Inverse of continuous bijective linear operator is continuous.












1












$begingroup$



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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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


















1












$begingroup$



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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















1












1








1





$begingroup$



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







share|cite|improve this question









$endgroup$





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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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




















  • $begingroup$
    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












2 Answers
2






active

oldest

votes


















1












$begingroup$

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 $TT^{-1}x=x$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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





















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    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 $TT^{-1}x=x$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      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


















    1












    $begingroup$

    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 $TT^{-1}x=x$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      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
















    1












    1








    1





    $begingroup$

    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 $TT^{-1}x=x$.






    share|cite|improve this answer









    $endgroup$



    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 $TT^{-1}x=x$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    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




















    • $begingroup$
      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













    0












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 24 at 8:40









        Lin XueleiLin Xuelei

        15510




        15510






























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            The Binding of Isaac: Rebirth/Afterbirth

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            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
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            What does “Dominus providebit” mean?

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            5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
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