analysis of $T : f to Tf$ with $[T(f)](x) = ie^{ipi x}(int_0^x e^{-ipi t}f(t)dt - int_x^1 e^{-ipi t}f(t)dt)$












1












$begingroup$


$, f in L^2([0,1],mathbb{C})$



show that $T : f to Tf, , f in L^2([0,1],mathbb{C})$ is continuous, $[T(f)](x) = ie^{ipi x}(int_0^x e^{-ipi t}f(t)dt - int_x^1 e^{-ipi t}f(t)dt)$



the operator is linear so it suffices to show the boundedness only



$[T(f)](x) overline{[T(f)](x) } = (int_0^x cos{(pi t)}f(t)dt - int_x^1 cos{(pi t)}f(t)dt)^2 + (int_0^x sin{(pi t)}f(t)dt - int_x^1 sin{(pi t)}f(t)dt)^2 $



then by Holder Inequality : $$|[T(f)](x) overline{[T(f)](x) }| leq 2x|f|^2_{L^2} +2(1-x)|f|^2_{L^2} + 4 sqrt{x(1-x)} |f|^2_{L^2} leq C |f|^2_{L^2} $$
and that implies $|[T(f)](x)||_{L^2}^2 leq C |f|^2_{L^2} $



now suppose $(f_n)_{n geq 0} subset L^2([0,1],mathbb{C}) $ such that $|f_n|_{L^2} leq 1$



also suppose that that sequence of functions converges $textbf{weakly}$ to some function in $L^2$ called $f$



from that we can say that $[T(f_n)](x) to [T(f)](x)$



now all the previous points are supposed to help me prove that $(T(f_n))_{n geq 0}$ converges $textbf{strongly}$ to $T(f)$



but I fail to see how, any help will be greatly appreciated.






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







share|cite|improve this question









$endgroup$












  • $begingroup$
    Do you know that operators of this type are compact operators? If you know this fact you can easily prove that $Tf_n to Tf$ in the norm.
    $endgroup$
    – Kavi Rama Murthy
    Jan 8 at 8:14
















1












$begingroup$


$, f in L^2([0,1],mathbb{C})$



show that $T : f to Tf, , f in L^2([0,1],mathbb{C})$ is continuous, $[T(f)](x) = ie^{ipi x}(int_0^x e^{-ipi t}f(t)dt - int_x^1 e^{-ipi t}f(t)dt)$



the operator is linear so it suffices to show the boundedness only



$[T(f)](x) overline{[T(f)](x) } = (int_0^x cos{(pi t)}f(t)dt - int_x^1 cos{(pi t)}f(t)dt)^2 + (int_0^x sin{(pi t)}f(t)dt - int_x^1 sin{(pi t)}f(t)dt)^2 $



then by Holder Inequality : $$|[T(f)](x) overline{[T(f)](x) }| leq 2x|f|^2_{L^2} +2(1-x)|f|^2_{L^2} + 4 sqrt{x(1-x)} |f|^2_{L^2} leq C |f|^2_{L^2} $$
and that implies $|[T(f)](x)||_{L^2}^2 leq C |f|^2_{L^2} $



now suppose $(f_n)_{n geq 0} subset L^2([0,1],mathbb{C}) $ such that $|f_n|_{L^2} leq 1$



also suppose that that sequence of functions converges $textbf{weakly}$ to some function in $L^2$ called $f$



from that we can say that $[T(f_n)](x) to [T(f)](x)$



now all the previous points are supposed to help me prove that $(T(f_n))_{n geq 0}$ converges $textbf{strongly}$ to $T(f)$



but I fail to see how, any help will be greatly appreciated.






