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General method to prove density, continuous and compact embedding of space into another
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We say that a set $X$ is dense into another one $X'$ if for any $x$ $in$ $X'$ there exists a sequence $x_n$ that is in $X$ such that $$limlimits_nto inftyx_n=x$$ we say that a set $X$ is compactly embedded into $Y$ if from any uniformly bounded sequence $x_n$ of $X$ one can extract a subsequence $x_varphi(n)$ that converges in $Y$. Finally a set $X$ is said to be continuously embedded into $Y$ if $$|x|_Yleq |x|_X$$ whenever $x$ belongs to $X$. Now, the question is: is there any general method or even a set of several methods to postulate whether a space is dense into another or not, continuously embedded or not compactly embedded or not?
general-topology functional-analysis compact-operators
asked Jul 23 at 19:36
AlphaXY
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We say that a set $X$ is dense into another one $X'$ if for any $x$ $in$ $X'$ there exists a sequence $x_n$ that is in $X$ such that $$limlimits_nto inftyx_n=x$$ we say that a set $X$ is compactly embedded into $Y$ if from any uniformly bounded sequence $x_n$ of $X$ one can extract a subsequence $x_varphi(n)$ that converges in $Y$. Finally a set $X$ is said to be continuously embedded into $Y$ if $$|x|_Yleq |x|_X$$ whenever $x$ belongs to $X$. Now, the question is: is there any general method or even a set of several methods to postulate whether a space is dense into another or not, continuously embedded or not compactly embedded or not?
general-topology functional-analysis compact-operators
asked Jul 23 at 19:36
AlphaXY
7282226
Is your question only about vector spaces?
– William Elliot
Jul 23 at 20:39
In fact, functional spaces: Sobolev spaces
– AlphaXY
Jul 23 at 20:42
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
We say that a set $X$ is dense into another one $X'$ if for any $x$ $in$ $X'$ there exists a sequence $x_n$ that is in $X$ such that $$limlimits_nto inftyx_n=x$$ we say that a set $X$ is compactly embedded into $Y$ if from any uniformly bounded sequence $x_n$ of $X$ one can extract a subsequence $x_varphi(n)$ that converges in $Y$. Finally a set $X$ is said to be continuously embedded into $Y$ if $$|x|_Yleq |x|_X$$ whenever $x$ belongs to $X$. Now, the question is: is there any general method or even a set of several methods to postulate whether a space is dense into another or not, continuously embedded or not compactly embedded or not?
general-topology functional-analysis compact-operators
asked Jul 23 at 19:36
AlphaXY
7282226
We say that a set $X$ is dense into another one $X'$ if for any $x$ $in$ $X'$ there exists a sequence $x_n$ that is in $X$ such that $$limlimits_nto inftyx_n=x$$ we say that a set $X$ is compactly embedded into $Y$ if from any uniformly bounded sequence $x_n$ of $X$ one can extract a subsequence $x_varphi(n)$ that converges in $Y$. Finally a set $X$ is said to be continuously embedded into $Y$ if $$|x|_Yleq |x|_X$$ whenever $x$ belongs to $X$. Now, the question is: is there any general method or even a set of several methods to postulate whether a space is dense into another or not, continuously embedded or not compactly embedded or not?
general-topology functional-analysis compact-operators
asked Jul 23 at 19:36
AlphaXY
7282226
edited yesterday
asked Jul 23 at 19:36
AlphaXY
7282226
asked Jul 23 at 19:36
AlphaXY
7282226
asked Jul 23 at 19:36
AlphaXY
7282226
7282226
Is your question only about vector spaces?
– William Elliot
Jul 23 at 20:39
In fact, functional spaces: Sobolev spaces
– AlphaXY
Jul 23 at 20:42
add a comment |Â
Is your question only about vector spaces?
– William Elliot
Jul 23 at 20:39
In fact, functional spaces: Sobolev spaces
– AlphaXY
Jul 23 at 20:42
Is your question only about vector spaces?
– William Elliot
Jul 23 at 20:39
Is your question only about vector spaces?
– William Elliot
Jul 23 at 20:39
In fact, functional spaces: Sobolev spaces
– AlphaXY
Jul 23 at 20:42
In fact, functional spaces: Sobolev spaces
– AlphaXY
Jul 23 at 20:42
add a comment |Â
1 Answer
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One typical pattern of events is that we start with a "small" space $D$ (e.g., test functions), and consider its completions $H^s$ with respect to norms $|cdot|_s$, where $|cdot|_sle |cdot|_t$ for $s<t$. Then $D$ is dense in every $H^s$, by construction, $H^t$ imbeds continuously in $H^s$ for $t>s$, and since $D$ is dense in $H^s$, certainly $H^t$ is dense in $H^s$.
