A non compact operator on $L^2[0,1]$












0












$begingroup$


Let $H$ be the Hilbert space $L^2[0,1]$.



and the operator $T : Hrightarrow H$, such as $T(f)(x)=x.f(x)$



Why $T$ isn't compact ?






functional-analysis operator-theory compact-operators







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Let $H$ be the Hilbert space $L^2[0,1]$.



    and the operator $T : Hrightarrow H$, such as $T(f)(x)=x.f(x)$



    Why $T$ isn't compact ?






    functional-analysis operator-theory compact-operators







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $H$ be the Hilbert space $L^2[0,1]$.



      and the operator $T : Hrightarrow H$, such as $T(f)(x)=x.f(x)$



      Why $T$ isn't compact ?






      functional-analysis operator-theory compact-operators







      share|cite|improve this question









      $endgroup$




      Let $H$ be the Hilbert space $L^2[0,1]$.



      and the operator $T : Hrightarrow H$, such as $T(f)(x)=x.f(x)$



      Why $T$ isn't compact ?






      functional-analysis operator-theory compact-operators




      functional-analysis operator-theory compact-operators






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 22 '18 at 8:23









      Anas BOUALIIAnas BOUALII

      1438




      1438






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Let $f_n(x) = sqrt{n}1_{[1-frac{1}{n},1]}(x)$. Then $||f_n||_2 = 1$ for each $n$ but $(Tf_n)_n$ has no convergent subsequence.



          Indeed, suppose $Tf_{n_k} to g$ in $L^2$ for some $g in L^2$ and subsequence $(f_{n_k})_k$. Since $||f_{n_k}||_2 = 1$ for each $k$, $g$ cannot be $0$, so $int_0^1 |g(x)|^2dx > 0$. Then $int_0^a |g(x)|^2dx =: epsilon > 0$ for some $a in (0,1)$. But then for large $k$, $$int_0^1 |g(x)-xf_{n_k}(x)|^2dx = int_0^a |g(x)|^2 +int_a^1 |g(x)-xf_{n_k}(x)|^2dx ge int_0^a |g(x)|^2dx ge epsilon,$$ which contradicts $f_{n_k} to g$ in $L^2$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I found dificulties in proving that $Tf_n=f_n$
            $endgroup$
            – Anas BOUALII
            Dec 22 '18 at 9:08










          • $begingroup$
            @AnasBOUALII I have added details
            $endgroup$
            – mathworker21
            Dec 22 '18 at 9:18










          • $begingroup$
            I see, thank you.
            $endgroup$
            – Anas BOUALII
            Dec 22 '18 at 9:24



















          2












          $begingroup$

          Observe that $T$ is a self-adjoint operator. And also $$
          ker (T-lambda)=(0)
          $$           holds for all $lambdain mathbb{C}$. If $T$ is compact, then by the spectral theorem of compact operators, it follows that
          $$
          T=0.
          $$           This leads to an obvious contradiction.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            By the theorem it follows that the spectrum is $mathbb{C}$, which is a contradiction with the spectrum is compact. Right ?
            $endgroup$
            – Anas BOUALII
            Dec 22 '18 at 8:56










          • $begingroup$
            I'll elaborate on this. In fact, $ker(T-lambda)=(0)$ means that the point spectrum of $T$ is empty. For a compact operator $T$, it holds generally that $sigma_p(T)setminus{0} = sigma(T)setminus{0}$, where $sigma_p(T) := {lambda;|;ker(T-lambda) neq (0)}$ is the point spectrum of $T$. For the given $T$, we have $sigma_p(T) =varnothing$ and hence it follows $sigma(T) subset {0}$. However, we can show that $sigma(T) = [0,1]$ holds, leading to a contradiction!
            $endgroup$
            – Song
            Dec 22 '18 at 9:03












          • $begingroup$
            I agree with you. so since $T$ is self-adjoint operator then $sigma(T) subset [m,M]$ with $M=sup_{||f||=1}<Tf,f>$, $m=inf_{||f||=1}<Tf,f>$, but how to prove that $m=0$ and $M=1$?
            $endgroup$
            – Anas BOUALII
            Dec 22 '18 at 9:22










