If $f$ is a Lebesgue integrable function on $[0,1]$ , find polynomials ${f_n }_{n=1}^{infty}$ such that...












0












$begingroup$


If $fin L^p$ is supported on $[0,1]$ , can we find a sequence of polynomials ${f_n }_{n=1}^{infty}$ such that $||f-f_n||_{L^1_{[0,1]}}to 0$ ?

My attempt:

We use $L^p$ denote $L^p_{[0,1]}$ for simplicity . If $fin L^2$ , then let $$S_Nf(x)=sum_{-N}^N a_n e^{- inx}$$ and by Weierstrass's approximation theorem we can find $f_n$ such that $||f_n-S_nf||_{L^{infty}} le frac 1n$ , then by Holder inequality we get $$||f-f_n||_{L^1} le ||f-f_n||_{L^2}||1||_{L^2} le (||f-S_nf||_{L^2}+||S_nf-f_n||_{L^2})||1||_{L^2}to 0$$ Since $[0,1]$ has finite measure , if $p ge2$ then $f in L^p$ implies $f in L^2$ , so we only need to check the condition when $p lt 2$ . But for $p neq 2$ , $L^p$ is not a Hilbert space , we can not get $$||f-S_Nf||_{L^p}to 0$$ Does the conclusion still valid for $p lt 2$ ?






real-analysis functional-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    How do you choose the $a_n$?
    $endgroup$
    – Severin Schraven
    Jan 14 at 20:41










  • $begingroup$
    @Severin Schraven $a_n =int_0^1 f(x) e^{-2 pi inx} , dx$ , the fourier coefficient of $f$ .
    $endgroup$
    – J.Guo
    Jan 14 at 20:44






  • 1




    $begingroup$
    Ah ok. Your claim still holds true for $pgeq 1$. You have to approximate by first by a sequence of continuous functions (if you have not seen this before: first approximate by simple functions, the simple functions by simple functions with open support (for this you have to use the regularity of the Lebesgue measure) and then those by continuous functions). Then you can use Stone-Weierstrass to approximate your continuous functions by polynomials.
    $endgroup$
    – Severin Schraven
    Jan 14 at 20:49










  • $begingroup$
    Thanks for the help , but what about the condition when $p lt 1$ ?
    $endgroup$
    – J.Guo
    Jan 14 at 20:54










  • $begingroup$
    The proof goes through for $0<p<1$. You have just to keep in mind that you do not have the triangle inequality, but $$ Vert f + g Vert_{p} leq 2^{frac{1}{p-1}} (Vert f Vert_p + Vert g Vert_p $$ there is no other point where you need anything about the $p$.
    $endgroup$
    – Severin Schraven
    Jan 14 at 22:02


















0












$begingroup$


If $fin L^p$ is supported on $[0,1]$ , can we find a sequence of polynomials ${f_n }_{n=1}^{infty}$ such that $||f-f_n||_{L^1_{[0,1]}}to 0$ ?

My attempt:

We use $L^p$ denote $L^p_{[0,1]}$ for simplicity . If $fin L^2$ , then let $$S_Nf(x)=sum_{-N}^N a_n e^{- inx}$$ and by Weierstrass's approximation theorem we can find $f_n$ such that $||f_n-S_nf||_{L^{infty}} le frac 1n$ , then by Holder inequality we get $$||f-f_n||_{L^1} le ||f-f_n||_{L^2}||1||_{L^2} le (||f-S_nf||_{L^2}+||S_nf-f_n||_{L^2})||1||_{L^2}to 0$$ Since $[0,1]$ has finite measure , if $p ge2$ then $f in L^p$ implies $f in L^2$ , so we only need to check the condition when $p lt 2$ . But for $p neq 2$ , $L^p$ is not a Hilbert space , we can not get $$||f-S_Nf||_{L^p}to 0$$ Does the conclusion still valid for $p lt 2$ ?






real-analysis functional-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    How do you choose the $a_n$?
    $endgroup$
    – Severin Schraven
    Jan 14 at 20:41










  • $begingroup$
    @Severin Schraven $a_n =int_0^1 f(x) e^{-2 pi inx} , dx$ , the fourier coefficient of $f$ .
    $endgroup$
    – J.Guo
    Jan 14 at 20:44