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







share|cite|improve this question









$endgroup$












  • $begingroup$
    Do you know that operators of this type are compact operators? If you know this fact you can easily prove that $Tf_n to Tf$ in the norm.
    $endgroup$
    – Kavi Rama Murthy
    Jan 8 at 8:14














1












1








1





$begingroup$


$, f in L^2([0,1],mathbb{C})$



show that $T : f to Tf, , f in L^2([0,1],mathbb{C})$ is continuous, $[T(f)](x) = ie^{ipi x}(int_0^x e^{-ipi t}f(t)dt - int_x^1 e^{-ipi t}f(t)dt)$



the operator is linear so it suffices to show the boundedness only



$[T(f)](x) overline{[T(f)](x) } = (int_0^x cos{(pi t)}f(t)dt - int_x^1 cos{(pi t)}f(t)dt)^2 + (int_0^x sin{(pi t)}f(t)dt - int_x^1 sin{(pi t)}f(t)dt)^2 $



then by Holder Inequality : $$|[T(f)](x) overline{[T(f)](x) }| leq 2x|f|^2_{L^2} +2(1-x)|f|^2_{L^2} + 4 sqrt{x(1-x)} |f|^2_{L^2} leq C |f|^2_{L^2} $$
and that implies $|[T(f)](x)||_{L^2}^2 leq C |f|^2_{L^2} $



now suppose $(f_n)_{n geq 0} subset L^2([0,1],mathbb{C}) $ such that $|f_n|_{L^2} leq 1$



also suppose that that sequence of functions converges $textbf{weakly}$ to some function in $L^2$ called $f$



from that we can say that $[T(f_n)](x) to [T(f)](x)$



now all the previous points are supposed to help me prove that $(T(f_n))_{n geq 0}$ converges $textbf{strongly}$ to $T(f)$



but I fail to see how, any help will be greatly appreciated.






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







share|cite|improve this question









$endgroup$




$, f in L^2([0,1],mathbb{C})$



show that $T : f to Tf, , f in L^2([0,1],mathbb{C})$ is continuous, $[T(f)](x) = ie^{ipi x}(int_0^x e^{-ipi t}f(t)dt - int_x^1 e^{-ipi t}f(t)dt)$



the operator is linear so it suffices to show the boundedness only



$[T(f)](x) overline{[T(f)](x) } = (int_0^x cos{(pi t)}f(t)dt - int_x^1 cos{(pi t)}f(t)dt)^2 + (int_0^x sin{(pi t)}f(t)dt - int_x^1 sin{(pi t)}f(t)dt)^2 $



then by Holder Inequality : $$|[T(f)](x) overline{[T(f)](x) }| leq 2x|f|^2_{L^2} +2(1-x)|f|^2_{L^2} + 4 sqrt{x(1-x)} |f|^2_{L^2} leq C |f|^2_{L^2} $$
and that implies $|[T(f)](x)||_{L^2}^2 leq C |f|^2_{L^2} $



now suppose $(f_n)_{n geq 0} subset L^2([0,1],mathbb{C}) $ such that $|f_n|_{L^2} leq 1$



also suppose that that sequence of functions converges $textbf{weakly}$ to some function in $L^2$ called $f$



from that we can say that $[T(f_n)](x) to [T(f)](x)$



now all the previous points are supposed to help me prove that $(T(f_n))_{n geq 0}$ converges $textbf{strongly}$ to $T(f)$



but I fail to see how, any help will be greatly appreciated.






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




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






share|cite|improve this question













share|cite|improve this question











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asked Jan 8 at 7:54









rapidracimrapidracim

1,5741319




1,5741319












  • $begingroup$
    Do you know that operators of this type are compact operators? If you know this fact you can easily prove that $Tf_n to Tf$ in the norm.
    $endgroup$
    – Kavi Rama Murthy
    Jan 8 at 8:14


















  • $begingroup$
    Do you know that operators of this type are compact operators? If you know this fact you can easily prove that $Tf_n to Tf$ in the norm.
    $endgroup$
    – Kavi Rama Murthy
    Jan 8 at 8:14
















$begingroup$
Do you know that operators of this type are compact operators? If you know this fact you can easily prove that $Tf_n to Tf$ in the norm.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 8:14