          • $begingroup$
            Well, $langle f,Tfranglesubset [0,1]$ is almost direct, and $f_n = sqrt{n}1_{[0,frac{1}{n}]}$ is an infimum-approaching sequence (note that $f_n$ is getting concentrated on $x=0$). In a similar way, $f_n =sqrt{n}1_{[1-frac{1}{n},1]}$ is a supremum-approaching sequence. But in fact, $sigma(T) =[0,1]$ can be seen in a more direct way. Since for $lambda in [0,1]$, $xmapsto frac{1}{x-lambda}$ is not bounded on $[0,1]$, it follows that $(T-lambda)^{-1}f(x) =frac{f(x)}{x-lambda}$ is not a bounded inverse in $L^2$ .
            $endgroup$
            – Song
            Dec 22 '18 at 9:34













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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Let $f_n(x) = sqrt{n}1_{[1-frac{1}{n},1]}(x)$. Then $||f_n||_2 = 1$ for each $n$ but $(Tf_n)_n$ has no convergent subsequence.



          Indeed, suppose $Tf_{n_k} to g$ in $L^2$ for some $g in L^2$ and subsequence $(f_{n_k})_k$. Since $||f_{n_k}||_2 = 1$ for each $k$, $g$ cannot be $0$, so $int_0^1 |g(x)|^2dx > 0$. Then $int_0^a |g(x)|^2dx =: epsilon > 0$ for some $a in (0,1)$. But then for large $k$, $$int_0^1 |g(x)-xf_{n_k}(x)|^2dx = int_0^a |g(x)|^2 +int_a^1 |g(x)-xf_{n_k}(x)|^2dx ge int_0^a |g(x)|^2dx ge epsilon,$$ which contradicts $f_{n_k} to g$ in $L^2$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I found dificulties in proving that $Tf_n=f_n$
            $endgroup$
            – Anas BOUALII
            Dec 22 '18 at 9:08










          • $begingroup$
            @AnasBOUALII I have added details
            $endgroup$
            – mathworker21
            Dec 22 '18 at 9:18










          • $begingroup$
            I see, thank you.
            $endgroup$
            – Anas BOUALII
            Dec 22 '18 at 9:24
















          1












          $begingroup$

          Let $f_n(x) = sqrt{n}1_{[1-frac{1}{n},1]}(x)$. Then $||f_n||_2 = 1$ for each $n$ but $(Tf_n)_n$ has no convergent subsequence.



          Indeed, suppose $Tf_{n_k} to g$ in $L^2$ for some $g in L^2$ and subsequence $(f_{n_k})_k$. Since $||f_{n_k}||_2 = 1$ for each $k$, $g$ cannot be $0$, so $int_0^1 |g(x)|^2dx > 0$. Then $int_0^a |g(x)|^2dx =: epsilon > 0$ for some $a in (0,1)$. But then for large $k$, $$int_0^1 |g(x)-xf_{n_k}(x)|^2dx = int_0^a |g(x)|^2 +int_a^1 |g(x)-xf_{n_k}(x)|^2dx ge int_0^a |g(x)|^2dx ge epsilon,$$ which contradicts $f_{n_k} to g$ in $L^2$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I found dificulties in proving that $Tf_n=f_n$
            $endgroup$
            – Anas BOUALII
            Dec 22 '18 at 9:08










          • $begingroup$
            @AnasBOUALII I have added details
            $endgroup$
            – mathworker21
            Dec 22 '18 at 9:18










          • $begingroup$
            I see, thank you.
            $endgroup$
            – Anas BOUALII
            Dec 22 '18 at 9:24














          1












          1








          1





          $begingroup$

          Let $f_n(x) = sqrt{n}1_{[1-frac{1}{n},1]}(x)$. Then $||f_n||_2 = 1$ for each $n$ but $(Tf_n)_n$ has no convergent subsequence.