  • 1




    $begingroup$
    Ah ok. Your claim still holds true for $pgeq 1$. You have to approximate by first by a sequence of continuous functions (if you have not seen this before: first approximate by simple functions, the simple functions by simple functions with open support (for this you have to use the regularity of the Lebesgue measure) and then those by continuous functions). Then you can use Stone-Weierstrass to approximate your continuous functions by polynomials.
    $endgroup$
    – Severin Schraven
    Jan 14 at 20:49










  • $begingroup$
    Thanks for the help , but what about the condition when $p lt 1$ ?
    $endgroup$
    – J.Guo
    Jan 14 at 20:54










  • $begingroup$
    The proof goes through for $0<p<1$. You have just to keep in mind that you do not have the triangle inequality, but $$ Vert f + g Vert_{p} leq 2^{frac{1}{p-1}} (Vert f Vert_p + Vert g Vert_p $$ there is no other point where you need anything about the $p$.
    $endgroup$
    – Severin Schraven
    Jan 14 at 22:02
















0












0








0





$begingroup$


If $fin L^p$ is supported on $[0,1]$ , can we find a sequence of polynomials ${f_n }_{n=1}^{infty}$ such that $||f-f_n||_{L^1_{[0,1]}}to 0$ ?

My attempt:

We use $L^p$ denote $L^p_{[0,1]}$ for simplicity . If $fin L^2$ , then let $$S_Nf(x)=sum_{-N}^N a_n e^{- inx}$$ and by Weierstrass's approximation theorem we can find $f_n$ such that $||f_n-S_nf||_{L^{infty}} le frac 1n$ , then by Holder inequality we get $$||f-f_n||_{L^1} le ||f-f_n||_{L^2}||1||_{L^2} le (||f-S_nf||_{L^2}+||S_nf-f_n||_{L^2})||1||_{L^2}to 0$$ Since $[0,1]$ has finite measure , if $p ge2$ then $f in L^p$ implies $f in L^2$ , so we only need to check the condition when $p lt 2$ . But for $p neq 2$ , $L^p$ is not a Hilbert space , we can not get $$||f-S_Nf||_{L^p}to 0$$ Does the conclusion still valid for $p lt 2$ ?






real-analysis functional-analysis







share|cite|improve this question











$endgroup$




If $fin L^p$ is supported on $[0,1]$ , can we find a sequence of polynomials ${f_n }_{n=1}^{infty}$ such that $||f-f_n||_{L^1_{[0,1]}}to 0$ ?

My attempt:

We use $L^p$ denote $L^p_{[0,1]}$ for simplicity . If $fin L^2$ , then let $$S_Nf(x)=sum_{-N}^N a_n e^{- inx}$$ and by Weierstrass's approximation theorem we can find $f_n$ such that $||f_n-S_nf||_{L^{infty}} le frac 1n$ , then by Holder inequality we get $$||f-f_n||_{L^1} le ||f-f_n||_{L^2}||1||_{L^2} le (||f-S_nf||_{L^2}+||S_nf-f_n||_{L^2})||1||_{L^2}to 0$$ Since $[0,1]$ has finite measure , if $p ge2$ then $f in L^p$ implies $f in L^2$ , so we only need to check the condition when $p lt 2$ . But for $p neq 2$ , $L^p$ is not a Hilbert space , we can not get $$||f-S_Nf||_{L^p}to 0$$ Does the conclusion still valid for $p lt 2$ ?






real-analysis functional-analysis




real-analysis functional-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 14 at 22:26









Bernard

120k740113




120k740113










asked Jan 14 at 20:36









J.GuoJ.Guo

3429




3429












  • $begingroup$
    How do you choose the $a_n$?
    $endgroup$
    – Severin Schraven
    Jan 14 at 20:41










  • $begingroup$
    @Severin Schraven $a_n =int_0^1 f(x) e^{-2 pi inx} , dx$ , the fourier coefficient of $f$ .
    $endgroup$
    – J.Guo
    Jan 14 at 20:44






