Strict Inequality for Fourier Coefficients
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I've been trying to solve this inequality but I only get the obvious part which is the $leq$ part. I need the $<$. The problem is the following:
Given a subset $Asubset [0,1)$ of measure not $0$, prove the following strict inequality:
$$
|widehat{chi_A}(n)| = bigg|int_A e^{2pi i n x} d x bigg| < |A|
$$
where $widehat{chi_A}(n)$ denotes the $n$-th Fourier coefficient of the function $chi_A$, that is, $1$ in $A$ and $0$ outside.
Thank you very much.
real-analysis fourier-analysis fourier-series fourier-transform
asked Dec 8 '18 at 10:27
Rubén FernándezRubén Fernández
51
$endgroup$
add a comment |
$begingroup$
I've been trying to solve this inequality but I only get the obvious part which is the $leq$ part. I need the $<$. The problem is the following:
Given a subset $Asubset [0,1)$ of measure not $0$, prove the following strict inequality:
$$
|widehat{chi_A}(n)| = bigg|int_A e^{2pi i n x} d x bigg| < |A|
$$
where $widehat{chi_A}(n)$ denotes the $n$-th Fourier coefficient of the function $chi_A$, that is, $1$ in $A$ and $0$ outside.
Thank you very much.
real-analysis fourier-analysis fourier-series fourier-transform
asked Dec 8 '18 at 10:27
Rubén FernándezRubén Fernández
51
$endgroup$
$begingroup$
You need to assume $nne0$. It's obvious that in fact $widehat{chi_A}(0)=|A|$.
$endgroup$
– David C. Ullrich
Dec 8 '18 at 14:18
add a comment |
$begingroup$
I've been trying to solve this inequality but I only get the obvious part which is the $leq$ part. I need the $<$. The problem is the following:
Given a subset $Asubset [0,1)$ of measure not $0$, prove the following strict inequality:
$$
|widehat{chi_A}(n)| = bigg|int_A e^{2pi i n x} d x bigg| < |A|
$$
where $widehat{chi_A}(n)$ denotes the $n$-th Fourier coefficient of the function $chi_A$, that is, $1$ in $A$ and $0$ outside.
Thank you very much.
real-analysis fourier-analysis fourier-series fourier-transform
asked Dec 8 '18 at 10:27
Rubén FernándezRubén Fernández
51
$endgroup$
I've been trying to solve this inequality but I only get the obvious part which is the $leq$ part. I need the $<$. The problem is the following:
Given a subset $Asubset [0,1)$ of measure not $0$, prove the following strict inequality:
$$
|widehat{chi_A}(n)| = bigg|int_A e^{2pi i n x} d x bigg| < |A|
$$
where $widehat{chi_A}(n)$ denotes the $n$-th Fourier coefficient of the function $chi_A$, that is, $1$ in $A$ and $0$ outside.
Thank you very much.
real-analysis fourier-analysis fourier-series fourier-transform
real-analysis fourier-analysis fourier-series fourier-transform
asked Dec 8 '18 at 10:27
Rubén FernándezRubén Fernández
51
asked Dec 8 '18 at 10:27
Rubén FernándezRubén Fernández
51
asked Dec 8 '18 at 10:27
Rubén FernándezRubén Fernández
51
asked Dec 8 '18 at 10:27
Rubén FernándezRubén Fernández
51
asked Dec 8 '18 at 10:27
Rubén FernándezRubén Fernández
51
51
$begingroup$
You need to assume $nne0$. It's obvious that in fact $widehat{chi_A}(0)=|A|$.
$endgroup$
– David C. Ullrich
Dec 8 '18 at 14:18
add a comment |
$begingroup$
You need to assume $nne0$. It's obvious that in fact $widehat{chi_A}(0)=|A|$.
$endgroup$
– David C. Ullrich
Dec 8 '18 at 14:18
$begingroup$
You need to assume $nne0$. It's obvious that in fact $widehat{chi_A}(0)=|A|$.
