$||T-T_n|| rightarrow 0$ and $T_n$ are compact but $T$ is not a compact operator.












1












$begingroup$


It is a result that if $||T-T_n|| rightarrow 0$ in the norm operator an that the $T_n in mathcal{L}(X,Y)$ (were $Y$ is a Banach space) are compact operators, then $T$ is compact. I found from here that pointwise convergence is not enough. But is there an easy counter-example for the case when $||T-T_n|| rightarrow 0$ but $Y$ is not Banach and when then $T$ is not compact?






functional-analysis compact-operators







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is that norm if $Y$ is not Banach?
    $endgroup$
    – lcv
    Jan 6 at 14:47










  • $begingroup$
    $||T||=sup_{||x le 1||}||Tx||_Y$ no?
    $endgroup$
    – roi_saumon
    Jan 6 at 15:02










  • $begingroup$
    If you redefine compact operator so that the image of the unit ball is precompact (instead of relatively compact, for Banach spaces this is equivalent) then the limit is indeed compact.
    $endgroup$
    – Jochen
    Jan 7 at 8:17
















1












$begingroup$


It is a result that if $||T-T_n|| rightarrow 0$ in the norm operator an that the $T_n in mathcal{L}(X,Y)$ (were $Y$ is a Banach space) are compact operators, then $T$ is compact. I found from here that pointwise convergence is not enough. But is there an easy counter-example for the case when $||T-T_n|| rightarrow 0$ but $Y$ is not Banach and when then $T$ is not compact?






functional-analysis compact-operators







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is that norm if $Y$ is not Banach?
    $endgroup$
    – lcv
    Jan 6 at 14:47










  • $begingroup$
    $||T||=sup_{||x le 1||}||Tx||_Y$ no?
    $endgroup$
    – roi_saumon
    Jan 6 at 15:02










  • $begingroup$
    If you redefine compact operator so that the image of the unit ball is precompact (instead of relatively compact, for Banach spaces this is equivalent) then the limit is indeed compact.
    $endgroup$
    – Jochen
    Jan 7 at 8:17














1












1








1





$begingroup$


It is a result that if $||T-T_n|| rightarrow 0$ in the norm operator an that the $T_n in mathcal{L}(X,Y)$ (were $Y$ is a Banach space) are compact operators, then $T$ is compact. I found from here that pointwise convergence is not enough. But is there an easy counter-example for the case when $||T-T_n|| rightarrow 0$ but $Y$ is not Banach and when then $T$ is not compact?






functional-analysis compact-operators







share|cite|improve this question











$endgroup$




It is a result that if $||T-T_n|| rightarrow 0$ in the norm operator an that the $T_n in mathcal{L}(X,Y)$ (were $Y$ is a Banach space) are compact operators, then $T$ is compact. I found from here that pointwise convergence is not enough. But is there an easy counter-example for the case when $||T-T_n|| rightarrow 0$ but $Y$ is not Banach and when then $T$ is not compact?






functional-analysis compact-operators




functional-analysis compact-operators






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 14:35









Paul Frost

9,9553932




9,9553932










asked Jan 6 at 13:41









roi_saumonroi_saumon

47328




47328












  • $begingroup$
    What is that norm if $Y$ is not Banach?
    $endgroup$
    – lcv
    Jan 6 at 14:47










  • $begingroup$
    $||T||=sup_{||x le 1||}||Tx||_Y$ no?
    $endgroup$
    – roi_saumon
    Jan 6 at 15:02










  • $begingroup$
    If you redefine compact operator so that the image of the unit ball is precompact (instead of relatively compact, for Banach spaces this is equivalent) then the limit is indeed compact.
    $endgroup$
    – Jochen
    Jan 7 at 8:17


















  • $begingroup$
    What is that norm if $Y$ is not Banach?
    $endgroup$
    – lcv
    Jan 6 at 14:47










  • $begingroup$
    $||T||=sup_{||x le 1||}||Tx||_Y$ no?
    $endgroup$
    – roi_saumon
    Jan 6 at 15:02










  • $begingroup$
    If you redefine compact operator so that the image of the unit ball is precompact (instead of relatively compact, for Banach spaces this is equivalent) then the limit is indeed compact.
    $endgroup$
    – Jochen
    Jan 7 at 8:17
















