Is the Schwartz topologically emebedded in space of tempered distributions?
3
$begingroup$
Let $gmapsto ( cdot, g)_2$ denote the map from the Schwartz space $S$ into its dual space $S'$ where $(f,g)_2$ is the inner product in $L^2$. Then is this a linear topological embedding ($S'$ is endowed with the weak* topology)? Can anyone provide a reference or a simple proof?
functional-analysis schwartz-space
asked Dec 1 '18 at 23:23
Andrew YuanAndrew Yuan
1508
$endgroup$
add a comment |
3
$begingroup$
Let $gmapsto ( cdot, g)_2$ denote the map from the Schwartz space $S$ into its dual space $S'$ where $(f,g)_2$ is the inner product in $L^2$. Then is this a linear topological embedding ($S'$ is endowed with the weak* topology)? Can anyone provide a reference or a simple proof?
functional-analysis schwartz-space
asked Dec 1 '18 at 23:23
Andrew YuanAndrew Yuan
1508
$endgroup$
add a comment |
3
3
3
1
$begingroup$
Let $gmapsto ( cdot, g)_2$ denote the map from the Schwartz space $S$ into its dual space $S'$ where $(f,g)_2$ is the inner product in $L^2$. Then is this a linear topological embedding ($S'$ is endowed with the weak* topology)? Can anyone provide a reference or a simple proof?
functional-analysis schwartz-space
asked Dec 1 '18 at 23:23
Andrew YuanAndrew Yuan
1508
$endgroup$
Let $gmapsto ( cdot, g)_2$ denote the map from the Schwartz space $S$ into its dual space $S'$ where $(f,g)_2$ is the inner product in $L^2$. Then is this a linear topological embedding ($S'$ is endowed with the weak* topology)? Can anyone provide a reference or a simple proof?
functional-analysis schwartz-space
functional-analysis schwartz-space
asked Dec 1 '18 at 23:23
Andrew YuanAndrew Yuan
1508
asked Dec 1 '18 at 23:23
Andrew YuanAndrew Yuan
1508
asked Dec 1 '18 at 23:23
Andrew YuanAndrew Yuan
1508
asked Dec 1 '18 at 23:23
Andrew YuanAndrew Yuan
1508
asked Dec 1 '18 at 23:23
Andrew YuanAndrew Yuan
1508
1508
add a comment |
add a comment |
1 Answer
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$$left|int_{mathbb{R}} f(x)g(x)dx right| le left|frac{f}{1+x^2}right|_{L^1} |(1+x^2)g|_{L^infty}le pi |f|_{L^infty}(|g|_{L^infty}+|x^2g|_{L^infty}) $$
$|.|_{L^infty}$,$|x^2.|_{L^infty}$ are in the semi-norms used to construct the Schwartz space.
answered Dec 2 '18 at 0:50
reunsreuns
19.9k21147
$endgroup$
$begingroup$
But why is the inverse continuous? Shouldn't a topological embedding be a homeomorphism onto its image?
$endgroup$
– Andrew Yuan
Dec 3 '18 at 4:00
$begingroup$
@AndrewYuan Inverse of what ? I you don't define the semi-norms, topology, dual, you won't see that it is really trivial. The Schwartz space is a subspace of the Banach space with norm $|.|_{L^infty}+|x^2.|_{L^infty}$ and as I shown that one embeds in its strong dual.
$endgroup$
– reuns
Dec 3 '18 at 4:09
$begingroup$
I mean the inverse of $gmapsto (cdot,g)_2$ on its image in $S'$, which is well-defined since its injective. And maybe I'm missing something, but I don't think the Schwartz space (with its usual topology) is a subspace of a Banach space or else it would be normable.
$endgroup$
– Andrew Yuan
Dec 3 '18 at 6:09
$begingroup$
@AndrewYuan A non-closed subspace. The Schwartz space is the intersection of infinitely many Banach spaces (with norms $sum_{|alpha| le K} | (1+|x|^M) partial_alpha f|_{L^infty}$) that's its definition. Its open sets are intersection of open sets in each of those Banach spaces.
