Showing any linear operator $T : X to Y$ is bounded, where $X$ is a finite dimensional normed vector space,...












4












$begingroup$


Let $X$ be a finite dimensional normed vector space and $Y$ an arbitrary normed vector space. Show that any linear operator $T : X to Y$ is bounded.



I got the hint to first show that $| x|_0 := | x | + | Tx|$, $x in X$, defines a norm on $X$, but I do not know how this should help me.



Further I should calculate $|T|$ for where $X = K^n$, equipped with the Euclidean norm $|cdot|_2$, $Y := ell_1(mathbb{N})$ and $Tx := (x_1,ldots,x_n,0,0,ldots) in ell_1(mathbb{N})$, for all $x = (x_1,ldots,x_n) in K^n$.



Can please someone help?






functional-analysis vector-spaces normed-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint: What theorem do you know about norms on a finite-dimensional vector space?
    $endgroup$
    – Mindlack
    Dec 16 '18 at 11:02










  • $begingroup$
    Norms on a finite-dimensional vector space are equivalent?
    $endgroup$
    – Anna Schmitz
    Dec 16 '18 at 11:14










  • $begingroup$
    Yes! So what can you infer, if you know that the original norn and the one you defined are equivalent?
    $endgroup$
    – Mindlack
    Dec 16 '18 at 11:16










  • $begingroup$
    $∥Tx∥ leq ∥x∥ + ∥Tx∥ = ∥x∥_0$
    $endgroup$
    – Anna Schmitz
    Dec 16 '18 at 11:25






  • 1




    $begingroup$
    I am sorry, I must have misunderstood your first comment and confused you further. You do have $|Tx| leq |x|_0$, thus the operator is bounded.
    $endgroup$
    – Mindlack
    Dec 16 '18 at 11:41
















4












$begingroup$


Let $X$ be a finite dimensional normed vector space and $Y$ an arbitrary normed vector space. Show that any linear operator $T : X to Y$ is bounded.



I got the hint to first show that $| x|_0 := | x | + | Tx|$, $x in X$, defines a norm on $X$, but I do not know how this should help me.



Further I should calculate $|T|$ for where $X = K^n$, equipped with the Euclidean norm $|cdot|_2$, $Y := ell_1(mathbb{N})$ and $Tx := (x_1,ldots,x_n,0,0,ldots) in ell_1(mathbb{N})$, for all $x = (x_1,ldots,x_n) in K^n$.



Can please someone help?






functional-analysis vector-spaces normed-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint: What theorem do you know about norms on a finite-dimensional vector space?
    $endgroup$
    – Mindlack
    Dec 16 '18 at 11:02










  • $begingroup$
    Norms on a finite-dimensional vector space are equivalent?
    $endgroup$
    – Anna Schmitz
    Dec 16 '18 at 11:14










  • $begingroup$
    Yes! So what can you infer, if you know that the original norn and the one you defined are equivalent?
    $endgroup$
    – Mindlack
    Dec 16 '18 at 11:16










  • $begingroup$
    $∥Tx∥ leq ∥x∥ + ∥Tx∥ = ∥x∥_0$
    $endgroup$
    – Anna Schmitz
    Dec 16 '18 at 11:25






  • 1




    $begingroup$
    I am sorry, I must have misunderstood your first comment and confused you further. You do have $|Tx| leq |x|_0$, thus the operator is bounded.
    $endgroup$
    – Mindlack
    Dec 16 '18 at 11:41














4












4








4





$begingroup$


Let $X$ be a finite dimensional normed vector space and $Y$ an arbitrary normed vector space. Show that any linear operator $T : X to Y$ is bounded.



I got the hint to first show that $| x|_0 := | x | + | Tx|$, $x in X$, defines a norm on $X$, but I do not know how this should help me.



Further I should calculate $|T|$ for where $X = K^n$, equipped with the Euclidean norm $|cdot|_2$, $Y := ell_1(mathbb{N})$ and $Tx := (x_1,ldots,x_n,0,0,ldots) in ell_1(mathbb{N})$, for all $x = (x_1,ldots,x_n) in K^n$.



