Lower semi continuous envelope is lower semi continuous












2












$begingroup$


Let X be a topological space and $F:X rightarrow overline{mathbb{R}}$. The lower semi continuous envelope of $F$ is defined by $sc^-F(u)=sup{phi(u);|; phi :X rightarrow overline{mathbb{R}} ; mathrm{is ; l.s.c. ; and} ; phi leq F}$. Show that $sc^-F$ is lower semi continuous.



Here is my proof, but it seems a bit too easy, so it would be great if somebody could check if it is correct.



Let $u_k rightarrow u$ in $X$. The set $M:={phi(u);|; phi :X rightarrow overline{mathbb{R}} ; mathrm{is ; l.s.c. ; and} ; phi leq F}$ is a family of l.s.c. functions, because every $phi in M$ is l.s.c. Then for every $phi in M$ we obtain: $phi(u) leq liminf ; phi(u_k) \
leq liminf ; sup {phi(u_k);|; phi :X rightarrow overline{mathbb{R}} ; mathrm{is ; l.s.c. ; and} ; phi leq F} = liminf ; sc^-F(u_k).$



Because this is true for every $phi in M$ it is also true for the supremum of all $phi$ in $M$ (at this point I'm not really sure if it is correct) an we get



$sup{phi(u);|; phi :X rightarrow overline{mathbb{R}} ; mathrm{is ; l.s.c. ; and} ; phi leq F} = sc^-F(u) leq liminf ; sc^-F(u_k)$.



So the envelope of $F$ is l.s.c.






real-analysis calculus-of-variations semicontinuous-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    If $X$ is not first-countable, it would be worth mentioning that $u_k$ is a net and how $liminf$ is understood here. Otherwise, it looks ok to me. Alternatively, a proof via epigraphs is more straightforward (as intersection of closed sets is closed).
    $endgroup$
    – A.Γ.
    Dec 6 '18 at 11:03










  • $begingroup$
    The set $M$ you define is not what you want it to be (it's a set of function values, not of functions).
    $endgroup$
    – MaoWao
    Dec 6 '18 at 11:15






  • 1




    $begingroup$
    We haven't mentioned epigraphs in the lecture so I don't know how to prove this assumption this way. Thanks for the comments, I'll add it to the proof.
    $endgroup$
    – mathstu
    Dec 7 '18 at 9:20
















2












$begingroup$


Let X be a topological space and $F:X rightarrow overline{mathbb{R}}$. The lower semi continuous envelope of $F$ is defined by $sc^-F(u)=sup{phi(u);|; phi :X rightarrow overline{mathbb{R}} ; mathrm{is ; l.s.c. ; and} ; phi leq F}$. Show that $sc^-F$ is lower semi continuous.



Here is my proof, but it seems a bit too easy, so it would be great if somebody could check if it is correct.



Let $u_k rightarrow u$ in $X$. The set $M:={phi(u);|; phi :X rightarrow overline{mathbb{R}} ; mathrm{is ; l.s.c. ; and} ; phi leq F}$ is a family of l.s.c. functions, because every $phi in M$ is l.s.c. Then for every $phi in M$ we obtain: $phi(u) leq liminf ; phi(u_k) \
leq liminf ; sup {phi(u_k);|; phi :X rightarrow overline{mathbb{R}} ; mathrm{is ; l.s.c. ; and} ; phi leq F} = liminf ; sc^-F(u_k).$



Because this is true for every $phi in M$ it is also true for the supremum of all $phi$ in $M$ (at this point I'm not really sure if it is correct) an we get



$sup{phi(u);|; phi :X rightarrow overline{mathbb{R}} ; mathrm{is ; l.s.c. ; and} ; phi leq F} = sc^-F(u) leq liminf ; sc^-F(u_k)$.



So the envelope of $F$ is l.s.c.






real-analysis calculus-of-variations semicontinuous-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    If $X$ is not first-countable, it would be worth mentioning that $u_k$ is a net and how $liminf$ is understood here. Otherwise, it looks ok to me. Alternatively, a proof via epigraphs is more straightforward (as intersection of closed sets is closed).
    $endgroup$
    – A.Γ.
    Dec 6 '18 at 11:03










  • $begingroup$
    The set $M$ you define is not what you want it to be (it's a set of function values, not of functions).
    $endgroup$
    – MaoWao
    Dec 6 '18 at 11:15






  • 1




    $begingroup$
    We haven't mentioned epigraphs in the lecture so I don't know how to prove this assumption this way. Thanks for the comments, I'll add it to the proof.
    $endgroup$
    – mathstu
    Dec 7 '18 at 9:20














2












2








2





$begingroup$


Let X be a topological space and $F:X rightarrow overline{mathbb{R}}$. The lower semi continuous envelope of $F$ is defined by $sc^-F(u)=sup{phi(u);|; phi :X rightarrow overline{mathbb{R}} ; mathrm{is ; l.s.c. ; and} ; phi leq F}$. Show that $sc^-F$ is lower semi continuous.



