Mixing
Show compactness of an operator
$begingroup$
Let $T: C^0[0,1] rightarrow l^1$, $(Tf)_n=a_n int_0^{1/n} f(x)dx$, for an $f$ in $C^0[0,1]$. Prove that $T$ is compact when ${frac{a_n}{n} }_n in l^1$.
I know the definitin of compact operator, but in this case I'd like to state that if ${ frac{a_n}{n} } in l^1$, then $T$ is a finite rank operator. Indeed,
$||(Tf)_n||_{l^1} = sum_n |frac{a_n}{n} int_0^1 f(frac{t}{n})dt|$. But $f$ is continuous in $[0,1]$, then that integral is finite, and, particularly, it's bounded from $||f||_{infty}$, thus the last sum is less or equal than $||f ||_{infty} sum_n |frac{a_n}{n}|$, thus it's convergent, and then $T$ has finite rank.
Is it okay?
EDIT
Set $T_m(f)=(a_1 int_0^1f(x)dx,ldots,a_m int_0^{1/m}f(x)dx, 0, 0, ldots)$.
I want to show that $|| T_m-T|| rightarrow _m 0$ in the op. norm. I have to compute $sup { ||T_m(f) - T(f)||_{l^1}: ||f||_{infty} leq 1 }=sup { sum_{k=m+1} |a_k int_0^frac{1}{k}f(x)dx|: ||f||_{infty} leq 1 }$
Now, $sum_{k=m+1} |a_k int_0^frac{1}{k}f(x)dx| leq sum_{k=1}^{infty} frac|{a_k}{k} int_0^1 f(frac{t}{n})dt| leq ||f||_{infty} sum_k |frac{a_k}{k}|$.
Thus, by taking the supremum I have that $||T-T_m|| = sum_{k=m+1}^infty |frac{a_k}{k}|$. But for ${ frac{a_k}{k} } in l^1$, this is the remainder of a convergent series, and then the limit over $m$ goes to $0$. So $T$ is compact, since it's limit of compact (they're finite rank) operators
functional-analysis operator-theory compact-operators
asked Jan 28 at 16:21
VoBVoB
761513
$endgroup$
add a comment |
$begingroup$
Let $T: C^0[0,1] rightarrow l^1$, $(Tf)_n=a_n int_0^{1/n} f(x)dx$, for an $f$ in $C^0[0,1]$. Prove that $T$ is compact when ${frac{a_n}{n} }_n in l^1$.
I know the definitin of compact operator, but in this case I'd like to state that if ${ frac{a_n}{n} } in l^1$, then $T$ is a finite rank operator. Indeed,
$||(Tf)_n||_{l^1} = sum_n |frac{a_n}{n} int_0^1 f(frac{t}{n})dt|$. But $f$ is continuous in $[0,1]$, then that integral is finite, and, particularly, it's bounded from $||f||_{infty}$, thus the last sum is less or equal than $||f ||_{infty} sum_n |frac{a_n}{n}|$, thus it's convergent, and then $T$ has finite rank.
Is it okay?
EDIT
Set $T_m(f)=(a_1 int_0^1f(x)dx,ldots,a_m int_0^{1/m}f(x)dx, 0, 0, ldots)$.
I want to show that $|| T_m-T|| rightarrow _m 0$ in the op. norm. I have to compute $sup { ||T_m(f) - T(f)||_{l^1}: ||f||_{infty} leq 1 }=sup { sum_{k=m+1} |a_k int_0^frac{1}{k}f(x)dx|: ||f||_{infty} leq 1 }$
Now, $sum_{k=m+1} |a_k int_0^frac{1}{k}f(x)dx| leq sum_{k=1}^{infty} frac|{a_k}{k} int_0^1 f(frac{t}{n})dt| leq ||f||_{infty} sum_k |frac{a_k}{k}|$.
