Closedness of Bounded and Continuous Function Space












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I haven't been able to do much progress on the following question:



Show that for any $f , g in mathcal { B } ( X ; Y )$ and $x in X$, there holds



$omega _ { f } ( x ) leq 2 rho _ { text { sup } } ( f , g ) + omega _ { g } ( x )$



Deduce that $mathcal { B C } ( X ; Y )$ is closed with respect to $rho_{sup}$.



Definitions:



For $E subseteq X$, $Omega _ { f } ( E ) = sup left{ d _ { Y } left( f ( x ) , f left( x ^ { prime } right) right) | x , x ^ { prime } in E right}$



$begin{array} { c } { mathcal { B C } ( X ; Y ) = mathcal { B } ( X ; Y ) cap mathcal { C } ( X ; Y )} \ { rho _ { text { sup } } ( f , g ) = sup _ { x in X } d _ { Y } ( f ( x ) , g ( x ) ) } end{array}$



$mathcal{B}(X,Y)$ and $mathcal{C}(X,Y)$ are respectively the set of bounded and continuous functions from $X$ to $Y$.



$omega _ { f } ( x ) = inf _ { r > 0 } Omega _ { f } left( N _ { r } ( x ) right)$, where $N_r(x)$ is a neighborhood of radius $r$ around $x$



What I did:



I wasn't able to do much. It seems to me that the boundedness comes into play when using/coming up with the supremum as it guarantees it's not infinity. I also tried to arrive at the inequality by the means of the triangle inequality, but I haven't been successful.










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  • Also define $omega_f(x)$ for a bounded function $f$?
    – Henno Brandsma
    Dec 1 '18 at 22:11










  • Just editted it to correct this, thanks. $omega _ { f } ( x ) = inf _ { r > 0 } Omega _ { f } left( N _ { r } ( x ) right)$
    – Fhoenix
    Dec 1 '18 at 23:08
















0














I haven't been able to do much progress on the following question:



Show that for any $f , g in mathcal { B } ( X ; Y )$ and $x in X$, there holds



$omega _ { f } ( x ) leq 2 rho _ { text { sup } } ( f , g ) + omega _ { g } ( x )$



Deduce that $mathcal { B C } ( X ; Y )$ is closed with respect to $rho_{sup}$.



Definitions:



For $E subseteq X$, $Omega _ { f } ( E ) = sup left{ d _ { Y } left( f ( x ) , f left( x ^ { prime } right) right) | x , x ^ { prime } in E right}$



$begin{array} { c } { mathcal { B C } ( X ; Y ) = mathcal { B } ( X ; Y ) cap mathcal { C } ( X ; Y )} \ { rho _ { text { sup } } ( f , g ) = sup _ { x in X } d _ { Y } ( f ( x ) , g ( x ) ) } end{array}$



$mathcal{B}(X,Y)$ and $mathcal{C}(X,Y)$ are respectively the set of bounded and continuous functions from $X$ to $Y$.



$omega _ { f } ( x ) = inf _ { r > 0 } Omega _ { f } left( N _ { r } ( x ) right)$, where $N_r(x)$ is a neighborhood of radius $r$ around $x$



What I did:



I wasn't able to do much. It seems to me that the boundedness comes into play when using/coming up with the supremum as it guarantees it's not infinity. I also tried to arrive at the inequality by the means of the triangle inequality, but I haven't been successful.










share|cite|improve this question
























  • Also define $omega_f(x)$ for a bounded function $f$?
    – Henno Brandsma
    Dec 1 '18 at 22:11










  • Just editted it to correct this, thanks. $omega _ { f } ( x ) = inf _ { r > 0 } Omega _ { f } left( N _ { r } ( x ) right)$
    – Fhoenix
    Dec 1 '18 at 23:08














0












0








0







I haven't been able to do much progress on the following question:



Show that for any $f , g in mathcal { B } ( X ; Y )$ and $x in X$, there holds



$omega _ { f } ( x ) leq 2 rho _ { text { sup } } ( f , g ) + omega _ { g } ( x )$



Deduce that $mathcal { B C } ( X ; Y )$ is closed with respect to $rho_{sup}$.