          Indeed, suppose $Tf_{n_k} to g$ in $L^2$ for some $g in L^2$ and subsequence $(f_{n_k})_k$. Since $||f_{n_k}||_2 = 1$ for each $k$, $g$ cannot be $0$, so $int_0^1 |g(x)|^2dx > 0$. Then $int_0^a |g(x)|^2dx =: epsilon > 0$ for some $a in (0,1)$. But then for large $k$, $$int_0^1 |g(x)-xf_{n_k}(x)|^2dx = int_0^a |g(x)|^2 +int_a^1 |g(x)-xf_{n_k}(x)|^2dx ge int_0^a |g(x)|^2dx ge epsilon,$$ which contradicts $f_{n_k} to g$ in $L^2$.






          share|cite|improve this answer











          $endgroup$



          Let $f_n(x) = sqrt{n}1_{[1-frac{1}{n},1]}(x)$. Then $||f_n||_2 = 1$ for each $n$ but $(Tf_n)_n$ has no convergent subsequence.



          Indeed, suppose $Tf_{n_k} to g$ in $L^2$ for some $g in L^2$ and subsequence $(f_{n_k})_k$. Since $||f_{n_k}||_2 = 1$ for each $k$, $g$ cannot be $0$, so $int_0^1 |g(x)|^2dx > 0$. Then $int_0^a |g(x)|^2dx =: epsilon > 0$ for some $a in (0,1)$. But then for large $k$, $$int_0^1 |g(x)-xf_{n_k}(x)|^2dx = int_0^a |g(x)|^2 +int_a^1 |g(x)-xf_{n_k}(x)|^2dx ge int_0^a |g(x)|^2dx ge epsilon,$$ which contradicts $f_{n_k} to g$ in $L^2$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 22 '18 at 9:18

























          answered Dec 22 '18 at 8:33









          mathworker21mathworker21

          8,9561928




          8,9561928












          • $begingroup$
            I found dificulties in proving that $Tf_n=f_n$
            $endgroup$
            – Anas BOUALII
            Dec 22 '18 at 9:08










          • $begingroup$
            @AnasBOUALII I have added details
            $endgroup$
            – mathworker21
            Dec 22 '18 at 9:18










          • $begingroup$
            I see, thank you.
            $endgroup$
            – Anas BOUALII
            Dec 22 '18 at 9:24


















          • $begingroup$
            I found dificulties in proving that $Tf_n=f_n$
            $endgroup$
            – Anas BOUALII
            Dec 22 '18 at 9:08










          • $begingroup$
            @AnasBOUALII I have added details
            $endgroup$
            – mathworker21
            Dec 22 '18 at 9:18










          • $begingroup$
            I see, thank you.
            $endgroup$
            – Anas BOUALII
            Dec 22 '18 at 9:24
















          $begingroup$
          I found dificulties in proving that $Tf_n=f_n$
          $endgroup$
          – Anas BOUALII
          Dec 22 '18 at 9:08




          $begingroup$
          I found dificulties in proving that $Tf_n=f_n$
          $endgroup$
          – Anas BOUALII
          Dec 22 '18 at 9:08












          $begingroup$
          @AnasBOUALII I have added details
          $endgroup$
          – mathworker21
          Dec 22 '18 at 9:18




          $begingroup$
          @AnasBOUALII I have added details
          $endgroup$
          – mathworker21
          Dec 22 '18 at 9:18












          $begingroup$
          I see, thank you.
          $endgroup$
          – Anas BOUALII
          Dec 22 '18 at 9:24




          $begingroup$
          I see, thank you.
          $endgroup$
          – Anas BOUALII
          Dec 22 '18 at 9:24











          2












          $begingroup$

          Observe that $T$ is a self-adjoint operator. And also $$
          ker (T-lambda)=(0)
          $$           holds for all $lambdain mathbb{C}$. If $T$ is compact, then by the spectral theorem of compact operators, it follows that
          $$
          T=0.
          $$           This leads to an obvious contradiction.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            By the theorem it follows that the spectrum is $mathbb{C}$, which is a contradiction with the spectrum is compact. Right ?
            $endgroup$
            – Anas BOUALII
            Dec 22 '18 at 8:56










          • $begingroup$
            I'll elaborate on this. In fact, $ker(T-lambda)=(0)$ means that the point spectrum of $T$ is empty. For a compact operator $T$, it holds generally that $sigma_p(T)setminus{0} = sigma(T)setminus{0}$, where $sigma_p(T) := {lambda;|;ker(T-lambda) neq (0)}$ is the point spectrum of $T$. For the given $T$, we have $sigma_p(T) =varnothing$ and hence it follows $sigma(T) subset {0}$. However, we can show that $sigma(T) = [0,1]$ holds, leading to a contradiction!
            $endgroup$
            – Song
            Dec 22 '18 at 9:03












