Operator norm $ ( ell_2 to ell_1)$












4












$begingroup$


Let $X$ be a finite dimensional normed vector space and $Y$ an arbitrary normed vector space. $ T:X→Y$.



I want to calculate $|T|$ for where $X = K^n$, equipped with the Euclidean norm $|cdot|_2$, $Y := ell_1(mathbb{N})$ and $Tx := (x_1,ldots,x_n,0,0,ldots) in ell_1(mathbb{N})$, for all $x = (x_1,ldots,x_n) in K^n$.



I do not know how to continue
$$ ||T∥_2 = sup limits_{x neq 0} frac{∥Tx∥_1}{∥x∥_2} = sup limits_{x neq 0} frac{∥(x_1,…,x_n,0,0,…)∥_1}{∥(x_1,…,x_n)∥_2} = sup limits_{x neq 0} frac{|x_1|+…+|x_n|}{(|x_1|^2+…+|x_n|^2)^{frac{1}{2}}}= ? $$






functional-analysis operator-theory norm normed-spaces lp-spaces







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$endgroup$












  • $begingroup$
    I don't think that it's a good approach, but you can use the fact that $||T|| = sup_{x: ||x||_{2} = 1} {||Tx||_{1}}$ and apply Lagrange multipliers then. In other words you shall find $|x_{1}| + ldots + |x_{n}| rightarrow text{max}$, while $|x_{1}|^{2} + ldots |x_{n}|^{2} = 1$
    $endgroup$
    – hyperkahler
    Dec 16 '18 at 20:30












  • $begingroup$
    I am tried around a lot. I have never worked with Lagrange multiplier, but $||T∥_2 = sup limits_{∥x∥_2 = 1} ∥Tx ||_1 $.
    $endgroup$
    – Anna Schmitz
    Dec 16 '18 at 20:38


















4












$begingroup$


Let $X$ be a finite dimensional normed vector space and $Y$ an arbitrary normed vector space. $ T:X→Y$.



I want to calculate $|T|$ for where $X = K^n$, equipped with the Euclidean norm $|cdot|_2$, $Y := ell_1(mathbb{N})$ and $Tx := (x_1,ldots,x_n,0,0,ldots) in ell_1(mathbb{N})$, for all $x = (x_1,ldots,x_n) in K^n$.



I do not know how to continue
$$ ||T∥_2 = sup limits_{x neq 0} frac{∥Tx∥_1}{∥x∥_2} = sup limits_{x neq 0} frac{∥(x_1,…,x_n,0,0,…)∥_1}{∥(x_1,…,x_n)∥_2} = sup limits_{x neq 0} frac{|x_1|+…+|x_n|}{(|x_1|^2+…+|x_n|^2)^{frac{1}{2}}}= ? $$






functional-analysis operator-theory norm normed-spaces lp-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    I don't think that it's a good approach, but you can use the fact that $||T|| = sup_{x: ||x||_{2} = 1} {||Tx||_{1}}$ and apply Lagrange multipliers then. In other words you shall find $|x_{1}| + ldots + |x_{n}| rightarrow text{max}$, while $|x_{1}|^{2} + ldots |x_{n}|^{2} = 1$
    $endgroup$
    – hyperkahler
    Dec 16 '18 at 20:30












  • $begingroup$
    I am tried around a lot. I have never worked with Lagrange multiplier, but $||T∥_2 = sup limits_{∥x∥_2 = 1} ∥Tx ||_1 $.
    $endgroup$
    – Anna Schmitz
    Dec 16 '18 at 20:38
















4












4








4





$begingroup$


Let $X$ be a finite dimensional normed vector space and $Y$ an arbitrary normed vector space. $ T:X→Y$.



I want to calculate $|T|$ for where $X = K^n$, equipped with the Euclidean norm $|cdot|_2$, $Y := ell_1(mathbb{N})$ and $Tx := (x_1,ldots,x_n,0,0,ldots) in ell_1(mathbb{N})$, for all $x = (x_1,ldots,x_n) in K^n$.



