Mixing
analysis of $T : f to Tf$ with $[T(f)](x) = ie^{ipi x}(int_0^x e^{-ipi t}f(t)dt - int_x^1 e^{-ipi t}f(t)dt)$
$begingroup$
$, f in L^2([0,1],mathbb{C})$
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
asked Jan 8 at 7:54
rapidracimrapidracim
1,5741319
$endgroup$
add a comment |
$begingroup$
$, f in L^2([0,1],mathbb{C})$
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
asked Jan 8 at 7:54
rapidracimrapidracim
1,5741319
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
$, f in L^2([0,1],mathbb{C})$
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
asked Jan 8 at 7:54
rapidracimrapidracim
1,5741319
$endgroup$
$, f in L^2([0,1],mathbb{C})$
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
asked Jan 8 at 7:54
rapidracimrapidracim
1,5741319
asked Jan 8 at 7:54
rapidracimrapidracim
1,5741319
asked Jan 8 at 7:54
rapidracimrapidracim
1,5741319
asked Jan 8 at 7:54
rapidracimrapidracim
1,5741319
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $Tf$.
answered Jan 8 at 8:22
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
$endgroup$
add a comment |
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1 Answer
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1 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 $Tf$.
answered Jan 8 at 8:22
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
$endgroup$
add a comment |
$begingroup$
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 $Tf$.
answered Jan 8 at 8:22
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
$endgroup$
add a comment |
$begingroup$
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 $Tf$.
answered Jan 8 at 8:22
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
$endgroup$
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 $Tf$.
answered Jan 8 at 8:22
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
answered Jan 8 at 8:22
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
answered Jan 8 at 8:22
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
answered Jan 8 at 8:22
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
56.6k42159
add a comment |
add a comment |
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$begingroup$
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