Can please someone help?






functional-analysis vector-spaces normed-spaces







share|cite|improve this question











$endgroup$




Let $X$ be a finite dimensional normed vector space and $Y$ an arbitrary normed vector space. Show that any linear operator $T : X to Y$ is bounded.



I got the hint to first show that $| x|_0 := | x | + | Tx|$, $x in X$, defines a norm on $X$, but I do not know how this should help me.



Further I should calculate $|T|$ for where $X = K^n$, equipped with the Euclidean norm $|cdot|_2$, $Y := ell_1(mathbb{N})$ and $Tx := (x_1,ldots,x_n,0,0,ldots) in ell_1(mathbb{N})$, for all $x = (x_1,ldots,x_n) in K^n$.



Can please someone help?






functional-analysis vector-spaces normed-spaces




functional-analysis vector-spaces normed-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 12:17









amWhy

1




1










asked Dec 16 '18 at 10:45









Anna SchmitzAnna Schmitz

917




917












  • $begingroup$
    Hint: What theorem do you know about norms on a finite-dimensional vector space?
    $endgroup$
    – Mindlack
    Dec 16 '18 at 11:02










  • $begingroup$
    Norms on a finite-dimensional vector space are equivalent?
    $endgroup$
    – Anna Schmitz
    Dec 16 '18 at 11:14










  • $begingroup$
    Yes! So what can you infer, if you know that the original norn and the one you defined are equivalent?
    $endgroup$
    – Mindlack
    Dec 16 '18 at 11:16










  • $begingroup$
    $∥Tx∥ leq ∥x∥ + ∥Tx∥ = ∥x∥_0$
    $endgroup$
    – Anna Schmitz
    Dec 16 '18 at 11:25






  • 1




    $begingroup$
    I am sorry, I must have misunderstood your first comment and confused you further. You do have $|Tx| leq |x|_0$, thus the operator is bounded.
    $endgroup$
    – Mindlack
    Dec 16 '18 at 11:41


















  • $begingroup$
    Hint: What theorem do you know about norms on a finite-dimensional vector space?
    $endgroup$
    – Mindlack
    Dec 16 '18 at 11:02










  • $begingroup$
    Norms on a finite-dimensional vector space are equivalent?
    $endgroup$
    – Anna Schmitz
    Dec 16 '18 at 11:14










  • $begingroup$
    Yes! So what can you infer, if you know that the original norn and the one you defined are equivalent?
    $endgroup$
    – Mindlack
    Dec 16 '18 at 11:16










  • $begingroup$
    $∥Tx∥ leq ∥x∥ + ∥Tx∥ = ∥x∥_0$
    $endgroup$
    – Anna Schmitz
    Dec 16 '18 at 11:25






  • 1




    $begingroup$
    I am sorry, I must have misunderstood your first comment and confused you further. You do have $|Tx| leq |x|_0$, thus the operator is bounded.
    $endgroup$
    – Mindlack
    Dec 16 '18 at 11:41
















$begingroup$
Hint: What theorem do you know about norms on a finite-dimensional vector space?
$endgroup$
– Mindlack
Dec 16 '18 at 11:02




$begingroup$
Hint: What theorem do you know about norms on a finite-dimensional vector space?
$endgroup$
– Mindlack
Dec 16 '18 at 11:02












$begingroup$
Norms on a finite-dimensional vector space are equivalent?
$endgroup$
– Anna Schmitz
Dec 16 '18 at 11:14




$begingroup$
Norms on a finite-dimensional vector space are equivalent?
$endgroup$
– Anna Schmitz
Dec 16 '18 at 11:14












$begingroup$
Yes! So what can you infer, if you know that the original norn and the one you defined are equivalent?
$endgroup$
– Mindlack
Dec 16 '18 at 11:16




$begingroup$
Yes! So what can you infer, if you know that the original norn and the one you defined are equivalent?
$endgroup$
– Mindlack
Dec 16 '18 at 11:16












