Operator norm on Lebesgue integrable functions
$begingroup$
Let $L_1([0,1],m)$ be the Banach space of $mathbb{K}$-valued integrable functions with respect to Lebesgue measure $m$, where $mathbb{K}$ is either $mathbb{R}$ or $mathbb{C}$. The norm on this space is defined like this: $||f||_1=int_{[0,1]}|f| dm$. I have to show that:
$a)$ For $n geq 2$ the operator $varphi_n(f)=int_{[0,1]} f g_n dm$, where $g_n(x)=n sin(n^2x)$ for $x in [0,1]$ is bounded with $||varphi_n||=n$.
$b)$ Show that there exists $f in L_1([0,1],m)$ such that $lim_{n to infty} |varphi_n( f)|=infty$.
MY ATTEMPT:
$g_n$ is Lebesgue integrable on $[0,1]$ since it's Riemann integrable. Hence $fg_n in L_1([0,1],m)$ and $|int_{[0,1]}fg_n dm| leq int_{[0,1]}|fg_n| dm$. We also have that $||fg_n||_1 leq ||f||_1||g_n||_infty$.
Thus $||varphi_n(f)||=|int_{[0,1]}fg_n dm| leq int_{[0,1]}|fg_n| dm=||fg_n||_1 leq ||f||_1 ||g_n||_infty$, i.e. $varphi_n$ is bounded. Now $||g_n||_infty=n$ since it's continuous on a bounded interval and the $essential$ $supremum$ is the same as the $max$. Now I would like to attain the equality with some function, and once that I find it I can use in part $b)$. Any ideas on the function?
functional-analysis lebesgue-integral lebesgue-measure norm
asked Dec 16 '18 at 22:00
user289143user289143
1,000313
$endgroup$
add a comment |
$begingroup$
Let $L_1([0,1],m)$ be the Banach space of $mathbb{K}$-valued integrable functions with respect to Lebesgue measure $m$, where $mathbb{K}$ is either $mathbb{R}$ or $mathbb{C}$. The norm on this space is defined like this: $||f||_1=int_{[0,1]}|f| dm$. I have to show that:
$a)$ For $n geq 2$ the operator $varphi_n(f)=int_{[0,1]} f g_n dm$, where $g_n(x)=n sin(n^2x)$ for $x in [0,1]$ is bounded with $||varphi_n||=n$.
$b)$ Show that there exists $f in L_1([0,1],m)$ such that $lim_{n to infty} |varphi_n( f)|=infty$.
MY ATTEMPT:
$g_n$ is Lebesgue integrable on $[0,1]$ since it's Riemann integrable. Hence $fg_n in L_1([0,1],m)$ and $|int_{[0,1]}fg_n dm| leq int_{[0,1]}|fg_n| dm$. We also have that $||fg_n||_1 leq ||f||_1||g_n||_infty$.
Thus $||varphi_n(f)||=|int_{[0,1]}fg_n dm| leq int_{[0,1]}|fg_n| dm=||fg_n||_1 leq ||f||_1 ||g_n||_infty$, i.e. $varphi_n$ is bounded. Now $||g_n||_infty=n$ since it's continuous on a bounded interval and the $essential$ $supremum$ is the same as the $max$. Now I would like to attain the equality with some function, and once that I find it I can use in part $b)$. Any ideas on the function?
functional-analysis lebesgue-integral lebesgue-measure norm
asked Dec 16 '18 at 22:00
user289143user289143
1,000313
$endgroup$
add a comment |
$begingroup$
Let $L_1([0,1],m)$ be the Banach space of $mathbb{K}$-valued integrable functions with respect to Lebesgue measure $m$, where $mathbb{K}$ is either $mathbb{R}$ or $mathbb{C}$. The norm on this space is defined like this: $||f||_1=int_{[0,1]}|f| dm$. I have to show that:
$a)$ For $n geq 2$ the operator $varphi_n(f)=int_{[0,1]} f g_n dm$, where $g_n(x)=n sin(n^2x)$ for $x in [0,1]$ is bounded with $||varphi_n||=n$.
