Norm limit of sequence of orthogonal projections on Hilbert space “contractive”












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Let $(P_n)_{n in mathbb N} subseteq B(cal H)$ be a sequence of (orthogonal) projections on a (separable) Hilbert space such that $leftVert P_{n}xirightVert rightarrow CleftVert xirightVert $ for every $xi in cal H$, where $C leq 1$. It feels like $C$ should be $1$ since $leftVert P_{n}rightVert =1$ for every $n$, but I don't see how to prove this. Is my guess even right?






operator-theory hilbert-spaces projection







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    $begingroup$


    Let $(P_n)_{n in mathbb N} subseteq B(cal H)$ be a sequence of (orthogonal) projections on a (separable) Hilbert space such that $leftVert P_{n}xirightVert rightarrow CleftVert xirightVert $ for every $xi in cal H$, where $C leq 1$. It feels like $C$ should be $1$ since $leftVert P_{n}rightVert =1$ for every $n$, but I don't see how to prove this. Is my guess even right?






    operator-theory hilbert-spaces projection







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Let $(P_n)_{n in mathbb N} subseteq B(cal H)$ be a sequence of (orthogonal) projections on a (separable) Hilbert space such that $leftVert P_{n}xirightVert rightarrow CleftVert xirightVert $ for every $xi in cal H$, where $C leq 1$. It feels like $C$ should be $1$ since $leftVert P_{n}rightVert =1$ for every $n$, but I don't see how to prove this. Is my guess even right?






      operator-theory hilbert-spaces projection







      share|cite|improve this question









      $endgroup$




      Let $(P_n)_{n in mathbb N} subseteq B(cal H)$ be a sequence of (orthogonal) projections on a (separable) Hilbert space such that $leftVert P_{n}xirightVert rightarrow CleftVert xirightVert $ for every $xi in cal H$, where $C leq 1$. It feels like $C$ should be $1$ since $leftVert P_{n}rightVert =1$ for every $n$, but I don't see how to prove this. Is my guess even right?






      operator-theory hilbert-spaces projection




      operator-theory hilbert-spaces projection






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      asked Dec 11 '18 at 8:55









      worldreporter14worldreporter14

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          If ${e_n}$ is an orthonormal basis for a Hilbert space and $P_n$ is the projection on the closed subspace spanned by ${e_n,e_{n+1},cdots }$ then $|P_nx| to 0$ for every $x$. So $C$ can be $0$.






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          • $begingroup$
            Thanks for the answer! Can we at least say that $C notin (0, 1)$?
            $endgroup$
            – worldreporter14
            Dec 11 '18 at 11:01











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          $begingroup$

          If ${e_n}$ is an orthonormal basis for a Hilbert space and $P_n$ is the projection on the closed subspace spanned by ${e_n,e_{n+1},cdots }$ then $|P_nx| to 0$ for every $x$. So $C$ can be $0$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for the answer! Can we at least say that $C notin (0, 1)$?
            $endgroup$
            – worldreporter14
            Dec 11 '18 at 11:01
















          1












          $begingroup$

          If ${e_n}$ is an orthonormal basis for a Hilbert space and $P_n$ is the projection on the closed subspace spanned by ${e_n,e_{n+1},cdots }$ then $|P_nx| to 0$ for every $x$. So $C$ can be $0$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for the answer! Can we at least say that $C notin (0, 1)$?
            $endgroup$
            – worldreporter14
            Dec 11 '18 at 11:01














          1












          1








          1





          $begingroup$

          If ${e_n}$ is an orthonormal basis for a Hilbert space and $P_n$ is the projection on the closed subspace spanned by ${e_n,e_{n+1},cdots }$ then $|P_nx| to 0$ for every $x$. So $C$ can be $0$.






          share|cite|improve this answer











          $endgroup$



          If ${e_n}$ is an orthonormal basis for a Hilbert space and $P_n$ is the projection on the closed subspace spanned by ${e_n,e_{n+1},cdots }$ then $|P_nx| to 0$ for every $x$. So $C$ can be $0$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 11 '18 at 9:15

























          answered Dec 11 '18 at 8:58









          Kavi Rama MurthyKavi Rama Murthy

          59k42161




          59k42161












          • $begingroup$
            Thanks for the answer! Can we at least say that $C notin (0, 1)$?
            $endgroup$
            – worldreporter14
            Dec 11 '18 at 11:01


















          • $begingroup$
            Thanks for the answer! Can we at least say that $C notin (0, 1)$?
            $endgroup$
            – worldreporter14
            Dec 11 '18 at 11:01
















          $begingroup$
          Thanks for the answer! Can we at least say that $C notin (0, 1)$?
          $endgroup$
          – worldreporter14
          Dec 11 '18 at 11:01




          $begingroup$
          Thanks for the answer! Can we at least say that $C notin (0, 1)$?
          $endgroup$
          – worldreporter14
          Dec 11 '18 at 11:01


















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