Relationship between compact operators and coefficients tending to zero
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While reading the solutions of an exercise, I found the following statement:
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
asked Dec 19 '18 at 14:55
Euler_SalterEuler_Salter
2,0671336
$endgroup$
add a comment |
$begingroup$
While reading the solutions of an exercise, I found the following statement:
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
asked Dec 19 '18 at 14:55
Euler_SalterEuler_Salter
2,0671336
$endgroup$
$begingroup$
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
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@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
add a comment |
$begingroup$
While reading the solutions of an exercise, I found the following statement:
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
asked Dec 19 '18 at 14:55
Euler_SalterEuler_Salter
2,0671336
$endgroup$
While reading the solutions of an exercise, I found the following statement:
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
asked Dec 19 '18 at 14:55
Euler_SalterEuler_Salter
2,0671336
asked Dec 19 '18 at 14:55
Euler_SalterEuler_Salter
2,0671336
asked Dec 19 '18 at 14:55
Euler_SalterEuler_Salter
2,0671336
asked Dec 19 '18 at 14:55
Euler_SalterEuler_Salter
2,0671336
asked Dec 19 '18 at 14:55
Euler_SalterEuler_Salter
2,0671336
2,0671336
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
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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.
answered Dec 21 '18 at 3:37
DisintegratingByPartsDisintegratingByParts
59.4k42580
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add a comment |
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1 Answer
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1 Answer
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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.
answered Dec 21 '18 at 3:37
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 21 '18 at 3:37
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 21 '18 at 3:37
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
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.
answered Dec 21 '18 at 3:37
DisintegratingByPartsDisintegratingByParts
59.4k42580
answered Dec 21 '18 at 3:37
DisintegratingByPartsDisintegratingByParts
59.4k42580
answered Dec 21 '18 at 3:37
DisintegratingByPartsDisintegratingByParts
59.4k42580
answered Dec 21 '18 at 3:37
DisintegratingByPartsDisintegratingByParts
59.4k42580
59.4k42580
add a comment |
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$begingroup$
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