$A$ is convex $iff$ for all $s,t gt 0$ , $(s+t)A = sA + tA$












0














Let $X$ be a TVS ,and $A subset X$.



Then $A$ is convex $iff$ for all $s,t gt 0$ , $(s+t)A = sA + tA$.



If for all $s,tgt 0$ we have $(s+t)A = sA+tA$ then of course $A$ is convex.



I'm not sure about the other direction. It is always true that $(s+t)A subset sA+tA$ ,so assuming $A$ is convex I need to show the other inclusion.



Any ideas?



Thanks for helping!










share|cite|improve this question



























    0














    Let $X$ be a TVS ,and $A subset X$.



    Then $A$ is convex $iff$ for all $s,t gt 0$ , $(s+t)A = sA + tA$.



    If for all $s,tgt 0$ we have $(s+t)A = sA+tA$ then of course $A$ is convex.



    I'm not sure about the other direction. It is always true that $(s+t)A subset sA+tA$ ,so assuming $A$ is convex I need to show the other inclusion.



    Any ideas?



    Thanks for helping!










    share|cite|improve this question

























      0












      0








      0







      Let $X$ be a TVS ,and $A subset X$.



      Then $A$ is convex $iff$ for all $s,t gt 0$ , $(s+t)A = sA + tA$.



      If for all $s,tgt 0$ we have $(s+t)A = sA+tA$ then of course $A$ is convex.



      I'm not sure about the other direction. It is always true that $(s+t)A subset sA+tA$ ,so assuming $A$ is convex I need to show the other inclusion.



      Any ideas?



      Thanks for helping!










      share|cite|improve this question













      Let $X$ be a TVS ,and $A subset X$.



      Then $A$ is convex $iff$ for all $s,t gt 0$ , $(s+t)A = sA + tA$.



      If for all $s,tgt 0$ we have $(s+t)A = sA+tA$ then of course $A$ is convex.



      I'm not sure about the other direction. It is always true that $(s+t)A subset sA+tA$ ,so assuming $A$ is convex I need to show the other inclusion.



      Any ideas?



      Thanks for helping!







      functional-analysis






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      share|cite|improve this question










      asked Jan 4 at 21:46









      user123user123

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          Let $sx+tyin sA+tA$. Then $$(s+t)left(frac{s}{s+t}x+frac{t}{s+t}yright)in(s+t)A$$






          share|cite|improve this answer





















          • So easy. Thanks!
            – user123
            Jan 4 at 21:56











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          1 Answer
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          active

          oldest

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          0














          Let $sx+tyin sA+tA$. Then $$(s+t)left(frac{s}{s+t}x+frac{t}{s+t}yright)in(s+t)A$$






          share|cite|improve this answer





















          • So easy. Thanks!
            – user123
            Jan 4 at 21:56
















          0














          Let $sx+tyin sA+tA$. Then $$(s+t)left(frac{s}{s+t}x+frac{t}{s+t}yright)in(s+t)A$$






          share|cite|improve this answer





















          • So easy. Thanks!
            – user123
            Jan 4 at 21:56














          0












          0








          0






          Let $sx+tyin sA+tA$. Then $$(s+t)left(frac{s}{s+t}x+frac{t}{s+t}yright)in(s+t)A$$






          share|cite|improve this answer












          Let $sx+tyin sA+tA$. Then $$(s+t)left(frac{s}{s+t}x+frac{t}{s+t}yright)in(s+t)A$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 4 at 21:53









          SmileyCraftSmileyCraft

          3,321516




          3,321516












          • So easy. Thanks!
            – user123
            Jan 4 at 21:56


















          • So easy. Thanks!
            – user123
            Jan 4 at 21:56
















          So easy. Thanks!
          – user123
          Jan 4 at 21:56




          So easy. Thanks!
          – user123
          Jan 4 at 21:56


















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