“algebraic openness” of the epigraph of a sublinear function












3












$begingroup$


Let $X$ be a real linear space, $p:X rightarrow mathbb{R}$ a sub-linear function, i.e:
$$forall x,yin X;p(x)+p(y) geq p(x+y)$$
$$forall lambda geq 0,x in X ; p(lambda x)=lambda p(x)$$
Consider the strict upper epigraph:
$$Gamma^+(p)={(x,y) ,vert, x in X,y > f(x)} subseteq Xtimesmathbb{R}$$
One sees that $p$ is convex and thus $Gamma^+(p)$ is convex.
My question is about it's "algebraic openness", formally define that $v in Gamma^+(p)$ is an internal point if for all $u in Xtimes mathbb{R}$ there exists $delta_0 >0$ so that for all $vert delta vert leq delta_0$ we have $v+delta u in Gamma^+(p)$. My Question is whether it is true that every point in $Gamma^+(p)$ is an internal point. For those who are curious this is for an application of the Geometric version of the Hahn Banach Theorem.



Any help is appreciated. Thanks



Note: an affirmative answer is tempting since for $X=mathbb{R^n}$, $p$ is continues and thus it's strict epigraph is (topologically and thus "algebraically") open






functional-analysis convex-analysis







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    Let $X$ be a real linear space, $p:X rightarrow mathbb{R}$ a sub-linear function, i.e:
    $$forall x,yin X;p(x)+p(y) geq p(x+y)$$
    $$forall lambda geq 0,x in X ; p(lambda x)=lambda p(x)$$
    Consider the strict upper epigraph:
    $$Gamma^+(p)={(x,y) ,vert, x in X,y > f(x)} subseteq Xtimesmathbb{R}$$
    One sees that $p$ is convex and thus $Gamma^+(p)$ is convex.
    My question is about it's "algebraic openness", formally define that $v in Gamma^+(p)$ is an internal point if for all $u in Xtimes mathbb{R}$ there exists $delta_0 >0$ so that for all $vert delta vert leq delta_0$ we have $v+delta u in Gamma^+(p)$. My Question is whether it is true that every point in $Gamma^+(p)$ is an internal point. For those who are curious this is for an application of the Geometric version of the Hahn Banach Theorem.



    Any help is appreciated. Thanks



    Note: an affirmative answer is tempting since for $X=mathbb{R^n}$, $p$ is continues and thus it's strict epigraph is (topologically and thus "algebraically") open






    functional-analysis convex-analysis







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      2



      $begingroup$


      Let $X$ be a real linear space, $p:X rightarrow mathbb{R}$ a sub-linear function, i.e:
      $$forall x,yin X;p(x)+p(y) geq p(x+y)$$
      $$forall lambda geq 0,x in X ; p(lambda x)=lambda p(x)$$
      Consider the strict upper epigraph:
      $$Gamma^+(p)={(x,y) ,vert, x in X,y > f(x)} subseteq Xtimesmathbb{R}$$
      One sees that $p$ is convex and thus $Gamma^+(p)$ is convex.
      My question is about it's "algebraic openness", formally define that $v in Gamma^+(p)$ is an internal point if for all $u in Xtimes mathbb{R}$ there exists $delta_0 >0$ so that for all $vert delta vert leq delta_0$ we have $v+delta u in Gamma^+(p)$. My Question is whether it is true that every point in $Gamma^+(p)$ is an internal point. For those who are curious this is for an application of the Geometric version of the Hahn Banach Theorem.



