Inverse of continuous bijective linear operator is continuous.
$begingroup$
If $T: X to Y$ is a continuous bijective linear operator where $X$ and $Y$ are Banach spaces, show that $T^{-1}$ is continuous.
Attempt:
Suppose that $T^{-1}$ is not continuous. Then, $T^{-1}$ is not bounded, so we have that $$||T^{-1}y_0||_X geq C||y_0||_Y$$ for all $C in mathbb{R}$. However, since $y_0 in Y$ and $T$ is surjective, we have that $T(x_0) = y_0$ for some $x_0 in X.$ Then,
$$||T^{-1}(Tx_0)||_X geq C||T(x_0)||_Y$$
$$||x_0||_X geq C||T(x_0)||_Y$$
$$||Tx_0||_Y leq frac{1}{C}||x_0||_X.$$
Since this is true for all $C in mathbb{R}$, we have that
$$||Tx_0||_Y = 0.$$
Since we can repeat this process for every $y in Y$, we get that
$$||Tx||_Y = 0, forall x in X.$$
Then, we must have that $Tx = 0$ for all $x$, because in general, $||x|| = 0$ iff $x=0$. So, this $T$ isn't injective, a contradiction. Thus, $T^{-1}$ must be bounded, and hence continuous.
Could someone please verify if I'm doing this properly? Linear operators are very new to me, and slightly uncomfortable. In particular, when we say a linear operator is bounded, I'm not sure if that means at a single $x_0$, or for all $x in X$. We've also never discussed the composition of linear operators, so I'm not 100% sure that $T^{-1}T(x_0) = x_0$, although I assume it should work just like function composition. Thanks for any help in advance.
real-analysis functional-analysis proof-verification
asked Sep 24 '17 at 19:03
user389056user389056
381112
$endgroup$
add a comment |
$begingroup$
If $T: X to Y$ is a continuous bijective linear operator where $X$ and $Y$ are Banach spaces, show that $T^{-1}$ is continuous.
Attempt:
Suppose that $T^{-1}$ is not continuous. Then, $T^{-1}$ is not bounded, so we have that $$||T^{-1}y_0||_X geq C||y_0||_Y$$ for all $C in mathbb{R}$. However, since $y_0 in Y$ and $T$ is surjective, we have that $T(x_0) = y_0$ for some $x_0 in X.$ Then,
$$||T^{-1}(Tx_0)||_X geq C||T(x_0)||_Y$$
$$||x_0||_X geq C||T(x_0)||_Y$$
$$||Tx_0||_Y leq frac{1}{C}||x_0||_X.$$
Since this is true for all $C in mathbb{R}$, we have that
$$||Tx_0||_Y = 0.$$
Since we can repeat this process for every $y in Y$, we get that
$$||Tx||_Y = 0, forall x in X.$$
Then, we must have that $Tx = 0$ for all $x$, because in general, $||x|| = 0$ iff $x=0$. So, this $T$ isn't injective, a contradiction. Thus, $T^{-1}$ must be bounded, and hence continuous.
Could someone please verify if I'm doing this properly? Linear operators are very new to me, and slightly uncomfortable. In particular, when we say a linear operator is bounded, I'm not sure if that means at a single $x_0$, or for all $x in X$. We've also never discussed the composition of linear operators, so I'm not 100% sure that $T^{-1}T(x_0) = x_0$, although I assume it should work just like function composition. Thanks for any help in advance.
real-analysis functional-analysis proof-verification
asked Sep 24 '17 at 19:03
user389056user389056
381112
$endgroup$
$begingroup$
This isn't right. Note your first display says $T^{-1}y_0$ has infinite norm... But, try using the Open Mapping Theorem to show $(T^{-1})^{-1} U$ is open for any $U$ open in $X$.
$endgroup$
– David Mitra
Sep 24 '17 at 20:23
$begingroup$
Oh. It's literally just, when $U subset X$ is open, then since $T$ is a continuous surjective linear operator, by the Open Mapping Theorem, $T(U)$ is open, which is precisely what $T^{-1}$ being continuous means? So... $T^{-1}y_0$ can't have infinite norm? Isn't this just the negation of being bounded?
$endgroup$
– user389056
Sep 24 '17 at 23:02
$begingroup$
If $y_0$ is fixed, no; $T^{-1}y_0$ is some element of $X$, it has finite norm. But $sup_{Vert y Vert=1} Vert T^{-1} yVert$ can be infinite. Different $C$ correspond to different $y$.