          • $begingroup$
            I agree with you. so since $T$ is self-adjoint operator then $sigma(T) subset [m,M]$ with $M=sup_{||f||=1}<Tf,f>$, $m=inf_{||f||=1}<Tf,f>$, but how to prove that $m=0$ and $M=1$?
            $endgroup$
            – Anas BOUALII
            Dec 22 '18 at 9:22










          • $begingroup$
            Well, $langle f,Tfranglesubset [0,1]$ is almost direct, and $f_n = sqrt{n}1_{[0,frac{1}{n}]}$ is an infimum-approaching sequence (note that $f_n$ is getting concentrated on $x=0$). In a similar way, $f_n =sqrt{n}1_{[1-frac{1}{n},1]}$ is a supremum-approaching sequence. But in fact, $sigma(T) =[0,1]$ can be seen in a more direct way. Since for $lambda in [0,1]$, $xmapsto frac{1}{x-lambda}$ is not bounded on $[0,1]$, it follows that $(T-lambda)^{-1}f(x) =frac{f(x)}{x-lambda}$ is not a bounded inverse in $L^2$ .
            $endgroup$
            – Song
            Dec 22 '18 at 9:34


















          2












          $begingroup$

          Observe that $T$ is a self-adjoint operator. And also $$
          ker (T-lambda)=(0)
          $$           holds for all $lambdain mathbb{C}$. If $T$ is compact, then by the spectral theorem of compact operators, it follows that
          $$
          T=0.
          $$           This leads to an obvious contradiction.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            By the theorem it follows that the spectrum is $mathbb{C}$, which is a contradiction with the spectrum is compact. Right ?
            $endgroup$
            – Anas BOUALII
            Dec 22 '18 at 8:56










          • $begingroup$
            I'll elaborate on this. In fact, $ker(T-lambda)=(0)$ means that the point spectrum of $T$ is empty. For a compact operator $T$, it holds generally that $sigma_p(T)setminus{0} = sigma(T)setminus{0}$, where $sigma_p(T) := {lambda;|;ker(T-lambda) neq (0)}$ is the point spectrum of $T$. For the given $T$, we have $sigma_p(T) =varnothing$ and hence it follows $sigma(T) subset {0}$. However, we can show that $sigma(T) = [0,1]$ holds, leading to a contradiction!
            $endgroup$
            – Song
            Dec 22 '18 at 9:03












          • $begingroup$
            I agree with you. so since $T$ is self-adjoint operator then $sigma(T) subset [m,M]$ with $M=sup_{||f||=1}<Tf,f>$, $m=inf_{||f||=1}<Tf,f>$, but how to prove that $m=0$ and $M=1$?
            $endgroup$
            – Anas BOUALII
            Dec 22 '18 at 9:22










          • $begingroup$
            Well, $langle f,Tfranglesubset [0,1]$ is almost direct, and $f_n = sqrt{n}1_{[0,frac{1}{n}]}$ is an infimum-approaching sequence (note that $f_n$ is getting concentrated on $x=0$). In a similar way, $f_n =sqrt{n}1_{[1-frac{1}{n},1]}$ is a supremum-approaching sequence. But in fact, $sigma(T) =[0,1]$ can be seen in a more direct way. Since for $lambda in [0,1]$, $xmapsto frac{1}{x-lambda}$ is not bounded on $[0,1]$, it follows that $(T-lambda)^{-1}f(x) =frac{f(x)}{x-lambda}$ is not a bounded inverse in $L^2$ .
            $endgroup$
            – Song
            Dec 22 '18 at 9:34
















          2












          2








          2





          $begingroup$

          Observe that $T$ is a self-adjoint operator. And also $$
          ker (T-lambda)=(0)
          $$           holds for all $lambdain mathbb{C}$. If $T$ is compact, then by the spectral theorem of compact operators, it follows that
          $$
          T=0.
          $$           This leads to an obvious contradiction.






          share|cite|improve this answer









          $endgroup$



          Observe that $T$ is a self-adjoint operator. And also $$
          ker (T-lambda)=(0)
          $$           holds for all $lambdain mathbb{C}$. If $T$ is compact, then by the spectral theorem of compact operators, it follows that
          $$
          T=0.
          $$           This leads to an obvious contradiction.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 22 '18 at 8:51