$endgroup$
– reuns
Dec 3 '18 at 6:32
$begingroup$
Define precisely the topologies in $S,S'$ then $(g,g) = |g|_{L^2}^2 > ...$ will make it obvious that the map $g to iota(g) in S'$ is bi-continuous.
$endgroup$
– reuns
Dec 3 '18 at 6:42
add a comment |
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$begingroup$
$$left|int_{mathbb{R}} f(x)g(x)dx right| le left|frac{f}{1+x^2}right|_{L^1} |(1+x^2)g|_{L^infty}le pi |f|_{L^infty}(|g|_{L^infty}+|x^2g|_{L^infty}) $$
$|.|_{L^infty}$,$|x^2.|_{L^infty}$ are in the semi-norms used to construct the Schwartz space.
answered Dec 2 '18 at 0:50
reunsreuns
19.9k21147
$endgroup$
$begingroup$
But why is the inverse continuous? Shouldn't a topological embedding be a homeomorphism onto its image?
$endgroup$
– Andrew Yuan
Dec 3 '18 at 4:00
$begingroup$
@AndrewYuan Inverse of what ? I you don't define the semi-norms, topology, dual, you won't see that it is really trivial. The Schwartz space is a subspace of the Banach space with norm $|.|_{L^infty}+|x^2.|_{L^infty}$ and as I shown that one embeds in its strong dual.
$endgroup$
– reuns
Dec 3 '18 at 4:09
$begingroup$
I mean the inverse of $gmapsto (cdot,g)_2$ on its image in $S'$, which is well-defined since its injective. And maybe I'm missing something, but I don't think the Schwartz space (with its usual topology) is a subspace of a Banach space or else it would be normable.
$endgroup$
– Andrew Yuan
Dec 3 '18 at 6:09
$begingroup$
@AndrewYuan A non-closed subspace. The Schwartz space is the intersection of infinitely many Banach spaces (with norms $sum_{|alpha| le K} | (1+|x|^M) partial_alpha f|_{L^infty}$) that's its definition. Its open sets are intersection of open sets in each of those Banach spaces.
$endgroup$
– reuns
Dec 3 '18 at 6:32
$begingroup$
Define precisely the topologies in $S,S'$ then $(g,g) = |g|_{L^2}^2 > ...$ will make it obvious that the map $g to iota(g) in S'$ is bi-continuous.
$endgroup$
– reuns
Dec 3 '18 at 6:42
add a comment |
0
$begingroup$
$$left|int_{mathbb{R}} f(x)g(x)dx right| le left|frac{f}{1+x^2}right|_{L^1} |(1+x^2)g|_{L^infty}le pi |f|_{L^infty}(|g|_{L^infty}+|x^2g|_{L^infty}) $$
$|.|_{L^infty}$,$|x^2.|_{L^infty}$ are in the semi-norms used to construct the Schwartz space.
answered Dec 2 '18 at 0:50
reunsreuns
19.9k21147
$endgroup$
$begingroup$
But why is the inverse continuous? Shouldn't a topological embedding be a homeomorphism onto its image?
$endgroup$
– Andrew Yuan
Dec 3 '18 at 4:00
$begingroup$
@AndrewYuan Inverse of what ? I you don't define the semi-norms, topology, dual, you won't see that it is really trivial. The Schwartz space is a subspace of the Banach space with norm $|.|_{L^infty}+|x^2.|_{L^infty}$ and as I shown that one embeds in its strong dual.
$endgroup$
– reuns
Dec 3 '18 at 4:09
$begingroup$
I mean the inverse of $gmapsto (cdot,g)_2$ on its image in $S'$, which is well-defined since its injective. And maybe I'm missing something, but I don't think the Schwartz space (with its usual topology) is a subspace of a Banach space or else it would be normable.
