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












1














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
























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










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










  • 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.
    – Jochen
    Jan 7 at 8:17
















1














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
























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










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










  • 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.
    – Jochen
    Jan 7 at 8:17














1












1








1







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















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

9,7203732




9,7203732










asked Jan 6 at 13:41









roi_saumonroi_saumon

43428




43428












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










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










  • 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.
    – Jochen
    Jan 7 at 8:17


















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










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










  • 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.
    – Jochen
    Jan 7 at 8:17
















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




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












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




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












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.
– Jochen
Jan 7 at 8:17




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.
– Jochen
Jan 7 at 8:17










1 Answer
1






active

oldest

votes


















2














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





















  • 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?
    – roi_saumon
    Jan 6 at 23:38












  • 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.)
    – SmileyCraft
    Jan 7 at 3:06












  • @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?
    – roi_saumon
    Jan 7 at 12:17












  • @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.
    – SmileyCraft
    yesterday











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1 Answer
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1 Answer
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active

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active

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votes









2














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





















  • 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?
    – roi_saumon
    Jan 6 at 23:38












  • 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.)
    – SmileyCraft
    Jan 7 at 3:06












  • @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?
    – roi_saumon
    Jan 7 at 12:17












  • @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.
    – SmileyCraft
    yesterday
















2














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





















  • 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?
    – roi_saumon
    Jan 6 at 23:38












  • 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.)
    – SmileyCraft
    Jan 7 at 3:06












  • @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?
    – roi_saumon
    Jan 7 at 12:17












  • @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.
    – SmileyCraft
    yesterday














2












2








2






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












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

3,261516




3,261516












  • 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?
    – roi_saumon
    Jan 6 at 23:38












  • 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.)
    – SmileyCraft
    Jan 7 at 3:06












  • @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?
    – roi_saumon
    Jan 7 at 12:17












  • @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.
    – SmileyCraft
    yesterday


















  • 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?
    – roi_saumon
    Jan 6 at 23:38












  • 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.)
    – SmileyCraft
    Jan 7 at 3:06












  • @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?
    – roi_saumon
    Jan 7 at 12:17












  • @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.
    – SmileyCraft
    yesterday
















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?
– roi_saumon
Jan 6 at 23:38






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?
– roi_saumon
Jan 6 at 23:38














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.)
– SmileyCraft
Jan 7 at 3:06






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.)
– SmileyCraft
Jan 7 at 3:06














@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?
– roi_saumon
Jan 7 at 12:17






@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?
– roi_saumon
Jan 7 at 12:17














@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.
– SmileyCraft
yesterday




@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.
– SmileyCraft
yesterday


















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