$begingroup$
What is that norm if $Y$ is not Banach?
$endgroup$
– lcv
Jan 6 at 14:47




$begingroup$
What is that norm if $Y$ is not Banach?
$endgroup$
– lcv
Jan 6 at 14:47












$begingroup$
$||T||=sup_{||x le 1||}||Tx||_Y$ no?
$endgroup$
– roi_saumon
Jan 6 at 15:02




$begingroup$
$||T||=sup_{||x le 1||}||Tx||_Y$ no?
$endgroup$
– roi_saumon
Jan 6 at 15:02












$begingroup$
If you redefine compact operator so that the image of the unit ball is precompact (instead of relatively compact, for Banach spaces this is equivalent) then the limit is indeed compact.
$endgroup$
– Jochen
Jan 7 at 8:17




$begingroup$
If you redefine compact operator so that the image of the unit ball is precompact (instead of relatively compact, for Banach spaces this is equivalent) then the limit is indeed compact.
$endgroup$
– Jochen
Jan 7 at 8:17










1 Answer
1






active

oldest

votes


















2












$begingroup$

Consider $X=Y=d$ the space of finite sequences with the supremum norm. Then consider $$T_n(x_1,x_2,...,x_n,x_{n+1},...)=left(frac{x_1}{1},frac{x_2}{2},...,frac{x_n}{n},0,...right).$$ This sequence of compact operators converges to $$T(x_1,x_2,...)=left(frac{x_1}{1},frac{x_2}{2},...right).$$ However this is not a compact operator since $T(B_1)$ is not complete.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I am not sure I understand the "However this is not a compact operator since T(B1) is not complete.". What would be a sequence for which the image has no converging subsequence?
    $endgroup$
    – roi_saumon
    Jan 6 at 23:38












  • $begingroup$
    Consider $x^{(n)}=(2^{-1},2^{-2},...,2^{-n},0,...)$. Then $T(x^{(n)})=(2^{-1}/1,2^{-2}/2,...,2^{-n}/n,0,...)$ is a non-convergent Cauchy sequence in $T(B_1)$. (Because $d$ only contains finite sequences.)
    $endgroup$
    – SmileyCraft
    Jan 7 at 3:06












  • $begingroup$
    @SimileyCraft I am not sure to understand why we have the $1/n$ and the $2^{-n}$. If I take $T_n in mathcal L(l^p)$ with $T_n : x mapsto (x_1, x_2, ..., x_n, 0, 0, ...)$ then $T_n$ is compact and $T_n rightarrow Id_{l^p}$ in the operator norm, no? But then $Id_{l^p}$ is not compact because the sequence $(1,0,0,0,...), (0,1,0,0,...),(0,0,1,0,0,...)$ has no converging subsequence even though it is bounded. Also, you conclude by saying that we have a Cauchy sequence in $T(B_1)$ that doesn't converge meaning that $T(B_1)$ is not complete. Does it imply that it is not relatively compact?
    $endgroup$
    – roi_saumon
    Jan 7 at 12:17












  • $begingroup$
    @roi_saumon Your $T_n$ does not converge to the identity operator. Remember we must use the operator norm. Compact implies sequentially compact. If a Cauchy sequence does not converge, then it also has no converging subsequence.
    $endgroup$
    – SmileyCraft
    Jan 8 at 14:05











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

Consider $X=Y=d$ the space of finite sequences with the supremum norm. Then consider $$T_n(x_1,x_2,...,x_n,x_{n+1},...)=left(frac{x_1}{1},frac{x_2}{2},...,frac{x_n}{n},0,...right).$$ This sequence of compact operators converges to $$T(x_1,x_2,...)=left(frac{x_1}{1},frac{x_2}{2},...right).$$ However this is not a compact operator since $T(B_1)$ is not complete.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I am not sure I understand the "However this is not a compact operator since T(B1) is not complete.". What would be a sequence for which the image has no converging subsequence?
    $endgroup$
    – roi_saumon
    Jan 6 at 23:38












