What is the operator norm of $Tf(x) = x^2f(x)$?












3












$begingroup$


Let $H = L^2([0,1],mathbb{R})$ and $T : H to H,, Tf(x) = x^2f(x) $.



$T$ is linear.



$$|Tf|_{L^2([0,1],mathbb{R})} = sqrt{int_0^1x^4f^2(x)dx} leqsqrt{int_0^1f^2(x)dx} = |f|_{L^2([0,1],mathbb{R})} $$



$T$ is linear and bounded therefore it's continuous. Also
$|T|| leq 1$.



I tried finding solution to $|Tf|_{L^2([0,1],mathbb{R})} = |f|_{L^2([0,1],mathbb{R})} $



I found $$f(x) = sqrt{frac{2x-1}{x^4-1}}$$



but it's not in $L^2$ so it doesn't work.



anyone knows an $f$ to reach $1$, I'm not even sure it's $1$.



any help will be greatly appreciated !






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







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hint: take $f=1_{[a,1]}$ where $a < 1$ is a positive real number close to $1$.
    $endgroup$
    – Mindlack
    Jan 7 at 18:16










  • $begingroup$
    The answers below are fine, but one may add, that the approach by the OP can't work since there is no $L^2$ function for which the norm is attained.
    $endgroup$
    – Dirk
    Jan 7 at 18:31
















3












$begingroup$


Let $H = L^2([0,1],mathbb{R})$ and $T : H to H,, Tf(x) = x^2f(x) $.



$T$ is linear.



$$|Tf|_{L^2([0,1],mathbb{R})} = sqrt{int_0^1x^4f^2(x)dx} leqsqrt{int_0^1f^2(x)dx} = |f|_{L^2([0,1],mathbb{R})} $$



$T$ is linear and bounded therefore it's continuous. Also
$|T|| leq 1$.



I tried finding solution to $|Tf|_{L^2([0,1],mathbb{R})} = |f|_{L^2([0,1],mathbb{R})} $



I found $$f(x) = sqrt{frac{2x-1}{x^4-1}}$$



but it's not in $L^2$ so it doesn't work.



anyone knows an $f$ to reach $1$, I'm not even sure it's $1$.



any help will be greatly appreciated !






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







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hint: take $f=1_{[a,1]}$ where $a < 1$ is a positive real number close to $1$.
    $endgroup$
    – Mindlack
    Jan 7 at 18:16










  • $begingroup$
    The answers below are fine, but one may add, that the approach by the OP can't work since there is no $L^2$ function for which the norm is attained.
    $endgroup$
    – Dirk
    Jan 7 at 18:31














3












3








3





$begingroup$


Let $H = L^2([0,1],mathbb{R})$ and $T : H to H,, Tf(x) = x^2f(x) $.



$T$ is linear.



$$|Tf|_{L^2([0,1],mathbb{R})} = sqrt{int_0^1x^4f^2(x)dx} leqsqrt{int_0^1f^2(x)dx} = |f|_{L^2([0,1],mathbb{R})} $$



$T$ is linear and bounded therefore it's continuous. Also
$|T|| leq 1$.



I tried finding solution to $|Tf|_{L^2([0,1],mathbb{R})} = |f|_{L^2([0,1],mathbb{R})} $



I found $$f(x) = sqrt{frac{2x-1}{x^4-1}}$$



but it's not in $L^2$ so it doesn't work.



anyone knows an $f$ to reach $1$, I'm not even sure it's $1$.



any help will be greatly appreciated !






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







share|cite|improve this question











$endgroup$




Let $H = L^2([0,1],mathbb{R})$ and $T : H to H,, Tf(x) = x^2f(x) $.



$T$ is linear.



$$|Tf|_{L^2([0,1],mathbb{R})} = sqrt{int_0^1x^4f^2(x)dx} leqsqrt{int_0^1f^2(x)dx} = |f|_{L^2([0,1],mathbb{R})} $$



$T$ is linear and bounded therefore it's continuous. Also
$|T|| leq 1$.