Thus, by taking the supremum I have that $||T-T_m|| = sum_{k=m+1}^infty |frac{a_k}{k}|$. But for ${ frac{a_k}{k} } in l^1$, this is the remainder of a convergent series, and then the limit over $m$ goes to $0$. So $T$ is compact, since it's limit of compact (they're finite rank) operators
functional-analysis operator-theory compact-operators
asked Jan 28 at 16:21
VoBVoB
761513
$endgroup$
$begingroup$
How did you leap from convergent to finite rank? That is the essence of the question and it is not true.
$endgroup$
– copper.hat
Jan 28 at 16:29
$begingroup$
Okay, now I see the problem. So I have to change strategy. Could I try with Ascoli Arzelà?
$endgroup$
– VoB
Jan 28 at 16:44
$begingroup$
Perhaps showing T to be the limit of finite rank operators would be less of a fishing trip?
$endgroup$
– copper.hat
Jan 28 at 16:51
$begingroup$
edited my original post !
$endgroup$
– VoB
Jan 28 at 17:25
add a comment |
$begingroup$
Let $T: C^0[0,1] rightarrow l^1$, $(Tf)_n=a_n int_0^{1/n} f(x)dx$, for an $f$ in $C^0[0,1]$. Prove that $T$ is compact when ${frac{a_n}{n} }_n in l^1$.
I know the definitin of compact operator, but in this case I'd like to state that if ${ frac{a_n}{n} } in l^1$, then $T$ is a finite rank operator. Indeed,
$||(Tf)_n||_{l^1} = sum_n |frac{a_n}{n} int_0^1 f(frac{t}{n})dt|$. But $f$ is continuous in $[0,1]$, then that integral is finite, and, particularly, it's bounded from $||f||_{infty}$, thus the last sum is less or equal than $||f ||_{infty} sum_n |frac{a_n}{n}|$, thus it's convergent, and then $T$ has finite rank.
Is it okay?
EDIT
Set $T_m(f)=(a_1 int_0^1f(x)dx,ldots,a_m int_0^{1/m}f(x)dx, 0, 0, ldots)$.
I want to show that $|| T_m-T|| rightarrow _m 0$ in the op. norm. I have to compute $sup { ||T_m(f) - T(f)||_{l^1}: ||f||_{infty} leq 1 }=sup { sum_{k=m+1} |a_k int_0^frac{1}{k}f(x)dx|: ||f||_{infty} leq 1 }$
Now, $sum_{k=m+1} |a_k int_0^frac{1}{k}f(x)dx| leq sum_{k=1}^{infty} frac|{a_k}{k} int_0^1 f(frac{t}{n})dt| leq ||f||_{infty} sum_k |frac{a_k}{k}|$.
Thus, by taking the supremum I have that $||T-T_m|| = sum_{k=m+1}^infty |frac{a_k}{k}|$. But for ${ frac{a_k}{k} } in l^1$, this is the remainder of a convergent series, and then the limit over $m$ goes to $0$. So $T$ is compact, since it's limit of compact (they're finite rank) operators
functional-analysis operator-theory compact-operators
asked Jan 28 at 16:21
VoBVoB
761513
$endgroup$
Let $T: C^0[0,1] rightarrow l^1$, $(Tf)_n=a_n int_0^{1/n} f(x)dx$, for an $f$ in $C^0[0,1]$. Prove that $T$ is compact when ${frac{a_n}{n} }_n in l^1$.
I know the definitin of compact operator, but in this case I'd like to state that if ${ frac{a_n}{n} } in l^1$, then $T$ is a finite rank operator. Indeed,
$||(Tf)_n||_{l^1} = sum_n |frac{a_n}{n} int_0^1 f(frac{t}{n})dt|$. But $f$ is continuous in $[0,1]$, then that integral is finite, and, particularly, it's bounded from $||f||_{infty}$, thus the last sum is less or equal than $||f ||_{infty} sum_n |frac{a_n}{n}|$, thus it's convergent, and then $T$ has finite rank.
Is it okay?
EDIT
Set $T_m(f)=(a_1 int_0^1f(x)dx,ldots,a_m int_0^{1/m}f(x)dx, 0, 0, ldots)$.