  • 1




    $begingroup$
    Ah ok. Your claim still holds true for $pgeq 1$. You have to approximate by first by a sequence of continuous functions (if you have not seen this before: first approximate by simple functions, the simple functions by simple functions with open support (for this you have to use the regularity of the Lebesgue measure) and then those by continuous functions). Then you can use Stone-Weierstrass to approximate your continuous functions by polynomials.
    $endgroup$
    – Severin Schraven
    Jan 14 at 20:49










  • $begingroup$
    Thanks for the help , but what about the condition when $p lt 1$ ?
    $endgroup$
    – J.Guo
    Jan 14 at 20:54










  • $begingroup$
    The proof goes through for $0<p<1$. You have just to keep in mind that you do not have the triangle inequality, but $$ Vert f + g Vert_{p} leq 2^{frac{1}{p-1}} (Vert f Vert_p + Vert g Vert_p $$ there is no other point where you need anything about the $p$.
    $endgroup$
    – Severin Schraven
    Jan 14 at 22:02




















  • $begingroup$
    How do you choose the $a_n$?
    $endgroup$
    – Severin Schraven
    Jan 14 at 20:41










  • $begingroup$
    @Severin Schraven $a_n =int_0^1 f(x) e^{-2 pi inx} , dx$ , the fourier coefficient of $f$ .
    $endgroup$
    – J.Guo
    Jan 14 at 20:44






  • 1




    $begingroup$
    Ah ok. Your claim still holds true for $pgeq 1$. You have to approximate by first by a sequence of continuous functions (if you have not seen this before: first approximate by simple functions, the simple functions by simple functions with open support (for this you have to use the regularity of the Lebesgue measure) and then those by continuous functions). Then you can use Stone-Weierstrass to approximate your continuous functions by polynomials.
    $endgroup$
    – Severin Schraven
    Jan 14 at 20:49










  • $begingroup$
    Thanks for the help , but what about the condition when $p lt 1$ ?
    $endgroup$
    – J.Guo
    Jan 14 at 20:54










  • $begingroup$
    The proof goes through for $0<p<1$. You have just to keep in mind that you do not have the triangle inequality, but $$ Vert f + g Vert_{p} leq 2^{frac{1}{p-1}} (Vert f Vert_p + Vert g Vert_p $$ there is no other point where you need anything about the $p$.
    $endgroup$
    – Severin Schraven
    Jan 14 at 22:02


















$begingroup$
How do you choose the $a_n$?
$endgroup$
– Severin Schraven
Jan 14 at 20:41




$begingroup$
How do you choose the $a_n$?
$endgroup$
– Severin Schraven
Jan 14 at 20:41












$begingroup$
@Severin Schraven $a_n =int_0^1 f(x) e^{-2 pi inx} , dx$ , the fourier coefficient of $f$ .
$endgroup$
– J.Guo
Jan 14 at 20:44




$begingroup$
@Severin Schraven $a_n =int_0^1 f(x) e^{-2 pi inx} , dx$ , the fourier coefficient of $f$ .
$endgroup$
– J.Guo
Jan 14 at 20:44




1




1




$begingroup$
Ah ok. Your claim still holds true for $pgeq 1$. You have to approximate by first by a sequence of continuous functions (if you have not seen this before: first approximate by simple functions, the simple functions by simple functions with open support (for this you have to use the regularity of the Lebesgue measure) and then those by continuous functions). Then you can use Stone-Weierstrass to approximate your continuous functions by polynomials.
$endgroup$
– Severin Schraven
Jan 14 at 20:49




$begingroup$
Ah ok. Your claim still holds true for $pgeq 1$. You have to approximate by first by a sequence of continuous functions (if you have not seen this before: first approximate by simple functions, the simple functions by simple functions with open support (for this you have to use the regularity of the Lebesgue measure) and then those by continuous functions). Then you can use Stone-Weierstrass to approximate your continuous functions by polynomials.
$endgroup$
– Severin Schraven
Jan 14 at 20:49












$begingroup$
Thanks for the help , but what about the condition when $p lt 1$ ?
$endgroup$
– J.Guo
Jan 14 at 20:54




$begingroup$
Thanks for the help , but what about the condition when $p lt 1$ ?
$endgroup$
– J.Guo
Jan 14 at 20:54