I tried finding solution to $|Tf|_{L^2([0,1],mathbb{R})} = |f|_{L^2([0,1],mathbb{R})} $



I found $$f(x) = sqrt{frac{2x-1}{x^4-1}}$$



but it's not in $L^2$ so it doesn't work.



anyone knows an $f$ to reach $1$, I'm not even sure it's $1$.



any help will be greatly appreciated !






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




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 9:30









Davide Giraudo

127k16151263




127k16151263










asked Jan 7 at 18:09









rapidracimrapidracim

1,7191419




1,7191419








  • 1




    $begingroup$
    Hint: take $f=1_{[a,1]}$ where $a < 1$ is a positive real number close to $1$.
    $endgroup$
    – Mindlack
    Jan 7 at 18:16










  • $begingroup$
    The answers below are fine, but one may add, that the approach by the OP can't work since there is no $L^2$ function for which the norm is attained.
    $endgroup$
    – Dirk
    Jan 7 at 18:31














  • 1




    $begingroup$
    Hint: take $f=1_{[a,1]}$ where $a < 1$ is a positive real number close to $1$.
    $endgroup$
    – Mindlack
    Jan 7 at 18:16










  • $begingroup$
    The answers below are fine, but one may add, that the approach by the OP can't work since there is no $L^2$ function for which the norm is attained.
    $endgroup$
    – Dirk
    Jan 7 at 18:31








1




1




$begingroup$
Hint: take $f=1_{[a,1]}$ where $a < 1$ is a positive real number close to $1$.
$endgroup$
– Mindlack
Jan 7 at 18:16




$begingroup$
Hint: take $f=1_{[a,1]}$ where $a < 1$ is a positive real number close to $1$.
$endgroup$
– Mindlack
Jan 7 at 18:16












$begingroup$
The answers below are fine, but one may add, that the approach by the OP can't work since there is no $L^2$ function for which the norm is attained.
$endgroup$
– Dirk
Jan 7 at 18:31




$begingroup$
The answers below are fine, but one may add, that the approach by the OP can't work since there is no $L^2$ function for which the norm is attained.
$endgroup$
– Dirk
Jan 7 at 18:31










2 Answers
2






active

oldest

votes


















5












$begingroup$

Consider $f_n(x)=x^n$ for $ninBbb N$.



Then $$|f_n|_2=sqrt{frac1{2n+1}}$$
and
$$|Tf_n|_2=sqrt{frac1{2n+5}}$$
Since
$$lim_{ntoinfty}frac{|Tf_n|_2}{|f_n|_2}=1$$
we have that $|T|ge1$.






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    Multiplication by $x^2$ on $[0,1]$. $T$ is a self-adjoint operator. The essential range of the multiplier $x^2$ is $[0,1]$, so the spectrum of $T$ is $[0,1]$. The spectral radius of $T$ is $1$, and (since $T$ is self-adjoint), the norm of $T$ is $1$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      As Dirk notes, there is no $f$ where the norm is attained. Here, we can see: $1$ is in the spectrum of $T$, but is not an eigenvalue of $T$.
      $endgroup$
      – GEdgar
      Jan 8 at 14:24











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    Consider $f_n(x)=x^n$ for $ninBbb N$.



    Then $$|f_n|_2=sqrt{frac1{2n+1}}$$
    and
    $$|Tf_n|_2=sqrt{frac1{2n+5}}$$
    Since
    $$lim_{ntoinfty}frac{|Tf_n|_2}{|f_n|_2}=1$$
    we have that $|T|ge1$.






    share|cite|improve this answer











    $endgroup$


















      5












      $begingroup$

      Consider $f_n(x)=x^n$ for $ninBbb N$.



      Then $$|f_n|_2=sqrt{frac1{2n+1}}$$
      and
      $$|Tf_n|_2=sqrt{frac1{2n+5}}$$
      Since
      $$lim_{ntoinfty}frac{|Tf_n|_2}{|f_n|_2}=1$$
      we have that $|T|ge1$.






      share|cite|improve this answer











      $endgroup$
















        5












        5








        5





        $begingroup$

        Consider $f_n(x)=x^n$ for $ninBbb N$.



