Why a set is weakly sequentially precompact set?

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I'm struggling to understand the following assertion:




Since $lim_trightarrow infty|x(t)-y_0|$ exists, $x(t):tgeq 0$ is weakly sequentially precompact.




Please give thorough explanations because my knowledge of weak topology is very weak (pun intended ;) )



If something is unclear, please refer to this paper: Asymptotic convergence of nonlinear contraction semigroups, (theorem 1 -- at the end of page 17 and beginning of page 18).




general-topology functional-analysis hilbert-spaces weak-convergence




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    up vote
    0
    down vote

    favorite












    I'm struggling to understand the following assertion:




    Since $lim_trightarrow infty|x(t)-y_0|$ exists, $x(t):tgeq 0$ is weakly sequentially precompact.




    Please give thorough explanations because my knowledge of weak topology is very weak (pun intended ;) )



    If something is unclear, please refer to this paper: Asymptotic convergence of nonlinear contraction semigroups, (theorem 1 -- at the end of page 17 and beginning of page 18).




    general-topology functional-analysis hilbert-spaces weak-convergence




    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I'm struggling to understand the following assertion:




      Since $lim_trightarrow infty|x(t)-y_0|$ exists, $x(t):tgeq 0$ is weakly sequentially precompact.




      Please give thorough explanations because my knowledge of weak topology is very weak (pun intended ;) )



      If something is unclear, please refer to this paper: Asymptotic convergence of nonlinear contraction semigroups, (theorem 1 -- at the end of page 17 and beginning of page 18).




      general-topology functional-analysis hilbert-spaces weak-convergence




      share|cite|improve this question














      I'm struggling to understand the following assertion:




      Since $lim_trightarrow infty|x(t)-y_0|$ exists, $x(t):tgeq 0$ is weakly sequentially precompact.




      Please give thorough explanations because my knowledge of weak topology is very weak (pun intended ;) )



      If something is unclear, please refer to this paper: Asymptotic convergence of nonlinear contraction semigroups, (theorem 1 -- at the end of page 17 and beginning of page 18).





      general-topology functional-analysis hilbert-spaces weak-convergence





      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 15 at 9:55









      Robert Z

      84.5k955123




      84.5k955123










      asked Aug 15 at 9:52









      Paolo

      356




      356




















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          Since $x$ is continuous the image of $[0,N]$ is is norm bounded for each positive integer $N$. Since $lim_t to infty |x(t)-y_0|$ exists it follows that $x(t): t geq 0$ lies in a norm bounded subset of $H$. In a separable Hilbert space any norm bounded set is weakly sequentially precompact.






          share|cite|improve this answer




















          • Where can I find that any norm bounded set is weakly sequentially precompact? Sorry I'm being pedantic but as I said I know very little about weak topology.
            – Paolo
            Aug 15 at 10:27










          • Actually this requires some knowledge of weak and weak* topologies. The Banach-Alaoglu Theorem (which can be found in many texts in Functional Analysis says that the closed unit ball of a dual space is compact in the weak* topology. Since Hilbert spaces are reflexive the weak* topology coincides with the weak topology and separability guarantees existence of weakly convergent. subsequence . Lot of reading to do but you can start with Banach-Alaoglu Theorem .
            – Kavi Rama Murthy
            Aug 15 at 10:33










          • Thank you, very much appreciated.
            – Paolo
            Aug 15 at 10:39










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          Since $x$ is continuous the image of $[0,N]$ is is norm bounded for each positive integer $N$. Since $lim_t to infty |x(t)-y_0|$ exists it follows that $x(t): t geq 0$ lies in a norm bounded subset of $H$. In a separable Hilbert space any norm bounded set is weakly sequentially precompact.






          share|cite|improve this answer




















          • Where can I find that any norm bounded set is weakly sequentially precompact? Sorry I'm being pedantic but as I said I know very little about weak topology.
            – Paolo
            Aug 15 at 10:27










          • Actually this requires some knowledge of weak and weak* topologies. The Banach-Alaoglu Theorem (which can be found in many texts in Functional Analysis says that the closed unit ball of a dual space is compact in the weak* topology. Since Hilbert spaces are reflexive the weak* topology coincides with the weak topology and separability guarantees existence of weakly convergent. subsequence . Lot of reading to do but you can start with Banach-Alaoglu Theorem .
            – Kavi Rama Murthy
            Aug 15 at 10:33










          • Thank you, very much appreciated.
            – Paolo
            Aug 15 at 10:39














          up vote
          1
          down vote



          accepted










          Since $x$ is continuous the image of $[0,N]$ is is norm bounded for each positive integer $N$. Since $lim_t to infty |x(t)-y_0|$ exists it follows that $x(t): t geq 0$ lies in a norm bounded subset of $H$. In a separable Hilbert space any norm bounded set is weakly sequentially precompact.






