Showing that the equation $x_i - sum_{j=1}^infty a_{ij}x_j = b_i$ has a unique solution.












6












$begingroup$


Exercise :




Consider the infinite-dimensional system of equations :
$$x_i - sum_{j=1}^infty a_{ij}x_j = b_i, quad i=1,2,3,dots$$
We suppose that $b=(b_1,b_2,dots) in ell^infty$ and that it exists $0<theta<1$ such that :
$$sup_i sum_{j=1}^infty |a_{ij}|leq theta$$
Show that the system of equations given has a unique solution $x=(x_1,x_2,dots) in ell^infty$.




Attempt :



In previous exercises and lessons, I have proved the following lemma :




Let $X$ be a Banach space and $T in B(X)$ with $|T| leq theta < 1$. Then, if $y in X$, the equation $x = y + Tx$ has a unique solution $x in X$.




My initial idea is to work over manipulating the exercise to reach the statements of the lemma above.



The equation for the infinite-dimensional system of equations given, can be rewritten as :



$$x_i - sum_{j=1}^infty a_{ij}x_j = b_i Rightarrow x = b + Tx$$



where $x = (x_1,x_2,dots)$ and $T$ be an operator, such that :



$$Tx = sum_{j=1}^infty a_{ij}x_j$$



Now, I observe that $ell^infty$ is a Banach space and since $b in ell^infty$ then $x - Tx in ell^infty$ which means that $x in ell^infty$ and $T$ is an operator defined over $ell^infty$.



Now, I need to prove that $T in B(ell^infty)$, which means that $T$ is a bounded linear operator $T : ell^infty to ell^infty$ and that $|T| < 1$.



For the case of linearity, let $x,y in ell^infty$ and $lambda in mathbb R$. Then, it is



$$T(lambda x + y) = sum_{j=1}^infty a_{ij}(lambda x_j + y_j) = sum_{j=1}^infty lambda a_{ij} x_j + sum_{j=1}^infty a_{ij}y_j$$



$$=$$



$$lambda sum_{j=1}^infty a_{ij}x_j + sum_{j=1}^infty a_{ij}y_j = lambda T x + Ty$$



which proves that $T$ is a linear operator.



Now, for the case of $T$ being bounded :



$$|Tx|_infty = bigg|sum_{j=1}^infty a_{ij}x_j bigg|_infty leq sum_{j=1}^infty |a_{ij}x_j|_infty leq sum_{j=1}^infty |a_{ij}| |x|_infty$$



But, it is



$$sup_i sum_{j=1}^infty |a_{ij}|leq theta < 1$$



thus, by combining the two last results above, we get :



$$|Tx|_infty leq theta |x|_infty$$



which proves that $T$ is a bounded linear operator $T in B(ell^infty)$ and thus the equation we transformed has a unique solution $x in ell^infty$ which means that the initial infinite-dimensional system of equations has a unique solution $x=(x_1,x2,dots) in ell^infty$.



Question : Is my approach correct and rigorous enough ? Any recommendations on what I can improve or any mistakes ? I would appreciate any input or approvement of my approach.






real-analysis functional-analysis proof-verification operator-theory banach-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    You are right. Your argument contains every necessary element.
    $endgroup$
    – Song
    Nov 25 '18 at 13:41
















6












$begingroup$


Exercise :




Consider the infinite-dimensional system of equations :
$$x_i - sum_{j=1}^infty a_{ij}x_j = b_i, quad i=1,2,3,dots$$
We suppose that $b=(b_1,b_2,dots) in ell^infty$ and that it exists $0<theta<1$ such that :
$$sup_i sum_{j=1}^infty |a_{ij}|leq theta$$
Show that the system of equations given has a unique solution $x=(x_1,x_2,dots) in ell^infty$.




Attempt :



In previous exercises and lessons, I have proved the following lemma :




Let $X$ be a Banach space and $T in B(X)$ with $|T| leq theta < 1$. Then, if $y in X$, the equation $x = y + Tx$ has a unique solution $x in X$.




My initial idea is to work over manipulating the exercise to reach the statements of the lemma above.



The equation for the infinite-dimensional system of equations given, can be rewritten as :



$$x_i - sum_{j=1}^infty a_{ij}x_j = b_i Rightarrow x = b + Tx$$



where $x = (x_1,x_2,dots)$ and $T$ be an operator, such that :



$$Tx = sum_{j=1}^infty a_{ij}x_j$$



Now, I observe that $ell^infty$ is a Banach space and since $b in ell^infty$ then $x - Tx in ell^infty$ which means that $x in ell^infty$ and $T$ is an operator defined over $ell^infty$.