          SongSong

          14.1k1633




          14.1k1633












          • $begingroup$
            By the theorem it follows that the spectrum is $mathbb{C}$, which is a contradiction with the spectrum is compact. Right ?
            $endgroup$
            – Anas BOUALII
            Dec 22 '18 at 8:56










          • $begingroup$
            I'll elaborate on this. In fact, $ker(T-lambda)=(0)$ means that the point spectrum of $T$ is empty. For a compact operator $T$, it holds generally that $sigma_p(T)setminus{0} = sigma(T)setminus{0}$, where $sigma_p(T) := {lambda;|;ker(T-lambda) neq (0)}$ is the point spectrum of $T$. For the given $T$, we have $sigma_p(T) =varnothing$ and hence it follows $sigma(T) subset {0}$. However, we can show that $sigma(T) = [0,1]$ holds, leading to a contradiction!
            $endgroup$
            – Song
            Dec 22 '18 at 9:03












          • $begingroup$
            I agree with you. so since $T$ is self-adjoint operator then $sigma(T) subset [m,M]$ with $M=sup_{||f||=1}<Tf,f>$, $m=inf_{||f||=1}<Tf,f>$, but how to prove that $m=0$ and $M=1$?
            $endgroup$
            – Anas BOUALII
            Dec 22 '18 at 9:22










          • $begingroup$
            Well, $langle f,Tfranglesubset [0,1]$ is almost direct, and $f_n = sqrt{n}1_{[0,frac{1}{n}]}$ is an infimum-approaching sequence (note that $f_n$ is getting concentrated on $x=0$). In a similar way, $f_n =sqrt{n}1_{[1-frac{1}{n},1]}$ is a supremum-approaching sequence. But in fact, $sigma(T) =[0,1]$ can be seen in a more direct way. Since for $lambda in [0,1]$, $xmapsto frac{1}{x-lambda}$ is not bounded on $[0,1]$, it follows that $(T-lambda)^{-1}f(x) =frac{f(x)}{x-lambda}$ is not a bounded inverse in $L^2$ .
            $endgroup$
            – Song
            Dec 22 '18 at 9:34




















          • $begingroup$
            By the theorem it follows that the spectrum is $mathbb{C}$, which is a contradiction with the spectrum is compact. Right ?
            $endgroup$
            – Anas BOUALII
            Dec 22 '18 at 8:56










          • $begingroup$
            I'll elaborate on this. In fact, $ker(T-lambda)=(0)$ means that the point spectrum of $T$ is empty. For a compact operator $T$, it holds generally that $sigma_p(T)setminus{0} = sigma(T)setminus{0}$, where $sigma_p(T) := {lambda;|;ker(T-lambda) neq (0)}$ is the point spectrum of $T$. For the given $T$, we have $sigma_p(T) =varnothing$ and hence it follows $sigma(T) subset {0}$. However, we can show that $sigma(T) = [0,1]$ holds, leading to a contradiction!
            $endgroup$
            – Song
            Dec 22 '18 at 9:03












          • $begingroup$
            I agree with you. so since $T$ is self-adjoint operator then $sigma(T) subset [m,M]$ with $M=sup_{||f||=1}<Tf,f>$, $m=inf_{||f||=1}<Tf,f>$, but how to prove that $m=0$ and $M=1$?
            $endgroup$
            – Anas BOUALII
            Dec 22 '18 at 9:22










          • $begingroup$
            Well, $langle f,Tfranglesubset [0,1]$ is almost direct, and $f_n = sqrt{n}1_{[0,frac{1}{n}]}$ is an infimum-approaching sequence (note that $f_n$ is getting concentrated on $x=0$). In a similar way, $f_n =sqrt{n}1_{[1-frac{1}{n},1]}$ is a supremum-approaching sequence. But in fact, $sigma(T) =[0,1]$ can be seen in a more direct way. Since for $lambda in [0,1]$, $xmapsto frac{1}{x-lambda}$ is not bounded on $[0,1]$, it follows that $(T-lambda)^{-1}f(x) =frac{f(x)}{x-lambda}$ is not a bounded inverse in $L^2$ .
            $endgroup$
            – Song
            Dec 22 '18 at 9:34


