Here is my proof, but it seems a bit too easy, so it would be great if somebody could check if it is correct.



Let $u_k rightarrow u$ in $X$. The set $M:={phi(u);|; phi :X rightarrow overline{mathbb{R}} ; mathrm{is ; l.s.c. ; and} ; phi leq F}$ is a family of l.s.c. functions, because every $phi in M$ is l.s.c. Then for every $phi in M$ we obtain: $phi(u) leq liminf ; phi(u_k) \
leq liminf ; sup {phi(u_k);|; phi :X rightarrow overline{mathbb{R}} ; mathrm{is ; l.s.c. ; and} ; phi leq F} = liminf ; sc^-F(u_k).$



Because this is true for every $phi in M$ it is also true for the supremum of all $phi$ in $M$ (at this point I'm not really sure if it is correct) an we get



$sup{phi(u);|; phi :X rightarrow overline{mathbb{R}} ; mathrm{is ; l.s.c. ; and} ; phi leq F} = sc^-F(u) leq liminf ; sc^-F(u_k)$.



So the envelope of $F$ is l.s.c.






real-analysis calculus-of-variations semicontinuous-functions







share|cite|improve this question











$endgroup$




Let X be a topological space and $F:X rightarrow overline{mathbb{R}}$. The lower semi continuous envelope of $F$ is defined by $sc^-F(u)=sup{phi(u);|; phi :X rightarrow overline{mathbb{R}} ; mathrm{is ; l.s.c. ; and} ; phi leq F}$. Show that $sc^-F$ is lower semi continuous.



Here is my proof, but it seems a bit too easy, so it would be great if somebody could check if it is correct.



Let $u_k rightarrow u$ in $X$. The set $M:={phi(u);|; phi :X rightarrow overline{mathbb{R}} ; mathrm{is ; l.s.c. ; and} ; phi leq F}$ is a family of l.s.c. functions, because every $phi in M$ is l.s.c. Then for every $phi in M$ we obtain: $phi(u) leq liminf ; phi(u_k) \
leq liminf ; sup {phi(u_k);|; phi :X rightarrow overline{mathbb{R}} ; mathrm{is ; l.s.c. ; and} ; phi leq F} = liminf ; sc^-F(u_k).$



Because this is true for every $phi in M$ it is also true for the supremum of all $phi$ in $M$ (at this point I'm not really sure if it is correct) an we get



$sup{phi(u);|; phi :X rightarrow overline{mathbb{R}} ; mathrm{is ; l.s.c. ; and} ; phi leq F} = sc^-F(u) leq liminf ; sc^-F(u_k)$.



So the envelope of $F$ is l.s.c.






real-analysis calculus-of-variations semicontinuous-functions




real-analysis calculus-of-variations semicontinuous-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 10:26









A.Γ.

22.7k32656




22.7k32656










asked Dec 6 '18 at 10:05









mathstumathstu

315




315












  • $begingroup$
    If $X$ is not first-countable, it would be worth mentioning that $u_k$ is a net and how $liminf$ is understood here. Otherwise, it looks ok to me. Alternatively, a proof via epigraphs is more straightforward (as intersection of closed sets is closed).
    $endgroup$
    – A.Γ.
    Dec 6 '18 at 11:03










  • $begingroup$
    The set $M$ you define is not what you want it to be (it's a set of function values, not of functions).
    $endgroup$
    – MaoWao
    Dec 6 '18 at 11:15






  • 1




    $begingroup$
    We haven't mentioned epigraphs in the lecture so I don't know how to prove this assumption this way. Thanks for the comments, I'll add it to the proof.
    $endgroup$
    – mathstu
    Dec 7 '18 at 9:20


















  • $begingroup$
    If $X$ is not first-countable, it would be worth mentioning that $u_k$ is a net and how $liminf$ is understood here. Otherwise, it looks ok to me. Alternatively, a proof via epigraphs is more straightforward (as intersection of closed sets is closed).
    $endgroup$
    – A.Γ.
    Dec 6 '18 at 11:03










  • $begingroup$
    The set $M$ you define is not what you want it to be (it's a set of function values, not of functions).
    $endgroup$
    – MaoWao
    Dec 6 '18 at 11:15






  • 1




    $begingroup$
    We haven't mentioned epigraphs in the lecture so I don't know how to prove this assumption this way. Thanks for the comments, I'll add it to the proof.
    $endgroup$
    – mathstu
    Dec 7 '18 at 9:20
















$begingroup$
If $X$ is not first-countable, it would be worth mentioning that $u_k$ is a net and how $liminf$ is understood here. Otherwise, it looks ok to me. Alternatively, a proof via epigraphs is more straightforward (as intersection of closed sets is closed).
$endgroup$
– A.Γ.
Dec 6 '18 at 11:03