I do not know how to continue
$$ ||T∥_2 = sup limits_{x neq 0} frac{∥Tx∥_1}{∥x∥_2} = sup limits_{x neq 0} frac{∥(x_1,…,x_n,0,0,…)∥_1}{∥(x_1,…,x_n)∥_2} = sup limits_{x neq 0} frac{|x_1|+…+|x_n|}{(|x_1|^2+…+|x_n|^2)^{frac{1}{2}}}= ? $$






functional-analysis operator-theory norm normed-spaces lp-spaces







share|cite|improve this question











$endgroup$




Let $X$ be a finite dimensional normed vector space and $Y$ an arbitrary normed vector space. $ T:X→Y$.



I want to calculate $|T|$ for where $X = K^n$, equipped with the Euclidean norm $|cdot|_2$, $Y := ell_1(mathbb{N})$ and $Tx := (x_1,ldots,x_n,0,0,ldots) in ell_1(mathbb{N})$, for all $x = (x_1,ldots,x_n) in K^n$.



I do not know how to continue
$$ ||T∥_2 = sup limits_{x neq 0} frac{∥Tx∥_1}{∥x∥_2} = sup limits_{x neq 0} frac{∥(x_1,…,x_n,0,0,…)∥_1}{∥(x_1,…,x_n)∥_2} = sup limits_{x neq 0} frac{|x_1|+…+|x_n|}{(|x_1|^2+…+|x_n|^2)^{frac{1}{2}}}= ? $$






functional-analysis operator-theory norm normed-spaces lp-spaces




functional-analysis operator-theory norm normed-spaces lp-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 15:34







user593746

















asked Dec 16 '18 at 20:16









Anna SchmitzAnna Schmitz

917




917












  • $begingroup$
    I don't think that it's a good approach, but you can use the fact that $||T|| = sup_{x: ||x||_{2} = 1} {||Tx||_{1}}$ and apply Lagrange multipliers then. In other words you shall find $|x_{1}| + ldots + |x_{n}| rightarrow text{max}$, while $|x_{1}|^{2} + ldots |x_{n}|^{2} = 1$
    $endgroup$
    – hyperkahler
    Dec 16 '18 at 20:30












  • $begingroup$
    I am tried around a lot. I have never worked with Lagrange multiplier, but $||T∥_2 = sup limits_{∥x∥_2 = 1} ∥Tx ||_1 $.
    $endgroup$
    – Anna Schmitz
    Dec 16 '18 at 20:38




















  • $begingroup$
    I don't think that it's a good approach, but you can use the fact that $||T|| = sup_{x: ||x||_{2} = 1} {||Tx||_{1}}$ and apply Lagrange multipliers then. In other words you shall find $|x_{1}| + ldots + |x_{n}| rightarrow text{max}$, while $|x_{1}|^{2} + ldots |x_{n}|^{2} = 1$
    $endgroup$
    – hyperkahler
    Dec 16 '18 at 20:30












  • $begingroup$
    I am tried around a lot. I have never worked with Lagrange multiplier, but $||T∥_2 = sup limits_{∥x∥_2 = 1} ∥Tx ||_1 $.
    $endgroup$
    – Anna Schmitz
    Dec 16 '18 at 20:38


















$begingroup$
I don't think that it's a good approach, but you can use the fact that $||T|| = sup_{x: ||x||_{2} = 1} {||Tx||_{1}}$ and apply Lagrange multipliers then. In other words you shall find $|x_{1}| + ldots + |x_{n}| rightarrow text{max}$, while $|x_{1}|^{2} + ldots |x_{n}|^{2} = 1$
$endgroup$
– hyperkahler
Dec 16 '18 at 20:30






$begingroup$
I don't think that it's a good approach, but you can use the fact that $||T|| = sup_{x: ||x||_{2} = 1} {||Tx||_{1}}$ and apply Lagrange multipliers then. In other words you shall find $|x_{1}| + ldots + |x_{n}| rightarrow text{max}$, while $|x_{1}|^{2} + ldots |x_{n}|^{2} = 1$
$endgroup$
– hyperkahler
Dec 16 '18 at 20:30














$begingroup$
I am tried around a lot. I have never worked with Lagrange multiplier, but $||T∥_2 = sup limits_{∥x∥_2 = 1} ∥Tx ||_1 $.
$endgroup$
– Anna Schmitz
Dec 16 '18 at 20:38






$begingroup$
I am tried around a lot. I have never worked with Lagrange multiplier, but $||T∥_2 = sup limits_{∥x∥_2 = 1} ∥Tx ||_1 $.
$endgroup$
– Anna Schmitz
Dec 16 '18 at 20:38












