General question on norm and distance in a normed vector space E and a continuous linear form u.












0












$begingroup$


Let E be a normed vector space and u be a linear, continuous form on E with kernel H.



We show that: $forall x in E setminus H, | u | le frac{|u(x)|}{d(x, H)}.$



$forall lambda in mathbb{R} setminus {0}, forall y in H,$



$frac{|u(lambda x + y)|}{|lambda x + y|} = frac{|lambda||u(x)|}{|lambda||x+frac{1}{lambda}y|} = frac{|u(x)|}{|x+frac{1}{lambda}y|} le frac{|u(x)|}{d(x,H)}.$



My question is on the last inequality (in particular, the denominator), comparing the norm on E and the distance d(x,H):



So, if I am correct:



$d(x,H) le d(x,frac{1}{lambda}y) = |x-frac{1}{lambda}y| le |x+frac{1}{lambda}y|$



So here, to compare the distance and the norm on E, we assume that $|w |:= d(0,w)$, $forall w in E$ ; that is the norm on the normed vector space E is the distance from a point w and the origin 0? (My confusion is E is a general normed vector space). Can someone please elaborate, thanks.






functional-analysis normed-spaces







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Let E be a normed vector space and u be a linear, continuous form on E with kernel H.



    We show that: $forall x in E setminus H, | u | le frac{|u(x)|}{d(x, H)}.$



    $forall lambda in mathbb{R} setminus {0}, forall y in H,$



    $frac{|u(lambda x + y)|}{|lambda x + y|} = frac{|lambda||u(x)|}{|lambda||x+frac{1}{lambda}y|} = frac{|u(x)|}{|x+frac{1}{lambda}y|} le frac{|u(x)|}{d(x,H)}.$



    My question is on the last inequality (in particular, the denominator), comparing the norm on E and the distance d(x,H):



    So, if I am correct:



    $d(x,H) le d(x,frac{1}{lambda}y) = |x-frac{1}{lambda}y| le |x+frac{1}{lambda}y|$



    So here, to compare the distance and the norm on E, we assume that $|w |:= d(0,w)$, $forall w in E$ ; that is the norm on the normed vector space E is the distance from a point w and the origin 0? (My confusion is E is a general normed vector space). Can someone please elaborate, thanks.






    functional-analysis normed-spaces







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let E be a normed vector space and u be a linear, continuous form on E with kernel H.



      We show that: $forall x in E setminus H, | u | le frac{|u(x)|}{d(x, H)}.$



      $forall lambda in mathbb{R} setminus {0}, forall y in H,$



      $frac{|u(lambda x + y)|}{|lambda x + y|} = frac{|lambda||u(x)|}{|lambda||x+frac{1}{lambda}y|} = frac{|u(x)|}{|x+frac{1}{lambda}y|} le frac{|u(x)|}{d(x,H)}.$



      My question is on the last inequality (in particular, the denominator), comparing the norm on E and the distance d(x,H):



      So, if I am correct:



      $d(x,H) le d(x,frac{1}{lambda}y) = |x-frac{1}{lambda}y| le |x+frac{1}{lambda}y|$



      So here, to compare the distance and the norm on E, we assume that $|w |:= d(0,w)$, $forall w in E$ ; that is the norm on the normed vector space E is the distance from a point w and the origin 0? (My confusion is E is a general normed vector space). Can someone please elaborate, thanks.






      functional-analysis normed-spaces







      share|cite|improve this question









      $endgroup$




      Let E be a normed vector space and u be a linear, continuous form on E with kernel H.



      We show that: $forall x in E setminus H, | u | le frac{|u(x)|}{d(x, H)}.$



      $forall lambda in mathbb{R} setminus {0}, forall y in H,$



      $frac{|u(lambda x + y)|}{|lambda x + y|} = frac{|lambda||u(x)|}{|lambda||x+frac{1}{lambda}y|} = frac{|u(x)|}{|x+frac{1}{lambda}y|} le frac{|u(x)|}{d(x,H)}.$



      My question is on the last inequality (in particular, the denominator), comparing the norm on E and the distance d(x,H):



      So, if I am correct:



      $d(x,H) le d(x,frac{1}{lambda}y) = |x-frac{1}{lambda}y| le |x+frac{1}{lambda}y|$



      So here, to compare the distance and the norm on E, we assume that $|w |:= d(0,w)$, $forall w in E$ ; that is the norm on the normed vector space E is the distance from a point w and the origin 0? (My confusion is E is a general normed vector space). Can someone please elaborate, thanks.






      functional-analysis normed-spaces




      functional-analysis normed-spaces






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 28 '18 at 16:06









      metalder9metalder9

      647




      647






















          1 Answer
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          1












          $begingroup$

          You are definitely on the right track, simply adding a minus will solve the problem:
          $$d(x,H)leq d(x,-frac1{lambda}y)=|x+frac1{lambda}y|$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I see, the way you did it is more direct. So, does the distance between two vectors in E equal the norm of the difference between the two vectors? How do we know this when the norm is not specified in the question (E a normed vector space)? I was interested if you could talk about the connection between the distance as shown and the norm on E.
            $endgroup$
            – metalder9
            Dec 28 '18 at 20:25