  • $begingroup$
    Consider $x^{(n)}=(2^{-1},2^{-2},...,2^{-n},0,...)$. Then $T(x^{(n)})=(2^{-1}/1,2^{-2}/2,...,2^{-n}/n,0,...)$ is a non-convergent Cauchy sequence in $T(B_1)$. (Because $d$ only contains finite sequences.)
    $endgroup$
    – SmileyCraft
    Jan 7 at 3:06












  • $begingroup$
    @SimileyCraft I am not sure to understand why we have the $1/n$ and the $2^{-n}$. If I take $T_n in mathcal L(l^p)$ with $T_n : x mapsto (x_1, x_2, ..., x_n, 0, 0, ...)$ then $T_n$ is compact and $T_n rightarrow Id_{l^p}$ in the operator norm, no? But then $Id_{l^p}$ is not compact because the sequence $(1,0,0,0,...), (0,1,0,0,...),(0,0,1,0,0,...)$ has no converging subsequence even though it is bounded. Also, you conclude by saying that we have a Cauchy sequence in $T(B_1)$ that doesn't converge meaning that $T(B_1)$ is not complete. Does it imply that it is not relatively compact?
    $endgroup$
    – roi_saumon
    Jan 7 at 12:17












  • $begingroup$
    @roi_saumon Your $T_n$ does not converge to the identity operator. Remember we must use the operator norm. Compact implies sequentially compact. If a Cauchy sequence does not converge, then it also has no converging subsequence.
    $endgroup$
    – SmileyCraft
    Jan 8 at 14:05
















2












$begingroup$

Consider $X=Y=d$ the space of finite sequences with the supremum norm. Then consider $$T_n(x_1,x_2,...,x_n,x_{n+1},...)=left(frac{x_1}{1},frac{x_2}{2},...,frac{x_n}{n},0,...right).$$ This sequence of compact operators converges to $$T(x_1,x_2,...)=left(frac{x_1}{1},frac{x_2}{2},...right).$$ However this is not a compact operator since $T(B_1)$ is not complete.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I am not sure I understand the "However this is not a compact operator since T(B1) is not complete.". What would be a sequence for which the image has no converging subsequence?
    $endgroup$
    – roi_saumon
    Jan 6 at 23:38












  • $begingroup$
    Consider $x^{(n)}=(2^{-1},2^{-2},...,2^{-n},0,...)$. Then $T(x^{(n)})=(2^{-1}/1,2^{-2}/2,...,2^{-n}/n,0,...)$ is a non-convergent Cauchy sequence in $T(B_1)$. (Because $d$ only contains finite sequences.)
    $endgroup$
    – SmileyCraft
    Jan 7 at 3:06












  • $begingroup$
    @SimileyCraft I am not sure to understand why we have the $1/n$ and the $2^{-n}$. If I take $T_n in mathcal L(l^p)$ with $T_n : x mapsto (x_1, x_2, ..., x_n, 0, 0, ...)$ then $T_n$ is compact and $T_n rightarrow Id_{l^p}$ in the operator norm, no? But then $Id_{l^p}$ is not compact because the sequence $(1,0,0,0,...), (0,1,0,0,...),(0,0,1,0,0,...)$ has no converging subsequence even though it is bounded. Also, you conclude by saying that we have a Cauchy sequence in $T(B_1)$ that doesn't converge meaning that $T(B_1)$ is not complete. Does it imply that it is not relatively compact?
    $endgroup$
    – roi_saumon
    Jan 7 at 12:17












  • $begingroup$
    @roi_saumon Your $T_n$ does not converge to the identity operator. Remember we must use the operator norm. Compact implies sequentially compact. If a Cauchy sequence does not converge, then it also has no converging subsequence.
    $endgroup$
    – SmileyCraft
    Jan 8 at 14:05














2












2








2





$begingroup$

Consider $X=Y=d$ the space of finite sequences with the supremum norm. Then consider $$T_n(x_1,x_2,...,x_n,x_{n+1},...)=left(frac{x_1}{1},frac{x_2}{2},...,frac{x_n}{n},0,...right).$$ This sequence of compact operators converges to $$T(x_1,x_2,...)=left(frac{x_1}{1},frac{x_2}{2},...right).$$ However this is not a compact operator since $T(B_1)$ is not complete.






