If $T$ is compact such as $T(e_n)=lambda_n e_n$ then $lim_nlambda_n=0$












2












$begingroup$


Let $H$ be a Hilbert separable space. with the orthonormal bais $(e_n)_{nin mathbb{N}}$, and $(lambda_n)_{nin mathbb{N}}$ a bounded sequence in $ mathbb{C}$.



We define the operator $T : H rightarrow H$, as $T(e_n)=lambda_n e_n$



We need to prove that $T$ is compact if and only if $lim_nlambda_n=0.$



I only did the reverse :



Suppose that $(lambda_n)_n $ converge to $0$.



We define the finite rank operators $T_k$ defined as : $T_k(x)=(lambda_1x_1,..,lambda_kx_k,0,0,...)$.



And now we prove that $||T_k-T||rightarrow 0.$



We have : $T(x)-T_k(x)=(0,0,...,lambda_{k+1}x_{k+1},...)$



then : $||T_k-T||leq sup_{n>k+1}|lambda_n|rightarrow0.$



then $T$ is compact as limit of finite rank operators.



[But I'm always stuck with the direct implication]






functional-analysis operator-theory hilbert-spaces







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Let $H$ be a Hilbert separable space. with the orthonormal bais $(e_n)_{nin mathbb{N}}$, and $(lambda_n)_{nin mathbb{N}}$ a bounded sequence in $ mathbb{C}$.



    We define the operator $T : H rightarrow H$, as $T(e_n)=lambda_n e_n$



    We need to prove that $T$ is compact if and only if $lim_nlambda_n=0.$



    I only did the reverse :



    Suppose that $(lambda_n)_n $ converge to $0$.



    We define the finite rank operators $T_k$ defined as : $T_k(x)=(lambda_1x_1,..,lambda_kx_k,0,0,...)$.



    And now we prove that $||T_k-T||rightarrow 0.$



    We have : $T(x)-T_k(x)=(0,0,...,lambda_{k+1}x_{k+1},...)$



    then : $||T_k-T||leq sup_{n>k+1}|lambda_n|rightarrow0.$



    then $T$ is compact as limit of finite rank operators.



    [But I'm always stuck with the direct implication]






    functional-analysis operator-theory hilbert-spaces







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Let $H$ be a Hilbert separable space. with the orthonormal bais $(e_n)_{nin mathbb{N}}$, and $(lambda_n)_{nin mathbb{N}}$ a bounded sequence in $ mathbb{C}$.



      We define the operator $T : H rightarrow H$, as $T(e_n)=lambda_n e_n$



      We need to prove that $T$ is compact if and only if $lim_nlambda_n=0.$



      I only did the reverse :



      Suppose that $(lambda_n)_n $ converge to $0$.



      We define the finite rank operators $T_k$ defined as : $T_k(x)=(lambda_1x_1,..,lambda_kx_k,0,0,...)$.



      And now we prove that $||T_k-T||rightarrow 0.$



      We have : $T(x)-T_k(x)=(0,0,...,lambda_{k+1}x_{k+1},...)$



      then : $||T_k-T||leq sup_{n>k+1}|lambda_n|rightarrow0.$



      then $T$ is compact as limit of finite rank operators.



      [But I'm always stuck with the direct implication]






      functional-analysis operator-theory hilbert-spaces







      share|cite|improve this question









      $endgroup$




      Let $H$ be a Hilbert separable space. with the orthonormal bais $(e_n)_{nin mathbb{N}}$, and $(lambda_n)_{nin mathbb{N}}$ a bounded sequence in $ mathbb{C}$.



      We define the operator $T : H rightarrow H$, as $T(e_n)=lambda_n e_n$



      We need to prove that $T$ is compact if and only if $lim_nlambda_n=0.$



      I only did the reverse :



      Suppose that $(lambda_n)_n $ converge to $0$.



      We define the finite rank operators $T_k$ defined as : $T_k(x)=(lambda_1x_1,..,lambda_kx_k,0,0,...)$.



      And now we prove that $||T_k-T||rightarrow 0.$



      We have : $T(x)-T_k(x)=(0,0,...,lambda_{k+1}x_{k+1},...)$



      then : $||T_k-T||leq sup_{n>k+1}|lambda_n|rightarrow0.$



      then $T$ is compact as limit of finite rank operators.