$b)$ Show that there exists $f in L_1([0,1],m)$ such that $lim_{n to infty} |varphi_n( f)|=infty$.
MY ATTEMPT:
$g_n$ is Lebesgue integrable on $[0,1]$ since it's Riemann integrable. Hence $fg_n in L_1([0,1],m)$ and $|int_{[0,1]}fg_n dm| leq int_{[0,1]}|fg_n| dm$. We also have that $||fg_n||_1 leq ||f||_1||g_n||_infty$.
Thus $||varphi_n(f)||=|int_{[0,1]}fg_n dm| leq int_{[0,1]}|fg_n| dm=||fg_n||_1 leq ||f||_1 ||g_n||_infty$, i.e. $varphi_n$ is bounded. Now $||g_n||_infty=n$ since it's continuous on a bounded interval and the $essential$ $supremum$ is the same as the $max$. Now I would like to attain the equality with some function, and once that I find it I can use in part $b)$. Any ideas on the function?
functional-analysis lebesgue-integral lebesgue-measure norm
asked Dec 16 '18 at 22:00
user289143user289143
1,000313
$endgroup$
Let $L_1([0,1],m)$ be the Banach space of $mathbb{K}$-valued integrable functions with respect to Lebesgue measure $m$, where $mathbb{K}$ is either $mathbb{R}$ or $mathbb{C}$. The norm on this space is defined like this: $||f||_1=int_{[0,1]}|f| dm$. I have to show that:
$a)$ For $n geq 2$ the operator $varphi_n(f)=int_{[0,1]} f g_n dm$, where $g_n(x)=n sin(n^2x)$ for $x in [0,1]$ is bounded with $||varphi_n||=n$.
$b)$ Show that there exists $f in L_1([0,1],m)$ such that $lim_{n to infty} |varphi_n( f)|=infty$.
MY ATTEMPT:
$g_n$ is Lebesgue integrable on $[0,1]$ since it's Riemann integrable. Hence $fg_n in L_1([0,1],m)$ and $|int_{[0,1]}fg_n dm| leq int_{[0,1]}|fg_n| dm$. We also have that $||fg_n||_1 leq ||f||_1||g_n||_infty$.
Thus $||varphi_n(f)||=|int_{[0,1]}fg_n dm| leq int_{[0,1]}|fg_n| dm=||fg_n||_1 leq ||f||_1 ||g_n||_infty$, i.e. $varphi_n$ is bounded. Now $||g_n||_infty=n$ since it's continuous on a bounded interval and the $essential$ $supremum$ is the same as the $max$. Now I would like to attain the equality with some function, and once that I find it I can use in part $b)$. Any ideas on the function?
functional-analysis lebesgue-integral lebesgue-measure norm
functional-analysis lebesgue-integral lebesgue-measure norm
asked Dec 16 '18 at 22:00
user289143user289143
1,000313
asked Dec 16 '18 at 22:00
user289143user289143
1,000313
asked Dec 16 '18 at 22:00
user289143user289143
1,000313
asked Dec 16 '18 at 22:00
user289143user289143
1,000313
asked Dec 16 '18 at 22:00
user289143user289143
1,000313
1,000313
add a comment |
add a comment |
1 Answer
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Answer to part a): it is not necessary to get equality in $|int_{[0,1]} f(x)nsin(n^{2}x),dx | leq |f|_1|g|_{infty}$. Instead, we get an 'approximate equality' as follows: let $epsilon >0$. Choose $delta >0$ such that $sin, x>1-epsilon$ for $frac {pi} 2 -delta <x <frac {pi} 2 +delta $. Let $f=frac {n^{2}} {2delta} I_A$ where $A=(frac {pi} {2n^{2}} -frac {delta} {n^{2}}, frac {pi} {2n^{2}} +frac {delta} {n^{2}})$. Simple calculations show that $|f|_1=1$ and $phi_n(f) >n(1-epsilon)$. Hence $|phi_n| geq n(1-epsilon)$ for all $epsilon >0$. Hence $|phi_n|geq n$.