$begingroup$
$∥Tx∥ leq ∥x∥ + ∥Tx∥ = ∥x∥_0$
$endgroup$
– Anna Schmitz
Dec 16 '18 at 11:25




$begingroup$
$∥Tx∥ leq ∥x∥ + ∥Tx∥ = ∥x∥_0$
$endgroup$
– Anna Schmitz
Dec 16 '18 at 11:25




1




1




$begingroup$
I am sorry, I must have misunderstood your first comment and confused you further. You do have $|Tx| leq |x|_0$, thus the operator is bounded.
$endgroup$
– Mindlack
Dec 16 '18 at 11:41




$begingroup$
I am sorry, I must have misunderstood your first comment and confused you further. You do have $|Tx| leq |x|_0$, thus the operator is bounded.
$endgroup$
– Mindlack
Dec 16 '18 at 11:41










1 Answer
1






active

oldest

votes


















1












$begingroup$

Regarding the operator norm:
I was thinking



$ ||T∥_2 = sup limits_{x neq 0} frac{∥Tx∥_1}{∥x∥_1} = sup limits_{x neq0} frac{∥(
x_1,…,x_n,0,0,…)∥_1}{∥(x_1,…,x_n)∥_1} = sup limits_{x neq0} frac{|x_1|+…+|x_n|}{|x_1|+…+|x_n|}= 1 $






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Seems correct !
    $endgroup$
    – Viktor Glombik
    Dec 16 '18 at 12:18










  • $begingroup$
    Why are you using the $1$ norm for $x$ if $X$ is equipped with the Euclidean norm?
    $endgroup$
    – user289143
    Dec 16 '18 at 16:37










  • $begingroup$
    Because $Y := ell_1(mathbb{N})$
    $endgroup$
    – Anna Schmitz
    Dec 16 '18 at 16:39










  • $begingroup$
    But for $x$ you have to look at the norm of $X$. You should use the norm of $Y$ just for $Tx$
    $endgroup$
    – user289143
    Dec 16 '18 at 16:42






  • 1




    $begingroup$
    True. So $ ||T∥_2 = sup limits_{x neq 0} frac{∥Tx∥_1}{∥x∥_2} = sup limits_{x neq0} frac{∥( x_1,…,x_n,0,0,…)∥_1}{∥(x_1,…,x_n)∥_2} = sup limits_{x neq0} frac{|x_1|+…+|x_n|}{(|x_1|^2+…+|x_n|^2)^{frac{1}{2}}}= ? $ and I thought I had figured it out
    $endgroup$
    – Anna Schmitz
    Dec 16 '18 at 16:50













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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









1












$begingroup$

Regarding the operator norm:
I was thinking



$ ||T∥_2 = sup limits_{x neq 0} frac{∥Tx∥_1}{∥x∥_1} = sup limits_{x neq0} frac{∥(
x_1,…,x_n,0,0,…)∥_1}{∥(x_1,…,x_n)∥_1} = sup limits_{x neq0} frac{|x_1|+…+|x_n|}{|x_1|+…+|x_n|}= 1 $






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Seems correct !
    $endgroup$
    – Viktor Glombik
    Dec 16 '18 at 12:18










  • $begingroup$
    Why are you using the $1$ norm for $x$ if $X$ is equipped with the Euclidean norm?
    $endgroup$
    – user289143
    Dec 16 '18 at 16:37










  • $begingroup$
    Because $Y := ell_1(mathbb{N})$
    $endgroup$
    – Anna Schmitz
    Dec 16 '18 at 16:39










  • $begingroup$
    But for $x$ you have to look at the norm of $X$. You should use the norm of $Y$ just for $Tx$
    $endgroup$
    – user289143
    Dec 16 '18 at 16:42






  • 1




    $begingroup$
    True. So $ ||T∥_2 = sup limits_{x neq 0} frac{∥Tx∥_1}{∥x∥_2} = sup limits_{x neq0} frac{∥( x_1,…,x_n,0,0,…)∥_1}{∥(x_1,…,x_n)∥_2} = sup limits_{x neq0} frac{|x_1|+…+|x_n|}{(|x_1|^2+…+|x_n|^2)^{frac{1}{2}}}= ? $ and I thought I had figured it out
    $endgroup$
    – Anna Schmitz
    Dec 16 '18 at 16:50


