      Any help is appreciated. Thanks



      Note: an affirmative answer is tempting since for $X=mathbb{R^n}$, $p$ is continues and thus it's strict epigraph is (topologically and thus "algebraically") open






      functional-analysis convex-analysis







      share|cite|improve this question











      $endgroup$




      Let $X$ be a real linear space, $p:X rightarrow mathbb{R}$ a sub-linear function, i.e:
      $$forall x,yin X;p(x)+p(y) geq p(x+y)$$
      $$forall lambda geq 0,x in X ; p(lambda x)=lambda p(x)$$
      Consider the strict upper epigraph:
      $$Gamma^+(p)={(x,y) ,vert, x in X,y > f(x)} subseteq Xtimesmathbb{R}$$
      One sees that $p$ is convex and thus $Gamma^+(p)$ is convex.
      My question is about it's "algebraic openness", formally define that $v in Gamma^+(p)$ is an internal point if for all $u in Xtimes mathbb{R}$ there exists $delta_0 >0$ so that for all $vert delta vert leq delta_0$ we have $v+delta u in Gamma^+(p)$. My Question is whether it is true that every point in $Gamma^+(p)$ is an internal point. For those who are curious this is for an application of the Geometric version of the Hahn Banach Theorem.



      Any help is appreciated. Thanks



      Note: an affirmative answer is tempting since for $X=mathbb{R^n}$, $p$ is continues and thus it's strict epigraph is (topologically and thus "algebraically") open






      functional-analysis convex-analysis




      functional-analysis convex-analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 26 '18 at 5:58







      pitariver

















      asked Dec 25 '18 at 11:26









      pitariverpitariver

      469213




      469213






















          1 Answer
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          $begingroup$

          Turns out this questions isn't so hard, I solved it after having another go at it.
          As I mentioned the topology here is a starting pont. But since we are only talking internal points, we can use the topology on one dimension at a time.



          Let $(x,y)in Gamma^+(p),; y>p(x)$. Let $(w,r) in X times mathbb{R}$ be some vector. Note that $p(x+tw)$ is a continues function of $t$ at $0$ since:
          $$p(x+tw) leq p(x) + p(tw) = p(x) + vert t vert p(sign(t)x) underset{t rightarrow 0}{rightarrow} p(x)$$
          Similarly,
          $$p(x+tw) geq p(x) - p(tw) = p(x) - vert t vert p(sign(t)x) underset{t rightarrow 0}{rightarrow} p(x)$$
          and so $p(x+tw) underset{t rightarrow 0}{rightarrow} p(x)$. Since $y > p(x)$ we can
          take $delta_0$ small enough so that for all $vert delta vert < delta_0$ we have $p(x+delta w) < y$. Now let's check the definition of $(x,y)$ being internal: we want:
          $$p(x+ delta w) underset{?}{<} y + delta r$$
          But for $delta_0$ even smaller we can guarantee it, since $p(x+delta w) < y$ and $r$ is constant. $blacksquare$






          share|cite|improve this answer









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            1 Answer
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            active

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            1












            $begingroup$

            Turns out this questions isn't so hard, I solved it after having another go at it.
            As I mentioned the topology here is a starting pont. But since we are only talking internal points, we can use the topology on one dimension at a time.



            Let $(x,y)in Gamma^+(p),; y>p(x)$. Let $(w,r) in X times mathbb{R}$ be some vector. Note that $p(x+tw)$ is a continues function of $t$ at $0$ since:
            $$p(x+tw) leq p(x) + p(tw) = p(x) + vert t vert p(sign(t)x) underset{t rightarrow 0}{rightarrow} p(x)$$
            Similarly,
            $$p(x+tw) geq p(x) - p(tw) = p(x) - vert t vert p(sign(t)x) underset{t rightarrow 0}{rightarrow} p(x)$$
            and so $p(x+tw) underset{t rightarrow 0}{rightarrow} p(x)$. Since $y > p(x)$ we can
            take $delta_0$ small enough so that for all $vert delta vert < delta_0$ we have $p(x+delta w) < y$. Now let's check the definition of $(x,y)$ being internal: we want:
            $$p(x+ delta w) underset{?}{<} y + delta r$$
            But for $delta_0$ even smaller we can guarantee it, since $p(x+delta w) < y$ and $r$ is constant. $blacksquare$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Turns out this questions isn't so hard, I solved it after having another go at it.
              As I mentioned the topology here is a starting pont. But since we are only talking internal points, we can use the topology on one dimension at a time.



