The orthonormal basis $f_n(x) = e^{2pi i nx}$ for $L^2(mathbb R/mathbb Z)$ defining Fourier series
$begingroup$
I thought I understood this, but now I don't think I do.
So the functions $f_n(x) = e^{2 pi i n x}$ form an orthonormal basis for the complex Hilbert space $L^2(mathbb R/mathbb Z)$. This means for any square integrable function $f: mathbb R/mathbb Z rightarrow mathbb C$, I can find unique complex numbers $c_n in mathbb C$ such that
$$f = sumlimits_{n in mathbb Z} c_n f_n$$
Formally, all this means is that the partial sums $F_N = sumlimits_{n=-N}^N c_nf_n$ converge in the $L^2$ norm to $f$.
But if I take a random sequence ${c_n}$ of complex numbers, what is the sum
$$f:=sumlimits_{n in mathbb Z} c_nf_n?$$
It is supposed to be an element of $L^2(mathbb R/mathbb Z)$. But $f$ does not appear to define a function $mathbb R/mathbb Z rightarrow mathbb C$, or even an equivalence class of such functions (identifying those $f, g$ such that $||f-g||_2 = 0$). After all, the sum $sumlimits_n |c_n| |f_n| = sumlimits_n |c_n|$ need not converge absolutely.
real-analysis functional-analysis convergence fourier-analysis fourier-series
asked Jan 2 at 3:58
D_SD_S
14.3k61756
$endgroup$
add a comment |
$begingroup$
I thought I understood this, but now I don't think I do.
So the functions $f_n(x) = e^{2 pi i n x}$ form an orthonormal basis for the complex Hilbert space $L^2(mathbb R/mathbb Z)$. This means for any square integrable function $f: mathbb R/mathbb Z rightarrow mathbb C$, I can find unique complex numbers $c_n in mathbb C$ such that
$$f = sumlimits_{n in mathbb Z} c_n f_n$$
Formally, all this means is that the partial sums $F_N = sumlimits_{n=-N}^N c_nf_n$ converge in the $L^2$ norm to $f$.
But if I take a random sequence ${c_n}$ of complex numbers, what is the sum
$$f:=sumlimits_{n in mathbb Z} c_nf_n?$$
It is supposed to be an element of $L^2(mathbb R/mathbb Z)$. But $f$ does not appear to define a function $mathbb R/mathbb Z rightarrow mathbb C$, or even an equivalence class of such functions (identifying those $f, g$ such that $||f-g||_2 = 0$). After all, the sum $sumlimits_n |c_n| |f_n| = sumlimits_n |c_n|$ need not converge absolutely.
real-analysis functional-analysis convergence fourier-analysis fourier-series
asked Jan 2 at 3:58
D_SD_S
14.3k61756
$endgroup$
$begingroup$
To nitpick, the usual definition of $f = sumlimits_{n in mathbb Z} c_n f_n$ is not that the partial sums from $-N$ to $N$ $L^2$-converge to $f$ but that the partial sums from $-M$ to $N$ $L^2$-converge to $f$ as $(M,N)toinfty$. It turns out to not make a difference in this case (as a consequence of the $f_n$ being orthonormal) but it does make a difference for more general sums.
$endgroup$
– Eric Wofsey
Jan 2 at 4:08
$begingroup$
What do you mean by "random sequence"?
$endgroup$
– Adrián González-Pérez
Jan 2 at 12:48
add a comment |
$begingroup$
I thought I understood this, but now I don't think I do.
So the functions $f_n(x) = e^{2 pi i n x}$ form an orthonormal basis for the complex Hilbert space $L^2(mathbb R/mathbb Z)$. This means for any square integrable function $f: mathbb R/mathbb Z rightarrow mathbb C$, I can find unique complex numbers $c_n in mathbb C$ such that
$$f = sumlimits_{n in mathbb Z} c_n f_n$$
Formally, all this means is that the partial sums $F_N = sumlimits_{n=-N}^N c_nf_n$ converge in the $L^2$ norm to $f$.
But if I take a random sequence ${c_n}$ of complex numbers, what is the sum
$$f:=sumlimits_{n in mathbb Z} c_nf_n?$$
It is supposed to be an element of $L^2(mathbb R/mathbb Z)$. But $f$ does not appear to define a function $mathbb R/mathbb Z rightarrow mathbb C$, or even an equivalence class of such functions (identifying those $f, g$ such that $||f-g||_2 = 0$). After all, the sum $sumlimits_n |c_n| |f_n| = sumlimits_n |c_n|$ need not converge absolutely.
real-analysis functional-analysis convergence fourier-analysis fourier-series
asked Jan 2 at 3:58
D_SD_S
14.3k61756
$endgroup$
I thought I understood this, but now I don't think I do.