        Then $$|f_n|_2=sqrt{frac1{2n+1}}$$
        and
        $$|Tf_n|_2=sqrt{frac1{2n+5}}$$
        Since
        $$lim_{ntoinfty}frac{|Tf_n|_2}{|f_n|_2}=1$$
        we have that $|T|ge1$.






        share|cite|improve this answer











        $endgroup$



        Consider $f_n(x)=x^n$ for $ninBbb N$.



        Then $$|f_n|_2=sqrt{frac1{2n+1}}$$
        and
        $$|Tf_n|_2=sqrt{frac1{2n+5}}$$
        Since
        $$lim_{ntoinfty}frac{|Tf_n|_2}{|f_n|_2}=1$$
        we have that $|T|ge1$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 7 at 20:05

























        answered Jan 7 at 18:16









        ajotatxeajotatxe

        53.8k23890




        53.8k23890























            3












            $begingroup$

            Multiplication by $x^2$ on $[0,1]$. $T$ is a self-adjoint operator. The essential range of the multiplier $x^2$ is $[0,1]$, so the spectrum of $T$ is $[0,1]$. The spectral radius of $T$ is $1$, and (since $T$ is self-adjoint), the norm of $T$ is $1$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              As Dirk notes, there is no $f$ where the norm is attained. Here, we can see: $1$ is in the spectrum of $T$, but is not an eigenvalue of $T$.
              $endgroup$
              – GEdgar
              Jan 8 at 14:24
















            3












            $begingroup$

            Multiplication by $x^2$ on $[0,1]$. $T$ is a self-adjoint operator. The essential range of the multiplier $x^2$ is $[0,1]$, so the spectrum of $T$ is $[0,1]$. The spectral radius of $T$ is $1$, and (since $T$ is self-adjoint), the norm of $T$ is $1$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              As Dirk notes, there is no $f$ where the norm is attained. Here, we can see: $1$ is in the spectrum of $T$, but is not an eigenvalue of $T$.
              $endgroup$
              – GEdgar
              Jan 8 at 14:24














            3












            3








            3





            $begingroup$

            Multiplication by $x^2$ on $[0,1]$. $T$ is a self-adjoint operator. The essential range of the multiplier $x^2$ is $[0,1]$, so the spectrum of $T$ is $[0,1]$. The spectral radius of $T$ is $1$, and (since $T$ is self-adjoint), the norm of $T$ is $1$.






            share|cite|improve this answer









            $endgroup$



            Multiplication by $x^2$ on $[0,1]$. $T$ is a self-adjoint operator. The essential range of the multiplier $x^2$ is $[0,1]$, so the spectrum of $T$ is $[0,1]$. The spectral radius of $T$ is $1$, and (since $T$ is self-adjoint), the norm of $T$ is $1$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 7 at 18:17









            GEdgarGEdgar

            62.5k267171




            62.5k267171












            • $begingroup$
              As Dirk notes, there is no $f$ where the norm is attained. Here, we can see: $1$ is in the spectrum of $T$, but is not an eigenvalue of $T$.
              $endgroup$
              – GEdgar
              Jan 8 at 14:24


















            • $begingroup$
              As Dirk notes, there is no $f$ where the norm is attained. Here, we can see: $1$ is in the spectrum of $T$, but is not an eigenvalue of $T$.
              $endgroup$
              – GEdgar
              Jan 8 at 14:24
















            $begingroup$
            As Dirk notes, there is no $f$ where the norm is attained. Here, we can see: $1$ is in the spectrum of $T$, but is not an eigenvalue of $T$.
            $endgroup$
            – GEdgar
            Jan 8 at 14:24




            $begingroup$
            As Dirk notes, there is no $f$ where the norm is attained. Here, we can see: $1$ is in the spectrum of $T$, but is not an eigenvalue of $T$.
            $endgroup$
            – GEdgar
            Jan 8 at 14:24


















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