share|cite|improve this answer









$endgroup$



Consider $X=Y=d$ the space of finite sequences with the supremum norm. Then consider $$T_n(x_1,x_2,...,x_n,x_{n+1},...)=left(frac{x_1}{1},frac{x_2}{2},...,frac{x_n}{n},0,...right).$$ This sequence of compact operators converges to $$T(x_1,x_2,...)=left(frac{x_1}{1},frac{x_2}{2},...right).$$ However this is not a compact operator since $T(B_1)$ is not complete.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 6 at 15:07









SmileyCraftSmileyCraft

3,401516




3,401516












  • $begingroup$
    I am not sure I understand the "However this is not a compact operator since T(B1) is not complete.". What would be a sequence for which the image has no converging subsequence?
    $endgroup$
    – roi_saumon
    Jan 6 at 23:38












  • $begingroup$
    Consider $x^{(n)}=(2^{-1},2^{-2},...,2^{-n},0,...)$. Then $T(x^{(n)})=(2^{-1}/1,2^{-2}/2,...,2^{-n}/n,0,...)$ is a non-convergent Cauchy sequence in $T(B_1)$. (Because $d$ only contains finite sequences.)
    $endgroup$
    – SmileyCraft
    Jan 7 at 3:06












  • $begingroup$
    @SimileyCraft I am not sure to understand why we have the $1/n$ and the $2^{-n}$. If I take $T_n in mathcal L(l^p)$ with $T_n : x mapsto (x_1, x_2, ..., x_n, 0, 0, ...)$ then $T_n$ is compact and $T_n rightarrow Id_{l^p}$ in the operator norm, no? But then $Id_{l^p}$ is not compact because the sequence $(1,0,0,0,...), (0,1,0,0,...),(0,0,1,0,0,...)$ has no converging subsequence even though it is bounded. Also, you conclude by saying that we have a Cauchy sequence in $T(B_1)$ that doesn't converge meaning that $T(B_1)$ is not complete. Does it imply that it is not relatively compact?
    $endgroup$
    – roi_saumon
    Jan 7 at 12:17












  • $begingroup$
    @roi_saumon Your $T_n$ does not converge to the identity operator. Remember we must use the operator norm. Compact implies sequentially compact. If a Cauchy sequence does not converge, then it also has no converging subsequence.
    $endgroup$
    – SmileyCraft
    Jan 8 at 14:05


















  • $begingroup$
    I am not sure I understand the "However this is not a compact operator since T(B1) is not complete.". What would be a sequence for which the image has no converging subsequence?
    $endgroup$
    – roi_saumon
    Jan 6 at 23:38












  • $begingroup$
    Consider $x^{(n)}=(2^{-1},2^{-2},...,2^{-n},0,...)$. Then $T(x^{(n)})=(2^{-1}/1,2^{-2}/2,...,2^{-n}/n,0,...)$ is a non-convergent Cauchy sequence in $T(B_1)$. (Because $d$ only contains finite sequences.)
    $endgroup$
    – SmileyCraft
    Jan 7 at 3:06












  • $begingroup$
    @SimileyCraft I am not sure to understand why we have the $1/n$ and the $2^{-n}$. If I take $T_n in mathcal L(l^p)$ with $T_n : x mapsto (x_1, x_2, ..., x_n, 0, 0, ...)$ then $T_n$ is compact and $T_n rightarrow Id_{l^p}$ in the operator norm, no? But then $Id_{l^p}$ is not compact because the sequence $(1,0,0,0,...), (0,1,0,0,...),(0,0,1,0,0,...)$ has no converging subsequence even though it is bounded. Also, you conclude by saying that we have a Cauchy sequence in $T(B_1)$ that doesn't converge meaning that $T(B_1)$ is not complete. Does it imply that it is not relatively compact?
    $endgroup$
    – roi_saumon
    Jan 7 at 12:17












  • $begingroup$
    @roi_saumon Your $T_n$ does not converge to the identity operator. Remember we must use the operator norm. Compact implies sequentially compact. If a Cauchy sequence does not converge, then it also has no converging subsequence.
    $endgroup$
    – SmileyCraft
    Jan 8 at 14:05
