So the functions $f_n(x) = e^{2 pi i n x}$ form an orthonormal basis for the complex Hilbert space $L^2(mathbb R/mathbb Z)$. This means for any square integrable function $f: mathbb R/mathbb Z rightarrow mathbb C$, I can find unique complex numbers $c_n in mathbb C$ such that
$$f = sumlimits_{n in mathbb Z} c_n f_n$$
Formally, all this means is that the partial sums $F_N = sumlimits_{n=-N}^N c_nf_n$ converge in the $L^2$ norm to $f$.
But if I take a random sequence ${c_n}$ of complex numbers, what is the sum
$$f:=sumlimits_{n in mathbb Z} c_nf_n?$$
It is supposed to be an element of $L^2(mathbb R/mathbb Z)$. But $f$ does not appear to define a function $mathbb R/mathbb Z rightarrow mathbb C$, or even an equivalence class of such functions (identifying those $f, g$ such that $||f-g||_2 = 0$). After all, the sum $sumlimits_n |c_n| |f_n| = sumlimits_n |c_n|$ need not converge absolutely.
real-analysis functional-analysis convergence fourier-analysis fourier-series
real-analysis functional-analysis convergence fourier-analysis fourier-series
asked Jan 2 at 3:58
D_SD_S
14.3k61756
asked Jan 2 at 3:58
D_SD_S
14.3k61756
asked Jan 2 at 3:58
D_SD_S
14.3k61756
asked Jan 2 at 3:58
D_SD_S
14.3k61756
asked Jan 2 at 3:58
D_SD_S
14.3k61756
14.3k61756
$begingroup$
To nitpick, the usual definition of $f = sumlimits_{n in mathbb Z} c_n f_n$ is not that the partial sums from $-N$ to $N$ $L^2$-converge to $f$ but that the partial sums from $-M$ to $N$ $L^2$-converge to $f$ as $(M,N)toinfty$. It turns out to not make a difference in this case (as a consequence of the $f_n$ being orthonormal) but it does make a difference for more general sums.
$endgroup$
– Eric Wofsey
Jan 2 at 4:08
$begingroup$
What do you mean by "random sequence"?
$endgroup$
– Adrián González-Pérez
Jan 2 at 12:48
add a comment |
$begingroup$
To nitpick, the usual definition of $f = sumlimits_{n in mathbb Z} c_n f_n$ is not that the partial sums from $-N$ to $N$ $L^2$-converge to $f$ but that the partial sums from $-M$ to $N$ $L^2$-converge to $f$ as $(M,N)toinfty$. It turns out to not make a difference in this case (as a consequence of the $f_n$ being orthonormal) but it does make a difference for more general sums.
$endgroup$
– Eric Wofsey
Jan 2 at 4:08
$begingroup$
What do you mean by "random sequence"?
$endgroup$
– Adrián González-Pérez
Jan 2 at 12:48
$begingroup$
To nitpick, the usual definition of $f = sumlimits_{n in mathbb Z} c_n f_n$ is not that the partial sums from $-N$ to $N$ $L^2$-converge to $f$ but that the partial sums from $-M$ to $N$ $L^2$-converge to $f$ as $(M,N)toinfty$. It turns out to not make a difference in this case (as a consequence of the $f_n$ being orthonormal) but it does make a difference for more general sums.
$endgroup$
– Eric Wofsey
Jan 2 at 4:08
$begingroup$
To nitpick, the usual definition of $f = sumlimits_{n in mathbb Z} c_n f_n$ is not that the partial sums from $-N$ to $N$ $L^2$-converge to $f$ but that the partial sums from $-M$ to $N$ $L^2$-converge to $f$ as $(M,N)toinfty$. It turns out to not make a difference in this case (as a consequence of the $f_n$ being orthonormal) but it does make a difference for more general sums.
$endgroup$
– Eric Wofsey
Jan 2 at 4:08
$begingroup$
What do you mean by "random sequence"?
$endgroup$
– Adrián González-Pérez
Jan 2 at 12:48
$begingroup$
What do you mean by "random sequence"?
$endgroup$
– Adrián González-Pérez
Jan 2 at 12:48
add a comment |
1 Answer
1
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$begingroup$
It's not true that an arbitrary sequence $(c_n)$ represents an element of $L^2$. Indeed, the only sequences that represent elements of $L^2$ are those for which $sum c_nf_n$ actually does converge with respect to the $L^2$ norm to some function. It turns out that this convergence is equivalent to the sum $sum|c_n|^2$ converging.
answered Jan 2 at 4:03
Eric WofseyEric Wofsey
193k14222353
$endgroup$
$begingroup$
Okay that makes sense. I forgot that the Hilbert space direct sum is strictly smaller than the direct product. So if $f$ is continuous, does $f(x)$ equal its Fourier series almost everywhere?