answered Dec 17 '18 at 4:57
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Answer to part a): it is not necessary to get equality in $|int_{[0,1]} f(x)nsin(n^{2}x),dx | leq |f|_1|g|_{infty}$. Instead, we get an 'approximate equality' as follows: let $epsilon >0$. Choose $delta >0$ such that $sin, x>1-epsilon$ for $frac {pi} 2 -delta <x <frac {pi} 2 +delta $. Let $f=frac {n^{2}} {2delta} I_A$ where $A=(frac {pi} {2n^{2}} -frac {delta} {n^{2}}, frac {pi} {2n^{2}} +frac {delta} {n^{2}})$. Simple calculations show that $|f|_1=1$ and $phi_n(f) >n(1-epsilon)$. Hence $|phi_n| geq n(1-epsilon)$ for all $epsilon >0$. Hence $|phi_n|geq n$.
answered Dec 17 '18 at 4:57
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
$endgroup$
add a comment |
$begingroup$
Answer to part a): it is not necessary to get equality in $|int_{[0,1]} f(x)nsin(n^{2}x),dx | leq |f|_1|g|_{infty}$. Instead, we get an 'approximate equality' as follows: let $epsilon >0$. Choose $delta >0$ such that $sin, x>1-epsilon$ for $frac {pi} 2 -delta <x <frac {pi} 2 +delta $. Let $f=frac {n^{2}} {2delta} I_A$ where $A=(frac {pi} {2n^{2}} -frac {delta} {n^{2}}, frac {pi} {2n^{2}} +frac {delta} {n^{2}})$. Simple calculations show that $|f|_1=1$ and $phi_n(f) >n(1-epsilon)$. Hence $|phi_n| geq n(1-epsilon)$ for all $epsilon >0$. Hence $|phi_n|geq n$.
answered Dec 17 '18 at 4:57
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
$endgroup$
add a comment |
$begingroup$
Answer to part a): it is not necessary to get equality in $|int_{[0,1]} f(x)nsin(n^{2}x),dx | leq |f|_1|g|_{infty}$. Instead, we get an 'approximate equality' as follows: let $epsilon >0$. Choose $delta >0$ such that $sin, x>1-epsilon$ for $frac {pi} 2 -delta <x <frac {pi} 2 +delta $. Let $f=frac {n^{2}} {2delta} I_A$ where $A=(frac {pi} {2n^{2}} -frac {delta} {n^{2}}, frac {pi} {2n^{2}} +frac {delta} {n^{2}})$. Simple calculations show that $|f|_1=1$ and $phi_n(f) >n(1-epsilon)$. Hence $|phi_n| geq n(1-epsilon)$ for all $epsilon >0$. Hence $|phi_n|geq n$.
answered Dec 17 '18 at 4:57
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
$endgroup$
Answer to part a): it is not necessary to get equality in $|int_{[0,1]} f(x)nsin(n^{2}x),dx | leq |f|_1|g|_{infty}$. Instead, we get an 'approximate equality' as follows: let $epsilon >0$. Choose $delta >0$ such that $sin, x>1-epsilon$ for $frac {pi} 2 -delta <x <frac {pi} 2 +delta $. Let $f=frac {n^{2}} {2delta} I_A$ where $A=(frac {pi} {2n^{2}} -frac {delta} {n^{2}}, frac {pi} {2n^{2}} +frac {delta} {n^{2}})$. Simple calculations show that $|f|_1=1$ and $phi_n(f) >n(1-epsilon)$. Hence $|phi_n| geq n(1-epsilon)$ for all $epsilon >0$. Hence $|phi_n|geq n$.
answered Dec 17 '18 at 4:57
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
edited Dec 17 '18 at 6:41
answered Dec 17 '18 at 4:57
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
answered Dec 17 '18 at 4:57
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
answered Dec 17 '18 at 4:57
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
67.5k53067
add a comment |
add a comment |
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