      [But I'm always stuck with the direct implication]






      functional-analysis operator-theory hilbert-spaces




      functional-analysis operator-theory hilbert-spaces






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 21 '18 at 19:27









      Anas BOUALIIAnas BOUALII

      1397




      1397






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          By the equivalent characterization of compact operators on a Hilbert space as taking the unit ball to a compact set, we just need to find a sequence in the image with no convergent subsequence. Because the $lambda_i$ do not converge to $0$, pick a subsequence $lambda_{n_i}$ with $|lambda_{n_i}|>delta$. Then the vectors $lambda_{n_i}e_{n_i}$ are in the image of the unit ball, but they are all orthogonal, so the distance between each of them is at least $sqrt{2}delta$.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            Recall that the image of the unit ball under $T$ is relatively (sequentially) compact by definition of compact operator. So begin by assuming to the contrary that there is a subsequence $n(k)$ such that
            $$
            |lambda_{n(k)}|ge epsilon>0,
            $$            for some $epsilon >0$. Then, we have
            $$
            |Te_{n(k)}-Te_{n(r)}|^2 = |lambda_{n(k)}e_{n(k)} - lambda_{n(r)}e_{n(r)} |^2= |lambda_{n(k)}|^2 +|lambda_{n(r)}|^2geq 2epsilon^2.
            $$             This means $(Te_{n(k)})_{kge 1}$ cannot have a Cauchy subsequence. This contradicts relative sequential compactness, hence we get
            $$
            lim_{ntoinfty}|lambda_n| = 0.
            $$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              If a sequence in not a Cauchy sequence,then surely it has no subsequence convergent at all ?
              $endgroup$
              – Anas BOUALII
              Dec 21 '18 at 19:43










            • $begingroup$
              I've just corrected my answer. It was an error.
              $endgroup$
              – Song
              Dec 21 '18 at 19:45










            • $begingroup$
              Now, the proof is good.
              $endgroup$
              – Anas BOUALII
              Dec 21 '18 at 19:47










            • $begingroup$
              Actually more is true: if $T$ is compact, then it sends a weakly convergent sequence to a strongly convergent sequence. So, in any case $Te_n$ should converge to $0$ because $e_n$ weakly converges to $0$.
              $endgroup$
              – Song
              Dec 21 '18 at 19:51












            • $begingroup$
              Yes, I saw that proposition. Do you have/recommand a good book for operators and exercies ?
              $endgroup$
              – Anas BOUALII
              Dec 21 '18 at 19:58











            Your Answer





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            2 Answers
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            active

            oldest

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            2 Answers
            2






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            By the equivalent characterization of compact operators on a Hilbert space as taking the unit ball to a compact set, we just need to find a sequence in the image with no convergent subsequence. Because the $lambda_i$ do not converge to $0$, pick a subsequence $lambda_{n_i}$ with $|lambda_{n_i}|>delta$. Then the vectors $lambda_{n_i}e_{n_i}$ are in the image of the unit ball, but they are all orthogonal, so the distance between each of them is at least $sqrt{2}delta$.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              By the equivalent characterization of compact operators on a Hilbert space as taking the unit ball to a compact set, we just need to find a sequence in the image with no convergent subsequence. Because the $lambda_i$ do not converge to $0$, pick a subsequence $lambda_{n_i}$ with $|lambda_{n_i}|>delta$. Then the vectors $lambda_{n_i}e_{n_i}$ are in the image of the unit ball, but they are all orthogonal, so the distance between each of them is at least $sqrt{2}delta$.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                By the equivalent characterization of compact operators on a Hilbert space as taking the unit ball to a compact set, we just need to find a sequence in the image with no convergent subsequence. Because the $lambda_i$ do not converge to $0$, pick a subsequence $lambda_{n_i}$ with $|lambda_{n_i}|>delta$. Then the vectors $lambda_{n_i}e_{n_i}$ are in the image of the unit ball, but they are all orthogonal, so the distance between each of them is at least $sqrt{2}delta$.






                share|cite|improve this answer









                $endgroup$



                By the equivalent characterization of compact operators on a Hilbert space as taking the unit ball to a compact set, we just need to find a sequence in the image with no convergent subsequence. Because the $lambda_i$ do not converge to $0$, pick a subsequence $lambda_{n_i}$ with $|lambda_{n_i}|>delta$. Then the vectors $lambda_{n_i}e_{n_i}$ are in the image of the unit ball, but they are all orthogonal, so the distance between each of them is at least $sqrt{2}delta$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 21 '18 at 19:35









                Ashwin TrisalAshwin Trisal

                1,2891516




                1,2891516























                    2












                    $begingroup$

                    Recall that the image of the unit ball under $T$ is relatively (sequentially) compact by definition of compact operator. So begin by assuming to the contrary that there is a subsequence $n(k)$ such that
                    $$
                    |lambda_{n(k)}|ge epsilon>0,
                    $$                    for some $epsilon >0$. Then, we have
                    $$
                    |Te_{n(k)}-Te_{n(r)}|^2 = |lambda_{n(k)}e_{n(k)} - lambda_{n(r)}e_{n(r)} |^2= |lambda_{n(k)}|^2 +|lambda_{n(r)}|^2geq 2epsilon^2.
                    $$                     This means $(Te_{n(k)})_{kge 1}$ cannot have a Cauchy subsequence. This contradicts relative sequential compactness, hence we get
                    $$
                    lim_{ntoinfty}|lambda_n| = 0.
                    $$