$endgroup$
– D_S
Jan 2 at 4:07
$begingroup$
Yes but that is a deep theorem (much harder than the fact that its Fourier series converges to it in the $L^2$ norm).
$endgroup$
– Eric Wofsey
Jan 2 at 4:15
$begingroup$
Okay thank you. One more question, does everything you said also hold for continuous functions and their Fourier expansions on the quotient $mathbb A_k/k$? Or more generally for the orthonormal basis of matrix coefficients on $L^2$ of a compact group via Peter-Weyl?
$endgroup$
– D_S
Jan 2 at 4:20
$begingroup$
Everything I said in the answer applies to orthonormal bases for any Hilbert space. I don't know anything about almost everywhere convergence for Fourier series on more general groups.
$endgroup$
– Eric Wofsey
Jan 2 at 4:22
add a comment |
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1 Answer
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1 Answer
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oldest
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oldest
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$begingroup$
It's not true that an arbitrary sequence $(c_n)$ represents an element of $L^2$. Indeed, the only sequences that represent elements of $L^2$ are those for which $sum c_nf_n$ actually does converge with respect to the $L^2$ norm to some function. It turns out that this convergence is equivalent to the sum $sum|c_n|^2$ converging.
answered Jan 2 at 4:03
Eric WofseyEric Wofsey
193k14222353
$endgroup$
$begingroup$
Okay that makes sense. I forgot that the Hilbert space direct sum is strictly smaller than the direct product. So if $f$ is continuous, does $f(x)$ equal its Fourier series almost everywhere?
$endgroup$
– D_S
Jan 2 at 4:07
$begingroup$
Yes but that is a deep theorem (much harder than the fact that its Fourier series converges to it in the $L^2$ norm).
$endgroup$
– Eric Wofsey
Jan 2 at 4:15
$begingroup$
Okay thank you. One more question, does everything you said also hold for continuous functions and their Fourier expansions on the quotient $mathbb A_k/k$? Or more generally for the orthonormal basis of matrix coefficients on $L^2$ of a compact group via Peter-Weyl?
$endgroup$
– D_S
Jan 2 at 4:20
$begingroup$
Everything I said in the answer applies to orthonormal bases for any Hilbert space. I don't know anything about almost everywhere convergence for Fourier series on more general groups.
$endgroup$
– Eric Wofsey
Jan 2 at 4:22
add a comment |
$begingroup$
It's not true that an arbitrary sequence $(c_n)$ represents an element of $L^2$. Indeed, the only sequences that represent elements of $L^2$ are those for which $sum c_nf_n$ actually does converge with respect to the $L^2$ norm to some function. It turns out that this convergence is equivalent to the sum $sum|c_n|^2$ converging.
answered Jan 2 at 4:03
Eric WofseyEric Wofsey
193k14222353
$endgroup$
$begingroup$
Okay that makes sense. I forgot that the Hilbert space direct sum is strictly smaller than the direct product. So if $f$ is continuous, does $f(x)$ equal its Fourier series almost everywhere?
$endgroup$
– D_S
Jan 2 at 4:07
$begingroup$
Yes but that is a deep theorem (much harder than the fact that its Fourier series converges to it in the $L^2$ norm).
$endgroup$
– Eric Wofsey
Jan 2 at 4:15
$begingroup$
Okay thank you. One more question, does everything you said also hold for continuous functions and their Fourier expansions on the quotient $mathbb A_k/k$? Or more generally for the orthonormal basis of matrix coefficients on $L^2$ of a compact group via Peter-Weyl?
$endgroup$
– D_S
Jan 2 at 4:20
$begingroup$
Everything I said in the answer applies to orthonormal bases for any Hilbert space. I don't know anything about almost everywhere convergence for Fourier series on more general groups.
$endgroup$
– Eric Wofsey
Jan 2 at 4:22
add a comment |
$begingroup$
It's not true that an arbitrary sequence $(c_n)$ represents an element of $L^2$. Indeed, the only sequences that represent elements of $L^2$ are those for which $sum c_nf_n$ actually does converge with respect to the $L^2$ norm to some function. It turns out that this convergence is equivalent to the sum $sum|c_n|^2$ converging.
answered Jan 2 at 4:03
Eric WofseyEric Wofsey
193k14222353
$endgroup$
It's not true that an arbitrary sequence $(c_n)$ represents an element of $L^2$. Indeed, the only sequences that represent elements of $L^2$ are those for which $sum c_nf_n$ actually does converge with respect to the $L^2$ norm to some function. It turns out that this convergence is equivalent to the sum $sum|c_n|^2$ converging.
answered Jan 2 at 4:03
Eric WofseyEric Wofsey
193k14222353
answered Jan 2 at 4:03
Eric WofseyEric Wofsey
193k14222353
answered Jan 2 at 4:03
Eric WofseyEric Wofsey
193k14222353
answered Jan 2 at 4:03
Eric WofseyEric Wofsey
193k14222353
193k14222353
$begingroup$
Okay that makes sense. I forgot that the Hilbert space direct sum is strictly smaller than the direct product. So if $f$ is continuous, does $f(x)$ equal its Fourier series almost everywhere?