2 Answers
2






active

oldest

votes


















1












$begingroup$

I will elaborate on my comment above



Given an operator $T: X rightarrow Y$ where $$X = mathbb{K}^{n}$$
$$Y = l^{1}(mathbb{N})$$ its norm is given by
$$||T||_{op} = sup_{x neq 0}{frac{||Tx||_{1}}{||x||_{2}}} = sup_{||x||_{2} leq 1}{||Tx||_{1}} = sup_{||x||_{2} = 1}{||Tx||_{1}}$$



So we have to maximize $$||Tx||_{1} = |x_{1}| + ldots + |x_{n}|$$ given $$||x||_{2} := (x_{1}^{2} + ldots + x_{n}^{2})^{frac{1}{2}} = 1$$



Let $t_{i} := |x_{i}|$, then our problem reformulates as follows:
$$t_{1} + t_{2} + ldots + t_{n} rightarrow text{max}$$
$$t_{1}^{2} + t_{2}^{2} + ldots + t_{n}^{2} = 1$$
$$t_{i} geq 0, forall i = 1, ldots, n$$



The Lagrangian for the problem is:
$$L = (t_{1} + ldots + t_{n}) - lambda (t_{1}^{2} + ldots + t_{n}^{2} - 1)$$
and $lambda$ stays for the lagrange multiplier.



The necessary extremum condition implies:
$$frac{partial L}{partial t_{i}} := 1 - lambda(2t_{i}) = 0$$
thus $t_{i} = frac{1}{2 lambda}$. Since $t_{i} geq 0$ it follows that $lambda > 0$.



The stationary point is $x = (frac{1}{2 lambda}, ldots, frac{1}{2 lambda})$ and we shall find the lambda such that the solution satisfy the second condition in the problem above.



$||x||_{2} = 1$ is equivalent to
$$ n cdot frac{1}{4 lambda^{2}} = 1$$ from which we get
$lambda = frac{sqrt{n}}{2}$
(note that we shall choose it positive due to the restrictions mentioned above)



Thus $t_{i} = frac{1}{sqrt{n}}$ and the maximum equals $$text{max} = n cdot frac{1}{sqrt{n}} = sqrt{n}$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Hi, how does $n cdot 1/n = sqrt{n}$ ? Also, nice alternative thinking but that's a bit off the road of functinal analysis.
    $endgroup$
    – Rebellos
    Dec 16 '18 at 21:16










  • $begingroup$
    @Rebellos i've made a mistake, please, check the answer for the updates
    $endgroup$
    – hyperkahler
    Dec 16 '18 at 21:25



















1












$begingroup$

Using Cauchy-Schwarz inequality, we have
$$
sum_{i=1}^nleftlvert x_irightrvert=sum_{i=1}^nleftlvert x_irightrvertcdot 1leqslant left(sum_{i=1}^nx_i^2right)^{1/2}left(sum_{i=1}^n1right)^{1/2}=sqrt nleft(sum_{i=1}^nx_i^2right)^{1/2}
$$
hence $leftlVert TrightrVertleqslant sqrt n$.
For the opposite inequality, look at the case where $x_i=1$ for all $iin{1,dots,n}$.






share|cite|improve this answer











$endgroup$













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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    I will elaborate on my comment above



    Given an operator $T: X rightarrow Y$ where $$X = mathbb{K}^{n}$$
    $$Y = l^{1}(mathbb{N})$$ its norm is given by
    $$||T||_{op} = sup_{x neq 0}{frac{||Tx||_{1}}{||x||_{2}}} = sup_{||x||_{2} leq 1}{||Tx||_{1}} = sup_{||x||_{2} = 1}{||Tx||_{1}}$$



    So we have to maximize $$||Tx||_{1} = |x_{1}| + ldots + |x_{n}|$$ given $$||x||_{2} := (x_{1}^{2} + ldots + x_{n}^{2})^{frac{1}{2}} = 1$$



    Let $t_{i} := |x_{i}|$, then our problem reformulates as follows:
    $$t_{1} + t_{2} + ldots + t_{n} rightarrow text{max}$$
    $$t_{1}^{2} + t_{2}^{2} + ldots + t_{n}^{2} = 1$$
    $$t_{i} geq 0, forall i = 1, ldots, n$$



    The Lagrangian for the problem is:
    $$L = (t_{1} + ldots + t_{n}) - lambda (t_{1}^{2} + ldots + t_{n}^{2} - 1)$$
    and $lambda$ stays for the lagrange multiplier.