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      If a sequence in not a Cauchy sequence,then surely it has no subsequence convergent at all ?
                      $endgroup$
                      – Anas BOUALII
                      Dec 21 '18 at 19:43










                    • $begingroup$
                      I've just corrected my answer. It was an error.
                      $endgroup$
                      – Song
                      Dec 21 '18 at 19:45










                    • $begingroup$
                      Now, the proof is good.
                      $endgroup$
                      – Anas BOUALII
                      Dec 21 '18 at 19:47










                    • $begingroup$
                      Actually more is true: if $T$ is compact, then it sends a weakly convergent sequence to a strongly convergent sequence. So, in any case $Te_n$ should converge to $0$ because $e_n$ weakly converges to $0$.
                      $endgroup$
                      – Song
                      Dec 21 '18 at 19:51












                    • $begingroup$
                      Yes, I saw that proposition. Do you have/recommand a good book for operators and exercies ?
                      $endgroup$
                      – Anas BOUALII
                      Dec 21 '18 at 19:58
















                    2












                    $begingroup$

                    Recall that the image of the unit ball under $T$ is relatively (sequentially) compact by definition of compact operator. So begin by assuming to the contrary that there is a subsequence $n(k)$ such that
                    $$
                    |lambda_{n(k)}|ge epsilon>0,
                    $$                    for some $epsilon >0$. Then, we have
                    $$
                    |Te_{n(k)}-Te_{n(r)}|^2 = |lambda_{n(k)}e_{n(k)} - lambda_{n(r)}e_{n(r)} |^2= |lambda_{n(k)}|^2 +|lambda_{n(r)}|^2geq 2epsilon^2.
                    $$                     This means $(Te_{n(k)})_{kge 1}$ cannot have a Cauchy subsequence. This contradicts relative sequential compactness, hence we get
                    $$
                    lim_{ntoinfty}|lambda_n| = 0.
                    $$






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      If a sequence in not a Cauchy sequence,then surely it has no subsequence convergent at all ?
                      $endgroup$
                      – Anas BOUALII
                      Dec 21 '18 at 19:43










                    • $begingroup$
                      I've just corrected my answer. It was an error.
                      $endgroup$
                      – Song
                      Dec 21 '18 at 19:45










                    • $begingroup$
                      Now, the proof is good.
                      $endgroup$
                      – Anas BOUALII
                      Dec 21 '18 at 19:47










                    • $begingroup$
                      Actually more is true: if $T$ is compact, then it sends a weakly convergent sequence to a strongly convergent sequence. So, in any case $Te_n$ should converge to $0$ because $e_n$ weakly converges to $0$.
                      $endgroup$
                      – Song
                      Dec 21 '18 at 19:51












                    • $begingroup$
                      Yes, I saw that proposition. Do you have/recommand a good book for operators and exercies ?
                      $endgroup$
                      – Anas BOUALII
                      Dec 21 '18 at 19:58














                    2












                    2








                    2





                    $begingroup$

                    Recall that the image of the unit ball under $T$ is relatively (sequentially) compact by definition of compact operator. So begin by assuming to the contrary that there is a subsequence $n(k)$ such that
                    $$
                    |lambda_{n(k)}|ge epsilon>0,
                    $$                    for some $epsilon >0$. Then, we have
                    $$
                    |Te_{n(k)}-Te_{n(r)}|^2 = |lambda_{n(k)}e_{n(k)} - lambda_{n(r)}e_{n(r)} |^2= |lambda_{n(k)}|^2 +|lambda_{n(r)}|^2geq 2epsilon^2.
                    $$                     This means $(Te_{n(k)})_{kge 1}$ cannot have a Cauchy subsequence. This contradicts relative sequential compactness, hence we get
                    $$
                    lim_{ntoinfty}|lambda_n| = 0.
                    $$






                    share|cite|improve this answer











                    $endgroup$



                    Recall that the image of the unit ball under $T$ is relatively (sequentially) compact by definition of compact operator. So begin by assuming to the contrary that there is a subsequence $n(k)$ such that
                    $$
                    |lambda_{n(k)}|ge epsilon>0,
                    $$                    for some $epsilon >0$. Then, we have
                    $$
                    |Te_{n(k)}-Te_{n(r)}|^2 = |lambda_{n(k)}e_{n(k)} - lambda_{n(r)}e_{n(r)} |^2= |lambda_{n(k)}|^2 +|lambda_{n(r)}|^2geq 2epsilon^2.
                    $$                     This means $(Te_{n(k)})_{kge 1}$ cannot have a Cauchy subsequence. This contradicts relative sequential compactness, hence we get
                    $$
                    lim_{ntoinfty}|lambda_n| = 0.
                    $$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 21 '18 at 19:44

























