Isomorphism between vector spaces concerning kernel of a continuous application
How can I prove that if $T:X -> Y$ is continuous, $T_0:X/Z -> Y$ is also continuous, with $Z=Ker(T)$? Also, why $||T||=||T_0||$? Thanks in advance
functional-analysis
asked Jan 4 at 0:05
João Costa
32
add a comment |
How can I prove that if $T:X -> Y$ is continuous, $T_0:X/Z -> Y$ is also continuous, with $Z=Ker(T)$? Also, why $||T||=||T_0||$? Thanks in advance
functional-analysis
asked Jan 4 at 0:05
João Costa
32
add a comment |
How can I prove that if $T:X -> Y$ is continuous, $T_0:X/Z -> Y$ is also continuous, with $Z=Ker(T)$? Also, why $||T||=||T_0||$? Thanks in advance
functional-analysis
asked Jan 4 at 0:05
João Costa
32
How can I prove that if $T:X -> Y$ is continuous, $T_0:X/Z -> Y$ is also continuous, with $Z=Ker(T)$? Also, why $||T||=||T_0||$? Thanks in advance
functional-analysis
functional-analysis
asked Jan 4 at 0:05
João Costa
32
asked Jan 4 at 0:05
João Costa
32
asked Jan 4 at 0:05
João Costa
32
asked Jan 4 at 0:05
João Costa
32
asked Jan 4 at 0:05
João Costa
32
32
add a comment |
add a comment |
1 Answer
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$T_0(x+Z)=T(x)$. This is a well defined linear map. If $x+Z_n to x+Z$ in $X|Z$ then there exists a sequence ${z_n}$ in $Z$ such that $x_n+z_n to x+z$. [This follows from the definition of norm in the quotient space]. Since $T$ is continuous we get $T(x_n+z_n) to T(x+z)$ and hence $T(x_n) to Tx$. Thus $T_0$ is continuous. Note that $|T_0(x+Z)|=|Tx|=|T(x+z)| leq |T||x+z|$ for all $z in Z$. taking inf over $z$ we get $|T_0| leq |T|$. Finally, given $epsilon >0$ choose $x$ such that $|x| leq 1$ and $|Tx| >|T|-epsilon$. We have $|T_0(x+Z)|= |Tx| >|T|-epsilon$ and $|x+Z| leq |x| leq 1$. Hence $|T_0| geq |T|$.
answered Jan 4 at 0:16
Kavi Rama Murthy
51.4k31855
Thanks. That's all
– João Costa
Jan 4 at 0:40
add a comment |
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$T_0(x+Z)=T(x)$. This is a well defined linear map. If $x+Z_n to x+Z$ in $X|Z$ then there exists a sequence ${z_n}$ in $Z$ such that $x_n+z_n to x+z$. [This follows from the definition of norm in the quotient space]. Since $T$ is continuous we get $T(x_n+z_n) to T(x+z)$ and hence $T(x_n) to Tx$. Thus $T_0$ is continuous. Note that $|T_0(x+Z)|=|Tx|=|T(x+z)| leq |T||x+z|$ for all $z in Z$. taking inf over $z$ we get $|T_0| leq |T|$. Finally, given $epsilon >0$ choose $x$ such that $|x| leq 1$ and $|Tx| >|T|-epsilon$. We have $|T_0(x+Z)|= |Tx| >|T|-epsilon$ and $|x+Z| leq |x| leq 1$. Hence $|T_0| geq |T|$.
answered Jan 4 at 0:16
Kavi Rama Murthy
51.4k31855
Thanks. That's all
– João Costa
Jan 4 at 0:40
add a comment |
$T_0(x+Z)=T(x)$. This is a well defined linear map. If $x+Z_n to x+Z$ in $X|Z$ then there exists a sequence ${z_n}$ in $Z$ such that $x_n+z_n to x+z$. [This follows from the definition of norm in the quotient space]. Since $T$ is continuous we get $T(x_n+z_n) to T(x+z)$ and hence $T(x_n) to Tx$. Thus $T_0$ is continuous. Note that $|T_0(x+Z)|=|Tx|=|T(x+z)| leq |T||x+z|$ for all $z in Z$. taking inf over $z$ we get $|T_0| leq |T|$. Finally, given $epsilon >0$ choose $x$ such that $|x| leq 1$ and $|Tx| >|T|-epsilon$. We have $|T_0(x+Z)|= |Tx| >|T|-epsilon$ and $|x+Z| leq |x| leq 1$. Hence $|T_0| geq |T|$.
answered Jan 4 at 0:16
Kavi Rama Murthy
51.4k31855
Thanks. That's all
– João Costa
Jan 4 at 0:40
add a comment |
$T_0(x+Z)=T(x)$. This is a well defined linear map. If $x+Z_n to x+Z$ in $X|Z$ then there exists a sequence ${z_n}$ in $Z$ such that $x_n+z_n to x+z$. [This follows from the definition of norm in the quotient space]. Since $T$ is continuous we get $T(x_n+z_n) to T(x+z)$ and hence $T(x_n) to Tx$. Thus $T_0$ is continuous. Note that $|T_0(x+Z)|=|Tx|=|T(x+z)| leq |T||x+z|$ for all $z in Z$. taking inf over $z$ we get $|T_0| leq |T|$. Finally, given $epsilon >0$ choose $x$ such that $|x| leq 1$ and $|Tx| >|T|-epsilon$. We have $|T_0(x+Z)|= |Tx| >|T|-epsilon$ and $|x+Z| leq |x| leq 1$. Hence $|T_0| geq |T|$.
answered Jan 4 at 0:16
Kavi Rama Murthy
51.4k31855
$T_0(x+Z)=T(x)$. This is a well defined linear map. If $x+Z_n to x+Z$ in $X|Z$ then there exists a sequence ${z_n}$ in $Z$ such that $x_n+z_n to x+z$. [This follows from the definition of norm in the quotient space]. Since $T$ is continuous we get $T(x_n+z_n) to T(x+z)$ and hence $T(x_n) to Tx$. Thus $T_0$ is continuous. Note that $|T_0(x+Z)|=|Tx|=|T(x+z)| leq |T||x+z|$ for all $z in Z$. taking inf over $z$ we get $|T_0| leq |T|$. Finally, given $epsilon >0$ choose $x$ such that $|x| leq 1$ and $|Tx| >|T|-epsilon$. We have $|T_0(x+Z)|= |Tx| >|T|-epsilon$ and $|x+Z| leq |x| leq 1$. Hence $|T_0| geq |T|$.
answered Jan 4 at 0:16
Kavi Rama Murthy
51.4k31855
answered Jan 4 at 0:16
Kavi Rama Murthy
51.4k31855
answered Jan 4 at 0:16
Kavi Rama Murthy
51.4k31855
answered Jan 4 at 0:16
Kavi Rama Murthy
51.4k31855
51.4k31855
Thanks. That's all
– João Costa
Jan 4 at 0:40
add a comment |
Thanks. That's all
– João Costa
Jan 4 at 0:40
Thanks. That's all
– João Costa
Jan 4 at 0:40
Thanks. That's all
– João Costa
Jan 4 at 0:40
add a comment |
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