          • 1




            $begingroup$
            Every norm naturally induces a metric by $d(x,y):=|x-y|$. You can check that the defining properties of a norm imply the defining properties of a metric here. Therefore, whenever we talk about a normed vector space, the induced metric is always assumed, unless specified otherwise.
            $endgroup$
            – SmileyCraft
            Dec 28 '18 at 20:30











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          1 Answer
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          active

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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

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          1












          $begingroup$

          You are definitely on the right track, simply adding a minus will solve the problem:
          $$d(x,H)leq d(x,-frac1{lambda}y)=|x+frac1{lambda}y|$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I see, the way you did it is more direct. So, does the distance between two vectors in E equal the norm of the difference between the two vectors? How do we know this when the norm is not specified in the question (E a normed vector space)? I was interested if you could talk about the connection between the distance as shown and the norm on E.
            $endgroup$
            – metalder9
            Dec 28 '18 at 20:25








          • 1




            $begingroup$
            Every norm naturally induces a metric by $d(x,y):=|x-y|$. You can check that the defining properties of a norm imply the defining properties of a metric here. Therefore, whenever we talk about a normed vector space, the induced metric is always assumed, unless specified otherwise.
            $endgroup$
            – SmileyCraft
            Dec 28 '18 at 20:30
















          1












          $begingroup$

          You are definitely on the right track, simply adding a minus will solve the problem:
          $$d(x,H)leq d(x,-frac1{lambda}y)=|x+frac1{lambda}y|$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I see, the way you did it is more direct. So, does the distance between two vectors in E equal the norm of the difference between the two vectors? How do we know this when the norm is not specified in the question (E a normed vector space)? I was interested if you could talk about the connection between the distance as shown and the norm on E.
            $endgroup$
            – metalder9
            Dec 28 '18 at 20:25








          • 1




            $begingroup$
            Every norm naturally induces a metric by $d(x,y):=|x-y|$. You can check that the defining properties of a norm imply the defining properties of a metric here. Therefore, whenever we talk about a normed vector space, the induced metric is always assumed, unless specified otherwise.
            $endgroup$
            – SmileyCraft
            Dec 28 '18 at 20:30














          1












          1








          1





          $begingroup$

          You are definitely on the right track, simply adding a minus will solve the problem:
          $$d(x,H)leq d(x,-frac1{lambda}y)=|x+frac1{lambda}y|$$






          share|cite|improve this answer









          $endgroup$



          You are definitely on the right track, simply adding a minus will solve the problem:
          $$d(x,H)leq d(x,-frac1{lambda}y)=|x+frac1{lambda}y|$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 28 '18 at 19:37









          SmileyCraftSmileyCraft

          3,749519




          3,749519












          • $begingroup$
            I see, the way you did it is more direct. So, does the distance between two vectors in E equal the norm of the difference between the two vectors? How do we know this when the norm is not specified in the question (E a normed vector space)? I was interested if you could talk about the connection between the distance as shown and the norm on E.
            $endgroup$
            – metalder9
            Dec 28 '18 at 20:25








          • 1




            $begingroup$
            Every norm naturally induces a metric by $d(x,y):=|x-y|$. You can check that the defining properties of a norm imply the defining properties of a metric here. Therefore, whenever we talk about a normed vector space, the induced metric is always assumed, unless specified otherwise.
            $endgroup$
            – SmileyCraft
            Dec 28 '18 at 20:30


















          • $begingroup$
            I see, the way you did it is more direct. So, does the distance between two vectors in E equal the norm of the difference between the two vectors? How do we know this when the norm is not specified in the question (E a normed vector space)? I was interested if you could talk about the connection between the distance as shown and the norm on E.
            $endgroup$
            – metalder9
            Dec 28 '18 at 20:25








          • 1




            $begingroup$
            Every norm naturally induces a metric by $d(x,y):=|x-y|$. You can check that the defining properties of a norm imply the defining properties of a metric here. Therefore, whenever we talk about a normed vector space, the induced metric is always assumed, unless specified otherwise.
            $endgroup$
            – SmileyCraft
            Dec 28 '18 at 20:30
















          $begingroup$
          I see, the way you did it is more direct. So, does the distance between two vectors in E equal the norm of the difference between the two vectors? How do we know this when the norm is not specified in the question (E a normed vector space)? I was interested if you could talk about the connection between the distance as shown and the norm on E.
          $endgroup$
          – metalder9
          Dec 28 '18 at 20:25






          $begingroup$
          I see, the way you did it is more direct. So, does the distance between two vectors in E equal the norm of the difference between the two vectors? How do we know this when the norm is not specified in the question (E a normed vector space)? I was interested if you could talk about the connection between the distance as shown and the norm on E.
          $endgroup$
          – metalder9
          Dec 28 '18 at 20:25






          1




          1




          $begingroup$
          Every norm naturally induces a metric by $d(x,y):=|x-y|$. You can check that the defining properties of a norm imply the defining properties of a metric here. Therefore, whenever we talk about a normed vector space, the induced metric is always assumed, unless specified otherwise.
          $endgroup$
          – SmileyCraft
          Dec 28 '18 at 20:30




          $begingroup$
          Every norm naturally induces a metric by $d(x,y):=|x-y|$. You can check that the defining properties of a norm imply the defining properties of a metric here. Therefore, whenever we talk about a normed vector space, the induced metric is always assumed, unless specified otherwise.
          $endgroup$
          – SmileyCraft
          Dec 28 '18 at 20:30


















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