          share|cite|improve this answer




















          • Where can I find that any norm bounded set is weakly sequentially precompact? Sorry I'm being pedantic but as I said I know very little about weak topology.
            – Paolo
            Aug 15 at 10:27










          • Actually this requires some knowledge of weak and weak* topologies. The Banach-Alaoglu Theorem (which can be found in many texts in Functional Analysis says that the closed unit ball of a dual space is compact in the weak* topology. Since Hilbert spaces are reflexive the weak* topology coincides with the weak topology and separability guarantees existence of weakly convergent. subsequence . Lot of reading to do but you can start with Banach-Alaoglu Theorem .
            – Kavi Rama Murthy
            Aug 15 at 10:33










          • Thank you, very much appreciated.
            – Paolo
            Aug 15 at 10:39












          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Since $x$ is continuous the image of $[0,N]$ is is norm bounded for each positive integer $N$. Since $lim_t to infty |x(t)-y_0|$ exists it follows that $x(t): t geq 0$ lies in a norm bounded subset of $H$. In a separable Hilbert space any norm bounded set is weakly sequentially precompact.






          share|cite|improve this answer












          Since $x$ is continuous the image of $[0,N]$ is is norm bounded for each positive integer $N$. Since $lim_t to infty |x(t)-y_0|$ exists it follows that $x(t): t geq 0$ lies in a norm bounded subset of $H$. In a separable Hilbert space any norm bounded set is weakly sequentially precompact.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 15 at 10:02









          Kavi Rama Murthy

          22.5k2933




          22.5k2933











          • Where can I find that any norm bounded set is weakly sequentially precompact? Sorry I'm being pedantic but as I said I know very little about weak topology.
            – Paolo
            Aug 15 at 10:27










          • Actually this requires some knowledge of weak and weak* topologies. The Banach-Alaoglu Theorem (which can be found in many texts in Functional Analysis says that the closed unit ball of a dual space is compact in the weak* topology. Since Hilbert spaces are reflexive the weak* topology coincides with the weak topology and separability guarantees existence of weakly convergent. subsequence . Lot of reading to do but you can start with Banach-Alaoglu Theorem .
            – Kavi Rama Murthy
            Aug 15 at 10:33










          • Thank you, very much appreciated.
            – Paolo
            Aug 15 at 10:39
















          • Where can I find that any norm bounded set is weakly sequentially precompact? Sorry I'm being pedantic but as I said I know very little about weak topology.
            – Paolo
            Aug 15 at 10:27










          • Actually this requires some knowledge of weak and weak* topologies. The Banach-Alaoglu Theorem (which can be found in many texts in Functional Analysis says that the closed unit ball of a dual space is compact in the weak* topology. Since Hilbert spaces are reflexive the weak* topology coincides with the weak topology and separability guarantees existence of weakly convergent. subsequence . Lot of reading to do but you can start with Banach-Alaoglu Theorem .
            – Kavi Rama Murthy
            Aug 15 at 10:33










          • Thank you, very much appreciated.
            – Paolo
            Aug 15 at 10:39















          Where can I find that any norm bounded set is weakly sequentially precompact? Sorry I'm being pedantic but as I said I know very little about weak topology.
          – Paolo
          Aug 15 at 10:27




          Where can I find that any norm bounded set is weakly sequentially precompact? Sorry I'm being pedantic but as I said I know very little about weak topology.
          – Paolo
          Aug 15 at 10:27












          Actually this requires some knowledge of weak and weak* topologies. The Banach-Alaoglu Theorem (which can be found in many texts in Functional Analysis says that the closed unit ball of a dual space is compact in the weak* topology. Since Hilbert spaces are reflexive the weak* topology coincides with the weak topology and separability guarantees existence of weakly convergent. subsequence . Lot of reading to do but you can start with Banach-Alaoglu Theorem .
          – Kavi Rama Murthy
          Aug 15 at 10:33




          Actually this requires some knowledge of weak and weak* topologies. The Banach-Alaoglu Theorem (which can be found in many texts in Functional Analysis says that the closed unit ball of a dual space is compact in the weak* topology. Since Hilbert spaces are reflexive the weak* topology coincides with the weak topology and separability guarantees existence of weakly convergent. subsequence . Lot of reading to do but you can start with Banach-Alaoglu Theorem .
          – Kavi Rama Murthy
          Aug 15 at 10:33












          Thank you, very much appreciated.
          – Paolo
          Aug 15 at 10:39




          Thank you, very much appreciated.
          – Paolo
          Aug 15 at 10:39












           

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