Approximation of a lower semi continuous function by monotone sequence of continuous functions












0












$begingroup$


Let $ (X, d_X)$ and $ (Y, d_Y) $ be metric spaces and let
$$ c: X times Y to [0, +infty]$$
a lower semicontinuous function. Then define for each $k in mathbb{N}$ the functions
$$ c_k (x,y): = inf_{x' in X, y' in Y} { c(x',y') wedge k + kd_X(x,x')+kd_Y(y,y') } $$
Then $c_k$ are continuous functions, monotonically increasing converging to $c$.



Proof:




  1. Continuity: take $(x,y) in X times Y$ and a sequence ${ x_n,y_n }_{n ge 1} $ converging to $(x,y)$. Then
    $$ | c_k(x,y)-c_k(x_n,y_n) | le sup_{x' in X, y' in Y} | kd_X(x,x')+kd_Y(y,y') -kd_X(x_n,x')-kd_Y(y_n,y') | le sup_{x' in X, y' in Y} k(d_X(x_n, x)|+d_Y(y_n,y)) to 0$$


  2. Monotonically increasing: fix $(x,y) in X times Y$ and consider any $(x',y') in X times Y$, then
    $$ c(x',y') wedge k + kd_X(x',x)+ kd_Y(y,y') le c(x',y') wedge (k+1) + (k+1)d_X(x',x)+ (k+1)d_Y(y,y') $$
    and then
    $$ c_{k}(x,y) le c_{k+1}(x,y) $$


  3. Convergence: First I observe that if $ (x',y') = (x,y)$ and $c(x,y) = + infty$ I have
    $$c(x',y') wedge k + kd_X(x',x)+ kd_Y(y,y')= c(x,y) wedge k = k$$
    Then $c_k(x,y) ge k$ and then $c_k to + infty$ when $k to + infty$.
    From now on I only consider the case in which $c(x,y) < + infty$. Then if $ (x',y') = (x,y)$, I have
    $$c(x',y') wedge k + kd_X(x',x)+ kd_Y(y,y')= c(x,y) wedge k le c(x,y) $$
    which implies $c_k(x,y) le c(x,y)$. Then it is enough to prove that $c(x,y)$ is the supremum of the sequence $c_k(x,y)$. Fix $0< epsilon < c(x,y)/2$.
    I want to prove that there exists a $k_{epsilon} in mathbb{N}$ s.t.
    $$c_{k_{epsilon}}(x,y) > c(x,y) - 2epsilon $$
    To do that I find $k_{epsilon} in mathbb{N}$ s.t.
    $$ c(x',y') wedge k_{epsilon} + k_{epsilon}d_X(x',x)+ k_{epsilon}d_Y(y,y') > c(x,y) -epsilon $$
    for any $(x',y') in X times Y$.
    Since $c$ is l.s.c. there exists a $delta= delta(epsilon)>0$ s.t.
    $$ (x',y') in B_{delta}((x,y)) Rightarrow c(x',y') > c(x,y) - epsilon$$



Then if $(x',y') in B_{delta}((x,y))$ I take $k_{epsilon} > c(x,y)- epsilon$ and then
$$ c(x',y') wedge k_{epsilon} + k_{epsilon}d_X(x',x)+ k_{epsilon}d_Y(y,y') > c(x,y) - epsilon + 0 $$



If $(x',y') notin B_{delta}((x,y))$ then $d_X(x',x) +d(y',y) > c(epsilon) >0$ and then, taking
$$k_{epsilon} > frac{c(x,y)-epsilon}{c(epsilon)}$$
I obtain
$$ c(x',y') wedge k_{epsilon} + k_{epsilon}d_X(x',x)+ k_{epsilon}d_Y(y,y') > 0 + c(x,y) - epsilon$$
QED



I think this proof is too long and involved.
So



is this proof correct? Does there exist a shorter proof?






real-analysis proof-verification convergence







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Let $ (X, d_X)$ and $ (Y, d_Y) $ be metric spaces and let
    $$ c: X times Y to [0, +infty]$$
    a lower semicontinuous function. Then define for each $k in mathbb{N}$ the functions
    $$ c_k (x,y): = inf_{x' in X, y' in Y} { c(x',y') wedge k + kd_X(x,x')+kd_Y(y,y') } $$
    Then $c_k$ are continuous functions, monotonically increasing converging to $c$.