$endgroup$
– Andrew Yuan
Dec 3 '18 at 6:09
$begingroup$
@AndrewYuan A non-closed subspace. The Schwartz space is the intersection of infinitely many Banach spaces (with norms $sum_{|alpha| le K} | (1+|x|^M) partial_alpha f|_{L^infty}$) that's its definition. Its open sets are intersection of open sets in each of those Banach spaces.
$endgroup$
– reuns
Dec 3 '18 at 6:32
$begingroup$
Define precisely the topologies in $S,S'$ then $(g,g) = |g|_{L^2}^2 > ...$ will make it obvious that the map $g to iota(g) in S'$ is bi-continuous.
$endgroup$
– reuns
Dec 3 '18 at 6:42
add a comment |
0
0
0
$begingroup$
$$left|int_{mathbb{R}} f(x)g(x)dx right| le left|frac{f}{1+x^2}right|_{L^1} |(1+x^2)g|_{L^infty}le pi |f|_{L^infty}(|g|_{L^infty}+|x^2g|_{L^infty}) $$
$|.|_{L^infty}$,$|x^2.|_{L^infty}$ are in the semi-norms used to construct the Schwartz space.
answered Dec 2 '18 at 0:50
reunsreuns
19.9k21147
$endgroup$
$$left|int_{mathbb{R}} f(x)g(x)dx right| le left|frac{f}{1+x^2}right|_{L^1} |(1+x^2)g|_{L^infty}le pi |f|_{L^infty}(|g|_{L^infty}+|x^2g|_{L^infty}) $$
$|.|_{L^infty}$,$|x^2.|_{L^infty}$ are in the semi-norms used to construct the Schwartz space.
answered Dec 2 '18 at 0:50
reunsreuns
19.9k21147
answered Dec 2 '18 at 0:50
reunsreuns
19.9k21147
answered Dec 2 '18 at 0:50
reunsreuns
19.9k21147
answered Dec 2 '18 at 0:50
reunsreuns
19.9k21147
19.9k21147
$begingroup$
But why is the inverse continuous? Shouldn't a topological embedding be a homeomorphism onto its image?
$endgroup$
– Andrew Yuan
Dec 3 '18 at 4:00
$begingroup$
@AndrewYuan Inverse of what ? I you don't define the semi-norms, topology, dual, you won't see that it is really trivial. The Schwartz space is a subspace of the Banach space with norm $|.|_{L^infty}+|x^2.|_{L^infty}$ and as I shown that one embeds in its strong dual.
$endgroup$
– reuns
Dec 3 '18 at 4:09
$begingroup$
I mean the inverse of $gmapsto (cdot,g)_2$ on its image in $S'$, which is well-defined since its injective. And maybe I'm missing something, but I don't think the Schwartz space (with its usual topology) is a subspace of a Banach space or else it would be normable.
$endgroup$
– Andrew Yuan
Dec 3 '18 at 6:09
$begingroup$
@AndrewYuan A non-closed subspace. The Schwartz space is the intersection of infinitely many Banach spaces (with norms $sum_{|alpha| le K} | (1+|x|^M) partial_alpha f|_{L^infty}$) that's its definition. Its open sets are intersection of open sets in each of those Banach spaces.
$endgroup$
– reuns
Dec 3 '18 at 6:32
$begingroup$
Define precisely the topologies in $S,S'$ then $(g,g) = |g|_{L^2}^2 > ...$ will make it obvious that the map $g to iota(g) in S'$ is bi-continuous.
$endgroup$
– reuns
Dec 3 '18 at 6:42
add a comment |
$begingroup$
But why is the inverse continuous? Shouldn't a topological embedding be a homeomorphism onto its image?
$endgroup$
– Andrew Yuan
Dec 3 '18 at 4:00
$begingroup$
@AndrewYuan Inverse of what ? I you don't define the semi-norms, topology, dual, you won't see that it is really trivial. The Schwartz space is a subspace of the Banach space with norm $|.|_{L^infty}+|x^2.|_{L^infty}$ and as I shown that one embeds in its strong dual.
