Relationship between compact operators and coefficients tending to zero












2












$begingroup$


While reading the solutions of an exercise, I found the following statement:



proof



Where $d$ is some operator and $I$ is the identity operator from $L^2$ to $L^2$. I managed to get the formula for $(I-dd^*)g$ that is written below, however, I couldn't draw the same conclusion.
I checked in the lecture notes and I couldn't find any reference regarding the relationship between Fourier coefficients of a function in the image of an operator, and the operator being compact. If it helps, It is pretty easy to show that this operator is bounded.



Any ideas?






real-analysis complex-analysis functional-analysis operator-theory hilbert-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    There is quite a definition : $T$ is compact iff there is a sequence of finite dimensional operators $T_n$ such that $|T g - T_n g | < A_n|g|$ and $A_n to 0$. Now that $Tg = sum_k c_k langle g,e_k rangle e_k$ with $c_k to 0$ and $(e_k)$ orthonormal implies $|Tg - T_ng| = |sum_{|k| > n} c_k langle g,e_k rangle e_k | < ...$
    $endgroup$
    – reuns
    Dec 19 '18 at 14:59












  • $begingroup$
    @reuns thanks for your comment! Can you please cite any theorem and / or give me any link to it?
    $endgroup$
    – Euler_Salter
    Dec 19 '18 at 15:14






  • 1




    $begingroup$
    You can take it as the definition of compact operators. The other definitions are more abstract and the proof they are equivalent are not hard. en.wikipedia.org/wiki/Compact_operator_on_Hilbert_space
    $endgroup$
    – reuns
    Dec 19 '18 at 15:57












  • $begingroup$
    @reuns Thanks! Do you mind writing up an answer and expanding a bit more your solution?
    $endgroup$
    – Euler_Salter
    Dec 19 '18 at 16:05
















2












$begingroup$


While reading the solutions of an exercise, I found the following statement:



proof



Where $d$ is some operator and $I$ is the identity operator from $L^2$ to $L^2$. I managed to get the formula for $(I-dd^*)g$ that is written below, however, I couldn't draw the same conclusion.
I checked in the lecture notes and I couldn't find any reference regarding the relationship between Fourier coefficients of a function in the image of an operator, and the operator being compact. If it helps, It is pretty easy to show that this operator is bounded.



Any ideas?






real-analysis complex-analysis functional-analysis operator-theory hilbert-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    There is quite a definition : $T$ is compact iff there is a sequence of finite dimensional operators $T_n$ such that $|T g - T_n g | < A_n|g|$ and $A_n to 0$. Now that $Tg = sum_k c_k langle g,e_k rangle e_k$ with $c_k to 0$ and $(e_k)$ orthonormal implies $|Tg - T_ng| = |sum_{|k| > n} c_k langle g,e_k rangle e_k | < ...$
    $endgroup$
    – reuns
    Dec 19 '18 at 14:59












  • $begingroup$
    @reuns thanks for your comment! Can you please cite any theorem and / or give me any link to it?
    $endgroup$
    – Euler_Salter
    Dec 19 '18 at 15:14






  • 1




    $begingroup$
    You can take it as the definition of compact operators. The other definitions are more abstract and the proof they are equivalent are not hard. en.wikipedia.org/wiki/Compact_operator_on_Hilbert_space
    $endgroup$
    – reuns
    Dec 19 '18 at 15:57












  • $begingroup$
    @reuns Thanks! Do you mind writing up an answer and expanding a bit more your solution?
    $endgroup$
    – Euler_Salter
    Dec 19 '18 at 16:05














2












2








2


1



$begingroup$


While reading the solutions of an exercise, I found the following statement:



proof



Where $d$ is some operator and $I$ is the identity operator from $L^2$ to $L^2$. I managed to get the formula for $(I-dd^*)g$ that is written below, however, I couldn't draw the same conclusion.
I checked in the lecture notes and I couldn't find any reference regarding the relationship between Fourier coefficients of a function in the image of an operator, and the operator being compact. If it helps, It is pretty easy to show that this operator is bounded.



Any ideas?






real-analysis complex-analysis functional-analysis operator-theory hilbert-spaces







share|cite|improve this question









$endgroup$




While reading the solutions of an exercise, I found the following statement:



proof



Where $d$ is some operator and $I$ is the identity operator from $L^2$ to $L^2$. I managed to get the formula for $(I-dd^*)g$ that is written below, however, I couldn't draw the same conclusion.
I checked in the lecture notes and I couldn't find any reference regarding the relationship between Fourier coefficients of a function in the image of an operator, and the operator being compact. If it helps, It is pretty easy to show that this operator is bounded.



