Is the Schwartz topologically emebedded in space of tempered distributions?












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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







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    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







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      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







      share|cite|improve this question









      $endgroup$




      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






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      asked Dec 1 '18 at 23:23









      Andrew YuanAndrew Yuan

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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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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













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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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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


















          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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
















          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.






          share|cite|improve this answer









          $endgroup$



          $$left|int_{mathbb{R}} f(x)g(x)dx right| le left|frac{f}{1+x^2}right|_{L^1} |(1+x^2)g|_{L^infty}le pi |f|_{L^infty}(|g|_{L^infty}+|x^2g|_{L^infty}) $$



          $|.|_{L^infty}$,$|x^2.|_{L^infty}$ are in the semi-norms used to construct the Schwartz space.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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




















          • $begingroup$
            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




















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