Prove that any linear functional on $mathbb{R^{n}}$ with the Euclidean norm is bounded











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I'm not sure how to prove this without just using the fact that is equivalent to say that a linear functional is continous.










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    Linear functional from what real linear space to $;Bbb R;$ ? From $;Bbb R;$ to itself?:
    – DonAntonio
    Nov 17 at 17:03










  • Sorry I meant on $mathbb{R^{n}}$
    – Roger
    Nov 17 at 17:05










  • Assuming you are referring to $f:Bbb{R}^n longrightarrow Bbb{R}$. By Riesz representation theorem it would be of the form $f(x)=a cdot x$ for some vector $a$.
    – Anurag A
    Nov 17 at 17:06












  • @AnuragA Thanks
    – Roger
    Nov 17 at 17:08










  • @AnuragA I think you may have wanted to mean "for some scalar $;ainBbb R;$ ..."
    – DonAntonio
    Nov 17 at 17:15

















up vote
1
down vote

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1












I'm not sure how to prove this without just using the fact that is equivalent to say that a linear functional is continous.










share|cite|improve this question




















  • 1




    Linear functional from what real linear space to $;Bbb R;$ ? From $;Bbb R;$ to itself?:
    – DonAntonio
    Nov 17 at 17:03










  • Sorry I meant on $mathbb{R^{n}}$
    – Roger
    Nov 17 at 17:05










  • Assuming you are referring to $f:Bbb{R}^n longrightarrow Bbb{R}$. By Riesz representation theorem it would be of the form $f(x)=a cdot x$ for some vector $a$.
    – Anurag A
    Nov 17 at 17:06












  • @AnuragA Thanks
    – Roger
    Nov 17 at 17:08










  • @AnuragA I think you may have wanted to mean "for some scalar $;ainBbb R;$ ..."
    – DonAntonio
    Nov 17 at 17:15















up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





I'm not sure how to prove this without just using the fact that is equivalent to say that a linear functional is continous.










share|cite|improve this question















I'm not sure how to prove this without just using the fact that is equivalent to say that a linear functional is continous.







real-analysis functional-analysis






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edited Nov 17 at 17:05

























asked Nov 17 at 16:58









Roger

545




545








  • 1




    Linear functional from what real linear space to $;Bbb R;$ ? From $;Bbb R;$ to itself?:
    – DonAntonio
    Nov 17 at 17:03










  • Sorry I meant on $mathbb{R^{n}}$
    – Roger
    Nov 17 at 17:05










  • Assuming you are referring to $f:Bbb{R}^n longrightarrow Bbb{R}$. By Riesz representation theorem it would be of the form $f(x)=a cdot x$ for some vector $a$.
    – Anurag A
    Nov 17 at 17:06












  • @AnuragA Thanks
    – Roger
    Nov 17 at 17:08










  • @AnuragA I think you may have wanted to mean "for some scalar $;ainBbb R;$ ..."
    – DonAntonio
    Nov 17 at 17:15
















  • 1




    Linear functional from what real linear space to $;Bbb R;$ ? From $;Bbb R;$ to itself?:
    – DonAntonio
    Nov 17 at 17:03










  • Sorry I meant on $mathbb{R^{n}}$
    – Roger
    Nov 17 at 17:05










  • Assuming you are referring to $f:Bbb{R}^n longrightarrow Bbb{R}$. By Riesz representation theorem it would be of the form $f(x)=a cdot x$ for some vector $a$.
    – Anurag A
    Nov 17 at 17:06












  • @AnuragA Thanks
    – Roger
    Nov 17 at 17:08










  • @AnuragA I think you may have wanted to mean "for some scalar $;ainBbb R;$ ..."
    – DonAntonio
    Nov 17 at 17:15










1




1




Linear functional from what real linear space to $;Bbb R;$ ? From $;Bbb R;$ to itself?:
– DonAntonio
Nov 17 at 17:03




Linear functional from what real linear space to $;Bbb R;$ ? From $;Bbb R;$ to itself?:
– DonAntonio
Nov 17 at 17:03












Sorry I meant on $mathbb{R^{n}}$
– Roger
Nov 17 at 17:05




Sorry I meant on $mathbb{R^{n}}$
– Roger
Nov 17 at 17:05












Assuming you are referring to $f:Bbb{R}^n longrightarrow Bbb{R}$. By Riesz representation theorem it would be of the form $f(x)=a cdot x$ for some vector $a$.
– Anurag A
Nov 17 at 17:06






Assuming you are referring to $f:Bbb{R}^n longrightarrow Bbb{R}$. By Riesz representation theorem it would be of the form $f(x)=a cdot x$ for some vector $a$.
– Anurag A
Nov 17 at 17:06