Now, I need to prove that $T in B(ell^infty)$, which means that $T$ is a bounded linear operator $T : ell^infty to ell^infty$ and that $|T| < 1$.



For the case of linearity, let $x,y in ell^infty$ and $lambda in mathbb R$. Then, it is



$$T(lambda x + y) = sum_{j=1}^infty a_{ij}(lambda x_j + y_j) = sum_{j=1}^infty lambda a_{ij} x_j + sum_{j=1}^infty a_{ij}y_j$$



$$=$$



$$lambda sum_{j=1}^infty a_{ij}x_j + sum_{j=1}^infty a_{ij}y_j = lambda T x + Ty$$



which proves that $T$ is a linear operator.



Now, for the case of $T$ being bounded :



$$|Tx|_infty = bigg|sum_{j=1}^infty a_{ij}x_j bigg|_infty leq sum_{j=1}^infty |a_{ij}x_j|_infty leq sum_{j=1}^infty |a_{ij}| |x|_infty$$



But, it is



$$sup_i sum_{j=1}^infty |a_{ij}|leq theta < 1$$



thus, by combining the two last results above, we get :



$$|Tx|_infty leq theta |x|_infty$$



which proves that $T$ is a bounded linear operator $T in B(ell^infty)$ and thus the equation we transformed has a unique solution $x in ell^infty$ which means that the initial infinite-dimensional system of equations has a unique solution $x=(x_1,x2,dots) in ell^infty$.



Question : Is my approach correct and rigorous enough ? Any recommendations on what I can improve or any mistakes ? I would appreciate any input or approvement of my approach.






real-analysis functional-analysis proof-verification operator-theory banach-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    You are right. Your argument contains every necessary element.
    $endgroup$
    – Song
    Nov 25 '18 at 13:41














6












6








6


2



$begingroup$


Exercise :




Consider the infinite-dimensional system of equations :
$$x_i - sum_{j=1}^infty a_{ij}x_j = b_i, quad i=1,2,3,dots$$
We suppose that $b=(b_1,b_2,dots) in ell^infty$ and that it exists $0<theta<1$ such that :
$$sup_i sum_{j=1}^infty |a_{ij}|leq theta$$
Show that the system of equations given has a unique solution $x=(x_1,x_2,dots) in ell^infty$.




Attempt :



In previous exercises and lessons, I have proved the following lemma :




Let $X$ be a Banach space and $T in B(X)$ with $|T| leq theta < 1$. Then, if $y in X$, the equation $x = y + Tx$ has a unique solution $x in X$.




My initial idea is to work over manipulating the exercise to reach the statements of the lemma above.



The equation for the infinite-dimensional system of equations given, can be rewritten as :



$$x_i - sum_{j=1}^infty a_{ij}x_j = b_i Rightarrow x = b + Tx$$



where $x = (x_1,x_2,dots)$ and $T$ be an operator, such that :



$$Tx = sum_{j=1}^infty a_{ij}x_j$$



Now, I observe that $ell^infty$ is a Banach space and since $b in ell^infty$ then $x - Tx in ell^infty$ which means that $x in ell^infty$ and $T$ is an operator defined over $ell^infty$.



Now, I need to prove that $T in B(ell^infty)$, which means that $T$ is a bounded linear operator $T : ell^infty to ell^infty$ and that $|T| < 1$.



For the case of linearity, let $x,y in ell^infty$ and $lambda in mathbb R$. Then, it is



$$T(lambda x + y) = sum_{j=1}^infty a_{ij}(lambda x_j + y_j) = sum_{j=1}^infty lambda a_{ij} x_j + sum_{j=1}^infty a_{ij}y_j$$



$$=$$



$$lambda sum_{j=1}^infty a_{ij}x_j + sum_{j=1}^infty a_{ij}y_j = lambda T x + Ty$$



which proves that $T$ is a linear operator.