$endgroup$
– David C. Ullrich
Dec 8 '18 at 14:18
$begingroup$
You need to assume $nne0$. It's obvious that in fact $widehat{chi_A}(0)=|A|$.
$endgroup$
– David C. Ullrich
Dec 8 '18 at 14:18
add a comment |
1 Answer
1
active
oldest
votes
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Let $f(x)=e^{2pi inx}$. If equality holds then $int_A f(x)dx=|A|e^{it}$ for some real $t$. Hence $int_A [e^{-it}f(x)-1]dx=0$. Taking real part we get $int_A [cos (2pi nx-t)-1]dx=0$. The integrand is $leq 0$ so this implies $cos (2pi nx-t)-1=0$ for almost all $x in A$. In other words $2pi nx-t in 2pi mathbb Z$ for almost all $x$. can you get a contradiction from this? [Hint: take $x,x'$ with $x-x'$ irrational]. Works only if $n neq 0$. Thanks to David Ullrich for pointing out.
answered Dec 8 '18 at 11:54
Kavi Rama MurthyKavi Rama Murthy
59.8k42161
$endgroup$
$begingroup$
Got it! Thank you very much.
$endgroup$
– Rubén Fernández
Dec 8 '18 at 13:02
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Of course there's no contradiction if $n=0$.
$endgroup$
– David C. Ullrich
Dec 8 '18 at 14:17
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Yes, I forgot writing $nnot=0$ in the hypothesis.
$endgroup$
– Rubén Fernández
Dec 8 '18 at 16:11
add a comment |
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1 Answer
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active
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votes
1 Answer
1
active
oldest
votes
active
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active
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votes
$begingroup$
Let $f(x)=e^{2pi inx}$. If equality holds then $int_A f(x)dx=|A|e^{it}$ for some real $t$. Hence $int_A [e^{-it}f(x)-1]dx=0$. Taking real part we get $int_A [cos (2pi nx-t)-1]dx=0$. The integrand is $leq 0$ so this implies $cos (2pi nx-t)-1=0$ for almost all $x in A$. In other words $2pi nx-t in 2pi mathbb Z$ for almost all $x$. can you get a contradiction from this? [Hint: take $x,x'$ with $x-x'$ irrational]. Works only if $n neq 0$. Thanks to David Ullrich for pointing out.
answered Dec 8 '18 at 11:54
Kavi Rama MurthyKavi Rama Murthy
59.8k42161
$endgroup$
$begingroup$
Got it! Thank you very much.
$endgroup$
– Rubén Fernández
Dec 8 '18 at 13:02
$begingroup$
Of course there's no contradiction if $n=0$.
$endgroup$
– David C. Ullrich
Dec 8 '18 at 14:17
$begingroup$
Yes, I forgot writing $nnot=0$ in the hypothesis.
$endgroup$
– Rubén Fernández
Dec 8 '18 at 16:11
add a comment |
$begingroup$
Let $f(x)=e^{2pi inx}$. If equality holds then $int_A f(x)dx=|A|e^{it}$ for some real $t$. Hence $int_A [e^{-it}f(x)-1]dx=0$. Taking real part we get $int_A [cos (2pi nx-t)-1]dx=0$. The integrand is $leq 0$ so this implies $cos (2pi nx-t)-1=0$ for almost all $x in A$. In other words $2pi nx-t in 2pi mathbb Z$ for almost all $x$. can you get a contradiction from this? [Hint: take $x,x'$ with $x-x'$ irrational]. Works only if $n neq 0$. Thanks to David Ullrich for pointing out.
answered Dec 8 '18 at 11:54
Kavi Rama MurthyKavi Rama Murthy
59.8k42161
$endgroup$
$begingroup$
Got it! Thank you very much.
$endgroup$
– Rubén Fernández
Dec 8 '18 at 13:02
$begingroup$
Of course there's no contradiction if $n=0$.
$endgroup$
– David C. Ullrich
Dec 8 '18 at 14:17
$begingroup$
Yes, I forgot writing $nnot=0$ in the hypothesis.
