operator norm of $f$ defined as follows $(f(u))_n = v_n = u_{n+1} - u_n,$
$begingroup$
let $l^{infty}$ be the vector space of bounded sequences equipped with the norm $|u| = sup_{n in mathbb{N}} |u_n|$
and let $f$ be an endomorphism on $l^{infty}$ defined as follows $f(u) = v$
where $v = (v_n)_{n in mathbb{N}}, text{and}, v_n = u_{n+1} - u_n$
we have $|f(u)| = sup_{n in mathbb{N}} |u_{n+1} - u_n| leq 2 sup_{n in mathbb{N}} |u_n| = 2|u|$
therefore $f$ is continuous
we also have $|f| leq 2$
on the other hand, for $w_n = frac{1}{n^k}$ we have $frac{|f(w)|}{|w|} = frac{2^k-1}{2^{k}}$
which for large $k$ approaches $1$
therefore $1 leq |f| leq 2$
this is the best approximation of the norm that I could get, any tips or help for computing the exact value of the norm are welcome. thanks !
sequences-and-series functional-analysis operator-theory
asked Dec 26 '18 at 10:10
rapidracimrapidracim
1,7191419
$endgroup$
add a comment |
$begingroup$
let $l^{infty}$ be the vector space of bounded sequences equipped with the norm $|u| = sup_{n in mathbb{N}} |u_n|$
and let $f$ be an endomorphism on $l^{infty}$ defined as follows $f(u) = v$
where $v = (v_n)_{n in mathbb{N}}, text{and}, v_n = u_{n+1} - u_n$
we have $|f(u)| = sup_{n in mathbb{N}} |u_{n+1} - u_n| leq 2 sup_{n in mathbb{N}} |u_n| = 2|u|$
therefore $f$ is continuous
we also have $|f| leq 2$
on the other hand, for $w_n = frac{1}{n^k}$ we have $frac{|f(w)|}{|w|} = frac{2^k-1}{2^{k}}$
which for large $k$ approaches $1$
therefore $1 leq |f| leq 2$
this is the best approximation of the norm that I could get, any tips or help for computing the exact value of the norm are welcome. thanks !
sequences-and-series functional-analysis operator-theory
asked Dec 26 '18 at 10:10
rapidracimrapidracim
1,7191419
$endgroup$
add a comment |
$begingroup$
let $l^{infty}$ be the vector space of bounded sequences equipped with the norm $|u| = sup_{n in mathbb{N}} |u_n|$
and let $f$ be an endomorphism on $l^{infty}$ defined as follows $f(u) = v$
where $v = (v_n)_{n in mathbb{N}}, text{and}, v_n = u_{n+1} - u_n$
we have $|f(u)| = sup_{n in mathbb{N}} |u_{n+1} - u_n| leq 2 sup_{n in mathbb{N}} |u_n| = 2|u|$
therefore $f$ is continuous
we also have $|f| leq 2$
on the other hand, for $w_n = frac{1}{n^k}$ we have $frac{|f(w)|}{|w|} = frac{2^k-1}{2^{k}}$
which for large $k$ approaches $1$
therefore $1 leq |f| leq 2$
this is the best approximation of the norm that I could get, any tips or help for computing the exact value of the norm are welcome. thanks !
sequences-and-series functional-analysis operator-theory
asked Dec 26 '18 at 10:10
rapidracimrapidracim
1,7191419
$endgroup$
let $l^{infty}$ be the vector space of bounded sequences equipped with the norm $|u| = sup_{n in mathbb{N}} |u_n|$
and let $f$ be an endomorphism on $l^{infty}$ defined as follows $f(u) = v$
where $v = (v_n)_{n in mathbb{N}}, text{and}, v_n = u_{n+1} - u_n$
we have $|f(u)| = sup_{n in mathbb{N}} |u_{n+1} - u_n| leq 2 sup_{n in mathbb{N}} |u_n| = 2|u|$
therefore $f$ is continuous
we also have $|f| leq 2$
on the other hand, for $w_n = frac{1}{n^k}$ we have $frac{|f(w)|}{|w|} = frac{2^k-1}{2^{k}}$
which for large $k$ approaches $1$
therefore $1 leq |f| leq 2$
this is the best approximation of the norm that I could get, any tips or help for computing the exact value of the norm are welcome. thanks !
sequences-and-series functional-analysis operator-theory
sequences-and-series functional-analysis operator-theory
asked Dec 26 '18 at 10:10
rapidracimrapidracim
1,7191419
asked Dec 26 '18 at 10:10
rapidracimrapidracim
1,7191419
asked Dec 26 '18 at 10:10
rapidracimrapidracim
1,7191419
asked Dec 26 '18 at 10:10
rapidracimrapidracim
1,7191419
asked Dec 26 '18 at 10:10
rapidracimrapidracim
1,7191419
1,7191419
add a comment |
add a comment |
1 Answer
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$begingroup$
You should try the sequence $(-1,1,0,0,0,0,0,dots)$.
answered Dec 26 '18 at 10:17
PhoemueXPhoemueX
27.5k22457
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You should try the sequence $(-1,1,0,0,0,0,0,dots)$.
answered Dec 26 '18 at 10:17
PhoemueXPhoemueX
27.5k22457
$endgroup$
add a comment |
$begingroup$
You should try the sequence $(-1,1,0,0,0,0,0,dots)$.
answered Dec 26 '18 at 10:17
PhoemueXPhoemueX
27.5k22457
$endgroup$
add a comment |
$begingroup$
You should try the sequence $(-1,1,0,0,0,0,0,dots)$.
answered Dec 26 '18 at 10:17
PhoemueXPhoemueX
27.5k22457
$endgroup$
You should try the sequence $(-1,1,0,0,0,0,0,dots)$.
answered Dec 26 '18 at 10:17
PhoemueXPhoemueX
27.5k22457
answered Dec 26 '18 at 10:17
PhoemueXPhoemueX
27.5k22457
answered Dec 26 '18 at 10:17
PhoemueXPhoemueX
27.5k22457
answered Dec 26 '18 at 10:17
PhoemueXPhoemueX
27.5k22457
27.5k22457
add a comment |
add a comment |
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