$endgroup$
– David Mitra
Sep 25 '17 at 6:22
add a comment |
$begingroup$
If $T: X to Y$ is a continuous bijective linear operator where $X$ and $Y$ are Banach spaces, show that $T^{-1}$ is continuous.
Attempt:
Suppose that $T^{-1}$ is not continuous. Then, $T^{-1}$ is not bounded, so we have that $$||T^{-1}y_0||_X geq C||y_0||_Y$$ for all $C in mathbb{R}$. However, since $y_0 in Y$ and $T$ is surjective, we have that $T(x_0) = y_0$ for some $x_0 in X.$ Then,
$$||T^{-1}(Tx_0)||_X geq C||T(x_0)||_Y$$
$$||x_0||_X geq C||T(x_0)||_Y$$
$$||Tx_0||_Y leq frac{1}{C}||x_0||_X.$$
Since this is true for all $C in mathbb{R}$, we have that
$$||Tx_0||_Y = 0.$$
Since we can repeat this process for every $y in Y$, we get that
$$||Tx||_Y = 0, forall x in X.$$
Then, we must have that $Tx = 0$ for all $x$, because in general, $||x|| = 0$ iff $x=0$. So, this $T$ isn't injective, a contradiction. Thus, $T^{-1}$ must be bounded, and hence continuous.
Could someone please verify if I'm doing this properly? Linear operators are very new to me, and slightly uncomfortable. In particular, when we say a linear operator is bounded, I'm not sure if that means at a single $x_0$, or for all $x in X$. We've also never discussed the composition of linear operators, so I'm not 100% sure that $T^{-1}T(x_0) = x_0$, although I assume it should work just like function composition. Thanks for any help in advance.
real-analysis functional-analysis proof-verification
asked Sep 24 '17 at 19:03
user389056user389056
381112
$endgroup$
If $T: X to Y$ is a continuous bijective linear operator where $X$ and $Y$ are Banach spaces, show that $T^{-1}$ is continuous.
Attempt:
Suppose that $T^{-1}$ is not continuous. Then, $T^{-1}$ is not bounded, so we have that $$||T^{-1}y_0||_X geq C||y_0||_Y$$ for all $C in mathbb{R}$. However, since $y_0 in Y$ and $T$ is surjective, we have that $T(x_0) = y_0$ for some $x_0 in X.$ Then,
$$||T^{-1}(Tx_0)||_X geq C||T(x_0)||_Y$$
$$||x_0||_X geq C||T(x_0)||_Y$$
$$||Tx_0||_Y leq frac{1}{C}||x_0||_X.$$
Since this is true for all $C in mathbb{R}$, we have that
$$||Tx_0||_Y = 0.$$
Since we can repeat this process for every $y in Y$, we get that
$$||Tx||_Y = 0, forall x in X.$$
Then, we must have that $Tx = 0$ for all $x$, because in general, $||x|| = 0$ iff $x=0$. So, this $T$ isn't injective, a contradiction. Thus, $T^{-1}$ must be bounded, and hence continuous.
Could someone please verify if I'm doing this properly? Linear operators are very new to me, and slightly uncomfortable. In particular, when we say a linear operator is bounded, I'm not sure if that means at a single $x_0$, or for all $x in X$. We've also never discussed the composition of linear operators, so I'm not 100% sure that $T^{-1}T(x_0) = x_0$, although I assume it should work just like function composition. Thanks for any help in advance.
real-analysis functional-analysis proof-verification
real-analysis functional-analysis proof-verification
asked Sep 24 '17 at 19:03
user389056user389056
381112
asked Sep 24 '17 at 19:03
user389056user389056
381112
asked Sep 24 '17 at 19:03
user389056user389056
381112
asked Sep 24 '17 at 19:03
user389056user389056
381112
asked Sep 24 '17 at 19:03
user389056user389056
381112
381112
$begingroup$
This isn't right. Note your first display says $T^{-1}y_0$ has infinite norm... But, try using the Open Mapping Theorem to show $(T^{-1})^{-1} U$ is open for any $U$ open in $X$.