    Proof:




    1. Continuity: take $(x,y) in X times Y$ and a sequence ${ x_n,y_n }_{n ge 1} $ converging to $(x,y)$. Then
      $$ | c_k(x,y)-c_k(x_n,y_n) | le sup_{x' in X, y' in Y} | kd_X(x,x')+kd_Y(y,y') -kd_X(x_n,x')-kd_Y(y_n,y') | le sup_{x' in X, y' in Y} k(d_X(x_n, x)|+d_Y(y_n,y)) to 0$$


    2. Monotonically increasing: fix $(x,y) in X times Y$ and consider any $(x',y') in X times Y$, then
      $$ c(x',y') wedge k + kd_X(x',x)+ kd_Y(y,y') le c(x',y') wedge (k+1) + (k+1)d_X(x',x)+ (k+1)d_Y(y,y') $$
      and then
      $$ c_{k}(x,y) le c_{k+1}(x,y) $$


    3. Convergence: First I observe that if $ (x',y') = (x,y)$ and $c(x,y) = + infty$ I have
      $$c(x',y') wedge k + kd_X(x',x)+ kd_Y(y,y')= c(x,y) wedge k = k$$
      Then $c_k(x,y) ge k$ and then $c_k to + infty$ when $k to + infty$.
      From now on I only consider the case in which $c(x,y) < + infty$. Then if $ (x',y') = (x,y)$, I have
      $$c(x',y') wedge k + kd_X(x',x)+ kd_Y(y,y')= c(x,y) wedge k le c(x,y) $$
      which implies $c_k(x,y) le c(x,y)$. Then it is enough to prove that $c(x,y)$ is the supremum of the sequence $c_k(x,y)$. Fix $0< epsilon < c(x,y)/2$.
      I want to prove that there exists a $k_{epsilon} in mathbb{N}$ s.t.
      $$c_{k_{epsilon}}(x,y) > c(x,y) - 2epsilon $$
      To do that I find $k_{epsilon} in mathbb{N}$ s.t.
      $$ c(x',y') wedge k_{epsilon} + k_{epsilon}d_X(x',x)+ k_{epsilon}d_Y(y,y') > c(x,y) -epsilon $$
      for any $(x',y') in X times Y$.
      Since $c$ is l.s.c. there exists a $delta= delta(epsilon)>0$ s.t.
      $$ (x',y') in B_{delta}((x,y)) Rightarrow c(x',y') > c(x,y) - epsilon$$



    Then if $(x',y') in B_{delta}((x,y))$ I take $k_{epsilon} > c(x,y)- epsilon$ and then
    $$ c(x',y') wedge k_{epsilon} + k_{epsilon}d_X(x',x)+ k_{epsilon}d_Y(y,y') > c(x,y) - epsilon + 0 $$



    If $(x',y') notin B_{delta}((x,y))$ then $d_X(x',x) +d(y',y) > c(epsilon) >0$ and then, taking
    $$k_{epsilon} > frac{c(x,y)-epsilon}{c(epsilon)}$$
    I obtain
    $$ c(x',y') wedge k_{epsilon} + k_{epsilon}d_X(x',x)+ k_{epsilon}d_Y(y,y') > 0 + c(x,y) - epsilon$$
    QED



    I think this proof is too long and involved.
    So



    is this proof correct? Does there exist a shorter proof?






    real-analysis proof-verification convergence







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Let $ (X, d_X)$ and $ (Y, d_Y) $ be metric spaces and let
      $$ c: X times Y to [0, +infty]$$
      a lower semicontinuous function. Then define for each $k in mathbb{N}$ the functions
      $$ c_k (x,y): = inf_{x' in X, y' in Y} { c(x',y') wedge k + kd_X(x,x')+kd_Y(y,y') } $$
      Then $c_k$ are continuous functions, monotonically increasing converging to $c$.



