Balls in finite dimensional normed spaces
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Let $X$, $Y$ be normed linear spaces and $Y$ be finite dimensional. Suppose $T in B(X,Y)$ such that $T$ is surjective. Prove that there is some $delta > 0$ so that $B_delta(0_Y ) subseteq T(B_1(0_X))$.
Any help/hints on this problem would be greatly appreciated. If you are not familiar with $ B(X,Y) $, it is the set of all bounded linear operators from $ X $ to $ Y $.
functional-analysis normed-spaces
asked Nov 13 at 21:56
Chase Sariaslani
565
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up vote
0
down vote
favorite
Let $X$, $Y$ be normed linear spaces and $Y$ be finite dimensional. Suppose $T in B(X,Y)$ such that $T$ is surjective. Prove that there is some $delta > 0$ so that $B_delta(0_Y ) subseteq T(B_1(0_X))$.
Any help/hints on this problem would be greatly appreciated. If you are not familiar with $ B(X,Y) $, it is the set of all bounded linear operators from $ X $ to $ Y $.
functional-analysis normed-spaces
asked Nov 13 at 21:56
Chase Sariaslani
565
open mapping theorem
– mathworker21
Nov 13 at 21:59
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $X$, $Y$ be normed linear spaces and $Y$ be finite dimensional. Suppose $T in B(X,Y)$ such that $T$ is surjective. Prove that there is some $delta > 0$ so that $B_delta(0_Y ) subseteq T(B_1(0_X))$.
Any help/hints on this problem would be greatly appreciated. If you are not familiar with $ B(X,Y) $, it is the set of all bounded linear operators from $ X $ to $ Y $.
functional-analysis normed-spaces
asked Nov 13 at 21:56
Chase Sariaslani
565
Let $X$, $Y$ be normed linear spaces and $Y$ be finite dimensional. Suppose $T in B(X,Y)$ such that $T$ is surjective. Prove that there is some $delta > 0$ so that $B_delta(0_Y ) subseteq T(B_1(0_X))$.
Any help/hints on this problem would be greatly appreciated. If you are not familiar with $ B(X,Y) $, it is the set of all bounded linear operators from $ X $ to $ Y $.
functional-analysis normed-spaces
functional-analysis normed-spaces
asked Nov 13 at 21:56
Chase Sariaslani
565
asked Nov 13 at 21:56
Chase Sariaslani
565
asked Nov 13 at 21:56
Chase Sariaslani
565
asked Nov 13 at 21:56
Chase Sariaslani
565
asked Nov 13 at 21:56
Chase Sariaslani
565
565
open mapping theorem
– mathworker21
Nov 13 at 21:59
add a comment |
open mapping theorem
– mathworker21
Nov 13 at 21:59
open mapping theorem
– mathworker21
Nov 13 at 21:59
open mapping theorem
– mathworker21
Nov 13 at 21:59
add a comment |
1 Answer
1
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Since it is surjective, there exists a finite dimensional subspace $U$ of $X$ such that the restriction of $T$ to $U$ is an isomorphism, This implies that $T_{mid U}(B_U(0,1))$ is open and contains $B_Y(0,delta)$. The result follows from the fact that $T_{mid U}(B_U(0,1))subset T(B_X(0,1)$.
answered Nov 13 at 22:00
Tsemo Aristide
54.3k11344
1
isn't the statement equivalent to the open mapping theorem, up to translations?
– mathworker21
Nov 13 at 22:00
Yes you are correct. I omitted that piece of info.
– Chase Sariaslani
Nov 13 at 22:04
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Since it is surjective, there exists a finite dimensional subspace $U$ of $X$ such that the restriction of $T$ to $U$ is an isomorphism, This implies that $T_{mid U}(B_U(0,1))$ is open and contains $B_Y(0,delta)$. The result follows from the fact that $T_{mid U}(B_U(0,1))subset T(B_X(0,1)$.
answered Nov 13 at 22:00
Tsemo Aristide
54.3k11344
1
isn't the statement equivalent to the open mapping theorem, up to translations?
– mathworker21
Nov 13 at 22:00
Yes you are correct. I omitted that piece of info.
– Chase Sariaslani
Nov 13 at 22:04
add a comment |
up vote
0
down vote
accepted
Since it is surjective, there exists a finite dimensional subspace $U$ of $X$ such that the restriction of $T$ to $U$ is an isomorphism, This implies that $T_{mid U}(B_U(0,1))$ is open and contains $B_Y(0,delta)$. The result follows from the fact that $T_{mid U}(B_U(0,1))subset T(B_X(0,1)$.
answered Nov 13 at 22:00
Tsemo Aristide
54.3k11344
1
isn't the statement equivalent to the open mapping theorem, up to translations?
– mathworker21
Nov 13 at 22:00
Yes you are correct. I omitted that piece of info.
– Chase Sariaslani
Nov 13 at 22:04
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Since it is surjective, there exists a finite dimensional subspace $U$ of $X$ such that the restriction of $T$ to $U$ is an isomorphism, This implies that $T_{mid U}(B_U(0,1))$ is open and contains $B_Y(0,delta)$. The result follows from the fact that $T_{mid U}(B_U(0,1))subset T(B_X(0,1)$.
answered Nov 13 at 22:00
Tsemo Aristide
54.3k11344
Since it is surjective, there exists a finite dimensional subspace $U$ of $X$ such that the restriction of $T$ to $U$ is an isomorphism, This implies that $T_{mid U}(B_U(0,1))$ is open and contains $B_Y(0,delta)$. The result follows from the fact that $T_{mid U}(B_U(0,1))subset T(B_X(0,1)$.
answered Nov 13 at 22:00
Tsemo Aristide
54.3k11344
edited Nov 13 at 22:05
answered Nov 13 at 22:00
Tsemo Aristide
54.3k11344
answered Nov 13 at 22:00
Tsemo Aristide
54.3k11344
answered Nov 13 at 22:00
Tsemo Aristide
54.3k11344
54.3k11344
1
isn't the statement equivalent to the open mapping theorem, up to translations?
– mathworker21
Nov 13 at 22:00
Yes you are correct. I omitted that piece of info.
– Chase Sariaslani
Nov 13 at 22:04
add a comment |
1
isn't the statement equivalent to the open mapping theorem, up to translations?
– mathworker21
Nov 13 at 22:00
Yes you are correct. I omitted that piece of info.
– Chase Sariaslani
Nov 13 at 22:04
1
1
isn't the statement equivalent to the open mapping theorem, up to translations?
– mathworker21
Nov 13 at 22:00
isn't the statement equivalent to the open mapping theorem, up to translations?
– mathworker21
Nov 13 at 22:00
Yes you are correct. I omitted that piece of info.
– Chase Sariaslani
Nov 13 at 22:04
Yes you are correct. I omitted that piece of info.
– Chase Sariaslani
Nov 13 at 22:04
add a comment |
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open mapping theorem
– mathworker21
Nov 13 at 21:59