Proof of compactness of $H^tto H^s$ is more delicate, and is less widely true than the previous. Often, when it does hold, such as with the Hilbert-space Sobolev spaces on the circle, it's because $H^s$ and $H^t$ have a common orthogonal basis (e.g., exponentials), and the identity map can be examined explicitly as a map $ell^2to ell^2$ by multiplication by a sequence of reals $mu_n$. When the $mu_n$'s go to $0$, the map is compact.
answered Aug 3 at 21:46
paul garrett
30.8k360116
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
One typical pattern of events is that we start with a "small" space $D$ (e.g., test functions), and consider its completions $H^s$ with respect to norms $|cdot|_s$, where $|cdot|_sle |cdot|_t$ for $s<t$. Then $D$ is dense in every $H^s$, by construction, $H^t$ imbeds continuously in $H^s$ for $t>s$, and since $D$ is dense in $H^s$, certainly $H^t$ is dense in $H^s$.
Proof of compactness of $H^tto H^s$ is more delicate, and is less widely true than the previous. Often, when it does hold, such as with the Hilbert-space Sobolev spaces on the circle, it's because $H^s$ and $H^t$ have a common orthogonal basis (e.g., exponentials), and the identity map can be examined explicitly as a map $ell^2to ell^2$ by multiplication by a sequence of reals $mu_n$. When the $mu_n$'s go to $0$, the map is compact.
answered Aug 3 at 21:46
paul garrett
30.8k360116
add a comment |Â
up vote
0
down vote
One typical pattern of events is that we start with a "small" space $D$ (e.g., test functions), and consider its completions $H^s$ with respect to norms $|cdot|_s$, where $|cdot|_sle |cdot|_t$ for $s<t$. Then $D$ is dense in every $H^s$, by construction, $H^t$ imbeds continuously in $H^s$ for $t>s$, and since $D$ is dense in $H^s$, certainly $H^t$ is dense in $H^s$.
Proof of compactness of $H^tto H^s$ is more delicate, and is less widely true than the previous. Often, when it does hold, such as with the Hilbert-space Sobolev spaces on the circle, it's because $H^s$ and $H^t$ have a common orthogonal basis (e.g., exponentials), and the identity map can be examined explicitly as a map $ell^2to ell^2$ by multiplication by a sequence of reals $mu_n$. When the $mu_n$'s go to $0$, the map is compact.
answered Aug 3 at 21:46
paul garrett
30.8k360116
add a comment |Â
up vote
0
down vote
up vote
0
down vote
One typical pattern of events is that we start with a "small" space $D$ (e.g., test functions), and consider its completions $H^s$ with respect to norms $|cdot|_s$, where $|cdot|_sle |cdot|_t$ for $s<t$. Then $D$ is dense in every $H^s$, by construction, $H^t$ imbeds continuously in $H^s$ for $t>s$, and since $D$ is dense in $H^s$, certainly $H^t$ is dense in $H^s$.
Proof of compactness of $H^tto H^s$ is more delicate, and is less widely true than the previous. Often, when it does hold, such as with the Hilbert-space Sobolev spaces on the circle, it's because $H^s$ and $H^t$ have a common orthogonal basis (e.g., exponentials), and the identity map can be examined explicitly as a map $ell^2to ell^2$ by multiplication by a sequence of reals $mu_n$. When the $mu_n$'s go to $0$, the map is compact.
answered Aug 3 at 21:46
paul garrett
30.8k360116
One typical pattern of events is that we start with a "small" space $D$ (e.g., test functions), and consider its completions $H^s$ with respect to norms $|cdot|_s$, where $|cdot|_sle |cdot|_t$ for $s<t$. Then $D$ is dense in every $H^s$, by construction, $H^t$ imbeds continuously in $H^s$ for $t>s$, and since $D$ is dense in $H^s$, certainly $H^t$ is dense in $H^s$.
Proof of compactness of $H^tto H^s$ is more delicate, and is less widely true than the previous. Often, when it does hold, such as with the Hilbert-space Sobolev spaces on the circle, it's because $H^s$ and $H^t$ have a common orthogonal basis (e.g., exponentials), and the identity map can be examined explicitly as a map $ell^2to ell^2$ by multiplication by a sequence of reals $mu_n$. When the $mu_n$'s go to $0$, the map is compact.
answered Aug 3 at 21:46
paul garrett
30.8k360116
answered Aug 3 at 21:46
paul garrett
30.8k360116
answered Aug 3 at 21:46
paul garrett
30.8k360116
answered Aug 3 at 21:46
paul garrett
30.8k360116
30.8k360116
add a comment |Â
add a comment |Â
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Is your question only about vector spaces?
– William Elliot
Jul 23 at 20:39
In fact, functional spaces: Sobolev spaces
– AlphaXY
Jul 23 at 20:42