$begingroup$
The proof goes through for $0<p<1$. You have just to keep in mind that you do not have the triangle inequality, but $$ Vert f + g Vert_{p} leq 2^{frac{1}{p-1}} (Vert f Vert_p + Vert g Vert_p $$ there is no other point where you need anything about the $p$.
$endgroup$
– Severin Schraven
Jan 14 at 22:02






$begingroup$
The proof goes through for $0<p<1$. You have just to keep in mind that you do not have the triangle inequality, but $$ Vert f + g Vert_{p} leq 2^{frac{1}{p-1}} (Vert f Vert_p + Vert g Vert_p $$ there is no other point where you need anything about the $p$.
$endgroup$
– Severin Schraven
Jan 14 at 22:02












1 Answer
1






active

oldest

votes


















3












$begingroup$

The Stone-Weierstrass theorem states that any continuous function $g$ on $[0,1]$ may be approximated uniformly by a polynomial $p$. That is, given $newcommand{eps}{varepsilon}eps > 0$ there exists a polynomial $p$ such that $sup_{x in [0,1]} |g(x) - p(x)| < eps$.



Lusin's theorem tells you that continuous functions are dense in $L^1([0,1])$. That is, for any $f in L^1([0,1])$ and any $eps > 0$ there exists a continuous function $g$ defined on $[0,1]$ for which $|f-g|_{L^1} < eps$.



Start with $f in L^1([0,1])$ and $eps > 0$. Choose $g$ and then $p$ accordingly. Then
$$|f-p|_{L^1} le |f-g|_{L^1} + |g-p|_{L^1} le |f-g|_{L^1} + |g-p|_{L^infty} < 2eps.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the help . For the proof of Lusin's theorem used the inequality $||f+g||_{L^1} le ||f||_{L^1}+||g||_{L^1}$ , which is not true when $p lt 1$ . So how to deal with this ?
    $endgroup$
    – J.Guo
    Jan 14 at 21:01












  • $begingroup$
    Your question specifically asks for polynomials $f_n$ with $|f - f_n|_{L^1} to 0$. What is the role of $p < 1$?
    $endgroup$
    – Umberto P.
    Jan 14 at 21:07










  • $begingroup$
    Sorry , what I really what to ask is the condition $p lt 2$ .I
    $endgroup$
    – J.Guo
    Jan 14 at 21:19










  • $begingroup$
    Lusin's theorem works in $p=1$.
    $endgroup$
    – Umberto P.
    Jan 14 at 21:20










  • $begingroup$
    Yes , it works in $p ge 1$ , but I'm not sure when $p lt 1$ .
    $endgroup$
    – J.Guo
    Jan 14 at 21:24











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3073716%2fif-f-is-a-lebesgue-integrable-function-on-0-1-find-polynomials-f-n%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

The Stone-Weierstrass theorem states that any continuous function $g$ on $[0,1]$ may be approximated uniformly by a polynomial $p$. That is, given $newcommand{eps}{varepsilon}eps > 0$ there exists a polynomial $p$ such that $sup_{x in [0,1]} |g(x) - p(x)| < eps$.



Lusin's theorem tells you that continuous functions are dense in $L^1([0,1])$. That is, for any $f in L^1([0,1])$ and any $eps > 0$ there exists a continuous function $g$ defined on $[0,1]$ for which $|f-g|_{L^1} < eps$.



Start with $f in L^1([0,1])$ and $eps > 0$. Choose $g$ and then $p$ accordingly. Then
$$|f-p|_{L^1} le |f-g|_{L^1} + |g-p|_{L^1} le |f-g|_{L^1} + |g-p|_{L^infty} < 2eps.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the help . For the proof of Lusin's theorem used the inequality $||f+g||_{L^1} le ||f||_{L^1}+||g||_{L^1}$ , which is not true when $p lt 1$ . So how to deal with this ?
    $endgroup$
    – J.Guo
    Jan 14 at 21:01












  • $begingroup$
    Your question specifically asks for polynomials $f_n$ with $|f - f_n|_{L^1} to 0$. What is the role of $p < 1$?
    $endgroup$
    – Umberto P.
    Jan 14 at 21:07










  • $begingroup$
    Sorry , what I really what to ask is the condition $p lt 2$ .I
    $endgroup$
    – J.Guo
    Jan 14 at 21:19










