T is a bounded linear operator and surjective, prove that there exists $k>0$ such that $|x|leq k|y|$.
$begingroup$
$B(X,Y)$ denotes the space of bounded linear operators from $X$ to $Y$.
Question: Let $Tin B(X,Y)$ such that $T(X)=Y$. Show that there exists $k>0$ such that given $yin Y$, there is an $xin X$ such that $T(x)=y$ and $|x|leq k|y|$.
Clearly, it is surjective. So we just need to prove the second part, i.e. prove that $|x|leq k|y|$. I was thinking that I can use the corollary of Open Mapping Theorem to prove it:
Corollary: Let $X,Y$ be Banach spaces and let $Tin B(X,Y)$ be bijective. Then $T^{-1}in B(X,Y)$ and there exist $c_1,c_2>0$ such that
$|x|leq c_1|Tx|leq c_2|x|,;xin X.$
Applying this corollary, we get $|x|leq c_1|y|$.
But how can we prove $T$ is injective? Assuming that $Tx_1=Tx_2,;forall x_1,x_2in X.$ How can we get $x_1=x_2$?
Could you please give me some ideas? Thank you so much.
functional-analysis analysis
asked Dec 17 '18 at 12:42
RossRoss
1237
$endgroup$
add a comment |
$begingroup$
$B(X,Y)$ denotes the space of bounded linear operators from $X$ to $Y$.
Question: Let $Tin B(X,Y)$ such that $T(X)=Y$. Show that there exists $k>0$ such that given $yin Y$, there is an $xin X$ such that $T(x)=y$ and $|x|leq k|y|$.
Clearly, it is surjective. So we just need to prove the second part, i.e. prove that $|x|leq k|y|$. I was thinking that I can use the corollary of Open Mapping Theorem to prove it:
Corollary: Let $X,Y$ be Banach spaces and let $Tin B(X,Y)$ be bijective. Then $T^{-1}in B(X,Y)$ and there exist $c_1,c_2>0$ such that
$|x|leq c_1|Tx|leq c_2|x|,;xin X.$
Applying this corollary, we get $|x|leq c_1|y|$.
But how can we prove $T$ is injective? Assuming that $Tx_1=Tx_2,;forall x_1,x_2in X.$ How can we get $x_1=x_2$?
Could you please give me some ideas? Thank you so much.
functional-analysis analysis
asked Dec 17 '18 at 12:42
RossRoss
1237
$endgroup$
add a comment |
$begingroup$
$B(X,Y)$ denotes the space of bounded linear operators from $X$ to $Y$.
Question: Let $Tin B(X,Y)$ such that $T(X)=Y$. Show that there exists $k>0$ such that given $yin Y$, there is an $xin X$ such that $T(x)=y$ and $|x|leq k|y|$.
Clearly, it is surjective. So we just need to prove the second part, i.e. prove that $|x|leq k|y|$. I was thinking that I can use the corollary of Open Mapping Theorem to prove it:
Corollary: Let $X,Y$ be Banach spaces and let $Tin B(X,Y)$ be bijective. Then $T^{-1}in B(X,Y)$ and there exist $c_1,c_2>0$ such that
$|x|leq c_1|Tx|leq c_2|x|,;xin X.$
Applying this corollary, we get $|x|leq c_1|y|$.
But how can we prove $T$ is injective? Assuming that $Tx_1=Tx_2,;forall x_1,x_2in X.$ How can we get $x_1=x_2$?
Could you please give me some ideas? Thank you so much.
functional-analysis analysis
asked Dec 17 '18 at 12:42
RossRoss
1237
$endgroup$
$B(X,Y)$ denotes the space of bounded linear operators from $X$ to $Y$.
Question: Let $Tin B(X,Y)$ such that $T(X)=Y$. Show that there exists $k>0$ such that given $yin Y$, there is an $xin X$ such that $T(x)=y$ and $|x|leq k|y|$.
Clearly, it is surjective. So we just need to prove the second part, i.e. prove that $|x|leq k|y|$. I was thinking that I can use the corollary of Open Mapping Theorem to prove it:
Corollary: Let $X,Y$ be Banach spaces and let $Tin B(X,Y)$ be bijective. Then $T^{-1}in B(X,Y)$ and there exist $c_1,c_2>0$ such that
$|x|leq c_1|Tx|leq c_2|x|,;xin X.$
Applying this corollary, we get $|x|leq c_1|y|$.
But how can we prove $T$ is injective? Assuming that $Tx_1=Tx_2,;forall x_1,x_2in X.$ How can we get $x_1=x_2$?
Could you please give me some ideas? Thank you so much.
functional-analysis analysis
functional-analysis analysis
asked Dec 17 '18 at 12:42
RossRoss
1237
asked Dec 17 '18 at 12:42
RossRoss
1237
asked Dec 17 '18 at 12:42
RossRoss
1237
asked Dec 17 '18 at 12:42
RossRoss
1237
asked Dec 17 '18 at 12:42
RossRoss
1237
1237
add a comment |
add a comment |
3 Answers
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You can't prove that it is injective. The only thing that you know about $T$ is that it's surjective, and there are linear surjective bounded operators between Banach spaces which are not injective. Take$$begin{array}{ccc}ell^1(mathbb{N})&longrightarrow&mathbb R\displaystylesum_{n=1}^infty a_n&mapsto&a_1,end{array}$$for instance.
answered Dec 17 '18 at 12:48
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
add a comment |
$begingroup$
Hint: Consider the quotient space $overline{X}=X/ker(T)$ , with the quotient norm, and the mapping $overline{T}:overline{X} to Y$ defined by
$overline{T}(overline{x}):=Tx$, where $overline{x}=x+ker(T).$
Then $overline{T}$ is bounded and bijective.
