“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
          1






          active

          oldest

          votes


















          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









          $endgroup$














            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3052023%2falgebraic-openness-of-the-epigraph-of-a-sublinear-function%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            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






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3052023%2falgebraic-openness-of-the-epigraph-of-a-sublinear-function%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Bundesstraße 106

                    Image
                    Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
                    Read more

                    Le Mesnil-Réaume

                    Image
                    Le Mesnil-Réaume Region Normandie Département Seine-Maritime Arrondissement Dieppe Kanton Eu Gemeindeverband Villes Sœurs Koordinaten 49° 58′  N , 1° 27′  O 49.9725 1.4463888888889 Koordinaten: 49° 58′  N , 1° 27′  O Höhe 92–156 m Fläche 5,55 km 2 Einwohner 793 (1. Januar 2016) Bevölkerungsdichte 143 Einw./km 2 Postleitzahl 76260 INSEE-Code 76435 Website http://www.mesnil-reaume.net/ Rathaus (Mairie) und Schule von Le Mesnil-Réaume Le Mesnil-Réaume ist eine französische Gemeinde mit 793 Einwohnern (Stand: 1. Januar 2016) im Département Seine-Maritime in der Region Normandie (vor 2016 Haute-Normandie). Sie gehört zum Arrondissement Dieppe und zum Kanton Eu. Die Einwohner werden Mesnillais genannt. Inhaltsverzeichnis 1 Geographie 2 Bevölkerungsentwicklung 3 Sehenswürdigkeiten 4 Weblinks Geographie | Le Mesnil-Réaume liegt etwa 43 Kilomet
                    Read more

                    Ida-Boy-Ed-Garten

                    Image
                    Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
                    Read more