Any ideas?






real-analysis complex-analysis functional-analysis operator-theory hilbert-spaces




real-analysis complex-analysis 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 19 '18 at 14:55









Euler_SalterEuler_Salter

2,0601337




2,0601337












  • $begingroup$
    There is quite a definition : $T$ is compact iff there is a sequence of finite dimensional operators $T_n$ such that $|T g - T_n g | < A_n|g|$ and $A_n to 0$. Now that $Tg = sum_k c_k langle g,e_k rangle e_k$ with $c_k to 0$ and $(e_k)$ orthonormal implies $|Tg - T_ng| = |sum_{|k| > n} c_k langle g,e_k rangle e_k | < ...$
    $endgroup$
    – reuns
    Dec 19 '18 at 14:59












  • $begingroup$
    @reuns thanks for your comment! Can you please cite any theorem and / or give me any link to it?
    $endgroup$
    – Euler_Salter
    Dec 19 '18 at 15:14






  • 1




    $begingroup$
    You can take it as the definition of compact operators. The other definitions are more abstract and the proof they are equivalent are not hard. en.wikipedia.org/wiki/Compact_operator_on_Hilbert_space
    $endgroup$
    – reuns
    Dec 19 '18 at 15:57












  • $begingroup$
    @reuns Thanks! Do you mind writing up an answer and expanding a bit more your solution?
    $endgroup$
    – Euler_Salter
    Dec 19 '18 at 16:05


















  • $begingroup$
    There is quite a definition : $T$ is compact iff there is a sequence of finite dimensional operators $T_n$ such that $|T g - T_n g | < A_n|g|$ and $A_n to 0$. Now that $Tg = sum_k c_k langle g,e_k rangle e_k$ with $c_k to 0$ and $(e_k)$ orthonormal implies $|Tg - T_ng| = |sum_{|k| > n} c_k langle g,e_k rangle e_k | < ...$
    $endgroup$
    – reuns
    Dec 19 '18 at 14:59












  • $begingroup$
    @reuns thanks for your comment! Can you please cite any theorem and / or give me any link to it?
    $endgroup$
    – Euler_Salter
    Dec 19 '18 at 15:14






  • 1




    $begingroup$
    You can take it as the definition of compact operators. The other definitions are more abstract and the proof they are equivalent are not hard. en.wikipedia.org/wiki/Compact_operator_on_Hilbert_space
    $endgroup$
    – reuns
    Dec 19 '18 at 15:57












  • $begingroup$
    @reuns Thanks! Do you mind writing up an answer and expanding a bit more your solution?
    $endgroup$
    – Euler_Salter
    Dec 19 '18 at 16:05
















$begingroup$
There is quite a definition : $T$ is compact iff there is a sequence of finite dimensional operators $T_n$ such that $|T g - T_n g | < A_n|g|$ and $A_n to 0$. Now that $Tg = sum_k c_k langle g,e_k rangle e_k$ with $c_k to 0$ and $(e_k)$ orthonormal implies $|Tg - T_ng| = |sum_{|k| > n} c_k langle g,e_k rangle e_k | < ...$
$endgroup$
– reuns
Dec 19 '18 at 14:59






$begingroup$
There is quite a definition : $T$ is compact iff there is a sequence of finite dimensional operators $T_n$ such that $|T g - T_n g | < A_n|g|$ and $A_n to 0$. Now that $Tg = sum_k c_k langle g,e_k rangle e_k$ with $c_k to 0$ and $(e_k)$ orthonormal implies $|Tg - T_ng| = |sum_{|k| > n} c_k langle g,e_k rangle e_k | < ...$
$endgroup$
– reuns
Dec 19 '18 at 14:59














$begingroup$
@reuns thanks for your comment! Can you please cite any theorem and / or give me any link to it?
$endgroup$
– Euler_Salter
Dec 19 '18 at 15:14




$begingroup$
@reuns thanks for your comment! Can you please cite any theorem and / or give me any link to it?
$endgroup$
– Euler_Salter
Dec 19 '18 at 15:14




1




1




$begingroup$
You can take it as the definition of compact operators. The other definitions are more abstract and the proof they are equivalent are not hard. en.wikipedia.org/wiki/Compact_operator_on_Hilbert_space
$endgroup$
– reuns
Dec 19 '18 at 15:57






$begingroup$
You can take it as the definition of compact operators. The other definitions are more abstract and the proof they are equivalent are not hard. en.wikipedia.org/wiki/Compact_operator_on_Hilbert_space
$endgroup$
– reuns
Dec 19 '18 at 15:57














