Operator on $ell^1(mathbb{N})$ space
$begingroup$
Let $A = (A_{n,m})$ with $A_{n,m} in mathbb{C}, n,m in mathbb{N}$ be a matrix. Assume $|A| = sup_m sum_n|A_{n,m}| < infty$. Show that for $T:ell^1(mathbb{N}) to ell^1(mathbb{N}), (Tf)(n) = sum_mA_{n,m}f(m)$ we have $|T| = |A|$.
Showing $|T| leq |A|$ was easy. But how can I show $|T| geq |A|$?
functional-analysis norm
edited Dec 12 '18 at 4:09
davyjones
395213
asked Jul 30 '18 at 8:02
AM88AM88
97
$endgroup$
add a comment |
$begingroup$
Let $A = (A_{n,m})$ with $A_{n,m} in mathbb{C}, n,m in mathbb{N}$ be a matrix. Assume $|A| = sup_m sum_n|A_{n,m}| < infty$. Show that for $T:ell^1(mathbb{N}) to ell^1(mathbb{N}), (Tf)(n) = sum_mA_{n,m}f(m)$ we have $|T| = |A|$.
Showing $|T| leq |A|$ was easy. But how can I show $|T| geq |A|$?
functional-analysis norm
edited Dec 12 '18 at 4:09
davyjones
395213
asked Jul 30 '18 at 8:02
AM88AM88
97
$endgroup$
add a comment |
$begingroup$
Let $A = (A_{n,m})$ with $A_{n,m} in mathbb{C}, n,m in mathbb{N}$ be a matrix. Assume $|A| = sup_m sum_n|A_{n,m}| < infty$. Show that for $T:ell^1(mathbb{N}) to ell^1(mathbb{N}), (Tf)(n) = sum_mA_{n,m}f(m)$ we have $|T| = |A|$.
Showing $|T| leq |A|$ was easy. But how can I show $|T| geq |A|$?
functional-analysis norm
edited Dec 12 '18 at 4:09
davyjones
395213
asked Jul 30 '18 at 8:02
AM88AM88
97
$endgroup$
Let $A = (A_{n,m})$ with $A_{n,m} in mathbb{C}, n,m in mathbb{N}$ be a matrix. Assume $|A| = sup_m sum_n|A_{n,m}| < infty$. Show that for $T:ell^1(mathbb{N}) to ell^1(mathbb{N}), (Tf)(n) = sum_mA_{n,m}f(m)$ we have $|T| = |A|$.
Showing $|T| leq |A|$ was easy. But how can I show $|T| geq |A|$?
functional-analysis norm
functional-analysis norm
edited Dec 12 '18 at 4:09
davyjones
395213
asked Jul 30 '18 at 8:02
AM88AM88
97
edited Dec 12 '18 at 4:09
davyjones
395213
asked Jul 30 '18 at 8:02
AM88AM88
97
edited Dec 12 '18 at 4:09
davyjones
395213
edited Dec 12 '18 at 4:09
davyjones
395213
edited Dec 12 '18 at 4:09
davyjones
395213
395213
asked Jul 30 '18 at 8:02
AM88AM88
97
asked Jul 30 '18 at 8:02
AM88AM88
97
asked Jul 30 '18 at 8:02
AM88AM88
97
97
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $e_k$ be the sequence with $1$ at the $k$-th place and $0$ everywhere else. Then $|Te_k|=sum_n |T(e_k)(n)|= sum _n |A_{n,k}|$ so $|T| geq sum _n |A_{n,k}|$ for all $k$. Take sup over $k$.
edited Dec 12 '18 at 4:09
davyjones
395213
answered Jul 30 '18 at 8:10
Kavi Rama MurthyKavi Rama Murthy
62.1k42262
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $e_k$ be the sequence with $1$ at the $k$-th place and $0$ everywhere else. Then $|Te_k|=sum_n |T(e_k)(n)|= sum _n |A_{n,k}|$ so $|T| geq sum _n |A_{n,k}|$ for all $k$. Take sup over $k$.
edited Dec 12 '18 at 4:09
davyjones
395213
answered Jul 30 '18 at 8:10
Kavi Rama MurthyKavi Rama Murthy
62.1k42262
$endgroup$
add a comment |
$begingroup$
Let $e_k$ be the sequence with $1$ at the $k$-th place and $0$ everywhere else. Then $|Te_k|=sum_n |T(e_k)(n)|= sum _n |A_{n,k}|$ so $|T| geq sum _n |A_{n,k}|$ for all $k$. Take sup over $k$.
edited Dec 12 '18 at 4:09
davyjones
395213
answered Jul 30 '18 at 8:10
Kavi Rama MurthyKavi Rama Murthy
62.1k42262
$endgroup$
add a comment |
$begingroup$
Let $e_k$ be the sequence with $1$ at the $k$-th place and $0$ everywhere else. Then $|Te_k|=sum_n |T(e_k)(n)|= sum _n |A_{n,k}|$ so $|T| geq sum _n |A_{n,k}|$ for all $k$. Take sup over $k$.
edited Dec 12 '18 at 4:09
davyjones
395213
answered Jul 30 '18 at 8:10
Kavi Rama MurthyKavi Rama Murthy
62.1k42262
$endgroup$
Let $e_k$ be the sequence with $1$ at the $k$-th place and $0$ everywhere else. Then $|Te_k|=sum_n |T(e_k)(n)|= sum _n |A_{n,k}|$ so $|T| geq sum _n |A_{n,k}|$ for all $k$. Take sup over $k$.
edited Dec 12 '18 at 4:09
davyjones
395213
answered Jul 30 '18 at 8:10
Kavi Rama MurthyKavi Rama Murthy
62.1k42262
edited Dec 12 '18 at 4:09
davyjones
395213
edited Dec 12 '18 at 4:09
davyjones
395213
edited Dec 12 '18 at 4:09
davyjones
395213
395213
answered Jul 30 '18 at 8:10
Kavi Rama MurthyKavi Rama Murthy
62.1k42262
answered Jul 30 '18 at 8:10
Kavi Rama MurthyKavi Rama Murthy
62.1k42262
answered Jul 30 '18 at 8:10
Kavi Rama MurthyKavi Rama Murthy
62.1k42262
62.1k42262
add a comment |
add a comment |
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