1












$begingroup$

Regarding the operator norm:
I was thinking



$ ||T∥_2 = sup limits_{x neq 0} frac{∥Tx∥_1}{∥x∥_1} = sup limits_{x neq0} frac{∥(
x_1,…,x_n,0,0,…)∥_1}{∥(x_1,…,x_n)∥_1} = sup limits_{x neq0} frac{|x_1|+…+|x_n|}{|x_1|+…+|x_n|}= 1 $






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Seems correct !
    $endgroup$
    – Viktor Glombik
    Dec 16 '18 at 12:18










  • $begingroup$
    Why are you using the $1$ norm for $x$ if $X$ is equipped with the Euclidean norm?
    $endgroup$
    – user289143
    Dec 16 '18 at 16:37










  • $begingroup$
    Because $Y := ell_1(mathbb{N})$
    $endgroup$
    – Anna Schmitz
    Dec 16 '18 at 16:39










  • $begingroup$
    But for $x$ you have to look at the norm of $X$. You should use the norm of $Y$ just for $Tx$
    $endgroup$
    – user289143
    Dec 16 '18 at 16:42






  • 1




    $begingroup$
    True. So $ ||T∥_2 = sup limits_{x neq 0} frac{∥Tx∥_1}{∥x∥_2} = sup limits_{x neq0} frac{∥( x_1,…,x_n,0,0,…)∥_1}{∥(x_1,…,x_n)∥_2} = sup limits_{x neq0} frac{|x_1|+…+|x_n|}{(|x_1|^2+…+|x_n|^2)^{frac{1}{2}}}= ? $ and I thought I had figured it out
    $endgroup$
    – Anna Schmitz
    Dec 16 '18 at 16:50
















1












1








1





$begingroup$

Regarding the operator norm:
I was thinking



$ ||T∥_2 = sup limits_{x neq 0} frac{∥Tx∥_1}{∥x∥_1} = sup limits_{x neq0} frac{∥(
x_1,…,x_n,0,0,…)∥_1}{∥(x_1,…,x_n)∥_1} = sup limits_{x neq0} frac{|x_1|+…+|x_n|}{|x_1|+…+|x_n|}= 1 $






share|cite|improve this answer









$endgroup$



Regarding the operator norm:
I was thinking



$ ||T∥_2 = sup limits_{x neq 0} frac{∥Tx∥_1}{∥x∥_1} = sup limits_{x neq0} frac{∥(
x_1,…,x_n,0,0,…)∥_1}{∥(x_1,…,x_n)∥_1} = sup limits_{x neq0} frac{|x_1|+…+|x_n|}{|x_1|+…+|x_n|}= 1 $







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 16 '18 at 11:57









Anna SchmitzAnna Schmitz

917




917








  • 1




    $begingroup$
    Seems correct !
    $endgroup$
    – Viktor Glombik
    Dec 16 '18 at 12:18










  • $begingroup$
    Why are you using the $1$ norm for $x$ if $X$ is equipped with the Euclidean norm?
    $endgroup$
    – user289143
    Dec 16 '18 at 16:37










  • $begingroup$
    Because $Y := ell_1(mathbb{N})$
    $endgroup$
    – Anna Schmitz
    Dec 16 '18 at 16:39










  • $begingroup$
    But for $x$ you have to look at the norm of $X$. You should use the norm of $Y$ just for $Tx$
    $endgroup$
    – user289143
    Dec 16 '18 at 16:42






  • 1




    $begingroup$
    True. So $ ||T∥_2 = sup limits_{x neq 0} frac{∥Tx∥_1}{∥x∥_2} = sup limits_{x neq0} frac{∥( x_1,…,x_n,0,0,…)∥_1}{∥(x_1,…,x_n)∥_2} = sup limits_{x neq0} frac{|x_1|+…+|x_n|}{(|x_1|^2+…+|x_n|^2)^{frac{1}{2}}}= ? $ and I thought I had figured it out
    $endgroup$
    – Anna Schmitz
    Dec 16 '18 at 16:50
















