Show that the sum is the norm of $l_1$












2












$begingroup$


I have the space $mathbb E = l_1$ and $Tx = (x_1, 0, x_3, 0, ...).$
I need to show that $T$ is continuous.



The norm of $l_1$ is $$Vert x Vert = sum_{k=1}^∞ vert x_k vert.$$



My solution:



$Vert Tx Vert = sum_{k=1}^∞ vert (Tx)_k vert = vert 0 vert + sum_{k=0}^∞ vert x_{2k+1} vert $.



Question:



How to continue this equality to get $sum_{k=1}^infty vert x_k vert$ (the norm)?






functional-analysis continuity operator-theory norm lp-spaces







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You do not need to continue the equality. An inequality will do just as well (provided the norm is on the greater side)
    $endgroup$
    – Mindlack
    Dec 16 '18 at 11:18
















2












$begingroup$


I have the space $mathbb E = l_1$ and $Tx = (x_1, 0, x_3, 0, ...).$
I need to show that $T$ is continuous.



The norm of $l_1$ is $$Vert x Vert = sum_{k=1}^∞ vert x_k vert.$$



My solution:



$Vert Tx Vert = sum_{k=1}^∞ vert (Tx)_k vert = vert 0 vert + sum_{k=0}^∞ vert x_{2k+1} vert $.



Question:



How to continue this equality to get $sum_{k=1}^infty vert x_k vert$ (the norm)?






functional-analysis continuity operator-theory norm lp-spaces







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You do not need to continue the equality. An inequality will do just as well (provided the norm is on the greater side)
    $endgroup$
    – Mindlack
    Dec 16 '18 at 11:18














2












2








2


1



$begingroup$


I have the space $mathbb E = l_1$ and $Tx = (x_1, 0, x_3, 0, ...).$
I need to show that $T$ is continuous.



The norm of $l_1$ is $$Vert x Vert = sum_{k=1}^∞ vert x_k vert.$$



My solution:



$Vert Tx Vert = sum_{k=1}^∞ vert (Tx)_k vert = vert 0 vert + sum_{k=0}^∞ vert x_{2k+1} vert $.



Question:



How to continue this equality to get $sum_{k=1}^infty vert x_k vert$ (the norm)?






functional-analysis continuity operator-theory norm lp-spaces







share|cite|improve this question











$endgroup$




I have the space $mathbb E = l_1$ and $Tx = (x_1, 0, x_3, 0, ...).$
I need to show that $T$ is continuous.



The norm of $l_1$ is $$Vert x Vert = sum_{k=1}^∞ vert x_k vert.$$



My solution:



$Vert Tx Vert = sum_{k=1}^∞ vert (Tx)_k vert = vert 0 vert + sum_{k=0}^∞ vert x_{2k+1} vert $.



Question:



How to continue this equality to get $sum_{k=1}^infty vert x_k vert$ (the norm)?






functional-analysis continuity operator-theory norm lp-spaces




functional-analysis continuity operator-theory norm lp-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 12:36









Davide Giraudo

126k16150261




126k16150261










asked Dec 16 '18 at 11:06









PhilipPhilip

826




826








  • 2




    $begingroup$
    You do not need to continue the equality. An inequality will do just as well (provided the norm is on the greater side)
    $endgroup$
    – Mindlack
    Dec 16 '18 at 11:18














  • 2




    $begingroup$
    You do not need to continue the equality. An inequality will do just as well (provided the norm is on the greater side)
    $endgroup$
    – Mindlack
    Dec 16 '18 at 11:18








2




2




$begingroup$
You do not need to continue the equality. An inequality will do just as well (provided the norm is on the greater side)
$endgroup$
– Mindlack
Dec 16 '18 at 11:18




$begingroup$
You do not need to continue the equality. An inequality will do just as well (provided the norm is on the greater side)
$endgroup$
– Mindlack
Dec 16 '18 at 11:18










1 Answer
1






active

oldest

votes


















2












$begingroup$

It is :



$$|Tx| = |(x_1,0,x_3,0,dots)| = sum_{k=1}^infty|x_{2k+1}| leq sum_{k=1}^infty|x_k| = |x|.$$



since the absolute sum of the odd entries of the sequence is definitely smaller or equal to the total sum of absolute values of all entries.



