Operator norm of $T:l^{2}rightarrow l^{1}$ where $Tx=(x_{1},x_{2}/2,x_{3}/3,x_{4}/4,…)$












4












$begingroup$


As the title states, I need to compute the operator norm of a linear operator
$T:l^{2}rightarrow l^{1}$, where



$$Tx=left(x_{1},frac{x_{2}}{2},frac{x_{3}}{3},frac{x_{4}}{4},... right)$$



Using Holder's inequality for any sequence $(x_{i})_{igeq 1}in l^{2}$, we can show



begin{align}
|Tx|_{1}&=sum_{i=1}^{infty}=|x_{1}|+left|frac{x_{2}}{2}right|+left|frac{x_{3}}{3}right|+cdots\
&=|x_{1}||1|+|x_{2}|left|frac{1}{2}right|+|x_{3}|left|frac{1}{3}right|+cdots \
&leq left|x_{i}right|_{2}left|frac{1}{i}right|_{2} \
&=frac{pi}{sqrt{6}}|(x_{i})|_{2}
end{align}



Hence



$$displaystyle |Tx|_{1}leqfrac{pi}{sqrt{6}}|(x_{i})|_{2}implies||T||leqfrac{pi}{sqrt{6}}$$



However, I am unable to find a sequence in $l_{2}$ which has norm $|(x_{i})|leq 1$ so that I may use the property $||T||=text{sup}_{|(x_{i})|_{2}=1}|Tx|_{1}$.



Any help is appreciated. Thank you.






linear-algebra functional-analysis operator-theory normed-spaces







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$endgroup$












  • $begingroup$
    You can use |T| to get $|T|$. It's claimed that this is better than ||T||, which gives $||T||$.
    $endgroup$
    – DanielWainfleet
    Jan 11 at 6:16


















4












$begingroup$


As the title states, I need to compute the operator norm of a linear operator
$T:l^{2}rightarrow l^{1}$, where



$$Tx=left(x_{1},frac{x_{2}}{2},frac{x_{3}}{3},frac{x_{4}}{4},... right)$$



Using Holder's inequality for any sequence $(x_{i})_{igeq 1}in l^{2}$, we can show



begin{align}
|Tx|_{1}&=sum_{i=1}^{infty}=|x_{1}|+left|frac{x_{2}}{2}right|+left|frac{x_{3}}{3}right|+cdots\
&=|x_{1}||1|+|x_{2}|left|frac{1}{2}right|+|x_{3}|left|frac{1}{3}right|+cdots \
&leq left|x_{i}right|_{2}left|frac{1}{i}right|_{2} \
&=frac{pi}{sqrt{6}}|(x_{i})|_{2}
end{align}



Hence



$$displaystyle |Tx|_{1}leqfrac{pi}{sqrt{6}}|(x_{i})|_{2}implies||T||leqfrac{pi}{sqrt{6}}$$



However, I am unable to find a sequence in $l_{2}$ which has norm $|(x_{i})|leq 1$ so that I may use the property $||T||=text{sup}_{|(x_{i})|_{2}=1}|Tx|_{1}$.



Any help is appreciated. Thank you.






linear-algebra functional-analysis operator-theory normed-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    You can use |T| to get $|T|$. It's claimed that this is better than ||T||, which gives $||T||$.
    $endgroup$
    – DanielWainfleet
    Jan 11 at 6:16
















4












4








4


1



$begingroup$


As the title states, I need to compute the operator norm of a linear operator
$T:l^{2}rightarrow l^{1}$, where



$$Tx=left(x_{1},frac{x_{2}}{2},frac{x_{3}}{3},frac{x_{4}}{4},... right)$$



Using Holder's inequality for any sequence $(x_{i})_{igeq 1}in l^{2}$, we can show



begin{align}
|Tx|_{1}&=sum_{i=1}^{infty}=|x_{1}|+left|frac{x_{2}}{2}right|+left|frac{x_{3}}{3}right|+cdots\
&=|x_{1}||1|+|x_{2}|left|frac{1}{2}right|+|x_{3}|left|frac{1}{3}right|+cdots \
&leq left|x_{i}right|_{2}left|frac{1}{i}right|_{2} \
&=frac{pi}{sqrt{6}}|(x_{i})|_{2}
end{align}



Hence



$$displaystyle |Tx|_{1}leqfrac{pi}{sqrt{6}}|(x_{i})|_{2}implies||T||leqfrac{pi}{sqrt{6}}$$



However, I am unable to find a sequence in $l_{2}$ which has norm $|(x_{i})|leq 1$ so that I may use the property $||T||=text{sup}_{|(x_{i})|_{2}=1}|Tx|_{1}$.



