If $f$ is a Lebesgue integrable function on $[0,1]$ , find polynomials ${f_n }_{n=1}^{infty}$ such that...
$begingroup$
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
edited Jan 14 at 22:26
Bernard
120k740113
asked Jan 14 at 20:36
J.GuoJ.Guo
3429
$endgroup$
|
show 1 more comment
$begingroup$
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
edited Jan 14 at 22:26
Bernard
120k740113
asked Jan 14 at 20:36
J.GuoJ.Guo
3429
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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
edited Jan 14 at 22:26
Bernard
120k740113
asked Jan 14 at 20:36
J.GuoJ.Guo
3429
$endgroup$
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
edited Jan 14 at 22:26
Bernard
120k740113
asked Jan 14 at 20:36
J.GuoJ.Guo
3429
edited Jan 14 at 22:26
Bernard
120k740113
asked Jan 14 at 20:36
J.GuoJ.Guo
3429
edited Jan 14 at 22:26
Bernard
120k740113
edited Jan 14 at 22:26
Bernard
120k740113
edited Jan 14 at 22:26
Bernard
120k740113
120k740113
asked Jan 14 at 20:36
J.GuoJ.Guo
3429
asked Jan 14 at 20:36
J.GuoJ.Guo
3429
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
|
show 1 more comment
$begingroup$
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
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
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
$$|f-p|_{L^1} le |f-g|_{L^1} + |g-p|_{L^1} le |f-g|_{L^1} + |g-p|_{L^infty} < 2eps.$$
answered Jan 14 at 20:45
Umberto P.Umberto P.
39.1k13066
$endgroup$
$begingroup$
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
add a comment |
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$begingroup$
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
$$|f-p|_{L^1} le |f-g|_{L^1} + |g-p|_{L^1} le |f-g|_{L^1} + |g-p|_{L^infty} < 2eps.$$
answered Jan 14 at 20:45
Umberto P.Umberto P.
39.1k13066
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$$|f-p|_{L^1} le |f-g|_{L^1} + |g-p|_{L^1} le |f-g|_{L^1} + |g-p|_{L^infty} < 2eps.$$
answered Jan 14 at 20:45
Umberto P.Umberto P.
39.1k13066
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$$|f-p|_{L^1} le |f-g|_{L^1} + |g-p|_{L^1} le |f-g|_{L^1} + |g-p|_{L^infty} < 2eps.$$
answered Jan 14 at 20:45
Umberto P.Umberto P.
39.1k13066
$endgroup$
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
$$|f-p|_{L^1} le |f-g|_{L^1} + |g-p|_{L^1} le |f-g|_{L^1} + |g-p|_{L^infty} < 2eps.$$
answered Jan 14 at 20:45
Umberto P.Umberto P.
39.1k13066
answered Jan 14 at 20:45
Umberto P.Umberto P.
39.1k13066
answered Jan 14 at 20:45
Umberto P.Umberto P.
39.1k13066
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
add a comment |
$begingroup$
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
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Lusin's theorem works in $p=1$.
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– Umberto P.
Jan 14 at 21:20
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Yes , it works in $p ge 1$ , but I'm not sure when $p lt 1$ .
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– J.Guo
Jan 14 at 21:24
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Yes , it works in $p ge 1$ , but I'm not sure when $p lt 1$ .
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– J.Guo
Jan 14 at 21:24
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$begingroup$
How do you choose the $a_n$?
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– Severin Schraven
Jan 14 at 20:41
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@Severin Schraven $a_n =int_0^1 f(x) e^{-2 pi inx} , dx$ , the fourier coefficient of $f$ .
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– 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.
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– Severin Schraven
Jan 14 at 20:49
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Thanks for the help , but what about the condition when $p lt 1$ ?
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– J.Guo
Jan 14 at 20:54
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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 $p$.
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– Severin Schraven
Jan 14 at 22:02