          $begingroup$
          By the theorem it follows that the spectrum is $mathbb{C}$, which is a contradiction with the spectrum is compact. Right ?
          $endgroup$
          – Anas BOUALII
          Dec 22 '18 at 8:56




          $begingroup$
          By the theorem it follows that the spectrum is $mathbb{C}$, which is a contradiction with the spectrum is compact. Right ?
          $endgroup$
          – Anas BOUALII
          Dec 22 '18 at 8:56












          $begingroup$
          I'll elaborate on this. In fact, $ker(T-lambda)=(0)$ means that the point spectrum of $T$ is empty. For a compact operator $T$, it holds generally that $sigma_p(T)setminus{0} = sigma(T)setminus{0}$, where $sigma_p(T) := {lambda;|;ker(T-lambda) neq (0)}$ is the point spectrum of $T$. For the given $T$, we have $sigma_p(T) =varnothing$ and hence it follows $sigma(T) subset {0}$. However, we can show that $sigma(T) = [0,1]$ holds, leading to a contradiction!
          $endgroup$
          – Song
          Dec 22 '18 at 9:03






          $begingroup$
          I'll elaborate on this. In fact, $ker(T-lambda)=(0)$ means that the point spectrum of $T$ is empty. For a compact operator $T$, it holds generally that $sigma_p(T)setminus{0} = sigma(T)setminus{0}$, where $sigma_p(T) := {lambda;|;ker(T-lambda) neq (0)}$ is the point spectrum of $T$. For the given $T$, we have $sigma_p(T) =varnothing$ and hence it follows $sigma(T) subset {0}$. However, we can show that $sigma(T) = [0,1]$ holds, leading to a contradiction!
          $endgroup$
          – Song
          Dec 22 '18 at 9:03














          $begingroup$
          I agree with you. so since $T$ is self-adjoint operator then $sigma(T) subset [m,M]$ with $M=sup_{||f||=1}<Tf,f>$, $m=inf_{||f||=1}<Tf,f>$, but how to prove that $m=0$ and $M=1$?
          $endgroup$
          – Anas BOUALII
          Dec 22 '18 at 9:22




          $begingroup$
          I agree with you. so since $T$ is self-adjoint operator then $sigma(T) subset [m,M]$ with $M=sup_{||f||=1}<Tf,f>$, $m=inf_{||f||=1}<Tf,f>$, but how to prove that $m=0$ and $M=1$?
          $endgroup$
          – Anas BOUALII
          Dec 22 '18 at 9:22












          $begingroup$
          Well, $langle f,Tfranglesubset [0,1]$ is almost direct, and $f_n = sqrt{n}1_{[0,frac{1}{n}]}$ is an infimum-approaching sequence (note that $f_n$ is getting concentrated on $x=0$). In a similar way, $f_n =sqrt{n}1_{[1-frac{1}{n},1]}$ is a supremum-approaching sequence. But in fact, $sigma(T) =[0,1]$ can be seen in a more direct way. Since for $lambda in [0,1]$, $xmapsto frac{1}{x-lambda}$ is not bounded on $[0,1]$, it follows that $(T-lambda)^{-1}f(x) =frac{f(x)}{x-lambda}$ is not a bounded inverse in $L^2$ .
          $endgroup$
          – Song
          Dec 22 '18 at 9:34






          $begingroup$
          Well, $langle f,Tfranglesubset [0,1]$ is almost direct, and $f_n = sqrt{n}1_{[0,frac{1}{n}]}$ is an infimum-approaching sequence (note that $f_n$ is getting concentrated on $x=0$). In a similar way, $f_n =sqrt{n}1_{[1-frac{1}{n},1]}$ is a supremum-approaching sequence. But in fact, $sigma(T) =[0,1]$ can be seen in a more direct way. Since for $lambda in [0,1]$, $xmapsto frac{1}{x-lambda}$ is not bounded on $[0,1]$, it follows that $(T-lambda)^{-1}f(x) =frac{f(x)}{x-lambda}$ is not a bounded inverse in $L^2$ .
          $endgroup$
          – Song
          Dec 22 '18 at 9:34




















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