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

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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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Mont Emei

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Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
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