Thus :



$$|Tx| leq |x| implies |T| leq 1$$



Now, if you take the sequence $x = (x_n) in ell^1$ with $x = (1,0,dots,0,dots)$, then :



$$|Tx| = |(1,0,dots,0,dots)| = 1 implies |T| = 1$$



Thus the operator $T$ is continuous.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I would like to have an input from the person that downvoted.
    $endgroup$
    – Rebellos
    Dec 16 '18 at 14:10











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1 Answer
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1 Answer
1






active

oldest

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active

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active

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2












$begingroup$

It is :



$$|Tx| = |(x_1,0,x_3,0,dots)| = sum_{k=1}^infty|x_{2k+1}| leq sum_{k=1}^infty|x_k| = |x|.$$



since the absolute sum of the odd entries of the sequence is definitely smaller or equal to the total sum of absolute values of all entries.



Thus :



$$|Tx| leq |x| implies |T| leq 1$$



Now, if you take the sequence $x = (x_n) in ell^1$ with $x = (1,0,dots,0,dots)$, then :



$$|Tx| = |(1,0,dots,0,dots)| = 1 implies |T| = 1$$



Thus the operator $T$ is continuous.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I would like to have an input from the person that downvoted.
    $endgroup$
    – Rebellos
    Dec 16 '18 at 14:10
















2












$begingroup$

It is :



$$|Tx| = |(x_1,0,x_3,0,dots)| = sum_{k=1}^infty|x_{2k+1}| leq sum_{k=1}^infty|x_k| = |x|.$$



since the absolute sum of the odd entries of the sequence is definitely smaller or equal to the total sum of absolute values of all entries.



Thus :



$$|Tx| leq |x| implies |T| leq 1$$



Now, if you take the sequence $x = (x_n) in ell^1$ with $x = (1,0,dots,0,dots)$, then :



$$|Tx| = |(1,0,dots,0,dots)| = 1 implies |T| = 1$$



Thus the operator $T$ is continuous.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I would like to have an input from the person that downvoted.
    $endgroup$
    – Rebellos
    Dec 16 '18 at 14:10














2












2








2





$begingroup$

It is :



$$|Tx| = |(x_1,0,x_3,0,dots)| = sum_{k=1}^infty|x_{2k+1}| leq sum_{k=1}^infty|x_k| = |x|.$$



since the absolute sum of the odd entries of the sequence is definitely smaller or equal to the total sum of absolute values of all entries.



Thus :



$$|Tx| leq |x| implies |T| leq 1$$



Now, if you take the sequence $x = (x_n) in ell^1$ with $x = (1,0,dots,0,dots)$, then :



$$|Tx| = |(1,0,dots,0,dots)| = 1 implies |T| = 1$$



Thus the operator $T$ is continuous.






share|cite|improve this answer











$endgroup$



It is :



$$|Tx| = |(x_1,0,x_3,0,dots)| = sum_{k=1}^infty|x_{2k+1}| leq sum_{k=1}^infty|x_k| = |x|.$$



since the absolute sum of the odd entries of the sequence is definitely smaller or equal to the total sum of absolute values of all entries.



Thus :



$$|Tx| leq |x| implies |T| leq 1$$



Now, if you take the sequence $x = (x_n) in ell^1$ with $x = (1,0,dots,0,dots)$, then :



$$|Tx| = |(1,0,dots,0,dots)| = 1 implies |T| = 1$$



Thus the operator $T$ is continuous.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 16 '18 at 11:37

























answered Dec 16 '18 at 11:18









RebellosRebellos

14.6k31247




14.6k31247












  • $begingroup$
    I would like to have an input from the person that downvoted.
    $endgroup$
    – Rebellos
    Dec 16 '18 at 14:10


















  • $begingroup$
    I would like to have an input from the person that downvoted.
    $endgroup$
    – Rebellos
    Dec 16 '18 at 14:10
















$begingroup$
I would like to have an input from the person that downvoted.
$endgroup$
– Rebellos
Dec 16 '18 at 14:10




$begingroup$
I would like to have an input from the person that downvoted.
$endgroup$
– Rebellos
Dec 16 '18 at 14:10


















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