Norm of a vector with one non-zero element

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In a proof, it is assumed that for $x=(alpha, 0, ..., 0)'$ with $alpha in R_+$ that the norm $||x||leqalpha$



The norm is not further specified.



However, I can define a norm which is , for example $|| cdot||=2*|| cdot||_2$, where the above does not hold.



Is my reasoning correct? And if so, is there an assumption I can make (without specifying the norm) such that $||x||leqalpha$ holds?










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    up vote
    2
    down vote

    favorite












    In a proof, it is assumed that for $x=(alpha, 0, ..., 0)'$ with $alpha in R_+$ that the norm $||x||leqalpha$



    The norm is not further specified.



    However, I can define a norm which is , for example $|| cdot||=2*|| cdot||_2$, where the above does not hold.



    Is my reasoning correct? And if so, is there an assumption I can make (without specifying the norm) such that $||x||leqalpha$ holds?










    share|cite|improve this question

























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      In a proof, it is assumed that for $x=(alpha, 0, ..., 0)'$ with $alpha in R_+$ that the norm $||x||leqalpha$



      The norm is not further specified.



      However, I can define a norm which is , for example $|| cdot||=2*|| cdot||_2$, where the above does not hold.



      Is my reasoning correct? And if so, is there an assumption I can make (without specifying the norm) such that $||x||leqalpha$ holds?










      share|cite|improve this question















      In a proof, it is assumed that for $x=(alpha, 0, ..., 0)'$ with $alpha in R_+$ that the norm $||x||leqalpha$



      The norm is not further specified.



      However, I can define a norm which is , for example $|| cdot||=2*|| cdot||_2$, where the above does not hold.



      Is my reasoning correct? And if so, is there an assumption I can make (without specifying the norm) such that $||x||leqalpha$ holds?







      vectors norm






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      edited Sep 7 at 10:38









      José Carlos Santos

      123k17101186




      123k17101186










      asked Sep 7 at 10:27









      Chris tie

      1303




      1303




















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          Yes, you are right.



          Note that if assume that $|cdot|=|cdot|_p$ for some $pin[1,+infty]$, then that inequality holds.






          share|cite|improve this answer




















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            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            3
            down vote



            accepted










            Yes, you are right.



            Note that if assume that $|cdot|=|cdot|_p$ for some $pin[1,+infty]$, then that inequality holds.






            share|cite|improve this answer
























              up vote
              3
              down vote



              accepted










              Yes, you are right.



              Note that if assume that $|cdot|=|cdot|_p$ for some $pin[1,+infty]$, then that inequality holds.






              share|cite|improve this answer






















                up vote
                3
                down vote



                accepted







                up vote
                3
                down vote



                accepted






                Yes, you are right.



                Note that if assume that $|cdot|=|cdot|_p$ for some $pin[1,+infty]$, then that inequality holds.






                share|cite|improve this answer












                Yes, you are right.



                Note that if assume that $|cdot|=|cdot|_p$ for some $pin[1,+infty]$, then that inequality holds.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Sep 7 at 10:33









                José Carlos Santos

                123k17101186




                123k17101186



























                     

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