $ mathbb R $. There you use that $ w < 0 $ close to the boundary of $ U $, because this means that the set $ w >= 0 $ is contained in a closed set inside $ U $. That is, $ A $ is a closed set with respect to a closed set and hence closed in $ mathbb R $.
answered Mar 7 '18 at 15:12
David MoselerDavid Moseler
1314
edited Mar 7 '18 at 15:39
answered Mar 7 '18 at 15:12
David MoselerDavid Moseler
1314
answered Mar 7 '18 at 15:12
David MoselerDavid Moseler
1314
answered Mar 7 '18 at 15:12
David MoselerDavid Moseler
1314
1314
1
$begingroup$
We know that $w^{-1}[0,infty)$ is closed in the space $U$ because $w:Uto Bbb R$ is continuous. But you must show that it is closed in $Bbb R^n$ to conclude that it is compact.
$endgroup$
– DanielWainfleet
Mar 8 '18 at 1:53
1
$begingroup$
@DanielWainfleet probably he shows that $A$ is closed in $mathbb{R}^n$ at the end of the post (see the edit). Anyway thank you.
$endgroup$
– Jeji
Mar 8 '18 at 11:02
$begingroup$
To the proposer: The edit appeared after my comment.... I interpret "$; w<0$ near $partial U;$" to mean $(partial U) cap overline {w^{-1}[0,infty )}=emptyset.$.... So if $v$ is a limit point of $S=w^{-1}[0,infty)$ then $vin (overline U)setminus partial U=U$ (as $U$ is open), so $vin dom (w) cap overline S,$ so $vin S$ by continuity of $w$... (Closure bars denoting closure in $Bbb R^n.)$
$endgroup$
– DanielWainfleet
Mar 8 '18 at 21:42
$begingroup$
This is a nice way to put it, @DanielWainfleet. To exploit the argument of relative closedness, however, my idea would be to use the fact mentioned by you that, denoting $ S = w^{-1} [0, infty) $, $ (partial U) cap bar{S} = emptyset $ to take $ a = inf S $, $ b = sup S $, and note that there is some $ epsilon > 0 $ for which $ S subset [a - epsilon, a + epsilon] subset U $. S being closed with respect to U means S is closed with respect with this closed interval, which in turn is closed with respect to $ mathbb R $, making S closed in $ mathbb R $.
$endgroup$
– David Moseler
Mar 10 '18 at 4:55
add a comment |
1
$begingroup$
We know that $w^{-1}[0,infty)$ is closed in the space $U$ because $w:Uto Bbb R$ is continuous. But you must show that it is closed in $Bbb R^n$ to conclude that it is compact.
$endgroup$
– DanielWainfleet
Mar 8 '18 at 1:53
1
$begingroup$
@DanielWainfleet probably he shows that $A$ is closed in $mathbb{R}^n$ at the end of the post (see the edit). Anyway thank you.
$endgroup$
– Jeji
Mar 8 '18 at 11:02
$begingroup$
To the proposer: The edit appeared after my comment.... I interpret "$; w<0$ near $partial U;$" to mean $(partial U) cap overline {w^{-1}[0,infty )}=emptyset.$.... So if $v$ is a limit point of $S=w^{-1}[0,infty)$ then $vin (overline U)setminus partial U=U$ (as $U$ is open), so $vin dom (w) cap overline S,$ so $vin S$ by continuity of $w$... (Closure bars denoting closure in $Bbb R^n.)$
$endgroup$
– DanielWainfleet
Mar 8 '18 at 21:42
$begingroup$
This is a nice way to put it, @DanielWainfleet. To exploit the argument of relative closedness, however, my idea would be to use the fact mentioned by you that, denoting $ S = w^{-1} [0, infty) $, $ (partial U) cap bar{S} = emptyset $ to take $ a = inf S $, $ b = sup S $, and note that there is some $ epsilon > 0 $ for which $ S subset [a - epsilon, a + epsilon] subset U $. S being closed with respect to U means S is closed with respect with this closed interval, which in turn is closed with respect to $ mathbb R $, making S closed in $ mathbb R $.
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– David Moseler
Mar 10 '18 at 4:55
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We know that $w^{-1}[0,infty)$ is closed in the space $U$ because $w:Uto Bbb R$ is continuous. But you must show that it is closed in $Bbb R^n$ to conclude that it is compact.