$begingroup$
@reuns Thanks! Do you mind writing up an answer and expanding a bit more your solution?
$endgroup$
– Euler_Salter
Dec 19 '18 at 16:05




$begingroup$
@reuns Thanks! Do you mind writing up an answer and expanding a bit more your solution?
$endgroup$
– Euler_Salter
Dec 19 '18 at 16:05










1 Answer
1






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oldest

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1












$begingroup$

Let $mathcal{H}$ be a Hilbert space with orthonormal basis ${ e_k }_{k=-infty}^{infty}$. Your operator is the same as the diagonal operator $D$ given by
$$
Df = sum_{k=-infty}^{infty}frac{1}{1+k^2}langle f,e_krangle e_k.
$$



To show that $D$ is compact, let ${ f_n }_{n=1}^{infty}$ be bounded sequence in $mathcal{H}$ with bound $M$; it must be shown that ${ Df_n }_{n=1}^{infty}$ has a convergent subsequence.



Every coordinate sequence ${ langle f_n,e_krangle}_{n=1}^{infty}$ is bounded by $M$; this bound holds for all $k$. Using Cantor diagonalization, there is a subsequence ${ f_{n_m} }_{m=1}^{infty}$ such that $lim_{m}langle f_{n_m},e_krangle$ converges for every $k$ to some limit $alpha_{k}$. The sequence ${alpha_k}$ is uniformly bounded by $M$; therefore $y=sum_{k=-infty}^{infty}frac{alpha_k}{1+k^2} e_k in mathcal{H}$. And,
$$
left|Df_{n_m}-yright|^2
= sum_{k=-infty}^{infty}frac{1}{1+k^2}|langle f_{n_m},e_krangle-alpha_k|^2
$$
converges to $0$ as $mrightarrowinfty$. (You can apply the Lebesgue dominated convergence Theorem applied to a discrete measure, or you can argue directly.) Hence,
$$
lim_{mrightarrowinfty}Df_{n_m}=sum_{k=-infty}^{infty}frac{1}{1+k^2}alpha_k e_k
$$
So $D$ is compact.






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    $begingroup$

    Let $mathcal{H}$ be a Hilbert space with orthonormal basis ${ e_k }_{k=-infty}^{infty}$. Your operator is the same as the diagonal operator $D$ given by
    $$
    Df = sum_{k=-infty}^{infty}frac{1}{1+k^2}langle f,e_krangle e_k.
    $$



    To show that $D$ is compact, let ${ f_n }_{n=1}^{infty}$ be bounded sequence in $mathcal{H}$ with bound $M$; it must be shown that ${ Df_n }_{n=1}^{infty}$ has a convergent subsequence.



    Every coordinate sequence ${ langle f_n,e_krangle}_{n=1}^{infty}$ is bounded by $M$; this bound holds for all $k$. Using Cantor diagonalization, there is a subsequence ${ f_{n_m} }_{m=1}^{infty}$ such that $lim_{m}langle f_{n_m},e_krangle$ converges for every $k$ to some limit $alpha_{k}$. The sequence ${alpha_k}$ is uniformly bounded by $M$; therefore $y=sum_{k=-infty}^{infty}frac{alpha_k}{1+k^2} e_k in mathcal{H}$. And,
    $$
    left|Df_{n_m}-yright|^2
    = sum_{k=-infty}^{infty}frac{1}{1+k^2}|langle f_{n_m},e_krangle-alpha_k|^2
    $$
    converges to $0$ as $mrightarrowinfty$. (You can apply the Lebesgue dominated convergence Theorem applied to a discrete measure, or you can argue directly.) Hence,
    $$
    lim_{mrightarrowinfty}Df_{n_m}=sum_{k=-infty}^{infty}frac{1}{1+k^2}alpha_k e_k
    $$
    So $D$ is compact.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Let $mathcal{H}$ be a Hilbert space with orthonormal basis ${ e_k }_{k=-infty}^{infty}$. Your operator is the same as the diagonal operator $D$ given by
      $$
      Df = sum_{k=-infty}^{infty}frac{1}{1+k^2}langle f,e_krangle e_k.
      $$



      To show that $D$ is compact, let ${ f_n }_{n=1}^{infty}$ be bounded sequence in $mathcal{H}$ with bound $M$; it must be shown that ${ Df_n }_{n=1}^{infty}$ has a convergent subsequence.



