Boundedness of a linear operator
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Let $X$ be a real normed linear space of all real sequences which are eventually zero with the 'sup' norm and $T:X to X$ be a bijective linear operator defined by $$T(x_1,x_2,x_3,....)=left(x_1,frac{x_2}{2^2},frac{x_3}{3^2},....right)$$
How to check whether $T$ and $T^{-1}$ is bounded or not ?
$$leftlVert TxrightrVert=sup Big{vert x_1 vert,frac{vert x_2 vert}{2^2},...Big}=sup_nBig{frac{vert x_n vert}{n^2}Big} leq sup_nBig{frac{vert x_n vert}{n}Big}$$
How to make the RHS of above in the form $K vertvert x vert vert$ if possible ? Any hint ?
On the otherhand, $T^{-1}:X to X$ is a map by $$T^{-1}(x_1.x_2,...)=(x_1,2^2x_2,3^2x_3,...)$$
$$leftlVert T^{-1}xrightrVert=sup_nBig{n^2 vert x_n vertBig} geq n$$ so $T^{-1}$ is not bounded. Am I right? Any help ?
functional-analysis operator-theory
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add a comment |
$begingroup$
Let $X$ be a real normed linear space of all real sequences which are eventually zero with the 'sup' norm and $T:X to X$ be a bijective linear operator defined by $$T(x_1,x_2,x_3,....)=left(x_1,frac{x_2}{2^2},frac{x_3}{3^2},....right)$$
How to check whether $T$ and $T^{-1}$ is bounded or not ?
$$leftlVert TxrightrVert=sup Big{vert x_1 vert,frac{vert x_2 vert}{2^2},...Big}=sup_nBig{frac{vert x_n vert}{n^2}Big} leq sup_nBig{frac{vert x_n vert}{n}Big}$$
How to make the RHS of above in the form $K vertvert x vert vert$ if possible ? Any hint ?
On the otherhand, $T^{-1}:X to X$ is a map by $$T^{-1}(x_1.x_2,...)=(x_1,2^2x_2,3^2x_3,...)$$
$$leftlVert T^{-1}xrightrVert=sup_nBig{n^2 vert x_n vertBig} geq n$$ so $T^{-1}$ is not bounded. Am I right? Any help ?
functional-analysis operator-theory
$endgroup$
add a comment |
$begingroup$
Let $X$ be a real normed linear space of all real sequences which are eventually zero with the 'sup' norm and $T:X to X$ be a bijective linear operator defined by $$T(x_1,x_2,x_3,....)=left(x_1,frac{x_2}{2^2},frac{x_3}{3^2},....right)$$
How to check whether $T$ and $T^{-1}$ is bounded or not ?
$$leftlVert TxrightrVert=sup Big{vert x_1 vert,frac{vert x_2 vert}{2^2},...Big}=sup_nBig{frac{vert x_n vert}{n^2}Big} leq sup_nBig{frac{vert x_n vert}{n}Big}$$
How to make the RHS of above in the form $K vertvert x vert vert$ if possible ? Any hint ?
On the otherhand, $T^{-1}:X to X$ is a map by $$T^{-1}(x_1.x_2,...)=(x_1,2^2x_2,3^2x_3,...)$$
$$leftlVert T^{-1}xrightrVert=sup_nBig{n^2 vert x_n vertBig} geq n$$ so $T^{-1}$ is not bounded. Am I right? Any help ?
functional-analysis operator-theory
$endgroup$
Let $X$ be a real normed linear space of all real sequences which are eventually zero with the 'sup' norm and $T:X to X$ be a bijective linear operator defined by $$T(x_1,x_2,x_3,....)=left(x_1,frac{x_2}{2^2},frac{x_3}{3^2},....right)$$
How to check whether $T$ and $T^{-1}$ is bounded or not ?
$$leftlVert TxrightrVert=sup Big{vert x_1 vert,frac{vert x_2 vert}{2^2},...Big}=sup_nBig{frac{vert x_n vert}{n^2}Big} leq sup_nBig{frac{vert x_n vert}{n}Big}$$
How to make the RHS of above in the form $K vertvert x vert vert$ if possible ? Any hint ?