$begingroup$
If $X$ is not first-countable, it would be worth mentioning that $u_k$ is a net and how $liminf$ is understood here. Otherwise, it looks ok to me. Alternatively, a proof via epigraphs is more straightforward (as intersection of closed sets is closed).
$endgroup$
– A.Γ.
Dec 6 '18 at 11:03












$begingroup$
The set $M$ you define is not what you want it to be (it's a set of function values, not of functions).
$endgroup$
– MaoWao
Dec 6 '18 at 11:15




$begingroup$
The set $M$ you define is not what you want it to be (it's a set of function values, not of functions).
$endgroup$
– MaoWao
Dec 6 '18 at 11:15




1




1




$begingroup$
We haven't mentioned epigraphs in the lecture so I don't know how to prove this assumption this way. Thanks for the comments, I'll add it to the proof.
$endgroup$
– mathstu
Dec 7 '18 at 9:20




$begingroup$
We haven't mentioned epigraphs in the lecture so I don't know how to prove this assumption this way. Thanks for the comments, I'll add it to the proof.
$endgroup$
– mathstu
Dec 7 '18 at 9:20










1 Answer
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$begingroup$

It is correct but the notation can be improved:




Then for every $phi in M$ we obtain: $phi(u) leq liminf ; phi(u_k) \
leq liminf ; sup {phi(u_k);|; phi :X rightarrow overline{mathbb{R}} ; mathrm{is ; l.s.c. ; and} ; phi leq F} = liminf ; sc^-F(u_k).$




Here I strongly suggest to choose different variables for $phi$ inside the set (or outside).
For example:



$ liminf phi(u_k) leq liminf ; sup {psi(u_k);|; psi :X rightarrow overline{mathbb{R}} ; mathrm{is ; l.s.c. ; and} ; psi leq F} = liminf ; sc^-F(u_k).$






share|cite|improve this answer









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    $begingroup$

    It is correct but the notation can be improved:




    Then for every $phi in M$ we obtain: $phi(u) leq liminf ; phi(u_k) \
    leq liminf ; sup {phi(u_k);|; phi :X rightarrow overline{mathbb{R}} ; mathrm{is ; l.s.c. ; and} ; phi leq F} = liminf ; sc^-F(u_k).$




    Here I strongly suggest to choose different variables for $phi$ inside the set (or outside).
    For example:



    $ liminf phi(u_k) leq liminf ; sup {psi(u_k);|; psi :X rightarrow overline{mathbb{R}} ; mathrm{is ; l.s.c. ; and} ; psi leq F} = liminf ; sc^-F(u_k).$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      It is correct but the notation can be improved:




      Then for every $phi in M$ we obtain: $phi(u) leq liminf ; phi(u_k) \
      leq liminf ; sup {phi(u_k);|; phi :X rightarrow overline{mathbb{R}} ; mathrm{is ; l.s.c. ; and} ; phi leq F} = liminf ; sc^-F(u_k).$




      Here I strongly suggest to choose different variables for $phi$ inside the set (or outside).
      For example:



      $ liminf phi(u_k) leq liminf ; sup {psi(u_k);|; psi :X rightarrow overline{mathbb{R}} ; mathrm{is ; l.s.c. ; and} ; psi leq F} = liminf ; sc^-F(u_k).$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        It is correct but the notation can be improved:




        Then for every $phi in M$ we obtain: $phi(u) leq liminf ; phi(u_k) \
        leq liminf ; sup {phi(u_k);|; phi :X rightarrow overline{mathbb{R}} ; mathrm{is ; l.s.c. ; and} ; phi leq F} = liminf ; sc^-F(u_k).$




        Here I strongly suggest to choose different variables for $phi$ inside the set (or outside).
        For example:



        $ liminf phi(u_k) leq liminf ; sup {psi(u_k);|; psi :X rightarrow overline{mathbb{R}} ; mathrm{is ; l.s.c. ; and} ; psi leq F} = liminf ; sc^-F(u_k).$






        share|cite|improve this answer









        $endgroup$



        It is correct but the notation can be improved:




        Then for every $phi in M$ we obtain: $phi(u) leq liminf ; phi(u_k) \
        leq liminf ; sup {phi(u_k);|; phi :X rightarrow overline{mathbb{R}} ; mathrm{is ; l.s.c. ; and} ; phi leq F} = liminf ; sc^-F(u_k).$




        Here I strongly suggest to choose different variables for $phi$ inside the set (or outside).
        For example:



        $ liminf phi(u_k) leq liminf ; sup {psi(u_k);|; psi :X rightarrow overline{mathbb{R}} ; mathrm{is ; l.s.c. ; and} ; psi leq F} = liminf ; sc^-F(u_k).$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 6 '18 at 10:56









        supinfsupinf

        6,3261028




        6,3261028






























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