    The necessary extremum condition implies:
    $$frac{partial L}{partial t_{i}} := 1 - lambda(2t_{i}) = 0$$
    thus $t_{i} = frac{1}{2 lambda}$. Since $t_{i} geq 0$ it follows that $lambda > 0$.



    The stationary point is $x = (frac{1}{2 lambda}, ldots, frac{1}{2 lambda})$ and we shall find the lambda such that the solution satisfy the second condition in the problem above.



    $||x||_{2} = 1$ is equivalent to
    $$ n cdot frac{1}{4 lambda^{2}} = 1$$ from which we get
    $lambda = frac{sqrt{n}}{2}$
    (note that we shall choose it positive due to the restrictions mentioned above)



    Thus $t_{i} = frac{1}{sqrt{n}}$ and the maximum equals $$text{max} = n cdot frac{1}{sqrt{n}} = sqrt{n}$$






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Hi, how does $n cdot 1/n = sqrt{n}$ ? Also, nice alternative thinking but that's a bit off the road of functinal analysis.
      $endgroup$
      – Rebellos
      Dec 16 '18 at 21:16










    • $begingroup$
      @Rebellos i've made a mistake, please, check the answer for the updates
      $endgroup$
      – hyperkahler
      Dec 16 '18 at 21:25
















    1












    $begingroup$

    I will elaborate on my comment above



    Given an operator $T: X rightarrow Y$ where $$X = mathbb{K}^{n}$$
    $$Y = l^{1}(mathbb{N})$$ its norm is given by
    $$||T||_{op} = sup_{x neq 0}{frac{||Tx||_{1}}{||x||_{2}}} = sup_{||x||_{2} leq 1}{||Tx||_{1}} = sup_{||x||_{2} = 1}{||Tx||_{1}}$$



    So we have to maximize $$||Tx||_{1} = |x_{1}| + ldots + |x_{n}|$$ given $$||x||_{2} := (x_{1}^{2} + ldots + x_{n}^{2})^{frac{1}{2}} = 1$$



    Let $t_{i} := |x_{i}|$, then our problem reformulates as follows:
    $$t_{1} + t_{2} + ldots + t_{n} rightarrow text{max}$$
    $$t_{1}^{2} + t_{2}^{2} + ldots + t_{n}^{2} = 1$$
    $$t_{i} geq 0, forall i = 1, ldots, n$$



    The Lagrangian for the problem is:
    $$L = (t_{1} + ldots + t_{n}) - lambda (t_{1}^{2} + ldots + t_{n}^{2} - 1)$$
    and $lambda$ stays for the lagrange multiplier.



    The necessary extremum condition implies:
    $$frac{partial L}{partial t_{i}} := 1 - lambda(2t_{i}) = 0$$
    thus $t_{i} = frac{1}{2 lambda}$. Since $t_{i} geq 0$ it follows that $lambda > 0$.



    The stationary point is $x = (frac{1}{2 lambda}, ldots, frac{1}{2 lambda})$ and we shall find the lambda such that the solution satisfy the second condition in the problem above.



    $||x||_{2} = 1$ is equivalent to
    $$ n cdot frac{1}{4 lambda^{2}} = 1$$ from which we get
    $lambda = frac{sqrt{n}}{2}$
    (note that we shall choose it positive due to the restrictions mentioned above)



    Thus $t_{i} = frac{1}{sqrt{n}}$ and the maximum equals $$text{max} = n cdot frac{1}{sqrt{n}} = sqrt{n}$$






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Hi, how does $n cdot 1/n = sqrt{n}$ ? Also, nice alternative thinking but that's a bit off the road of functinal analysis.
      $endgroup$
      – Rebellos
      Dec 16 '18 at 21:16










    • $begingroup$
      @Rebellos i've made a mistake, please, check the answer for the updates
      $endgroup$
      – hyperkahler
      Dec 16 '18 at 21:25