$endgroup$
– Rubén Fernández
Dec 8 '18 at 16:11
add a comment |
$begingroup$
Let $f(x)=e^{2pi inx}$. If equality holds then $int_A f(x)dx=|A|e^{it}$ for some real $t$. Hence $int_A [e^{-it}f(x)-1]dx=0$. Taking real part we get $int_A [cos (2pi nx-t)-1]dx=0$. The integrand is $leq 0$ so this implies $cos (2pi nx-t)-1=0$ for almost all $x in A$. In other words $2pi nx-t in 2pi mathbb Z$ for almost all $x$. can you get a contradiction from this? [Hint: take $x,x'$ with $x-x'$ irrational]. Works only if $n neq 0$. Thanks to David Ullrich for pointing out.
answered Dec 8 '18 at 11:54
Kavi Rama MurthyKavi Rama Murthy
59.8k42161
$endgroup$
Let $f(x)=e^{2pi inx}$. If equality holds then $int_A f(x)dx=|A|e^{it}$ for some real $t$. Hence $int_A [e^{-it}f(x)-1]dx=0$. Taking real part we get $int_A [cos (2pi nx-t)-1]dx=0$. The integrand is $leq 0$ so this implies $cos (2pi nx-t)-1=0$ for almost all $x in A$. In other words $2pi nx-t in 2pi mathbb Z$ for almost all $x$. can you get a contradiction from this? [Hint: take $x,x'$ with $x-x'$ irrational]. Works only if $n neq 0$. Thanks to David Ullrich for pointing out.
answered Dec 8 '18 at 11:54
Kavi Rama MurthyKavi Rama Murthy
59.8k42161
edited Dec 8 '18 at 23:14
answered Dec 8 '18 at 11:54
Kavi Rama MurthyKavi Rama Murthy
59.8k42161
answered Dec 8 '18 at 11:54
Kavi Rama MurthyKavi Rama Murthy
59.8k42161
answered Dec 8 '18 at 11:54
Kavi Rama MurthyKavi Rama Murthy
59.8k42161
59.8k42161
$begingroup$
Got it! Thank you very much.
$endgroup$
– Rubén Fernández
Dec 8 '18 at 13:02
$begingroup$
Of course there's no contradiction if $n=0$.
$endgroup$
– David C. Ullrich
Dec 8 '18 at 14:17
$begingroup$
Yes, I forgot writing $nnot=0$ in the hypothesis.
$endgroup$
– Rubén Fernández
Dec 8 '18 at 16:11
add a comment |
$begingroup$
Got it! Thank you very much.
$endgroup$
– Rubén Fernández
Dec 8 '18 at 13:02
$begingroup$
Of course there's no contradiction if $n=0$.
$endgroup$
– David C. Ullrich
Dec 8 '18 at 14:17
$begingroup$
Yes, I forgot writing $nnot=0$ in the hypothesis.
$endgroup$
– Rubén Fernández
Dec 8 '18 at 16:11
$begingroup$
Got it! Thank you very much.
$endgroup$
– Rubén Fernández
Dec 8 '18 at 13:02
$begingroup$
Got it! Thank you very much.
$endgroup$
– Rubén Fernández
Dec 8 '18 at 13:02
$begingroup$
Of course there's no contradiction if $n=0$.
$endgroup$
– David C. Ullrich
Dec 8 '18 at 14:17
$begingroup$
Of course there's no contradiction if $n=0$.
$endgroup$
– David C. Ullrich
Dec 8 '18 at 14:17
$begingroup$
Yes, I forgot writing $nnot=0$ in the hypothesis.
$endgroup$
– Rubén Fernández
Dec 8 '18 at 16:11
$begingroup$
Yes, I forgot writing $nnot=0$ in the hypothesis.
$endgroup$
– Rubén Fernández
Dec 8 '18 at 16:11
add a comment |
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$begingroup$
You need to assume $nne0$. It's obvious that in fact $widehat{chi_A}(0)=|A|$.
$endgroup$
– David C. Ullrich
Dec 8 '18 at 14:18