$begingroup$
Do you know that operators of this type are compact operators? If you know this fact you can easily prove that $Tf_n to Tf$ in the norm.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 8:14










1 Answer
1






active

oldest

votes


















0












$begingroup$

Use Theorem 8.2.3 in
https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&ved=2ahUKEwjWt7j83t3fAhXKso8KHSR3BecQFjACegQIBRAC&url=http%3A%2F%2Fpeople.math.gatech.edu%2F~heil%2Fmetricnote%2Fchap8.pdf&usg=AOvVaw3hD60cdJ-gOIAiuX4Jr5hh



to conclude that $T$ is a compact operator. Hence any subsequence of ${Tf_n}$ has a further subsequence converging in the norm. The limit has to be $Tf$ because $f_n to f$ weakly implies $Tf_n to Tf$ weakly. Hence the entire sequence ${Tf_n}$ converges in the norm to $Tf$.






share|cite|improve this answer









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    $begingroup$

    Use Theorem 8.2.3 in
    https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&ved=2ahUKEwjWt7j83t3fAhXKso8KHSR3BecQFjACegQIBRAC&url=http%3A%2F%2Fpeople.math.gatech.edu%2F~heil%2Fmetricnote%2Fchap8.pdf&usg=AOvVaw3hD60cdJ-gOIAiuX4Jr5hh



    to conclude that $T$ is a compact operator. Hence any subsequence of ${Tf_n}$ has a further subsequence converging in the norm. The limit has to be $Tf$ because $f_n to f$ weakly implies $Tf_n to Tf$ weakly. Hence the entire sequence ${Tf_n}$ converges in the norm to $Tf$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Use Theorem 8.2.3 in
      https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&ved=2ahUKEwjWt7j83t3fAhXKso8KHSR3BecQFjACegQIBRAC&url=http%3A%2F%2Fpeople.math.gatech.edu%2F~heil%2Fmetricnote%2Fchap8.pdf&usg=AOvVaw3hD60cdJ-gOIAiuX4Jr5hh



      to conclude that $T$ is a compact operator. Hence any subsequence of ${Tf_n}$ has a further subsequence converging in the norm. The limit has to be $Tf$ because $f_n to f$ weakly implies $Tf_n to Tf$ weakly. Hence the entire sequence ${Tf_n}$ converges in the norm to $Tf$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Use Theorem 8.2.3 in
        https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&ved=2ahUKEwjWt7j83t3fAhXKso8KHSR3BecQFjACegQIBRAC&url=http%3A%2F%2Fpeople.math.gatech.edu%2F~heil%2Fmetricnote%2Fchap8.pdf&usg=AOvVaw3hD60cdJ-gOIAiuX4Jr5hh



        to conclude that $T$ is a compact operator. Hence any subsequence of ${Tf_n}$ has a further subsequence converging in the norm. The limit has to be $Tf$ because $f_n to f$ weakly implies $Tf_n to Tf$ weakly. Hence the entire sequence ${Tf_n}$ converges in the norm to $Tf$.






        share|cite|improve this answer









        $endgroup$



        Use Theorem 8.2.3 in
        https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&ved=2ahUKEwjWt7j83t3fAhXKso8KHSR3BecQFjACegQIBRAC&url=http%3A%2F%2Fpeople.math.gatech.edu%2F~heil%2Fmetricnote%2Fchap8.pdf&usg=AOvVaw3hD60cdJ-gOIAiuX4Jr5hh



        to conclude that $T$ is a compact operator. Hence any subsequence of ${Tf_n}$ has a further subsequence converging in the norm. The limit has to be $Tf$ because $f_n to f$ weakly implies $Tf_n to Tf$ weakly. Hence the entire sequence ${Tf_n}$ converges in the norm to $Tf$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 8 at 8:22









        Kavi Rama MurthyKavi Rama Murthy

        56.6k42159




        56.6k42159






























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