Any help is appreciated. Thank you.






linear-algebra functional-analysis operator-theory normed-spaces







share|cite|improve this question











$endgroup$




As the title states, I need to compute the operator norm of a linear operator
$T:l^{2}rightarrow l^{1}$, where



$$Tx=left(x_{1},frac{x_{2}}{2},frac{x_{3}}{3},frac{x_{4}}{4},... right)$$



Using Holder's inequality for any sequence $(x_{i})_{igeq 1}in l^{2}$, we can show



begin{align}
|Tx|_{1}&=sum_{i=1}^{infty}=|x_{1}|+left|frac{x_{2}}{2}right|+left|frac{x_{3}}{3}right|+cdots\
&=|x_{1}||1|+|x_{2}|left|frac{1}{2}right|+|x_{3}|left|frac{1}{3}right|+cdots \
&leq left|x_{i}right|_{2}left|frac{1}{i}right|_{2} \
&=frac{pi}{sqrt{6}}|(x_{i})|_{2}
end{align}



Hence



$$displaystyle |Tx|_{1}leqfrac{pi}{sqrt{6}}|(x_{i})|_{2}implies||T||leqfrac{pi}{sqrt{6}}$$



However, I am unable to find a sequence in $l_{2}$ which has norm $|(x_{i})|leq 1$ so that I may use the property $||T||=text{sup}_{|(x_{i})|_{2}=1}|Tx|_{1}$.



Any help is appreciated. Thank you.






linear-algebra functional-analysis operator-theory normed-spaces




linear-algebra functional-analysis operator-theory normed-spaces






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share|cite|improve this question













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edited Jan 11 at 4:03









Mattos

2,73721321




2,73721321










asked Jan 10 at 6:47









JackJack

283




283












  • $begingroup$
    You can use |T| to get $|T|$. It's claimed that this is better than ||T||, which gives $||T||$.
    $endgroup$
    – DanielWainfleet
    Jan 11 at 6:16




















  • $begingroup$
    You can use |T| to get $|T|$. It's claimed that this is better than ||T||, which gives $||T||$.
    $endgroup$
    – DanielWainfleet
    Jan 11 at 6:16


















$begingroup$
You can use |T| to get $|T|$. It's claimed that this is better than ||T||, which gives $||T||$.
$endgroup$
– DanielWainfleet
Jan 11 at 6:16






$begingroup$
You can use |T| to get $|T|$. It's claimed that this is better than ||T||, which gives $||T||$.
$endgroup$
– DanielWainfleet
Jan 11 at 6:16












1 Answer
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$begingroup$

Let $x_i=frac c i, i=1,2cdots$ where $c$ is such that $csumlimits_{i=1}^{infty} x_i^{2}=1$. In other words, $c=frac {sqrt 6} {pi}$. Then $|T(x_i)|=sumlimits_{i=1}^{infty} frac c {i^{2}}$ which is exactly $frac {pi} {sqrt 6}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Wow, sometimes when you look at a problem for so long continuously, you don't even realise the solution is right in front of you. Thank you so much!
    $endgroup$
    – Jack
    Jan 10 at 7:51











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Let $x_i=frac c i, i=1,2cdots$ where $c$ is such that $csumlimits_{i=1}^{infty} x_i^{2}=1$. In other words, $c=frac {sqrt 6} {pi}$. Then $|T(x_i)|=sumlimits_{i=1}^{infty} frac c {i^{2}}$ which is exactly $frac {pi} {sqrt 6}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Wow, sometimes when you look at a problem for so long continuously, you don't even realise the solution is right in front of you. Thank you so much!
    $endgroup$
    – Jack
    Jan 10 at 7:51
















2












$begingroup$

Let $x_i=frac c i, i=1,2cdots$ where $c$ is such that $csumlimits_{i=1}^{infty} x_i^{2}=1$. In other words, $c=frac {sqrt 6} {pi}$. Then $|T(x_i)|=sumlimits_{i=1}^{infty} frac c {i^{2}}$ which is exactly $frac {pi} {sqrt 6}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Wow, sometimes when you look at a problem for so long continuously, you don't even realise the solution is right in front of you. Thank you so much!
    $endgroup$
    – Jack
    Jan 10 at 7:51














2












2








2





$begingroup$

Let $x_i=frac c i, i=1,2cdots$ where $c$ is such that $csumlimits_{i=1}^{infty} x_i^{2}=1$. In other words, $c=frac {sqrt 6} {pi}$. Then $|T(x_i)|=sumlimits_{i=1}^{infty} frac c {i^{2}}$ which is exactly $frac {pi} {sqrt 6}$.






share|cite|improve this answer











$endgroup$



Let $x_i=frac c i, i=1,2cdots$ where $c$ is such that $csumlimits_{i=1}^{infty} x_i^{2}=1$. In other words, $c=frac {sqrt 6} {pi}$. Then $|T(x_i)|=sumlimits_{i=1}^{infty} frac c {i^{2}}$ which is exactly $frac {pi} {sqrt 6}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 10 at 7:32

























answered Jan 10 at 7:27









Kavi Rama MurthyKavi Rama Murthy

54.5k32055




54.5k32055












  • $begingroup$
    Wow, sometimes when you look at a problem for so long continuously, you don't even realise the solution is right in front of you. Thank you so much!
    $endgroup$
    – Jack
    Jan 10 at 7:51


















  • $begingroup$
    Wow, sometimes when you look at a problem for so long continuously, you don't even realise the solution is right in front of you. Thank you so much!
    $endgroup$
    – Jack
    Jan 10 at 7:51
















$begingroup$
Wow, sometimes when you look at a problem for so long continuously, you don't even realise the solution is right in front of you. Thank you so much!
$endgroup$
– Jack
Jan 10 at 7:51




$begingroup$
Wow, sometimes when you look at a problem for so long continuously, you don't even realise the solution is right in front of you. Thank you so much!
$endgroup$
– Jack
Jan 10 at 7:51


















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