  • 1




    $begingroup$
    Seems correct !
    $endgroup$
    – Viktor Glombik
    Dec 16 '18 at 12:18










  • $begingroup$
    Why are you using the $1$ norm for $x$ if $X$ is equipped with the Euclidean norm?
    $endgroup$
    – user289143
    Dec 16 '18 at 16:37










  • $begingroup$
    Because $Y := ell_1(mathbb{N})$
    $endgroup$
    – Anna Schmitz
    Dec 16 '18 at 16:39










  • $begingroup$
    But for $x$ you have to look at the norm of $X$. You should use the norm of $Y$ just for $Tx$
    $endgroup$
    – user289143
    Dec 16 '18 at 16:42






  • 1




    $begingroup$
    True. So $ ||T∥_2 = sup limits_{x neq 0} frac{∥Tx∥_1}{∥x∥_2} = sup limits_{x neq0} frac{∥( x_1,…,x_n,0,0,…)∥_1}{∥(x_1,…,x_n)∥_2} = sup limits_{x neq0} frac{|x_1|+…+|x_n|}{(|x_1|^2+…+|x_n|^2)^{frac{1}{2}}}= ? $ and I thought I had figured it out
    $endgroup$
    – Anna Schmitz
    Dec 16 '18 at 16:50










1




1




$begingroup$
Seems correct !
$endgroup$
– Viktor Glombik
Dec 16 '18 at 12:18




$begingroup$
Seems correct !
$endgroup$
– Viktor Glombik
Dec 16 '18 at 12:18












$begingroup$
Why are you using the $1$ norm for $x$ if $X$ is equipped with the Euclidean norm?
$endgroup$
– user289143
Dec 16 '18 at 16:37




$begingroup$
Why are you using the $1$ norm for $x$ if $X$ is equipped with the Euclidean norm?
$endgroup$
– user289143
Dec 16 '18 at 16:37












$begingroup$
Because $Y := ell_1(mathbb{N})$
$endgroup$
– Anna Schmitz
Dec 16 '18 at 16:39




$begingroup$
Because $Y := ell_1(mathbb{N})$
$endgroup$
– Anna Schmitz
Dec 16 '18 at 16:39












$begingroup$
But for $x$ you have to look at the norm of $X$. You should use the norm of $Y$ just for $Tx$
$endgroup$
– user289143
Dec 16 '18 at 16:42




$begingroup$
But for $x$ you have to look at the norm of $X$. You should use the norm of $Y$ just for $Tx$
$endgroup$
– user289143
Dec 16 '18 at 16:42




1




1




$begingroup$
True. So $ ||T∥_2 = sup limits_{x neq 0} frac{∥Tx∥_1}{∥x∥_2} = sup limits_{x neq0} frac{∥( x_1,…,x_n,0,0,…)∥_1}{∥(x_1,…,x_n)∥_2} = sup limits_{x neq0} frac{|x_1|+…+|x_n|}{(|x_1|^2+…+|x_n|^2)^{frac{1}{2}}}= ? $ and I thought I had figured it out
$endgroup$
– Anna Schmitz
Dec 16 '18 at 16:50






$begingroup$
True. So $ ||T∥_2 = sup limits_{x neq 0} frac{∥Tx∥_1}{∥x∥_2} = sup limits_{x neq0} frac{∥( x_1,…,x_n,0,0,…)∥_1}{∥(x_1,…,x_n)∥_2} = sup limits_{x neq0} frac{|x_1|+…+|x_n|}{(|x_1|^2+…+|x_n|^2)^{frac{1}{2}}}= ? $ and I thought I had figured it out
$endgroup$
– Anna Schmitz
Dec 16 '18 at 16:50




















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Cabo Verde

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  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
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Gyllenstierna

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