@AnuragA Thanks
– Roger
Nov 17 at 17:08




@AnuragA Thanks
– Roger
Nov 17 at 17:08












@AnuragA I think you may have wanted to mean "for some scalar $;ainBbb R;$ ..."
– DonAntonio
Nov 17 at 17:15






@AnuragA I think you may have wanted to mean "for some scalar $;ainBbb R;$ ..."
– DonAntonio
Nov 17 at 17:15












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For



$vec v in Bbb R^n, tag 1$



we may write



$vec v = displaystyle sum_1^n v_i vec e_i, tag 2$



where $vec e_i$, $1 le i le n$ is a basis of $Bbb R^n$, orthonormal in the Euclidean inner product $langle cdot, cdot rangle_2$ on $Bbb R^n$, which defines the Euclidean norm norm $Vert cdot Vert_2 = langle cdot, cdot rangle^{1/2}$ on $Bbb R^n$ (see note below); then with



$phi: Bbb R^n to Bbb R tag 3$



linear, we have



$phi(vec v) = phi left (displaystyle sum_1^n v_i vec e_i right ) = displaystyle sum_1^n v_i phi(vec e_i); tag 4$



thus



$vert phi(vec v) vert = left vert displaystyle sum_1^n v_i phi(vec e_i) right vert le displaystyle sum_1^n vert v_i phi(vec e_i) vert = sum_1^n vert v_i vert vert phi(vec e_i) vert; tag 5$



let



$M = max {vert phi(vec e_i) vert, ; 1 le i le n }; tag 6$



then



$vert phi(vec v) vert le displaystyle sum_1^n vert v_i vert vert phi(vec e_i) vert le M sum_1^n vert v_i vert; tag 7$



also,



$vert v_i vert le sqrt { displaystyle sum_1^n v_i^2 }, 1 le i le n; tag 8$



whence



$displaystyle sum_1^n vert v_i vert le n sqrt { displaystyle sum_1^n v_i^2 }, tag 9$



so that (7) becomes



$vert phi(vec v) vert le M n sqrt { displaystyle sum_1^n v_i^2 } = MnVert vec v Vert_2, tag {10}$



which shows that $phi$ is bounded in the Euclidean norm $Vert cdot Vert_2$ on $Bbb R^n$, with bound $Mn$.



Nota Bene: The definition of Euclidean norm $Vert cdot Vert_2$ on $Bbb R^n$ used here is the usual one defined in terms of the Euclidean inner product $langle cdot, cdot rangle_2$, where if



$vec w = displaystyle sum_1^n w_i vec e_i, tag{11}$



we have



$langle vec v, vec w rangle_2 = displaystyle sum_1^n v_i w_i, tag{12}$



from which



$Vert vec v Vert_2 = sqrt{langle vec v, vec v rangle_2} = sqrt {displaystyle sum_1^n v_i^2}; tag{13}$



the vectors $vec e_i$ are chosen orthonormal with respect to this basis:



$langle vec e_i, vec e_j rangle = delta_{ij}, ; 1 le i, j le n. tag{14}$



End of Note.






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    up vote
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    down vote



    accepted










    For



    $vec v in Bbb R^n, tag 1$



    we may write



    $vec v = displaystyle sum_1^n v_i vec e_i, tag 2$



    where $vec e_i$, $1 le i le n$ is a basis of $Bbb R^n$, orthonormal in the Euclidean inner product $langle cdot, cdot rangle_2$ on $Bbb R^n$, which defines the Euclidean norm norm $Vert cdot Vert_2 = langle cdot, cdot rangle^{1/2}$ on $Bbb R^n$ (see note below); then with



    $phi: Bbb R^n to Bbb R tag 3$



    linear, we have



    $phi(vec v) = phi left (displaystyle sum_1^n v_i vec e_i right ) = displaystyle sum_1^n v_i phi(vec e_i); tag 4$



    thus



    $vert phi(vec v) vert = left vert displaystyle sum_1^n v_i phi(vec e_i) right vert le displaystyle sum_1^n vert v_i phi(vec e_i) vert = sum_1^n vert v_i vert vert phi(vec e_i) vert; tag 5$



    let



    $M = max {vert phi(vec e_i) vert, ; 1 le i le n }; tag 6$



    then



    $vert phi(vec v) vert le displaystyle sum_1^n vert v_i vert vert phi(vec e_i) vert le M sum_1^n vert v_i vert; tag 7$



    also,



    $vert v_i vert le sqrt { displaystyle sum_1^n v_i^2 }, 1 le i le n; tag 8$



    whence



    $displaystyle sum_1^n vert v_i vert le n sqrt { displaystyle sum_1^n v_i^2 }, tag 9$



    so that (7) becomes



    $vert phi(vec v) vert le M n sqrt { displaystyle sum_1^n v_i^2 } = MnVert vec v Vert_2, tag {10}$



    which shows that $phi$ is bounded in the Euclidean norm $Vert cdot Vert_2$ on $Bbb R^n$, with bound $Mn$.



