Mixing
Particular continuous functions are dense on $L_p$
$begingroup$
I am trying to prove the next:
Let $finmathcal{L}_p(mathbb{R},mathcal{A}_{mathbb{R}}^*,overline{lambda)},$ $pin[1,+infty),$ and $epsilon>0$ given.
Then there is $A=A(epsilon)>0$ and $g:mathbb{R}rightarrowmathbb{R}$ continuous with $g(x)=0$ for all $xnotin[-A,A]$ such that $||f-g||<epsilon.$
Here $mathcal{A}_{mathbb{R}}^*$ is the $sigma-$algebra of Lebesgue and $overline{lambda}$ the Lebesgue measure on $mathbb{R}.$
I was using Lusin's theorem to give a continuous map $g_{epsilon}:mathbb{R}rightarrowmathbb{R}$ for each $epsilon>0$ such that $overline{lambda}({xinmathbb{R}:f(x)neq g_{epsilon}(x)})<epsilon,$ but I cannot see how to use this to give to $g$ the asked property.
I was thinking to use regularity of Lebesgue measure to give two disjoint closed subsets, one of them a compact set of the form $[-A,A]$ of $mathbb{R}$ and give a continuous function on $mathbb{R}$ such that $0leq f(x)leq 1$ for each $x inmathbb{R}$ and $f(x)=0 inmathbb{R}-[-A,A] $ and $f(x)=1 in [-A,A] $ but in this way I don't know how to ensure that $||f-g||<epsilon.$
Any kind of help is thanked in advanced.
measure-theory lebesgue-measure
asked Jan 29 at 7:26
Suiz96Suiz96
24717
$endgroup$
add a comment |
$begingroup$
I am trying to prove the next:
Let $finmathcal{L}_p(mathbb{R},mathcal{A}_{mathbb{R}}^*,overline{lambda)},$ $pin[1,+infty),$ and $epsilon>0$ given.
Then there is $A=A(epsilon)>0$ and $g:mathbb{R}rightarrowmathbb{R}$ continuous with $g(x)=0$ for all $xnotin[-A,A]$ such that $||f-g||<epsilon.$
Here $mathcal{A}_{mathbb{R}}^*$ is the $sigma-$algebra of Lebesgue and $overline{lambda}$ the Lebesgue measure on $mathbb{R}.$
I was using Lusin's theorem to give a continuous map $g_{epsilon}:mathbb{R}rightarrowmathbb{R}$ for each $epsilon>0$ such that $overline{lambda}({xinmathbb{R}:f(x)neq g_{epsilon}(x)})<epsilon,$ but I cannot see how to use this to give to $g$ the asked property.
I was thinking to use regularity of Lebesgue measure to give two disjoint closed subsets, one of them a compact set of the form $[-A,A]$ of $mathbb{R}$ and give a continuous function on $mathbb{R}$ such that $0leq f(x)leq 1$ for each $x inmathbb{R}$ and $f(x)=0 inmathbb{R}-[-A,A] $ and $f(x)=1 in [-A,A] $ but in this way I don't know how to ensure that $||f-g||<epsilon.$
Any kind of help is thanked in advanced.
measure-theory lebesgue-measure
asked Jan 29 at 7:26
Suiz96Suiz96
24717
$endgroup$
$begingroup$
this might be helpful. Although it is for the case $p=1$, it should be similar to other cases
$endgroup$
– supinf
Jan 29 at 8:39
$begingroup$
Thanks @supinf. I'm checking the link; it's so helpful.
$endgroup$
– Suiz96
Jan 29 at 18:48
add a comment |
$begingroup$
I am trying to prove the next:
Let $finmathcal{L}_p(mathbb{R},mathcal{A}_{mathbb{R}}^*,overline{lambda)},$ $pin[1,+infty),$ and $epsilon>0$ given.
Then there is $A=A(epsilon)>0$ and $g:mathbb{R}rightarrowmathbb{R}$ continuous with $g(x)=0$ for all $xnotin[-A,A]$ such that $||f-g||<epsilon.$
Here $mathcal{A}_{mathbb{R}}^*$ is the $sigma-$algebra of Lebesgue and $overline{lambda}$ the Lebesgue measure on $mathbb{R}.$
I was using Lusin's theorem to give a continuous map $g_{epsilon}:mathbb{R}rightarrowmathbb{R}$ for each $epsilon>0$ such that $overline{lambda}({xinmathbb{R}:f(x)neq g_{epsilon}(x)})<epsilon,$ but I cannot see how to use this to give to $g$ the asked property.