  • $begingroup$
    Lusin's theorem works in $p=1$.
    $endgroup$
    – Umberto P.
    Jan 14 at 21:20










  • $begingroup$
    Yes , it works in $p ge 1$ , but I'm not sure when $p lt 1$ .
    $endgroup$
    – J.Guo
    Jan 14 at 21:24
















3












$begingroup$

The Stone-Weierstrass theorem states that any continuous function $g$ on $[0,1]$ may be approximated uniformly by a polynomial $p$. That is, given $newcommand{eps}{varepsilon}eps > 0$ there exists a polynomial $p$ such that $sup_{x in [0,1]} |g(x) - p(x)| < eps$.



Lusin's theorem tells you that continuous functions are dense in $L^1([0,1])$. That is, for any $f in L^1([0,1])$ and any $eps > 0$ there exists a continuous function $g$ defined on $[0,1]$ for which $|f-g|_{L^1} < eps$.



Start with $f in L^1([0,1])$ and $eps > 0$. Choose $g$ and then $p$ accordingly. Then
$$|f-p|_{L^1} le |f-g|_{L^1} + |g-p|_{L^1} le |f-g|_{L^1} + |g-p|_{L^infty} < 2eps.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the help . For the proof of Lusin's theorem used the inequality $||f+g||_{L^1} le ||f||_{L^1}+||g||_{L^1}$ , which is not true when $p lt 1$ . So how to deal with this ?
    $endgroup$
    – J.Guo
    Jan 14 at 21:01












  • $begingroup$
    Your question specifically asks for polynomials $f_n$ with $|f - f_n|_{L^1} to 0$. What is the role of $p < 1$?
    $endgroup$
    – Umberto P.
    Jan 14 at 21:07










  • $begingroup$
    Sorry , what I really what to ask is the condition $p lt 2$ .I
    $endgroup$
    – J.Guo
    Jan 14 at 21:19










  • $begingroup$
    Lusin's theorem works in $p=1$.
    $endgroup$
    – Umberto P.
    Jan 14 at 21:20










  • $begingroup$
    Yes , it works in $p ge 1$ , but I'm not sure when $p lt 1$ .
    $endgroup$
    – J.Guo
    Jan 14 at 21:24














3












3








3





$begingroup$

The Stone-Weierstrass theorem states that any continuous function $g$ on $[0,1]$ may be approximated uniformly by a polynomial $p$. That is, given $newcommand{eps}{varepsilon}eps > 0$ there exists a polynomial $p$ such that $sup_{x in [0,1]} |g(x) - p(x)| < eps$.



Lusin's theorem tells you that continuous functions are dense in $L^1([0,1])$. That is, for any $f in L^1([0,1])$ and any $eps > 0$ there exists a continuous function $g$ defined on $[0,1]$ for which $|f-g|_{L^1} < eps$.



Start with $f in L^1([0,1])$ and $eps > 0$. Choose $g$ and then $p$ accordingly. Then
$$|f-p|_{L^1} le |f-g|_{L^1} + |g-p|_{L^1} le |f-g|_{L^1} + |g-p|_{L^infty} < 2eps.$$






share|cite|improve this answer









$endgroup$



The Stone-Weierstrass theorem states that any continuous function $g$ on $[0,1]$ may be approximated uniformly by a polynomial $p$. That is, given $newcommand{eps}{varepsilon}eps > 0$ there exists a polynomial $p$ such that $sup_{x in [0,1]} |g(x) - p(x)| < eps$.



Lusin's theorem tells you that continuous functions are dense in $L^1([0,1])$. That is, for any $f in L^1([0,1])$ and any $eps > 0$ there exists a continuous function $g$ defined on $[0,1]$ for which $|f-g|_{L^1} < eps$.



Start with $f in L^1([0,1])$ and $eps > 0$. Choose $g$ and then $p$ accordingly. Then
$$|f-p|_{L^1} le |f-g|_{L^1} + |g-p|_{L^1} le |f-g|_{L^1} + |g-p|_{L^infty} < 2eps.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 14 at 20:45









Umberto P.Umberto P.