Definitions:



For $E subseteq X$, $Omega _ { f } ( E ) = sup left{ d _ { Y } left( f ( x ) , f left( x ^ { prime } right) right) | x , x ^ { prime } in E right}$



$begin{array} { c } { mathcal { B C } ( X ; Y ) = mathcal { B } ( X ; Y ) cap mathcal { C } ( X ; Y )} \ { rho _ { text { sup } } ( f , g ) = sup _ { x in X } d _ { Y } ( f ( x ) , g ( x ) ) } end{array}$



$mathcal{B}(X,Y)$ and $mathcal{C}(X,Y)$ are respectively the set of bounded and continuous functions from $X$ to $Y$.



$omega _ { f } ( x ) = inf _ { r > 0 } Omega _ { f } left( N _ { r } ( x ) right)$, where $N_r(x)$ is a neighborhood of radius $r$ around $x$



What I did:



I wasn't able to do much. It seems to me that the boundedness comes into play when using/coming up with the supremum as it guarantees it's not infinity. I also tried to arrive at the inequality by the means of the triangle inequality, but I haven't been successful.










share|cite|improve this question















I haven't been able to do much progress on the following question:



Show that for any $f , g in mathcal { B } ( X ; Y )$ and $x in X$, there holds



$omega _ { f } ( x ) leq 2 rho _ { text { sup } } ( f , g ) + omega _ { g } ( x )$



Deduce that $mathcal { B C } ( X ; Y )$ is closed with respect to $rho_{sup}$.



Definitions:



For $E subseteq X$, $Omega _ { f } ( E ) = sup left{ d _ { Y } left( f ( x ) , f left( x ^ { prime } right) right) | x , x ^ { prime } in E right}$



$begin{array} { c } { mathcal { B C } ( X ; Y ) = mathcal { B } ( X ; Y ) cap mathcal { C } ( X ; Y )} \ { rho _ { text { sup } } ( f , g ) = sup _ { x in X } d _ { Y } ( f ( x ) , g ( x ) ) } end{array}$



$mathcal{B}(X,Y)$ and $mathcal{C}(X,Y)$ are respectively the set of bounded and continuous functions from $X$ to $Y$.



$omega _ { f } ( x ) = inf _ { r > 0 } Omega _ { f } left( N _ { r } ( x ) right)$, where $N_r(x)$ is a neighborhood of radius $r$ around $x$



What I did:



I wasn't able to do much. It seems to me that the boundedness comes into play when using/coming up with the supremum as it guarantees it's not infinity. I also tried to arrive at the inequality by the means of the triangle inequality, but I haven't been successful.







general-topology analysis






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edited Dec 2 '18 at 0:01

























asked Dec 1 '18 at 20:50









Fhoenix

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11211












  • Also define $omega_f(x)$ for a bounded function $f$?
    – Henno Brandsma
    Dec 1 '18 at 22:11










  • Just editted it to correct this, thanks. $omega _ { f } ( x ) = inf _ { r > 0 } Omega _ { f } left( N _ { r } ( x ) right)$
    – Fhoenix
    Dec 1 '18 at 23:08


















  • Also define $omega_f(x)$ for a bounded function $f$?
    – Henno Brandsma
    Dec 1 '18 at 22:11










  • Just editted it to correct this, thanks. $omega _ { f } ( x ) = inf _ { r > 0 } Omega _ { f } left( N _ { r } ( x ) right)$
    – Fhoenix
    Dec 1 '18 at 23:08
















Also define $omega_f(x)$ for a bounded function $f$?
– Henno Brandsma
Dec 1 '18 at 22:11