Now, for the case of $T$ being bounded :



$$|Tx|_infty = bigg|sum_{j=1}^infty a_{ij}x_j bigg|_infty leq sum_{j=1}^infty |a_{ij}x_j|_infty leq sum_{j=1}^infty |a_{ij}| |x|_infty$$



But, it is



$$sup_i sum_{j=1}^infty |a_{ij}|leq theta < 1$$



thus, by combining the two last results above, we get :



$$|Tx|_infty leq theta |x|_infty$$



which proves that $T$ is a bounded linear operator $T in B(ell^infty)$ and thus the equation we transformed has a unique solution $x in ell^infty$ which means that the initial infinite-dimensional system of equations has a unique solution $x=(x_1,x2,dots) in ell^infty$.



Question : Is my approach correct and rigorous enough ? Any recommendations on what I can improve or any mistakes ? I would appreciate any input or approvement of my approach.






real-analysis functional-analysis proof-verification operator-theory banach-spaces







share|cite|improve this question











$endgroup$




Exercise :




Consider the infinite-dimensional system of equations :
$$x_i - sum_{j=1}^infty a_{ij}x_j = b_i, quad i=1,2,3,dots$$
We suppose that $b=(b_1,b_2,dots) in ell^infty$ and that it exists $0<theta<1$ such that :
$$sup_i sum_{j=1}^infty |a_{ij}|leq theta$$
Show that the system of equations given has a unique solution $x=(x_1,x_2,dots) in ell^infty$.




Attempt :



In previous exercises and lessons, I have proved the following lemma :




Let $X$ be a Banach space and $T in B(X)$ with $|T| leq theta < 1$. Then, if $y in X$, the equation $x = y + Tx$ has a unique solution $x in X$.




My initial idea is to work over manipulating the exercise to reach the statements of the lemma above.



The equation for the infinite-dimensional system of equations given, can be rewritten as :



$$x_i - sum_{j=1}^infty a_{ij}x_j = b_i Rightarrow x = b + Tx$$



where $x = (x_1,x_2,dots)$ and $T$ be an operator, such that :



$$Tx = sum_{j=1}^infty a_{ij}x_j$$



Now, I observe that $ell^infty$ is a Banach space and since $b in ell^infty$ then $x - Tx in ell^infty$ which means that $x in ell^infty$ and $T$ is an operator defined over $ell^infty$.



Now, I need to prove that $T in B(ell^infty)$, which means that $T$ is a bounded linear operator $T : ell^infty to ell^infty$ and that $|T| < 1$.



For the case of linearity, let $x,y in ell^infty$ and $lambda in mathbb R$. Then, it is



$$T(lambda x + y) = sum_{j=1}^infty a_{ij}(lambda x_j + y_j) = sum_{j=1}^infty lambda a_{ij} x_j + sum_{j=1}^infty a_{ij}y_j$$



$$=$$



$$lambda sum_{j=1}^infty a_{ij}x_j + sum_{j=1}^infty a_{ij}y_j = lambda T x + Ty$$



which proves that $T$ is a linear operator.



Now, for the case of $T$ being bounded :



$$|Tx|_infty = bigg|sum_{j=1}^infty a_{ij}x_j bigg|_infty leq sum_{j=1}^infty |a_{ij}x_j|_infty leq sum_{j=1}^infty |a_{ij}| |x|_infty$$



But, it is



$$sup_i sum_{j=1}^infty |a_{ij}|leq theta < 1$$



thus, by combining the two last results above, we get :



$$|Tx|_infty leq theta |x|_infty$$



which proves that $T$ is a bounded linear operator $T in B(ell^infty)$ and thus the equation we transformed has a unique solution $x in ell^infty$ which means that the initial infinite-dimensional system of equations has a unique solution $x=(x_1,x2,dots) in ell^infty$.



Question : Is my approach correct and rigorous enough ? Any recommendations on what I can improve or any mistakes ? I would appreciate any input or approvement of my approach.






real-analysis functional-analysis proof-verification operator-theory banach-spaces




real-analysis functional-analysis proof-verification operator-theory banach-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 '18 at 15:02







Rebellos

















asked Nov 25 '18 at 12:25









RebellosRebellos

14.5k31246




14.5k31246












  • $begingroup$
    You are right. Your argument contains every necessary element.
    $endgroup$
    – Song
    Nov 25 '18 at 13:41


















  • $begingroup$
    You are right. Your argument contains every necessary element.
    $endgroup$
    – Song
    Nov 25 '18 at 13:41
















$begingroup$
You are right. Your argument contains every necessary element.
$endgroup$
– Song
Nov 25 '18 at 13:41




$begingroup$
You are right. Your argument contains every necessary element.
$endgroup$
– Song
Nov 25 '18 at 13:41










0






active

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