    1












    1








    1





    $begingroup$

    I will elaborate on my comment above



    Given an operator $T: X rightarrow Y$ where $$X = mathbb{K}^{n}$$
    $$Y = l^{1}(mathbb{N})$$ its norm is given by
    $$||T||_{op} = sup_{x neq 0}{frac{||Tx||_{1}}{||x||_{2}}} = sup_{||x||_{2} leq 1}{||Tx||_{1}} = sup_{||x||_{2} = 1}{||Tx||_{1}}$$



    So we have to maximize $$||Tx||_{1} = |x_{1}| + ldots + |x_{n}|$$ given $$||x||_{2} := (x_{1}^{2} + ldots + x_{n}^{2})^{frac{1}{2}} = 1$$



    Let $t_{i} := |x_{i}|$, then our problem reformulates as follows:
    $$t_{1} + t_{2} + ldots + t_{n} rightarrow text{max}$$
    $$t_{1}^{2} + t_{2}^{2} + ldots + t_{n}^{2} = 1$$
    $$t_{i} geq 0, forall i = 1, ldots, n$$



    The Lagrangian for the problem is:
    $$L = (t_{1} + ldots + t_{n}) - lambda (t_{1}^{2} + ldots + t_{n}^{2} - 1)$$
    and $lambda$ stays for the lagrange multiplier.



    The necessary extremum condition implies:
    $$frac{partial L}{partial t_{i}} := 1 - lambda(2t_{i}) = 0$$
    thus $t_{i} = frac{1}{2 lambda}$. Since $t_{i} geq 0$ it follows that $lambda > 0$.



    The stationary point is $x = (frac{1}{2 lambda}, ldots, frac{1}{2 lambda})$ and we shall find the lambda such that the solution satisfy the second condition in the problem above.



    $||x||_{2} = 1$ is equivalent to
    $$ n cdot frac{1}{4 lambda^{2}} = 1$$ from which we get
    $lambda = frac{sqrt{n}}{2}$
    (note that we shall choose it positive due to the restrictions mentioned above)



    Thus $t_{i} = frac{1}{sqrt{n}}$ and the maximum equals $$text{max} = n cdot frac{1}{sqrt{n}} = sqrt{n}$$






    share|cite|improve this answer











    $endgroup$



    I will elaborate on my comment above



    Given an operator $T: X rightarrow Y$ where $$X = mathbb{K}^{n}$$
    $$Y = l^{1}(mathbb{N})$$ its norm is given by
    $$||T||_{op} = sup_{x neq 0}{frac{||Tx||_{1}}{||x||_{2}}} = sup_{||x||_{2} leq 1}{||Tx||_{1}} = sup_{||x||_{2} = 1}{||Tx||_{1}}$$



    So we have to maximize $$||Tx||_{1} = |x_{1}| + ldots + |x_{n}|$$ given $$||x||_{2} := (x_{1}^{2} + ldots + x_{n}^{2})^{frac{1}{2}} = 1$$



    Let $t_{i} := |x_{i}|$, then our problem reformulates as follows:
    $$t_{1} + t_{2} + ldots + t_{n} rightarrow text{max}$$
    $$t_{1}^{2} + t_{2}^{2} + ldots + t_{n}^{2} = 1$$
    $$t_{i} geq 0, forall i = 1, ldots, n$$



    The Lagrangian for the problem is:
    $$L = (t_{1} + ldots + t_{n}) - lambda (t_{1}^{2} + ldots + t_{n}^{2} - 1)$$
    and $lambda$ stays for the lagrange multiplier.



    The necessary extremum condition implies:
    $$frac{partial L}{partial t_{i}} := 1 - lambda(2t_{i}) = 0$$
    thus $t_{i} = frac{1}{2 lambda}$. Since $t_{i} geq 0$ it follows that $lambda > 0$.



    The stationary point is $x = (frac{1}{2 lambda}, ldots, frac{1}{2 lambda})$ and we shall find the lambda such that the solution satisfy the second condition in the problem above.