answered Dec 17 '18 at 12:47
FredFred
46.9k1848
$endgroup$
add a comment |
$begingroup$
Hint: from the Open Mapping theorem, there exists some $c>0$ such that every element of $Y$ with norm at most $c$ has a pre-image with norm at most $1$.
answered Dec 17 '18 at 12:45
MindlackMindlack
4,740210
$endgroup$
add a comment |
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3 Answers
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3 Answers
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$begingroup$
You can't prove that it is injective. The only thing that you know about $T$ is that it's surjective, and there are linear surjective bounded operators between Banach spaces which are not injective. Take$$begin{array}{ccc}ell^1(mathbb{N})&longrightarrow&mathbb R\displaystylesum_{n=1}^infty a_n&mapsto&a_1,end{array}$$for instance.
answered Dec 17 '18 at 12:48
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
add a comment |
$begingroup$
You can't prove that it is injective. The only thing that you know about $T$ is that it's surjective, and there are linear surjective bounded operators between Banach spaces which are not injective. Take$$begin{array}{ccc}ell^1(mathbb{N})&longrightarrow&mathbb R\displaystylesum_{n=1}^infty a_n&mapsto&a_1,end{array}$$for instance.
answered Dec 17 '18 at 12:48
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
add a comment |
$begingroup$
You can't prove that it is injective. The only thing that you know about $T$ is that it's surjective, and there are linear surjective bounded operators between Banach spaces which are not injective. Take$$begin{array}{ccc}ell^1(mathbb{N})&longrightarrow&mathbb R\displaystylesum_{n=1}^infty a_n&mapsto&a_1,end{array}$$for instance.
answered Dec 17 '18 at 12:48
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
You can't prove that it is injective. The only thing that you know about $T$ is that it's surjective, and there are linear surjective bounded operators between Banach spaces which are not injective. Take$$begin{array}{ccc}ell^1(mathbb{N})&longrightarrow&mathbb R\displaystylesum_{n=1}^infty a_n&mapsto&a_1,end{array}$$for instance.
answered Dec 17 '18 at 12:48
José Carlos SantosJosé Carlos Santos
163k22131234
answered Dec 17 '18 at 12:48
José Carlos SantosJosé Carlos Santos
163k22131234
answered Dec 17 '18 at 12:48
José Carlos SantosJosé Carlos Santos
163k22131234
answered Dec 17 '18 at 12:48
José Carlos SantosJosé Carlos Santos
163k22131234
163k22131234
add a comment |
add a comment |
$begingroup$
Hint: Consider the quotient space $overline{X}=X/ker(T)$ , with the quotient norm, and the mapping $overline{T}:overline{X} to Y$ defined by
$overline{T}(overline{x}):=Tx$, where $overline{x}=x+ker(T).$
Then $overline{T}$ is bounded and bijective.
answered Dec 17 '18 at 12:47
FredFred
46.9k1848
$endgroup$
add a comment |
$begingroup$
Hint: Consider the quotient space $overline{X}=X/ker(T)$ , with the quotient norm, and the mapping $overline{T}:overline{X} to Y$ defined by
$overline{T}(overline{x}):=Tx$, where $overline{x}=x+ker(T).$
Then $overline{T}$ is bounded and bijective.
answered Dec 17 '18 at 12:47
FredFred
46.9k1848
$endgroup$
add a comment |
$begingroup$
Hint: Consider the quotient space $overline{X}=X/ker(T)$ , with the quotient norm, and the mapping $overline{T}:overline{X} to Y$ defined by
$overline{T}(overline{x}):=Tx$, where $overline{x}=x+ker(T).$
Then $overline{T}$ is bounded and bijective.
answered Dec 17 '18 at 12:47
FredFred
46.9k1848
$endgroup$
Hint: Consider the quotient space $overline{X}=X/ker(T)$ , with the quotient norm, and the mapping $overline{T}:overline{X} to Y$ defined by
$overline{T}(overline{x}):=Tx$, where $overline{x}=x+ker(T).$
Then $overline{T}$ is bounded and bijective.
answered Dec 17 '18 at 12:47
FredFred
46.9k1848
edited Dec 17 '18 at 14:08
answered Dec 17 '18 at 12:47
FredFred
46.9k1848
answered Dec 17 '18 at 12:47
FredFred
46.9k1848
answered Dec 17 '18 at 12:47
FredFred
46.9k1848
46.9k1848
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$begingroup$
Hint: from the Open Mapping theorem, there exists some $c>0$ such that every element of $Y$ with norm at most $c$ has a pre-image with norm at most $1$.
answered Dec 17 '18 at 12:45
MindlackMindlack
4,740210
$endgroup$
add a comment |
$begingroup$
Hint: from the Open Mapping theorem, there exists some $c>0$ such that every element of $Y$ with norm at most $c$ has a pre-image with norm at most $1$.
answered Dec 17 '18 at 12:45
MindlackMindlack
4,740210
$endgroup$
add a comment |
$begingroup$
Hint: from the Open Mapping theorem, there exists some $c>0$ such that every element of $Y$ with norm at most $c$ has a pre-image with norm at most $1$.
answered Dec 17 '18 at 12:45
MindlackMindlack
4,740210
$endgroup$
Hint: from the Open Mapping theorem, there exists some $c>0$ such that every element of $Y$ with norm at most $c$ has a pre-image with norm at most $1$.
answered Dec 17 '18 at 12:45
MindlackMindlack
4,740210
answered Dec 17 '18 at 12:45
MindlackMindlack
4,740210
answered Dec 17 '18 at 12:45
MindlackMindlack
4,740210
answered Dec 17 '18 at 12:45
MindlackMindlack
4,740210
4,740210
add a comment |
add a comment |
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