$endgroup$
– D_S
Jan 2 at 4:07
$begingroup$
Yes but that is a deep theorem (much harder than the fact that its Fourier series converges to it in the $L^2$ norm).
$endgroup$
– Eric Wofsey
Jan 2 at 4:15
$begingroup$
Okay thank you. One more question, does everything you said also hold for continuous functions and their Fourier expansions on the quotient $mathbb A_k/k$? Or more generally for the orthonormal basis of matrix coefficients on $L^2$ of a compact group via Peter-Weyl?
$endgroup$
– D_S
Jan 2 at 4:20
$begingroup$
Everything I said in the answer applies to orthonormal bases for any Hilbert space. I don't know anything about almost everywhere convergence for Fourier series on more general groups.
$endgroup$
– Eric Wofsey
Jan 2 at 4:22
add a comment |
$begingroup$
Okay that makes sense. I forgot that the Hilbert space direct sum is strictly smaller than the direct product. So if $f$ is continuous, does $f(x)$ equal its Fourier series almost everywhere?
$endgroup$
– D_S
Jan 2 at 4:07
$begingroup$
Yes but that is a deep theorem (much harder than the fact that its Fourier series converges to it in the $L^2$ norm).
$endgroup$
– Eric Wofsey
Jan 2 at 4:15
$begingroup$
Okay thank you. One more question, does everything you said also hold for continuous functions and their Fourier expansions on the quotient $mathbb A_k/k$? Or more generally for the orthonormal basis of matrix coefficients on $L^2$ of a compact group via Peter-Weyl?
$endgroup$
– D_S
Jan 2 at 4:20
$begingroup$
Everything I said in the answer applies to orthonormal bases for any Hilbert space. I don't know anything about almost everywhere convergence for Fourier series on more general groups.
$endgroup$
– Eric Wofsey
Jan 2 at 4:22
$begingroup$
Okay that makes sense. I forgot that the Hilbert space direct sum is strictly smaller than the direct product. So if $f$ is continuous, does $f(x)$ equal its Fourier series almost everywhere?
$endgroup$
– D_S
Jan 2 at 4:07
$begingroup$
Okay that makes sense. I forgot that the Hilbert space direct sum is strictly smaller than the direct product. So if $f$ is continuous, does $f(x)$ equal its Fourier series almost everywhere?
$endgroup$
– D_S
Jan 2 at 4:07
$begingroup$
Yes but that is a deep theorem (much harder than the fact that its Fourier series converges to it in the $L^2$ norm).
$endgroup$
– Eric Wofsey
Jan 2 at 4:15
$begingroup$
Yes but that is a deep theorem (much harder than the fact that its Fourier series converges to it in the $L^2$ norm).
$endgroup$
– Eric Wofsey
Jan 2 at 4:15
$begingroup$
Okay thank you. One more question, does everything you said also hold for continuous functions and their Fourier expansions on the quotient $mathbb A_k/k$? Or more generally for the orthonormal basis of matrix coefficients on $L^2$ of a compact group via Peter-Weyl?
$endgroup$
– D_S
Jan 2 at 4:20
$begingroup$
Okay thank you. One more question, does everything you said also hold for continuous functions and their Fourier expansions on the quotient $mathbb A_k/k$? Or more generally for the orthonormal basis of matrix coefficients on $L^2$ of a compact group via Peter-Weyl?
$endgroup$
– D_S
Jan 2 at 4:20
$begingroup$
Everything I said in the answer applies to orthonormal bases for any Hilbert space. I don't know anything about almost everywhere convergence for Fourier series on more general groups.
$endgroup$
– Eric Wofsey
Jan 2 at 4:22
$begingroup$
Everything I said in the answer applies to orthonormal bases for any Hilbert space. I don't know anything about almost everywhere convergence for Fourier series on more general groups.
$endgroup$
– Eric Wofsey
Jan 2 at 4:22
add a comment |
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$begingroup$
To nitpick, the usual definition of $f = sumlimits_{n in mathbb Z} c_n f_n$ is not that the partial sums from $-N$ to $N$ $L^2$-converge to $f$ but that the partial sums from $-M$ to $N$ $L^2$-converge to $f$ as $(M,N)toinfty$. It turns out to not make a difference in this case (as a consequence of the $f_n$ being orthonormal) but it does make a difference for more general sums.
$endgroup$
– Eric Wofsey
Jan 2 at 4:08
$begingroup$
What do you mean by "random sequence"?
$endgroup$
– Adrián González-Pérez
Jan 2 at 12:48