    Nota Bene: The definition of Euclidean norm $Vert cdot Vert_2$ on $Bbb R^n$ used here is the usual one defined in terms of the Euclidean inner product $langle cdot, cdot rangle_2$, where if



    $vec w = displaystyle sum_1^n w_i vec e_i, tag{11}$



    we have



    $langle vec v, vec w rangle_2 = displaystyle sum_1^n v_i w_i, tag{12}$



    from which



    $Vert vec v Vert_2 = sqrt{langle vec v, vec v rangle_2} = sqrt {displaystyle sum_1^n v_i^2}; tag{13}$



    the vectors $vec e_i$ are chosen orthonormal with respect to this basis:



    $langle vec e_i, vec e_j rangle = delta_{ij}, ; 1 le i, j le n. tag{14}$



    End of Note.






    share|cite|improve this answer

























      up vote
      1
      down vote



      accepted










      For



      $vec v in Bbb R^n, tag 1$



      we may write



      $vec v = displaystyle sum_1^n v_i vec e_i, tag 2$



      where $vec e_i$, $1 le i le n$ is a basis of $Bbb R^n$, orthonormal in the Euclidean inner product $langle cdot, cdot rangle_2$ on $Bbb R^n$, which defines the Euclidean norm norm $Vert cdot Vert_2 = langle cdot, cdot rangle^{1/2}$ on $Bbb R^n$ (see note below); then with



      $phi: Bbb R^n to Bbb R tag 3$



      linear, we have



      $phi(vec v) = phi left (displaystyle sum_1^n v_i vec e_i right ) = displaystyle sum_1^n v_i phi(vec e_i); tag 4$



      thus



      $vert phi(vec v) vert = left vert displaystyle sum_1^n v_i phi(vec e_i) right vert le displaystyle sum_1^n vert v_i phi(vec e_i) vert = sum_1^n vert v_i vert vert phi(vec e_i) vert; tag 5$



      let



      $M = max {vert phi(vec e_i) vert, ; 1 le i le n }; tag 6$



      then



      $vert phi(vec v) vert le displaystyle sum_1^n vert v_i vert vert phi(vec e_i) vert le M sum_1^n vert v_i vert; tag 7$



      also,



      $vert v_i vert le sqrt { displaystyle sum_1^n v_i^2 }, 1 le i le n; tag 8$



      whence



      $displaystyle sum_1^n vert v_i vert le n sqrt { displaystyle sum_1^n v_i^2 }, tag 9$



      so that (7) becomes



      $vert phi(vec v) vert le M n sqrt { displaystyle sum_1^n v_i^2 } = MnVert vec v Vert_2, tag {10}$



      which shows that $phi$ is bounded in the Euclidean norm $Vert cdot Vert_2$ on $Bbb R^n$, with bound $Mn$.



      Nota Bene: The definition of Euclidean norm $Vert cdot Vert_2$ on $Bbb R^n$ used here is the usual one defined in terms of the Euclidean inner product $langle cdot, cdot rangle_2$, where if



      $vec w = displaystyle sum_1^n w_i vec e_i, tag{11}$



      we have



      $langle vec v, vec w rangle_2 = displaystyle sum_1^n v_i w_i, tag{12}$



      from which



      $Vert vec v Vert_2 = sqrt{langle vec v, vec v rangle_2} = sqrt {displaystyle sum_1^n v_i^2}; tag{13}$



      the vectors $vec e_i$ are chosen orthonormal with respect to this basis:



      $langle vec e_i, vec e_j rangle = delta_{ij}, ; 1 le i, j le n. tag{14}$



      End of Note.






      share|cite|improve this answer























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        For



        $vec v in Bbb R^n, tag 1$



        we may write



        $vec v = displaystyle sum_1^n v_i vec e_i, tag 2$



        where $vec e_i$, $1 le i le n$ is a basis of $Bbb R^n$, orthonormal in the Euclidean inner product $langle cdot, cdot rangle_2$ on $Bbb R^n$, which defines the Euclidean norm norm $Vert cdot Vert_2 = langle cdot, cdot rangle^{1/2}$ on $Bbb R^n$ (see note below); then with