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– DanielWainfleet
Mar 8 '18 at 1:53
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We know that $w^{-1}[0,infty)$ is closed in the space $U$ because $w:Uto Bbb R$ is continuous. But you must show that it is closed in $Bbb R^n$ to conclude that it is compact.
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– DanielWainfleet
Mar 8 '18 at 1:53
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@DanielWainfleet probably he shows that $A$ is closed in $mathbb{R}^n$ at the end of the post (see the edit). Anyway thank you.
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– Jeji
Mar 8 '18 at 11:02
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@DanielWainfleet probably he shows that $A$ is closed in $mathbb{R}^n$ at the end of the post (see the edit). Anyway thank you.
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– Jeji
Mar 8 '18 at 11:02
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To the proposer: The edit appeared after my comment.... I interpret "$; w<0$ near $partial U;$" to mean $(partial U) cap overline {w^{-1}[0,infty )}=emptyset.$.... So if $v$ is a limit point of $S=w^{-1}[0,infty)$ then $vin (overline U)setminus partial U=U$ (as $U$ is open), so $vin dom (w) cap overline S,$ so $vin S$ by continuity of $w$... (Closure bars denoting closure in $Bbb R^n.)$
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– DanielWainfleet
Mar 8 '18 at 21:42
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To the proposer: The edit appeared after my comment.... I interpret "$; w<0$ near $partial U;$" to mean $(partial U) cap overline {w^{-1}[0,infty )}=emptyset.$.... So if $v$ is a limit point of $S=w^{-1}[0,infty)$ then $vin (overline U)setminus partial U=U$ (as $U$ is open), so $vin dom (w) cap overline S,$ so $vin S$ by continuity of $w$... (Closure bars denoting closure in $Bbb R^n.)$
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– DanielWainfleet
Mar 8 '18 at 21:42
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This is a nice way to put it, @DanielWainfleet. To exploit the argument of relative closedness, however, my idea would be to use the fact mentioned by you that, denoting $ S = w^{-1} [0, infty) $, $ (partial U) cap bar{S} = emptyset $ to take $ a = inf S $, $ b = sup S $, and note that there is some $ epsilon > 0 $ for which $ S subset [a - epsilon, a + epsilon] subset U $. S being closed with respect to U means S is closed with respect with this closed interval, which in turn is closed with respect to $ mathbb R $, making S closed in $ mathbb R $.
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– David Moseler
Mar 10 '18 at 4:55
$begingroup$
This is a nice way to put it, @DanielWainfleet. To exploit the argument of relative closedness, however, my idea would be to use the fact mentioned by you that, denoting $ S = w^{-1} [0, infty) $, $ (partial U) cap bar{S} = emptyset $ to take $ a = inf S $, $ b = sup S $, and note that there is some $ epsilon > 0 $ for which $ S subset [a - epsilon, a + epsilon] subset U $. S being closed with respect to U means S is closed with respect with this closed interval, which in turn is closed with respect to $ mathbb R $, making S closed in $ mathbb R $.
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– David Moseler
Mar 10 '18 at 4:55
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You should prove that $lbrace xin U: w(x) geq 0rbrace$ is closed in $U$ (and thus closed in $mathbb{R}^n$) and then use the existence of a maximum on a compact domain. (Also: Did you perhaps mean $w:Urightarrow mathbb{R}$?)
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– Max Freiburghaus
Mar 7 '18 at 14:50
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@MaxFreiburghaus Yes, sorry. It is $w:Urightarrowmathbb{R}$. I correct it now. But is the set ${w(x)geq0}$ automatically closed if $w$ is continuous in $U$?
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– Jeji
Mar 7 '18 at 14:54
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It's the pre-image of a closed set and therefore closed. But you need to be a bit careful in using the information about the boundary. Generally, "closed in open" is not necessarily closed. but here, the closed set is bounded away from the boundary of the open set within which it is contained.
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– Max Freiburghaus
Mar 7 '18 at 14:54
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@MaxFreiburghaus Ok, I understand that the set ${w(x)geq0}$ is closed. The information about the boundary says that this set is bounded away from the boundary. So this set is compact. Now should I apply the existence of a maximum on a compact domain?
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– Jeji
Mar 7 '18 at 15:09