$endgroup$
– reuns
Dec 3 '18 at 4:09
$begingroup$
I mean the inverse of $gmapsto (cdot,g)_2$ on its image in $S'$, which is well-defined since its injective. And maybe I'm missing something, but I don't think the Schwartz space (with its usual topology) is a subspace of a Banach space or else it would be normable.
$endgroup$
– Andrew Yuan
Dec 3 '18 at 6:09
$begingroup$
@AndrewYuan A non-closed subspace. The Schwartz space is the intersection of infinitely many Banach spaces (with norms $sum_{|alpha| le K} | (1+|x|^M) partial_alpha f|_{L^infty}$) that's its definition. Its open sets are intersection of open sets in each of those Banach spaces.
$endgroup$
– reuns
Dec 3 '18 at 6:32
$begingroup$
Define precisely the topologies in $S,S'$ then $(g,g) = |g|_{L^2}^2 > ...$ will make it obvious that the map $g to iota(g) in S'$ is bi-continuous.
$endgroup$
– reuns
Dec 3 '18 at 6:42
$begingroup$
But why is the inverse continuous? Shouldn't a topological embedding be a homeomorphism onto its image?
$endgroup$
– Andrew Yuan
Dec 3 '18 at 4:00
$begingroup$
But why is the inverse continuous? Shouldn't a topological embedding be a homeomorphism onto its image?
$endgroup$
– Andrew Yuan
Dec 3 '18 at 4:00
$begingroup$
@AndrewYuan Inverse of what ? I you don't define the semi-norms, topology, dual, you won't see that it is really trivial. The Schwartz space is a subspace of the Banach space with norm $|.|_{L^infty}+|x^2.|_{L^infty}$ and as I shown that one embeds in its strong dual.
$endgroup$
– reuns
Dec 3 '18 at 4:09
$begingroup$
@AndrewYuan Inverse of what ? I you don't define the semi-norms, topology, dual, you won't see that it is really trivial. The Schwartz space is a subspace of the Banach space with norm $|.|_{L^infty}+|x^2.|_{L^infty}$ and as I shown that one embeds in its strong dual.
$endgroup$
– reuns
Dec 3 '18 at 4:09
$begingroup$
I mean the inverse of $gmapsto (cdot,g)_2$ on its image in $S'$, which is well-defined since its injective. And maybe I'm missing something, but I don't think the Schwartz space (with its usual topology) is a subspace of a Banach space or else it would be normable.
$endgroup$
– Andrew Yuan
Dec 3 '18 at 6:09
$begingroup$
I mean the inverse of $gmapsto (cdot,g)_2$ on its image in $S'$, which is well-defined since its injective. And maybe I'm missing something, but I don't think the Schwartz space (with its usual topology) is a subspace of a Banach space or else it would be normable.
$endgroup$
– Andrew Yuan
Dec 3 '18 at 6:09
$begingroup$
@AndrewYuan A non-closed subspace. The Schwartz space is the intersection of infinitely many Banach spaces (with norms $sum_{|alpha| le K} | (1+|x|^M) partial_alpha f|_{L^infty}$) that's its definition. Its open sets are intersection of open sets in each of those Banach spaces.
$endgroup$
– reuns
Dec 3 '18 at 6:32
$begingroup$
@AndrewYuan A non-closed subspace. The Schwartz space is the intersection of infinitely many Banach spaces (with norms $sum_{|alpha| le K} | (1+|x|^M) partial_alpha f|_{L^infty}$) that's its definition. Its open sets are intersection of open sets in each of those Banach spaces.
$endgroup$
– reuns
Dec 3 '18 at 6:32
$begingroup$
Define precisely the topologies in $S,S'$ then $(g,g) = |g|_{L^2}^2 > ...$ will make it obvious that the map $g to iota(g) in S'$ is bi-continuous.
$endgroup$
– reuns
Dec 3 '18 at 6:42
$begingroup$
Define precisely the topologies in $S,S'$ then $(g,g) = |g|_{L^2}^2 > ...$ will make it obvious that the map $g to iota(g) in S'$ is bi-continuous.
$endgroup$
– reuns
Dec 3 '18 at 6:42
add a comment |
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