On the otherhand, $T^{-1}:X to X$ is a map by $$T^{-1}(x_1.x_2,...)=(x_1,2^2x_2,3^2x_3,...)$$
$$leftlVert T^{-1}xrightrVert=sup_nBig{n^2 vert x_n vertBig} geq n$$ so $T^{-1}$ is not bounded. Am I right? Any help ?
functional-analysis operator-theory
functional-analysis operator-theory
asked Jan 22 at 6:27
Chinnapparaj RChinnapparaj R
5,7032928
5,7032928
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1 Answer
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$begingroup$
- $sup_n {frac{|x_n|}{n}} le sup_n{|x_n|} =||x||.$
- Your considerations concerning $T^{-1}$ are correct.
$endgroup$
$begingroup$
Thanks! Is there any text book which deals with lot of this type of problems ?
$endgroup$
– Chinnapparaj R
Jan 22 at 6:43
$begingroup$
Why no response? Is I'm asking anything wrong?
$endgroup$
– Chinnapparaj R
Jan 22 at 7:06
add a comment |
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1 Answer
1
active
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1 Answer
1
active
oldest
votes
active
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active
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votes
$begingroup$
- $sup_n {frac{|x_n|}{n}} le sup_n{|x_n|} =||x||.$
- Your considerations concerning $T^{-1}$ are correct.
$endgroup$
$begingroup$
Thanks! Is there any text book which deals with lot of this type of problems ?
$endgroup$
– Chinnapparaj R
Jan 22 at 6:43
$begingroup$
Why no response? Is I'm asking anything wrong?
$endgroup$
– Chinnapparaj R
Jan 22 at 7:06
add a comment |
$begingroup$
- $sup_n {frac{|x_n|}{n}} le sup_n{|x_n|} =||x||.$
- Your considerations concerning $T^{-1}$ are correct.
$endgroup$
$begingroup$
Thanks! Is there any text book which deals with lot of this type of problems ?
$endgroup$
– Chinnapparaj R
Jan 22 at 6:43
$begingroup$
Why no response? Is I'm asking anything wrong?
$endgroup$
– Chinnapparaj R
Jan 22 at 7:06
add a comment |
$begingroup$
- $sup_n {frac{|x_n|}{n}} le sup_n{|x_n|} =||x||.$
- Your considerations concerning $T^{-1}$ are correct.
$endgroup$
- $sup_n {frac{|x_n|}{n}} le sup_n{|x_n|} =||x||.$
- Your considerations concerning $T^{-1}$ are correct.
answered Jan 22 at 6:37
FredFred
47.8k1849
47.8k1849
$begingroup$
Thanks! Is there any text book which deals with lot of this type of problems ?
$endgroup$
– Chinnapparaj R
Jan 22 at 6:43
$begingroup$
Why no response? Is I'm asking anything wrong?
$endgroup$
– Chinnapparaj R
Jan 22 at 7:06
add a comment |
$begingroup$
Thanks! Is there any text book which deals with lot of this type of problems ?
$endgroup$
– Chinnapparaj R
Jan 22 at 6:43
$begingroup$
Why no response? Is I'm asking anything wrong?
$endgroup$
– Chinnapparaj R
Jan 22 at 7:06
$begingroup$
Thanks! Is there any text book which deals with lot of this type of problems ?
$endgroup$
– Chinnapparaj R
Jan 22 at 6:43
$begingroup$
Thanks! Is there any text book which deals with lot of this type of problems ?
$endgroup$
– Chinnapparaj R
Jan 22 at 6:43
$begingroup$
Why no response? Is I'm asking anything wrong?
$endgroup$
– Chinnapparaj R
Jan 22 at 7:06
$begingroup$
Why no response? Is I'm asking anything wrong?
$endgroup$
– Chinnapparaj R
Jan 22 at 7:06
add a comment |
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