$begingroup$
I am not sure I understand the "However this is not a compact operator since T(B1) is not complete.". What would be a sequence for which the image has no converging subsequence?
$endgroup$
– roi_saumon
Jan 6 at 23:38






$begingroup$
I am not sure I understand the "However this is not a compact operator since T(B1) is not complete.". What would be a sequence for which the image has no converging subsequence?
$endgroup$
– roi_saumon
Jan 6 at 23:38














$begingroup$
Consider $x^{(n)}=(2^{-1},2^{-2},...,2^{-n},0,...)$. Then $T(x^{(n)})=(2^{-1}/1,2^{-2}/2,...,2^{-n}/n,0,...)$ is a non-convergent Cauchy sequence in $T(B_1)$. (Because $d$ only contains finite sequences.)
$endgroup$
– SmileyCraft
Jan 7 at 3:06






$begingroup$
Consider $x^{(n)}=(2^{-1},2^{-2},...,2^{-n},0,...)$. Then $T(x^{(n)})=(2^{-1}/1,2^{-2}/2,...,2^{-n}/n,0,...)$ is a non-convergent Cauchy sequence in $T(B_1)$. (Because $d$ only contains finite sequences.)
$endgroup$
– SmileyCraft
Jan 7 at 3:06














$begingroup$
@SimileyCraft I am not sure to understand why we have the $1/n$ and the $2^{-n}$. If I take $T_n in mathcal L(l^p)$ with $T_n : x mapsto (x_1, x_2, ..., x_n, 0, 0, ...)$ then $T_n$ is compact and $T_n rightarrow Id_{l^p}$ in the operator norm, no? But then $Id_{l^p}$ is not compact because the sequence $(1,0,0,0,...), (0,1,0,0,...),(0,0,1,0,0,...)$ has no converging subsequence even though it is bounded. Also, you conclude by saying that we have a Cauchy sequence in $T(B_1)$ that doesn't converge meaning that $T(B_1)$ is not complete. Does it imply that it is not relatively compact?
$endgroup$
– roi_saumon
Jan 7 at 12:17






$begingroup$
@SimileyCraft I am not sure to understand why we have the $1/n$ and the $2^{-n}$. If I take $T_n in mathcal L(l^p)$ with $T_n : x mapsto (x_1, x_2, ..., x_n, 0, 0, ...)$ then $T_n$ is compact and $T_n rightarrow Id_{l^p}$ in the operator norm, no? But then $Id_{l^p}$ is not compact because the sequence $(1,0,0,0,...), (0,1,0,0,...),(0,0,1,0,0,...)$ has no converging subsequence even though it is bounded. Also, you conclude by saying that we have a Cauchy sequence in $T(B_1)$ that doesn't converge meaning that $T(B_1)$ is not complete. Does it imply that it is not relatively compact?
$endgroup$
– roi_saumon
Jan 7 at 12:17














$begingroup$
@roi_saumon Your $T_n$ does not converge to the identity operator. Remember we must use the operator norm. Compact implies sequentially compact. If a Cauchy sequence does not converge, then it also has no converging subsequence.
$endgroup$
– SmileyCraft
Jan 8 at 14:05




$begingroup$
@roi_saumon Your $T_n$ does not converge to the identity operator. Remember we must use the operator norm. Compact implies sequentially compact. If a Cauchy sequence does not converge, then it also has no converging subsequence.
$endgroup$
– SmileyCraft
Jan 8 at 14:05


















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Display a custom attribute below product name in the front-end Magento 1.9.3.8

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0 I need to display a custom attribute bellow the product name and I don't know how can I call this attribute. I am editing the file /app/design/frontend/templatemela/MAG100219_1/template/catalog/product/view.phtml But how can I call this attribute in php? I tried calling a string echo with the name I put in the attribute, but it didn't work. Code is as below: <div class="page-title product-title"> <h1><?php echo $_helper->productAttribute($_product, $_product->getName(), 'name') ?></h1> </div> <div class="codigoinmetro"> <h2><?php echo $_helper->productAttribute($_product, $_product->getName(), 'registroinmetro') ?></h2> </div> php attrib...
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