I was thinking to use regularity of Lebesgue measure to give two disjoint closed subsets, one of them a compact set of the form $[-A,A]$ of $mathbb{R}$ and give a continuous function on $mathbb{R}$ such that $0leq f(x)leq 1$ for each $x inmathbb{R}$ and $f(x)=0 inmathbb{R}-[-A,A] $ and $f(x)=1 in [-A,A] $ but in this way I don't know how to ensure that $||f-g||<epsilon.$
Any kind of help is thanked in advanced.
measure-theory lebesgue-measure
asked Jan 29 at 7:26
Suiz96Suiz96
24717
$endgroup$
I am trying to prove the next:
Let $finmathcal{L}_p(mathbb{R},mathcal{A}_{mathbb{R}}^*,overline{lambda)},$ $pin[1,+infty),$ and $epsilon>0$ given.
Then there is $A=A(epsilon)>0$ and $g:mathbb{R}rightarrowmathbb{R}$ continuous with $g(x)=0$ for all $xnotin[-A,A]$ such that $||f-g||<epsilon.$
Here $mathcal{A}_{mathbb{R}}^*$ is the $sigma-$algebra of Lebesgue and $overline{lambda}$ the Lebesgue measure on $mathbb{R}.$
I was using Lusin's theorem to give a continuous map $g_{epsilon}:mathbb{R}rightarrowmathbb{R}$ for each $epsilon>0$ such that $overline{lambda}({xinmathbb{R}:f(x)neq g_{epsilon}(x)})<epsilon,$ but I cannot see how to use this to give to $g$ the asked property.
I was thinking to use regularity of Lebesgue measure to give two disjoint closed subsets, one of them a compact set of the form $[-A,A]$ of $mathbb{R}$ and give a continuous function on $mathbb{R}$ such that $0leq f(x)leq 1$ for each $x inmathbb{R}$ and $f(x)=0 inmathbb{R}-[-A,A] $ and $f(x)=1 in [-A,A] $ but in this way I don't know how to ensure that $||f-g||<epsilon.$
Any kind of help is thanked in advanced.
measure-theory lebesgue-measure
measure-theory lebesgue-measure
asked Jan 29 at 7:26
Suiz96Suiz96
24717
asked Jan 29 at 7:26
Suiz96Suiz96
24717
edited Jan 29 at 8:22
Suiz96
asked Jan 29 at 7:26
Suiz96Suiz96
24717
asked Jan 29 at 7:26
Suiz96Suiz96
24717
asked Jan 29 at 7:26
Suiz96Suiz96
24717
24717
$begingroup$
this might be helpful. Although it is for the case $p=1$, it should be similar to other cases
$endgroup$
– supinf
Jan 29 at 8:39
$begingroup$
Thanks @supinf. I'm checking the link; it's so helpful.
$endgroup$
– Suiz96
Jan 29 at 18:48
add a comment |
$begingroup$
this might be helpful. Although it is for the case $p=1$, it should be similar to other cases
$endgroup$
– supinf
Jan 29 at 8:39
$begingroup$
Thanks @supinf. I'm checking the link; it's so helpful.
$endgroup$
– Suiz96
Jan 29 at 18:48
$begingroup$
this might be helpful. Although it is for the case $p=1$, it should be similar to other cases
$endgroup$
– supinf
Jan 29 at 8:39
$begingroup$
this might be helpful. Although it is for the case $p=1$, it should be similar to other cases
$endgroup$
– supinf
Jan 29 at 8:39
$begingroup$
Thanks @supinf. I'm checking the link; it's so helpful.
$endgroup$
– Suiz96
Jan 29 at 18:48
$begingroup$
Thanks @supinf. I'm checking the link; it's so helpful.
$endgroup$
– Suiz96
Jan 29 at 18:48
add a comment |
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$begingroup$
this might be helpful. Although it is for the case $p=1$, it should be similar to other cases
$endgroup$
– supinf
Jan 29 at 8:39
$begingroup$
Thanks @supinf. I'm checking the link; it's so helpful.
$endgroup$
– Suiz96
Jan 29 at 18:48