$endgroup$
– David Mitra
Sep 24 '17 at 20:23
$begingroup$
Oh. It's literally just, when $U subset X$ is open, then since $T$ is a continuous surjective linear operator, by the Open Mapping Theorem, $T(U)$ is open, which is precisely what $T^{-1}$ being continuous means? So... $T^{-1}y_0$ can't have infinite norm? Isn't this just the negation of being bounded?
$endgroup$
– user389056
Sep 24 '17 at 23:02
$begingroup$
If $y_0$ is fixed, no; $T^{-1}y_0$ is some element of $X$, it has finite norm. But $sup_{Vert y Vert=1} Vert T^{-1} yVert$ can be infinite. Different $C$ correspond to different $y$.
$endgroup$
– David Mitra
Sep 25 '17 at 6:22
add a comment |
$begingroup$
This isn't right. Note your first display says $T^{-1}y_0$ has infinite norm... But, try using the Open Mapping Theorem to show $(T^{-1})^{-1} U$ is open for any $U$ open in $X$.
$endgroup$
– David Mitra
Sep 24 '17 at 20:23
$begingroup$
Oh. It's literally just, when $U subset X$ is open, then since $T$ is a continuous surjective linear operator, by the Open Mapping Theorem, $T(U)$ is open, which is precisely what $T^{-1}$ being continuous means? So... $T^{-1}y_0$ can't have infinite norm? Isn't this just the negation of being bounded?
$endgroup$
– user389056
Sep 24 '17 at 23:02
$begingroup$
If $y_0$ is fixed, no; $T^{-1}y_0$ is some element of $X$, it has finite norm. But $sup_{Vert y Vert=1} Vert T^{-1} yVert$ can be infinite. Different $C$ correspond to different $y$.
$endgroup$
– David Mitra
Sep 25 '17 at 6:22
$begingroup$
This isn't right. Note your first display says $T^{-1}y_0$ has infinite norm... But, try using the Open Mapping Theorem to show $(T^{-1})^{-1} U$ is open for any $U$ open in $X$.
$endgroup$
– David Mitra
Sep 24 '17 at 20:23
$begingroup$
This isn't right. Note your first display says $T^{-1}y_0$ has infinite norm... But, try using the Open Mapping Theorem to show $(T^{-1})^{-1} U$ is open for any $U$ open in $X$.
$endgroup$
– David Mitra
Sep 24 '17 at 20:23
$begingroup$
Oh. It's literally just, when $U subset X$ is open, then since $T$ is a continuous surjective linear operator, by the Open Mapping Theorem, $T(U)$ is open, which is precisely what $T^{-1}$ being continuous means? So... $T^{-1}y_0$ can't have infinite norm? Isn't this just the negation of being bounded?
$endgroup$
– user389056
Sep 24 '17 at 23:02
$begingroup$
Oh. It's literally just, when $U subset X$ is open, then since $T$ is a continuous surjective linear operator, by the Open Mapping Theorem, $T(U)$ is open, which is precisely what $T^{-1}$ being continuous means? So... $T^{-1}y_0$ can't have infinite norm? Isn't this just the negation of being bounded?
$endgroup$
– user389056
Sep 24 '17 at 23:02
$begingroup$
If $y_0$ is fixed, no; $T^{-1}y_0$ is some element of $X$, it has finite norm. But $sup_{Vert y Vert=1} Vert T^{-1} yVert$ can be infinite. Different $C$ correspond to different $y$.
$endgroup$
– David Mitra
Sep 25 '17 at 6:22
$begingroup$
If $y_0$ is fixed, no; $T^{-1}y_0$ is some element of $X$, it has finite norm. But $sup_{Vert y Vert=1} Vert T^{-1} yVert$ can be infinite. Different $C$ correspond to different $y$.
$endgroup$
– David Mitra
Sep 25 '17 at 6:22
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The proof looks good to me. To say it is ``bounded" means that there is a bound on how much a linear operator can stretch a vector.
It seems like you know the definition, but we want to just ensure that there exists some $c$ so that $|Tx| leq c|x|$ for all $x in X$.
Your second question is just the definition of an inverse, as in $TT^{-1}x=x$.
answered Sep 24 '17 at 19:09
Andres MejiaAndres Mejia
16.2k21548
$endgroup$
$begingroup$
So, the same $c$ for all $x$? Okay, thanks.