39.1k13066




39.1k13066












  • $begingroup$
    Thanks for the help . For the proof of Lusin's theorem used the inequality $||f+g||_{L^1} le ||f||_{L^1}+||g||_{L^1}$ , which is not true when $p lt 1$ . So how to deal with this ?
    $endgroup$
    – J.Guo
    Jan 14 at 21:01












  • $begingroup$
    Your question specifically asks for polynomials $f_n$ with $|f - f_n|_{L^1} to 0$. What is the role of $p < 1$?
    $endgroup$
    – Umberto P.
    Jan 14 at 21:07










  • $begingroup$
    Sorry , what I really what to ask is the condition $p lt 2$ .I
    $endgroup$
    – J.Guo
    Jan 14 at 21:19










  • $begingroup$
    Lusin's theorem works in $p=1$.
    $endgroup$
    – Umberto P.
    Jan 14 at 21:20










  • $begingroup$
    Yes , it works in $p ge 1$ , but I'm not sure when $p lt 1$ .
    $endgroup$
    – J.Guo
    Jan 14 at 21:24


















  • $begingroup$
    Thanks for the help . For the proof of Lusin's theorem used the inequality $||f+g||_{L^1} le ||f||_{L^1}+||g||_{L^1}$ , which is not true when $p lt 1$ . So how to deal with this ?
    $endgroup$
    – J.Guo
    Jan 14 at 21:01












  • $begingroup$
    Your question specifically asks for polynomials $f_n$ with $|f - f_n|_{L^1} to 0$. What is the role of $p < 1$?
    $endgroup$
    – Umberto P.
    Jan 14 at 21:07










  • $begingroup$
    Sorry , what I really what to ask is the condition $p lt 2$ .I
    $endgroup$
    – J.Guo
    Jan 14 at 21:19










  • $begingroup$
    Lusin's theorem works in $p=1$.
    $endgroup$
    – Umberto P.
    Jan 14 at 21:20










  • $begingroup$
    Yes , it works in $p ge 1$ , but I'm not sure when $p lt 1$ .
    $endgroup$
    – J.Guo
    Jan 14 at 21:24
















$begingroup$
Thanks for the help . For the proof of Lusin's theorem used the inequality $||f+g||_{L^1} le ||f||_{L^1}+||g||_{L^1}$ , which is not true when $p lt 1$ . So how to deal with this ?
$endgroup$
– J.Guo
Jan 14 at 21:01






$begingroup$
Thanks for the help . For the proof of Lusin's theorem used the inequality $||f+g||_{L^1} le ||f||_{L^1}+||g||_{L^1}$ , which is not true when $p lt 1$ . So how to deal with this ?
$endgroup$
– J.Guo
Jan 14 at 21:01














$begingroup$
Your question specifically asks for polynomials $f_n$ with $|f - f_n|_{L^1} to 0$. What is the role of $p < 1$?
$endgroup$
– Umberto P.
Jan 14 at 21:07




$begingroup$
Your question specifically asks for polynomials $f_n$ with $|f - f_n|_{L^1} to 0$. What is the role of $p < 1$?
$endgroup$
– Umberto P.
Jan 14 at 21:07












$begingroup$
Sorry , what I really what to ask is the condition $p lt 2$ .I
$endgroup$
– J.Guo
Jan 14 at 21:19




$begingroup$
Sorry , what I really what to ask is the condition $p lt 2$ .I
$endgroup$
– J.Guo
Jan 14 at 21:19












$begingroup$
Lusin's theorem works in $p=1$.
$endgroup$
– Umberto P.
Jan 14 at 21:20




$begingroup$
Lusin's theorem works in $p=1$.
$endgroup$
– Umberto P.
Jan 14 at 21:20












$begingroup$
Yes , it works in $p ge 1$ , but I'm not sure when $p lt 1$ .
$endgroup$
– J.Guo
Jan 14 at 21:24




$begingroup$
Yes , it works in $p ge 1$ , but I'm not sure when $p lt 1$ .
$endgroup$
– J.Guo
Jan 14 at 21:24


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3073716%2fif-f-is-a-lebesgue-integrable-function-on-0-1-find-polynomials-f-n%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Mario Kart Wii

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
Read more

The Binding of Isaac: Rebirth/Afterbirth

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
Read more

What does “Dominus providebit” mean?

Image
5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
Read more