        $phi: Bbb R^n to Bbb R tag 3$



        linear, we have



        $phi(vec v) = phi left (displaystyle sum_1^n v_i vec e_i right ) = displaystyle sum_1^n v_i phi(vec e_i); tag 4$



        thus



        $vert phi(vec v) vert = left vert displaystyle sum_1^n v_i phi(vec e_i) right vert le displaystyle sum_1^n vert v_i phi(vec e_i) vert = sum_1^n vert v_i vert vert phi(vec e_i) vert; tag 5$



        let



        $M = max {vert phi(vec e_i) vert, ; 1 le i le n }; tag 6$



        then



        $vert phi(vec v) vert le displaystyle sum_1^n vert v_i vert vert phi(vec e_i) vert le M sum_1^n vert v_i vert; tag 7$



        also,



        $vert v_i vert le sqrt { displaystyle sum_1^n v_i^2 }, 1 le i le n; tag 8$



        whence



        $displaystyle sum_1^n vert v_i vert le n sqrt { displaystyle sum_1^n v_i^2 }, tag 9$



        so that (7) becomes



        $vert phi(vec v) vert le M n sqrt { displaystyle sum_1^n v_i^2 } = MnVert vec v Vert_2, tag {10}$



        which shows that $phi$ is bounded in the Euclidean norm $Vert cdot Vert_2$ on $Bbb R^n$, with bound $Mn$.



        Nota Bene: The definition of Euclidean norm $Vert cdot Vert_2$ on $Bbb R^n$ used here is the usual one defined in terms of the Euclidean inner product $langle cdot, cdot rangle_2$, where if



        $vec w = displaystyle sum_1^n w_i vec e_i, tag{11}$



        we have



        $langle vec v, vec w rangle_2 = displaystyle sum_1^n v_i w_i, tag{12}$



        from which



        $Vert vec v Vert_2 = sqrt{langle vec v, vec v rangle_2} = sqrt {displaystyle sum_1^n v_i^2}; tag{13}$



        the vectors $vec e_i$ are chosen orthonormal with respect to this basis:



        $langle vec e_i, vec e_j rangle = delta_{ij}, ; 1 le i, j le n. tag{14}$



        End of Note.






        share|cite|improve this answer












        For



        $vec v in Bbb R^n, tag 1$



        we may write



        $vec v = displaystyle sum_1^n v_i vec e_i, tag 2$



        where $vec e_i$, $1 le i le n$ is a basis of $Bbb R^n$, orthonormal in the Euclidean inner product $langle cdot, cdot rangle_2$ on $Bbb R^n$, which defines the Euclidean norm norm $Vert cdot Vert_2 = langle cdot, cdot rangle^{1/2}$ on $Bbb R^n$ (see note below); then with



        $phi: Bbb R^n to Bbb R tag 3$



        linear, we have



        $phi(vec v) = phi left (displaystyle sum_1^n v_i vec e_i right ) = displaystyle sum_1^n v_i phi(vec e_i); tag 4$



        thus



        $vert phi(vec v) vert = left vert displaystyle sum_1^n v_i phi(vec e_i) right vert le displaystyle sum_1^n vert v_i phi(vec e_i) vert = sum_1^n vert v_i vert vert phi(vec e_i) vert; tag 5$



        let



        $M = max {vert phi(vec e_i) vert, ; 1 le i le n }; tag 6$



        then



        $vert phi(vec v) vert le displaystyle sum_1^n vert v_i vert vert phi(vec e_i) vert le M sum_1^n vert v_i vert; tag 7$



        also,



        $vert v_i vert le sqrt { displaystyle sum_1^n v_i^2 }, 1 le i le n; tag 8$



        whence



        $displaystyle sum_1^n vert v_i vert le n sqrt { displaystyle sum_1^n v_i^2 }, tag 9$



        so that (7) becomes



        $vert phi(vec v) vert le M n sqrt { displaystyle sum_1^n v_i^2 } = MnVert vec v Vert_2, tag {10}$



        which shows that $phi$ is bounded in the Euclidean norm $Vert cdot Vert_2$ on $Bbb R^n$, with bound $Mn$.



        Nota Bene: The definition of Euclidean norm $Vert cdot Vert_2$ on $Bbb R^n$ used here is the usual one defined in terms of the Euclidean inner product $langle cdot, cdot rangle_2$, where if



        $vec w = displaystyle sum_1^n w_i vec e_i, tag{11}$



        we have



        $langle vec v, vec w rangle_2 = displaystyle sum_1^n v_i w_i, tag{12}$



        from which



        $Vert vec v Vert_2 = sqrt{langle vec v, vec v rangle_2} = sqrt {displaystyle sum_1^n v_i^2}; tag{13}$



        the vectors $vec e_i$ are chosen orthonormal with respect to this basis:



        $langle vec e_i, vec e_j rangle = delta_{ij}, ; 1 le i, j le n. tag{14}$



        End of Note.







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        answered Nov 17 at 18:27









        Robert Lewis

        42k22760




        42k22760






























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