      Proof:




      1. Continuity: take $(x,y) in X times Y$ and a sequence ${ x_n,y_n }_{n ge 1} $ converging to $(x,y)$. Then
        $$ | c_k(x,y)-c_k(x_n,y_n) | le sup_{x' in X, y' in Y} | kd_X(x,x')+kd_Y(y,y') -kd_X(x_n,x')-kd_Y(y_n,y') | le sup_{x' in X, y' in Y} k(d_X(x_n, x)|+d_Y(y_n,y)) to 0$$


      2. Monotonically increasing: fix $(x,y) in X times Y$ and consider any $(x',y') in X times Y$, then
        $$ c(x',y') wedge k + kd_X(x',x)+ kd_Y(y,y') le c(x',y') wedge (k+1) + (k+1)d_X(x',x)+ (k+1)d_Y(y,y') $$
        and then
        $$ c_{k}(x,y) le c_{k+1}(x,y) $$


      3. Convergence: First I observe that if $ (x',y') = (x,y)$ and $c(x,y) = + infty$ I have
        $$c(x',y') wedge k + kd_X(x',x)+ kd_Y(y,y')= c(x,y) wedge k = k$$
        Then $c_k(x,y) ge k$ and then $c_k to + infty$ when $k to + infty$.
        From now on I only consider the case in which $c(x,y) < + infty$. Then if $ (x',y') = (x,y)$, I have
        $$c(x',y') wedge k + kd_X(x',x)+ kd_Y(y,y')= c(x,y) wedge k le c(x,y) $$
        which implies $c_k(x,y) le c(x,y)$. Then it is enough to prove that $c(x,y)$ is the supremum of the sequence $c_k(x,y)$. Fix $0< epsilon < c(x,y)/2$.
        I want to prove that there exists a $k_{epsilon} in mathbb{N}$ s.t.
        $$c_{k_{epsilon}}(x,y) > c(x,y) - 2epsilon $$
        To do that I find $k_{epsilon} in mathbb{N}$ s.t.
        $$ c(x',y') wedge k_{epsilon} + k_{epsilon}d_X(x',x)+ k_{epsilon}d_Y(y,y') > c(x,y) -epsilon $$
        for any $(x',y') in X times Y$.
        Since $c$ is l.s.c. there exists a $delta= delta(epsilon)>0$ s.t.
        $$ (x',y') in B_{delta}((x,y)) Rightarrow c(x',y') > c(x,y) - epsilon$$



      Then if $(x',y') in B_{delta}((x,y))$ I take $k_{epsilon} > c(x,y)- epsilon$ and then
      $$ c(x',y') wedge k_{epsilon} + k_{epsilon}d_X(x',x)+ k_{epsilon}d_Y(y,y') > c(x,y) - epsilon + 0 $$



      If $(x',y') notin B_{delta}((x,y))$ then $d_X(x',x) +d(y',y) > c(epsilon) >0$ and then, taking
      $$k_{epsilon} > frac{c(x,y)-epsilon}{c(epsilon)}$$
      I obtain
      $$ c(x',y') wedge k_{epsilon} + k_{epsilon}d_X(x',x)+ k_{epsilon}d_Y(y,y') > 0 + c(x,y) - epsilon$$
      QED



      I think this proof is too long and involved.
      So



      is this proof correct? Does there exist a shorter proof?






      real-analysis proof-verification convergence







      share|cite|improve this question











      $endgroup$




      Let $ (X, d_X)$ and $ (Y, d_Y) $ be metric spaces and let
      $$ c: X times Y to [0, +infty]$$
      a lower semicontinuous function. Then define for each $k in mathbb{N}$ the functions
      $$ c_k (x,y): = inf_{x' in X, y' in Y} { c(x',y') wedge k + kd_X(x,x')+kd_Y(y,y') } $$
      Then $c_k$ are continuous functions, monotonically increasing converging to $c$.