                    answered Dec 21 '18 at 19:40









                    SongSong

                    11.3k628




                    11.3k628












                    • $begingroup$
                      If a sequence in not a Cauchy sequence,then surely it has no subsequence convergent at all ?
                      $endgroup$
                      – Anas BOUALII
                      Dec 21 '18 at 19:43










                    • $begingroup$
                      I've just corrected my answer. It was an error.
                      $endgroup$
                      – Song
                      Dec 21 '18 at 19:45










                    • $begingroup$
                      Now, the proof is good.
                      $endgroup$
                      – Anas BOUALII
                      Dec 21 '18 at 19:47










                    • $begingroup$
                      Actually more is true: if $T$ is compact, then it sends a weakly convergent sequence to a strongly convergent sequence. So, in any case $Te_n$ should converge to $0$ because $e_n$ weakly converges to $0$.
                      $endgroup$
                      – Song
                      Dec 21 '18 at 19:51












                    • $begingroup$
                      Yes, I saw that proposition. Do you have/recommand a good book for operators and exercies ?
                      $endgroup$
                      – Anas BOUALII
                      Dec 21 '18 at 19:58


















                    • $begingroup$
                      If a sequence in not a Cauchy sequence,then surely it has no subsequence convergent at all ?
                      $endgroup$
                      – Anas BOUALII
                      Dec 21 '18 at 19:43










                    • $begingroup$
                      I've just corrected my answer. It was an error.
                      $endgroup$
                      – Song
                      Dec 21 '18 at 19:45










                    • $begingroup$
                      Now, the proof is good.
                      $endgroup$
                      – Anas BOUALII
                      Dec 21 '18 at 19:47










                    • $begingroup$
                      Actually more is true: if $T$ is compact, then it sends a weakly convergent sequence to a strongly convergent sequence. So, in any case $Te_n$ should converge to $0$ because $e_n$ weakly converges to $0$.
                      $endgroup$
                      – Song
                      Dec 21 '18 at 19:51












                    • $begingroup$
                      Yes, I saw that proposition. Do you have/recommand a good book for operators and exercies ?
                      $endgroup$
                      – Anas BOUALII
                      Dec 21 '18 at 19:58
















                    $begingroup$
                    If a sequence in not a Cauchy sequence,then surely it has no subsequence convergent at all ?
                    $endgroup$
                    – Anas BOUALII
                    Dec 21 '18 at 19:43




                    $begingroup$
                    If a sequence in not a Cauchy sequence,then surely it has no subsequence convergent at all ?
                    $endgroup$
                    – Anas BOUALII
                    Dec 21 '18 at 19:43












                    $begingroup$
                    I've just corrected my answer. It was an error.
                    $endgroup$
                    – Song
                    Dec 21 '18 at 19:45




                    $begingroup$
                    I've just corrected my answer. It was an error.
                    $endgroup$
                    – Song
                    Dec 21 '18 at 19:45












                    $begingroup$
                    Now, the proof is good.
                    $endgroup$
                    – Anas BOUALII
                    Dec 21 '18 at 19:47




                    $begingroup$
                    Now, the proof is good.
                    $endgroup$
                    – Anas BOUALII
                    Dec 21 '18 at 19:47












                    $begingroup$
                    Actually more is true: if $T$ is compact, then it sends a weakly convergent sequence to a strongly convergent sequence. So, in any case $Te_n$ should converge to $0$ because $e_n$ weakly converges to $0$.
                    $endgroup$
                    – Song
                    Dec 21 '18 at 19:51






                    $begingroup$
                    Actually more is true: if $T$ is compact, then it sends a weakly convergent sequence to a strongly convergent sequence. So, in any case $Te_n$ should converge to $0$ because $e_n$ weakly converges to $0$.
                    $endgroup$
                    – Song
                    Dec 21 '18 at 19:51














                    $begingroup$
                    Yes, I saw that proposition. Do you have/recommand a good book for operators and exercies ?
                    $endgroup$
                    – Anas BOUALII
                    Dec 21 '18 at 19:58




                    $begingroup$
                    Yes, I saw that proposition. Do you have/recommand a good book for operators and exercies ?
                    $endgroup$
                    – Anas BOUALII
                    Dec 21 '18 at 19:58


















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