    $||x||_{2} = 1$ is equivalent to
    $$ n cdot frac{1}{4 lambda^{2}} = 1$$ from which we get
    $lambda = frac{sqrt{n}}{2}$
    (note that we shall choose it positive due to the restrictions mentioned above)



    Thus $t_{i} = frac{1}{sqrt{n}}$ and the maximum equals $$text{max} = n cdot frac{1}{sqrt{n}} = sqrt{n}$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 16 '18 at 21:24

























    answered Dec 16 '18 at 20:52









    hyperkahlerhyperkahler

    1,483714




    1,483714








    • 1




      $begingroup$
      Hi, how does $n cdot 1/n = sqrt{n}$ ? Also, nice alternative thinking but that's a bit off the road of functinal analysis.
      $endgroup$
      – Rebellos
      Dec 16 '18 at 21:16










    • $begingroup$
      @Rebellos i've made a mistake, please, check the answer for the updates
      $endgroup$
      – hyperkahler
      Dec 16 '18 at 21:25














    • 1




      $begingroup$
      Hi, how does $n cdot 1/n = sqrt{n}$ ? Also, nice alternative thinking but that's a bit off the road of functinal analysis.
      $endgroup$
      – Rebellos
      Dec 16 '18 at 21:16










    • $begingroup$
      @Rebellos i've made a mistake, please, check the answer for the updates
      $endgroup$
      – hyperkahler
      Dec 16 '18 at 21:25








    1




    1




    $begingroup$
    Hi, how does $n cdot 1/n = sqrt{n}$ ? Also, nice alternative thinking but that's a bit off the road of functinal analysis.
    $endgroup$
    – Rebellos
    Dec 16 '18 at 21:16




    $begingroup$
    Hi, how does $n cdot 1/n = sqrt{n}$ ? Also, nice alternative thinking but that's a bit off the road of functinal analysis.
    $endgroup$
    – Rebellos
    Dec 16 '18 at 21:16












    $begingroup$
    @Rebellos i've made a mistake, please, check the answer for the updates
    $endgroup$
    – hyperkahler
    Dec 16 '18 at 21:25




    $begingroup$
    @Rebellos i've made a mistake, please, check the answer for the updates
    $endgroup$
    – hyperkahler
    Dec 16 '18 at 21:25











    1












    $begingroup$

    Using Cauchy-Schwarz inequality, we have
    $$
    sum_{i=1}^nleftlvert x_irightrvert=sum_{i=1}^nleftlvert x_irightrvertcdot 1leqslant left(sum_{i=1}^nx_i^2right)^{1/2}left(sum_{i=1}^n1right)^{1/2}=sqrt nleft(sum_{i=1}^nx_i^2right)^{1/2}
    $$
    hence $leftlVert TrightrVertleqslant sqrt n$.
    For the opposite inequality, look at the case where $x_i=1$ for all $iin{1,dots,n}$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Using Cauchy-Schwarz inequality, we have
      $$
      sum_{i=1}^nleftlvert x_irightrvert=sum_{i=1}^nleftlvert x_irightrvertcdot 1leqslant left(sum_{i=1}^nx_i^2right)^{1/2}left(sum_{i=1}^n1right)^{1/2}=sqrt nleft(sum_{i=1}^nx_i^2right)^{1/2}
      $$
      hence $leftlVert TrightrVertleqslant sqrt n$.
      For the opposite inequality, look at the case where $x_i=1$ for all $iin{1,dots,n}$.






      share|cite|improve this answer











      $endgroup$
















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        $begingroup$

        Using Cauchy-Schwarz inequality, we have
        $$
        sum_{i=1}^nleftlvert x_irightrvert=sum_{i=1}^nleftlvert x_irightrvertcdot 1leqslant left(sum_{i=1}^nx_i^2right)^{1/2}left(sum_{i=1}^n1right)^{1/2}=sqrt nleft(sum_{i=1}^nx_i^2right)^{1/2}
        $$
        hence $leftlVert TrightrVertleqslant sqrt n$.
        For the opposite inequality, look at the case where $x_i=1$ for all $iin{1,dots,n}$.






        share|cite|improve this answer











        $endgroup$



        Using Cauchy-Schwarz inequality, we have
        $$
        sum_{i=1}^nleftlvert x_irightrvert=sum_{i=1}^nleftlvert x_irightrvertcdot 1leqslant left(sum_{i=1}^nx_i^2right)^{1/2}left(sum_{i=1}^n1right)^{1/2}=sqrt nleft(sum_{i=1}^nx_i^2right)^{1/2}
        $$
        hence $leftlVert TrightrVertleqslant sqrt n$.
        For the opposite inequality, look at the case where $x_i=1$ for all $iin{1,dots,n}$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 18 '18 at 15:41

























        answered Dec 18 '18 at 10:56









        Davide GiraudoDavide Giraudo

        127k16151264




        127k16151264






























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