Also define $omega_f(x)$ for a bounded function $f$?
– Henno Brandsma
Dec 1 '18 at 22:11












Just editted it to correct this, thanks. $omega _ { f } ( x ) = inf _ { r > 0 } Omega _ { f } left( N _ { r } ( x ) right)$
– Fhoenix
Dec 1 '18 at 23:08




Just editted it to correct this, thanks. $omega _ { f } ( x ) = inf _ { r > 0 } Omega _ { f } left( N _ { r } ( x ) right)$
– Fhoenix
Dec 1 '18 at 23:08










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$d(f(x_1),f(x_2)) leq d(g(x_1),g(x_2)) +2sup d(f(x),g(x))$. Take sup over $x_1,x_2 in N_r(x)$ and let $r to 0$. Note that (by monotonicity) $omega_f(x)$ is also $lim_{r to 0} Omega_f(N_r(x))$. For the second part use the fa ct that $f$ is continuous at $f$ iff $omega_f(x)=0$. Thus, if $f_n$'s are continuous and $f_n to f$ w.r.t. $rho_{sup}$ then $omega_f(x) leq 2rho_{sup} (f_n,f) +0 to 0$ for all $x$ so $f$ is continuous.






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    $d(f(x_1),f(x_2)) leq d(g(x_1),g(x_2)) +2sup d(f(x),g(x))$. Take sup over $x_1,x_2 in N_r(x)$ and let $r to 0$. Note that (by monotonicity) $omega_f(x)$ is also $lim_{r to 0} Omega_f(N_r(x))$. For the second part use the fa ct that $f$ is continuous at $f$ iff $omega_f(x)=0$. Thus, if $f_n$'s are continuous and $f_n to f$ w.r.t. $rho_{sup}$ then $omega_f(x) leq 2rho_{sup} (f_n,f) +0 to 0$ for all $x$ so $f$ is continuous.






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      $d(f(x_1),f(x_2)) leq d(g(x_1),g(x_2)) +2sup d(f(x),g(x))$. Take sup over $x_1,x_2 in N_r(x)$ and let $r to 0$. Note that (by monotonicity) $omega_f(x)$ is also $lim_{r to 0} Omega_f(N_r(x))$. For the second part use the fa ct that $f$ is continuous at $f$ iff $omega_f(x)=0$. Thus, if $f_n$'s are continuous and $f_n to f$ w.r.t. $rho_{sup}$ then $omega_f(x) leq 2rho_{sup} (f_n,f) +0 to 0$ for all $x$ so $f$ is continuous.






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        $d(f(x_1),f(x_2)) leq d(g(x_1),g(x_2)) +2sup d(f(x),g(x))$. Take sup over $x_1,x_2 in N_r(x)$ and let $r to 0$. Note that (by monotonicity) $omega_f(x)$ is also $lim_{r to 0} Omega_f(N_r(x))$. For the second part use the fa ct that $f$ is continuous at $f$ iff $omega_f(x)=0$. Thus, if $f_n$'s are continuous and $f_n to f$ w.r.t. $rho_{sup}$ then $omega_f(x) leq 2rho_{sup} (f_n,f) +0 to 0$ for all $x$ so $f$ is continuous.






        share|cite|improve this answer












        $d(f(x_1),f(x_2)) leq d(g(x_1),g(x_2)) +2sup d(f(x),g(x))$. Take sup over $x_1,x_2 in N_r(x)$ and let $r to 0$. Note that (by monotonicity) $omega_f(x)$ is also $lim_{r to 0} Omega_f(N_r(x))$. For the second part use the fa ct that $f$ is continuous at $f$ iff $omega_f(x)=0$. Thus, if $f_n$'s are continuous and $f_n to f$ w.r.t. $rho_{sup}$ then $omega_f(x) leq 2rho_{sup} (f_n,f) +0 to 0$ for all $x$ so $f$ is continuous.







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        answered Dec 2 '18 at 0:02









        Kavi Rama Murthy

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