      Proof:




      1. Continuity: take $(x,y) in X times Y$ and a sequence ${ x_n,y_n }_{n ge 1} $ converging to $(x,y)$. Then
        $$ | c_k(x,y)-c_k(x_n,y_n) | le sup_{x' in X, y' in Y} | kd_X(x,x')+kd_Y(y,y') -kd_X(x_n,x')-kd_Y(y_n,y') | le sup_{x' in X, y' in Y} k(d_X(x_n, x)|+d_Y(y_n,y)) to 0$$


      2. Monotonically increasing: fix $(x,y) in X times Y$ and consider any $(x',y') in X times Y$, then
        $$ c(x',y') wedge k + kd_X(x',x)+ kd_Y(y,y') le c(x',y') wedge (k+1) + (k+1)d_X(x',x)+ (k+1)d_Y(y,y') $$
        and then
        $$ c_{k}(x,y) le c_{k+1}(x,y) $$


      3. Convergence: First I observe that if $ (x',y') = (x,y)$ and $c(x,y) = + infty$ I have
        $$c(x',y') wedge k + kd_X(x',x)+ kd_Y(y,y')= c(x,y) wedge k = k$$
        Then $c_k(x,y) ge k$ and then $c_k to + infty$ when $k to + infty$.
        From now on I only consider the case in which $c(x,y) < + infty$. Then if $ (x',y') = (x,y)$, I have
        $$c(x',y') wedge k + kd_X(x',x)+ kd_Y(y,y')= c(x,y) wedge k le c(x,y) $$
        which implies $c_k(x,y) le c(x,y)$. Then it is enough to prove that $c(x,y)$ is the supremum of the sequence $c_k(x,y)$. Fix $0< epsilon < c(x,y)/2$.
        I want to prove that there exists a $k_{epsilon} in mathbb{N}$ s.t.
        $$c_{k_{epsilon}}(x,y) > c(x,y) - 2epsilon $$
        To do that I find $k_{epsilon} in mathbb{N}$ s.t.
        $$ c(x',y') wedge k_{epsilon} + k_{epsilon}d_X(x',x)+ k_{epsilon}d_Y(y,y') > c(x,y) -epsilon $$
        for any $(x',y') in X times Y$.
        Since $c$ is l.s.c. there exists a $delta= delta(epsilon)>0$ s.t.
        $$ (x',y') in B_{delta}((x,y)) Rightarrow c(x',y') > c(x,y) - epsilon$$



      Then if $(x',y') in B_{delta}((x,y))$ I take $k_{epsilon} > c(x,y)- epsilon$ and then
      $$ c(x',y') wedge k_{epsilon} + k_{epsilon}d_X(x',x)+ k_{epsilon}d_Y(y,y') > c(x,y) - epsilon + 0 $$



      If $(x',y') notin B_{delta}((x,y))$ then $d_X(x',x) +d(y',y) > c(epsilon) >0$ and then, taking
      $$k_{epsilon} > frac{c(x,y)-epsilon}{c(epsilon)}$$
      I obtain
      $$ c(x',y') wedge k_{epsilon} + k_{epsilon}d_X(x',x)+ k_{epsilon}d_Y(y,y') > 0 + c(x,y) - epsilon$$
      QED



      I think this proof is too long and involved.
      So



      is this proof correct? Does there exist a shorter proof?






      real-analysis proof-verification convergence




      real-analysis proof-verification convergence






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 3 '18 at 18:10







      Bremen000

















      asked Dec 1 '18 at 23:44









      Bremen000Bremen000

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