      Every coordinate sequence ${ langle f_n,e_krangle}_{n=1}^{infty}$ is bounded by $M$; this bound holds for all $k$. Using Cantor diagonalization, there is a subsequence ${ f_{n_m} }_{m=1}^{infty}$ such that $lim_{m}langle f_{n_m},e_krangle$ converges for every $k$ to some limit $alpha_{k}$. The sequence ${alpha_k}$ is uniformly bounded by $M$; therefore $y=sum_{k=-infty}^{infty}frac{alpha_k}{1+k^2} e_k in mathcal{H}$. And,
      $$
      left|Df_{n_m}-yright|^2
      = sum_{k=-infty}^{infty}frac{1}{1+k^2}|langle f_{n_m},e_krangle-alpha_k|^2
      $$
      converges to $0$ as $mrightarrowinfty$. (You can apply the Lebesgue dominated convergence Theorem applied to a discrete measure, or you can argue directly.) Hence,
      $$
      lim_{mrightarrowinfty}Df_{n_m}=sum_{k=-infty}^{infty}frac{1}{1+k^2}alpha_k e_k
      $$
      So $D$ is compact.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Let $mathcal{H}$ be a Hilbert space with orthonormal basis ${ e_k }_{k=-infty}^{infty}$. Your operator is the same as the diagonal operator $D$ given by
        $$
        Df = sum_{k=-infty}^{infty}frac{1}{1+k^2}langle f,e_krangle e_k.
        $$



        To show that $D$ is compact, let ${ f_n }_{n=1}^{infty}$ be bounded sequence in $mathcal{H}$ with bound $M$; it must be shown that ${ Df_n }_{n=1}^{infty}$ has a convergent subsequence.



        Every coordinate sequence ${ langle f_n,e_krangle}_{n=1}^{infty}$ is bounded by $M$; this bound holds for all $k$. Using Cantor diagonalization, there is a subsequence ${ f_{n_m} }_{m=1}^{infty}$ such that $lim_{m}langle f_{n_m},e_krangle$ converges for every $k$ to some limit $alpha_{k}$. The sequence ${alpha_k}$ is uniformly bounded by $M$; therefore $y=sum_{k=-infty}^{infty}frac{alpha_k}{1+k^2} e_k in mathcal{H}$. And,
        $$
        left|Df_{n_m}-yright|^2
        = sum_{k=-infty}^{infty}frac{1}{1+k^2}|langle f_{n_m},e_krangle-alpha_k|^2
        $$
        converges to $0$ as $mrightarrowinfty$. (You can apply the Lebesgue dominated convergence Theorem applied to a discrete measure, or you can argue directly.) Hence,
        $$
        lim_{mrightarrowinfty}Df_{n_m}=sum_{k=-infty}^{infty}frac{1}{1+k^2}alpha_k e_k
        $$
        So $D$ is compact.






        share|cite|improve this answer









        $endgroup$



        Let $mathcal{H}$ be a Hilbert space with orthonormal basis ${ e_k }_{k=-infty}^{infty}$. Your operator is the same as the diagonal operator $D$ given by
        $$
        Df = sum_{k=-infty}^{infty}frac{1}{1+k^2}langle f,e_krangle e_k.
        $$



        To show that $D$ is compact, let ${ f_n }_{n=1}^{infty}$ be bounded sequence in $mathcal{H}$ with bound $M$; it must be shown that ${ Df_n }_{n=1}^{infty}$ has a convergent subsequence.



        Every coordinate sequence ${ langle f_n,e_krangle}_{n=1}^{infty}$ is bounded by $M$; this bound holds for all $k$. Using Cantor diagonalization, there is a subsequence ${ f_{n_m} }_{m=1}^{infty}$ such that $lim_{m}langle f_{n_m},e_krangle$ converges for every $k$ to some limit $alpha_{k}$. The sequence ${alpha_k}$ is uniformly bounded by $M$; therefore $y=sum_{k=-infty}^{infty}frac{alpha_k}{1+k^2} e_k in mathcal{H}$. And,
        $$
        left|Df_{n_m}-yright|^2
        = sum_{k=-infty}^{infty}frac{1}{1+k^2}|langle f_{n_m},e_krangle-alpha_k|^2
        $$
        converges to $0$ as $mrightarrowinfty$. (You can apply the Lebesgue dominated convergence Theorem applied to a discrete measure, or you can argue directly.) Hence,
        $$
        lim_{mrightarrowinfty}Df_{n_m}=sum_{k=-infty}^{infty}frac{1}{1+k^2}alpha_k e_k
        $$
        So $D$ is compact.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 21 '18 at 3:37









        DisintegratingByPartsDisintegratingByParts

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