I want to show that $|| T_m-T|| rightarrow _m 0$ in the op. norm. I have to compute $sup { ||T_m(f) - T(f)||_{l^1}: ||f||_{infty} leq 1 }=sup { sum_{k=m+1} |a_k int_0^frac{1}{k}f(x)dx|: ||f||_{infty} leq 1 }$
Now, $sum_{k=m+1} |a_k int_0^frac{1}{k}f(x)dx| leq sum_{k=1}^{infty} frac|{a_k}{k} int_0^1 f(frac{t}{n})dt| leq ||f||_{infty} sum_k |frac{a_k}{k}|$.
Thus, by taking the supremum I have that $||T-T_m|| = sum_{k=m+1}^infty |frac{a_k}{k}|$. But for ${ frac{a_k}{k} } in l^1$, this is the remainder of a convergent series, and then the limit over $m$ goes to $0$. So $T$ is compact, since it's limit of compact (they're finite rank) operators
functional-analysis operator-theory compact-operators
functional-analysis operator-theory compact-operators
asked Jan 28 at 16:21
VoBVoB
761513
asked Jan 28 at 16:21
VoBVoB
761513
edited Jan 28 at 17:16
VoB
asked Jan 28 at 16:21
VoBVoB
761513
asked Jan 28 at 16:21
VoBVoB
761513
asked Jan 28 at 16:21
VoBVoB
761513
761513
$begingroup$
How did you leap from convergent to finite rank? That is the essence of the question and it is not true.
$endgroup$
– copper.hat
Jan 28 at 16:29
$begingroup$
Okay, now I see the problem. So I have to change strategy. Could I try with Ascoli Arzelà?
$endgroup$
– VoB
Jan 28 at 16:44
$begingroup$
Perhaps showing T to be the limit of finite rank operators would be less of a fishing trip?
$endgroup$
– copper.hat
Jan 28 at 16:51
$begingroup$
edited my original post !
$endgroup$
– VoB
Jan 28 at 17:25
add a comment |
$begingroup$
How did you leap from convergent to finite rank? That is the essence of the question and it is not true.
$endgroup$
– copper.hat
Jan 28 at 16:29
$begingroup$
Okay, now I see the problem. So I have to change strategy. Could I try with Ascoli Arzelà?
$endgroup$
– VoB
Jan 28 at 16:44
$begingroup$
Perhaps showing T to be the limit of finite rank operators would be less of a fishing trip?
$endgroup$
– copper.hat
Jan 28 at 16:51
$begingroup$
edited my original post !
$endgroup$
– VoB
Jan 28 at 17:25
$begingroup$
How did you leap from convergent to finite rank? That is the essence of the question and it is not true.
$endgroup$
– copper.hat
Jan 28 at 16:29
$begingroup$
How did you leap from convergent to finite rank? That is the essence of the question and it is not true.
$endgroup$
– copper.hat
Jan 28 at 16:29
$begingroup$
Okay, now I see the problem. So I have to change strategy. Could I try with Ascoli Arzelà?
$endgroup$
– VoB
Jan 28 at 16:44
$begingroup$
Okay, now I see the problem. So I have to change strategy. Could I try with Ascoli Arzelà?
$endgroup$
– VoB
Jan 28 at 16:44
$begingroup$
Perhaps showing T to be the limit of finite rank operators would be less of a fishing trip?
$endgroup$
– copper.hat
Jan 28 at 16:51
$begingroup$
Perhaps showing T to be the limit of finite rank operators would be less of a fishing trip?
$endgroup$
– copper.hat
Jan 28 at 16:51
$begingroup$
edited my original post !
$endgroup$
– VoB
Jan 28 at 17:25
$begingroup$
edited my original post !