              Let $(x,y)in Gamma^+(p),; y>p(x)$. Let $(w,r) in X times mathbb{R}$ be some vector. Note that $p(x+tw)$ is a continues function of $t$ at $0$ since:
              $$p(x+tw) leq p(x) + p(tw) = p(x) + vert t vert p(sign(t)x) underset{t rightarrow 0}{rightarrow} p(x)$$
              Similarly,
              $$p(x+tw) geq p(x) - p(tw) = p(x) - vert t vert p(sign(t)x) underset{t rightarrow 0}{rightarrow} p(x)$$
              and so $p(x+tw) underset{t rightarrow 0}{rightarrow} p(x)$. Since $y > p(x)$ we can
              take $delta_0$ small enough so that for all $vert delta vert < delta_0$ we have $p(x+delta w) < y$. Now let's check the definition of $(x,y)$ being internal: we want:
              $$p(x+ delta w) underset{?}{<} y + delta r$$
              But for $delta_0$ even smaller we can guarantee it, since $p(x+delta w) < y$ and $r$ is constant. $blacksquare$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Turns out this questions isn't so hard, I solved it after having another go at it.
                As I mentioned the topology here is a starting pont. But since we are only talking internal points, we can use the topology on one dimension at a time.



                Let $(x,y)in Gamma^+(p),; y>p(x)$. Let $(w,r) in X times mathbb{R}$ be some vector. Note that $p(x+tw)$ is a continues function of $t$ at $0$ since:
                $$p(x+tw) leq p(x) + p(tw) = p(x) + vert t vert p(sign(t)x) underset{t rightarrow 0}{rightarrow} p(x)$$
                Similarly,
                $$p(x+tw) geq p(x) - p(tw) = p(x) - vert t vert p(sign(t)x) underset{t rightarrow 0}{rightarrow} p(x)$$
                and so $p(x+tw) underset{t rightarrow 0}{rightarrow} p(x)$. Since $y > p(x)$ we can
                take $delta_0$ small enough so that for all $vert delta vert < delta_0$ we have $p(x+delta w) < y$. Now let's check the definition of $(x,y)$ being internal: we want:
                $$p(x+ delta w) underset{?}{<} y + delta r$$
                But for $delta_0$ even smaller we can guarantee it, since $p(x+delta w) < y$ and $r$ is constant. $blacksquare$






                share|cite|improve this answer









                $endgroup$



                Turns out this questions isn't so hard, I solved it after having another go at it.
                As I mentioned the topology here is a starting pont. But since we are only talking internal points, we can use the topology on one dimension at a time.



                Let $(x,y)in Gamma^+(p),; y>p(x)$. Let $(w,r) in X times mathbb{R}$ be some vector. Note that $p(x+tw)$ is a continues function of $t$ at $0$ since:
                $$p(x+tw) leq p(x) + p(tw) = p(x) + vert t vert p(sign(t)x) underset{t rightarrow 0}{rightarrow} p(x)$$
                Similarly,
                $$p(x+tw) geq p(x) - p(tw) = p(x) - vert t vert p(sign(t)x) underset{t rightarrow 0}{rightarrow} p(x)$$
                and so $p(x+tw) underset{t rightarrow 0}{rightarrow} p(x)$. Since $y > p(x)$ we can
                take $delta_0$ small enough so that for all $vert delta vert < delta_0$ we have $p(x+delta w) < y$. Now let's check the definition of $(x,y)$ being internal: we want:
                $$p(x+ delta w) underset{?}{<} y + delta r$$
                But for $delta_0$ even smaller we can guarantee it, since $p(x+delta w) < y$ and $r$ is constant. $blacksquare$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 27 '18 at 6:20









                pitariverpitariver

                469213




                469213






























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