Linear transformation in space of infinite dimension
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Problem.
(a) Show that every linear transformation $A: mathbbR^n to mathbbR^m$ is continuous.
(b) If we change $mathbbR^n$ by a normed space $X$ and take $m=1$, the item (a) is true?
Solution. For item (a), is easy to show that $A$ is Lipschitz. For (b), is $dim X$ is finite, take an isomorphism $T: mathbbR^p to X$ where $dim X = p$ and $T(e_i) = u_i$ ($e_i,u_i$ are bases for $mathbbR^p,X$ respectively). My question is when $dim X = infty$. There is spaces of infinite dimension such that a linear transformation may not be continuous, but in this specific case, I don't know. Can someone help me?
real-analysis functional-analysis continuity
add a comment |Â
up vote
1
down vote
favorite
Problem.
(a) Show that every linear transformation $A: mathbbR^n to mathbbR^m$ is continuous.
(b) If we change $mathbbR^n$ by a normed space $X$ and take $m=1$, the item (a) is true?
Solution. For item (a), is easy to show that $A$ is Lipschitz. For (b), is $dim X$ is finite, take an isomorphism $T: mathbbR^p to X$ where $dim X = p$ and $T(e_i) = u_i$ ($e_i,u_i$ are bases for $mathbbR^p,X$ respectively). My question is when $dim X = infty$. There is spaces of infinite dimension such that a linear transformation may not be continuous, but in this specific case, I don't know. Can someone help me?
real-analysis functional-analysis continuity
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Problem.
(a) Show that every linear transformation $A: mathbbR^n to mathbbR^m$ is continuous.
(b) If we change $mathbbR^n$ by a normed space $X$ and take $m=1$, the item (a) is true?
Solution. For item (a), is easy to show that $A$ is Lipschitz. For (b), is $dim X$ is finite, take an isomorphism $T: mathbbR^p to X$ where $dim X = p$ and $T(e_i) = u_i$ ($e_i,u_i$ are bases for $mathbbR^p,X$ respectively). My question is when $dim X = infty$. There is spaces of infinite dimension such that a linear transformation may not be continuous, but in this specific case, I don't know. Can someone help me?
real-analysis functional-analysis continuity
Problem.
(a) Show that every linear transformation $A: mathbbR^n to mathbbR^m$ is continuous.
(b) If we change $mathbbR^n$ by a normed space $X$ and take $m=1$, the item (a) is true?
Solution. For item (a), is easy to show that $A$ is Lipschitz. For (b), is $dim X$ is finite, take an isomorphism $T: mathbbR^p to X$ where $dim X = p$ and $T(e_i) = u_i$ ($e_i,u_i$ are bases for $mathbbR^p,X$ respectively). My question is when $dim X = infty$. There is spaces of infinite dimension such that a linear transformation may not be continuous, but in this specific case, I don't know. Can someone help me?
real-analysis functional-analysis continuity
edited 12 hours ago
José Carlos Santos
111k1695171
111k1695171
asked 12 hours ago
Lucas Corrêa
1,041219
1,041219
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
A classical example consists in taking:
- $X=Cbigl([0,1]bigr)$;
- if $fin X$, $|f|=int_0^1|f|$;
- $psi(f)=f(1)$.
Then $psi$ is discontinuous: if $f_n(t)=t^n$, then $lim_ntoinftyf_n=0$, but $lim_ntoinftypsi(f_n)=1neqpsi(0)$.
Nice! Thank you!
â Lucas Corrêa
12 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
A classical example consists in taking:
- $X=Cbigl([0,1]bigr)$;
- if $fin X$, $|f|=int_0^1|f|$;
- $psi(f)=f(1)$.
Then $psi$ is discontinuous: if $f_n(t)=t^n$, then $lim_ntoinftyf_n=0$, but $lim_ntoinftypsi(f_n)=1neqpsi(0)$.
Nice! Thank you!
â Lucas Corrêa
12 hours ago
add a comment |Â
up vote
3
down vote
accepted
A classical example consists in taking:
- $X=Cbigl([0,1]bigr)$;
- if $fin X$, $|f|=int_0^1|f|$;
- $psi(f)=f(1)$.
Then $psi$ is discontinuous: if $f_n(t)=t^n$, then $lim_ntoinftyf_n=0$, but $lim_ntoinftypsi(f_n)=1neqpsi(0)$.
Nice! Thank you!
â Lucas Corrêa
12 hours ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
A classical example consists in taking:
- $X=Cbigl([0,1]bigr)$;
- if $fin X$, $|f|=int_0^1|f|$;
- $psi(f)=f(1)$.
Then $psi$ is discontinuous: if $f_n(t)=t^n$, then $lim_ntoinftyf_n=0$, but $lim_ntoinftypsi(f_n)=1neqpsi(0)$.
A classical example consists in taking:
- $X=Cbigl([0,1]bigr)$;
- if $fin X$, $|f|=int_0^1|f|$;
- $psi(f)=f(1)$.
Then $psi$ is discontinuous: if $f_n(t)=t^n$, then $lim_ntoinftyf_n=0$, but $lim_ntoinftypsi(f_n)=1neqpsi(0)$.
answered 12 hours ago
José Carlos Santos
111k1695171
111k1695171
Nice! Thank you!
â Lucas Corrêa
12 hours ago
add a comment |Â
Nice! Thank you!
â Lucas Corrêa
12 hours ago
Nice! Thank you!
â Lucas Corrêa
12 hours ago
Nice! Thank you!
â Lucas Corrêa
12 hours ago
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2873140%2flinear-transformation-in-space-of-infinite-dimension%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password