$endgroup$
– user389056
Sep 24 '17 at 19:11
$begingroup$
Follow up question, if you have a chance: If we say that a family $T_j$ is bounded, then $||T_j(x)|| < C||x||$ for all j and x, or is every j allowed to use a different C?
$endgroup$
– user389056
Sep 24 '17 at 19:21
$begingroup$
Look up the uniform boundedness principle. I think it depends on the context, but more likely than not, the former
$endgroup$
– Andres Mejia
Sep 24 '17 at 19:44
add a comment |
$begingroup$
By open mapping theorem, the preimage of open set in $X$ under $T^{-1}$ is still open in $Y$, which shows that $T^{-1}$ is continuous.
answered Jan 24 at 8:40
Lin XueleiLin Xuelei
15510
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
The proof looks good to me. To say it is ``bounded" means that there is a bound on how much a linear operator can stretch a vector.
It seems like you know the definition, but we want to just ensure that there exists some $c$ so that $|Tx| leq c|x|$ for all $x in X$.
Your second question is just the definition of an inverse, as in $TT^{-1}x=x$.
answered Sep 24 '17 at 19:09
Andres MejiaAndres Mejia
16.2k21548
$endgroup$
$begingroup$
So, the same $c$ for all $x$? Okay, thanks.
$endgroup$
– user389056
Sep 24 '17 at 19:11
$begingroup$
Follow up question, if you have a chance: If we say that a family $T_j$ is bounded, then $||T_j(x)|| < C||x||$ for all j and x, or is every j allowed to use a different C?
$endgroup$
– user389056
Sep 24 '17 at 19:21
$begingroup$
Look up the uniform boundedness principle. I think it depends on the context, but more likely than not, the former
$endgroup$
– Andres Mejia
Sep 24 '17 at 19:44
add a comment |
$begingroup$
The proof looks good to me. To say it is ``bounded" means that there is a bound on how much a linear operator can stretch a vector.
It seems like you know the definition, but we want to just ensure that there exists some $c$ so that $|Tx| leq c|x|$ for all $x in X$.
Your second question is just the definition of an inverse, as in $TT^{-1}x=x$.
answered Sep 24 '17 at 19:09
Andres MejiaAndres Mejia
16.2k21548
$endgroup$
$begingroup$
So, the same $c$ for all $x$? Okay, thanks.
$endgroup$
– user389056
Sep 24 '17 at 19:11
$begingroup$
Follow up question, if you have a chance: If we say that a family $T_j$ is bounded, then $||T_j(x)|| < C||x||$ for all j and x, or is every j allowed to use a different C?
$endgroup$
– user389056
Sep 24 '17 at 19:21
$begingroup$
Look up the uniform boundedness principle. I think it depends on the context, but more likely than not, the former
$endgroup$
– Andres Mejia
Sep 24 '17 at 19:44
add a comment |
$begingroup$
The proof looks good to me. To say it is ``bounded" means that there is a bound on how much a linear operator can stretch a vector.
It seems like you know the definition, but we want to just ensure that there exists some $c$ so that $|Tx| leq c|x|$ for all $x in X$.
Your second question is just the definition of an inverse, as in $TT^{-1}x=x$.
answered Sep 24 '17 at 19:09
Andres MejiaAndres Mejia
16.2k21548
$endgroup$
The proof looks good to me. To say it is ``bounded" means that there is a bound on how much a linear operator can stretch a vector.
It seems like you know the definition, but we want to just ensure that there exists some $c$ so that $|Tx| leq c|x|$ for all $x in X$.
Your second question is just the definition of an inverse, as in $TT^{-1}x=x$.
answered Sep 24 '17 at 19:09
Andres MejiaAndres Mejia
16.2k21548
answered Sep 24 '17 at 19:09
Andres MejiaAndres Mejia
16.2k21548
answered Sep 24 '17 at 19:09
Andres MejiaAndres Mejia
16.2k21548
answered Sep 24 '17 at 19:09
Andres MejiaAndres Mejia
16.2k21548
16.2k21548
$begingroup$
So, the same $c$ for all $x$? Okay, thanks.
$endgroup$
– user389056
Sep 24 '17 at 19:11
$begingroup$
Follow up question, if you have a chance: If we say that a family $T_j$ is bounded, then $||T_j(x)|| < C||x||$ for all j and x, or is every j allowed to use a different C?