$endgroup$
– VoB
Jan 28 at 17:25
add a comment |
1 Answer
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While this operator is not of finite rank, can you show it is a limit of finite rank operators? (In this case you're done since a limit of compact operators is compact).
answered Jan 28 at 16:51
bitesizebobitesizebo
1,63118
$endgroup$
$begingroup$
Okay, I know this result. I'm editing my original post
$endgroup$
– VoB
Jan 28 at 16:58
$begingroup$
Fixed. Let me know if it's alright
$endgroup$
– VoB
Jan 28 at 17:16
$begingroup$
Looks good to me
$endgroup$
– bitesizebo
Jan 28 at 17:38
add a comment |
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1 Answer
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1 Answer
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$begingroup$
While this operator is not of finite rank, can you show it is a limit of finite rank operators? (In this case you're done since a limit of compact operators is compact).
answered Jan 28 at 16:51
bitesizebobitesizebo
1,63118
$endgroup$
$begingroup$
Okay, I know this result. I'm editing my original post
$endgroup$
– VoB
Jan 28 at 16:58
$begingroup$
Fixed. Let me know if it's alright
$endgroup$
– VoB
Jan 28 at 17:16
$begingroup$
Looks good to me
$endgroup$
– bitesizebo
Jan 28 at 17:38
add a comment |
$begingroup$
While this operator is not of finite rank, can you show it is a limit of finite rank operators? (In this case you're done since a limit of compact operators is compact).
answered Jan 28 at 16:51
bitesizebobitesizebo
1,63118
$endgroup$
$begingroup$
Okay, I know this result. I'm editing my original post
$endgroup$
– VoB
Jan 28 at 16:58
$begingroup$
Fixed. Let me know if it's alright
$endgroup$
– VoB
Jan 28 at 17:16
$begingroup$
Looks good to me
$endgroup$
– bitesizebo
Jan 28 at 17:38
add a comment |
$begingroup$
While this operator is not of finite rank, can you show it is a limit of finite rank operators? (In this case you're done since a limit of compact operators is compact).
answered Jan 28 at 16:51
bitesizebobitesizebo
1,63118
$endgroup$
While this operator is not of finite rank, can you show it is a limit of finite rank operators? (In this case you're done since a limit of compact operators is compact).
answered Jan 28 at 16:51
bitesizebobitesizebo
1,63118
answered Jan 28 at 16:51
bitesizebobitesizebo
1,63118
answered Jan 28 at 16:51
bitesizebobitesizebo
1,63118
answered Jan 28 at 16:51
bitesizebobitesizebo
1,63118
1,63118
$begingroup$
Okay, I know this result. I'm editing my original post
$endgroup$
– VoB
Jan 28 at 16:58
$begingroup$
Fixed. Let me know if it's alright
$endgroup$
– VoB
Jan 28 at 17:16
$begingroup$
Looks good to me
$endgroup$
– bitesizebo
Jan 28 at 17:38
add a comment |
$begingroup$
Okay, I know this result. I'm editing my original post
$endgroup$
– VoB
Jan 28 at 16:58
$begingroup$
Fixed. Let me know if it's alright
$endgroup$
– VoB
Jan 28 at 17:16
$begingroup$
Looks good to me
$endgroup$
– bitesizebo
Jan 28 at 17:38
$begingroup$
Okay, I know this result. I'm editing my original post
$endgroup$
– VoB
Jan 28 at 16:58
$begingroup$
Okay, I know this result. I'm editing my original post
$endgroup$
– VoB
Jan 28 at 16:58
$begingroup$
Fixed. Let me know if it's alright
$endgroup$
– VoB
Jan 28 at 17:16
$begingroup$
Fixed. Let me know if it's alright
$endgroup$
– VoB
Jan 28 at 17:16
$begingroup$
Looks good to me
$endgroup$
– bitesizebo
Jan 28 at 17:38
$begingroup$
Looks good to me
$endgroup$
– bitesizebo
Jan 28 at 17:38
add a comment |
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$begingroup$
How did you leap from convergent to finite rank? That is the essence of the question and it is not true.
$endgroup$
– copper.hat
Jan 28 at 16:29
$begingroup$
Okay, now I see the problem. So I have to change strategy. Could I try with Ascoli Arzelà?
$endgroup$
– VoB
Jan 28 at 16:44
$begingroup$
Perhaps showing T to be the limit of finite rank operators would be less of a fishing trip?
$endgroup$
– copper.hat
Jan 28 at 16:51
$begingroup$
edited my original post !
$endgroup$
– VoB
Jan 28 at 17:25