$endgroup$
– user389056
Sep 24 '17 at 19:21
$begingroup$
Look up the uniform boundedness principle. I think it depends on the context, but more likely than not, the former
$endgroup$
– Andres Mejia
Sep 24 '17 at 19:44
add a comment |
$begingroup$
So, the same $c$ for all $x$? Okay, thanks.
$endgroup$
– user389056
Sep 24 '17 at 19:11
$begingroup$
Follow up question, if you have a chance: If we say that a family $T_j$ is bounded, then $||T_j(x)|| < C||x||$ for all j and x, or is every j allowed to use a different C?
$endgroup$
– user389056
Sep 24 '17 at 19:21
$begingroup$
Look up the uniform boundedness principle. I think it depends on the context, but more likely than not, the former
$endgroup$
– Andres Mejia
Sep 24 '17 at 19:44
$begingroup$
So, the same $c$ for all $x$? Okay, thanks.
$endgroup$
– user389056
Sep 24 '17 at 19:11
$begingroup$
So, the same $c$ for all $x$? Okay, thanks.
$endgroup$
– user389056
Sep 24 '17 at 19:11
$begingroup$
Follow up question, if you have a chance: If we say that a family $T_j$ is bounded, then $||T_j(x)|| < C||x||$ for all j and x, or is every j allowed to use a different C?
$endgroup$
– user389056
Sep 24 '17 at 19:21
$begingroup$
Follow up question, if you have a chance: If we say that a family $T_j$ is bounded, then $||T_j(x)|| < C||x||$ for all j and x, or is every j allowed to use a different C?
$endgroup$
– user389056
Sep 24 '17 at 19:21
$begingroup$
Look up the uniform boundedness principle. I think it depends on the context, but more likely than not, the former
$endgroup$
– Andres Mejia
Sep 24 '17 at 19:44
$begingroup$
Look up the uniform boundedness principle. I think it depends on the context, but more likely than not, the former
$endgroup$
– Andres Mejia
Sep 24 '17 at 19:44
add a comment |
$begingroup$
By open mapping theorem, the preimage of open set in $X$ under $T^{-1}$ is still open in $Y$, which shows that $T^{-1}$ is continuous.
answered Jan 24 at 8:40
Lin XueleiLin Xuelei
15510
$endgroup$
add a comment |
$begingroup$
By open mapping theorem, the preimage of open set in $X$ under $T^{-1}$ is still open in $Y$, which shows that $T^{-1}$ is continuous.
answered Jan 24 at 8:40
Lin XueleiLin Xuelei
15510
$endgroup$
add a comment |
$begingroup$
By open mapping theorem, the preimage of open set in $X$ under $T^{-1}$ is still open in $Y$, which shows that $T^{-1}$ is continuous.
answered Jan 24 at 8:40
Lin XueleiLin Xuelei
15510
$endgroup$
By open mapping theorem, the preimage of open set in $X$ under $T^{-1}$ is still open in $Y$, which shows that $T^{-1}$ is continuous.
answered Jan 24 at 8:40
Lin XueleiLin Xuelei
15510
answered Jan 24 at 8:40
Lin XueleiLin Xuelei
15510
answered Jan 24 at 8:40
Lin XueleiLin Xuelei
15510
answered Jan 24 at 8:40
Lin XueleiLin Xuelei
15510
15510
add a comment |
add a comment |
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$begingroup$
This isn't right. Note your first display says $T^{-1}y_0$ has infinite norm... But, try using the Open Mapping Theorem to show $(T^{-1})^{-1} U$ is open for any $U$ open in $X$.
$endgroup$
– David Mitra
Sep 24 '17 at 20:23
$begingroup$
Oh. It's literally just, when $U subset X$ is open, then since $T$ is a continuous surjective linear operator, by the Open Mapping Theorem, $T(U)$ is open, which is precisely what $T^{-1}$ being continuous means? So... $T^{-1}y_0$ can't have infinite norm? Isn't this just the negation of being bounded?
$endgroup$
– user389056
Sep 24 '17 at 23:02
$begingroup$
If $y_0$ is fixed, no; $T^{-1}y_0$ is some element of $X$, it has finite norm. But $sup_{Vert y Vert=1} Vert T^{-1} yVert$ can be infinite. Different $C$ correspond to different $y$.
$endgroup$
– David Mitra
Sep 25 '17 at 6:22