Uniform convergence of parametrised Fourier series.
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Let $X$ be a compact topological space, and $f: S^1 times X rightarrow mathbb{C}$ be a continuous function. Then for every point $x in X$ we can compute the Fourier coefficients
$$c_k(x) = int_{S^1} f(z,x) z^k text{d}z,$$
and construct the Fourier series as the limit of the partial sums
$$s_n(z, x) = sum_{k=-n}^n c_k z^k.$$
This must not converge point-wise, but taking Cesàro means
$$ sigma_l(z,x) = frac 1 lsum_{n = 0}^{l-1} s_n(z,x) = frac 1 l sum_{n=0}^{l-1}sum_{k=-n}^{n}c_kz^k = sum_{k=-l}^lfrac{l - |k|}{l}c_kz^k,$$
we get uniform convergence in $z$ by Fejér's theorem.
Note that $sigma_l(z,x)$ is still continuous in $x$.
In his book K-Theory, Atiyah claims that the $sigma_l$ converge uniformly in $x$, that is given $epsilon > 0$ there exists $l_0 in mathbb{N}$, such that for all $(z,x) in S^1 times X$ and $l geq l_0$ we have
$$|f(z,x) - sigma_l(z,x)| < epsilon .$$
I don't have any idea why this should be true, but my knowledge about Fourier series and analysis is pretty limited. Any help would be appreciated.
Here are some thoughts I had:
One obstacle is that the $sigma_l$ do not form a true series, because of the factor $frac 1 l$. So the classical Weierstrass test does not even apply here. Is there another kind of Weierstrass test that one could apply to this situation?
Because $X$ is compact, it would suffice to show uniform convergence locally.
analysis convergence fourier-analysis fourier-series uniform-convergence
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up vote
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Let $X$ be a compact topological space, and $f: S^1 times X rightarrow mathbb{C}$ be a continuous function. Then for every point $x in X$ we can compute the Fourier coefficients
$$c_k(x) = int_{S^1} f(z,x) z^k text{d}z,$$
and construct the Fourier series as the limit of the partial sums
$$s_n(z, x) = sum_{k=-n}^n c_k z^k.$$
This must not converge point-wise, but taking Cesàro means
$$ sigma_l(z,x) = frac 1 lsum_{n = 0}^{l-1} s_n(z,x) = frac 1 l sum_{n=0}^{l-1}sum_{k=-n}^{n}c_kz^k = sum_{k=-l}^lfrac{l - |k|}{l}c_kz^k,$$
we get uniform convergence in $z$ by Fejér's theorem.
Note that $sigma_l(z,x)$ is still continuous in $x$.
In his book K-Theory, Atiyah claims that the $sigma_l$ converge uniformly in $x$, that is given $epsilon > 0$ there exists $l_0 in mathbb{N}$, such that for all $(z,x) in S^1 times X$ and $l geq l_0$ we have
$$|f(z,x) - sigma_l(z,x)| < epsilon .$$
I don't have any idea why this should be true, but my knowledge about Fourier series and analysis is pretty limited. Any help would be appreciated.
Here are some thoughts I had:
One obstacle is that the $sigma_l$ do not form a true series, because of the factor $frac 1 l$. So the classical Weierstrass test does not even apply here. Is there another kind of Weierstrass test that one could apply to this situation?
Because $X$ is compact, it would suffice to show uniform convergence locally.
analysis convergence fourier-analysis fourier-series uniform-convergence
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $X$ be a compact topological space, and $f: S^1 times X rightarrow mathbb{C}$ be a continuous function. Then for every point $x in X$ we can compute the Fourier coefficients
$$c_k(x) = int_{S^1} f(z,x) z^k text{d}z,$$
and construct the Fourier series as the limit of the partial sums
$$s_n(z, x) = sum_{k=-n}^n c_k z^k.$$
This must not converge point-wise, but taking Cesàro means
$$ sigma_l(z,x) = frac 1 lsum_{n = 0}^{l-1} s_n(z,x) = frac 1 l sum_{n=0}^{l-1}sum_{k=-n}^{n}c_kz^k = sum_{k=-l}^lfrac{l - |k|}{l}c_kz^k,$$
we get uniform convergence in $z$ by Fejér's theorem.
Note that $sigma_l(z,x)$ is still continuous in $x$.
In his book K-Theory, Atiyah claims that the $sigma_l$ converge uniformly in $x$, that is given $epsilon > 0$ there exists $l_0 in mathbb{N}$, such that for all $(z,x) in S^1 times X$ and $l geq l_0$ we have
$$|f(z,x) - sigma_l(z,x)| < epsilon .$$
I don't have any idea why this should be true, but my knowledge about Fourier series and analysis is pretty limited. Any help would be appreciated.
Here are some thoughts I had:
One obstacle is that the $sigma_l$ do not form a true series, because of the factor $frac 1 l$. So the classical Weierstrass test does not even apply here. Is there another kind of Weierstrass test that one could apply to this situation?
Because $X$ is compact, it would suffice to show uniform convergence locally.
analysis convergence fourier-analysis fourier-series uniform-convergence
Let $X$ be a compact topological space, and $f: S^1 times X rightarrow mathbb{C}$ be a continuous function. Then for every point $x in X$ we can compute the Fourier coefficients
$$c_k(x) = int_{S^1} f(z,x) z^k text{d}z,$$
and construct the Fourier series as the limit of the partial sums
$$s_n(z, x) = sum_{k=-n}^n c_k z^k.$$
This must not converge point-wise, but taking Cesàro means
$$ sigma_l(z,x) = frac 1 lsum_{n = 0}^{l-1} s_n(z,x) = frac 1 l sum_{n=0}^{l-1}sum_{k=-n}^{n}c_kz^k = sum_{k=-l}^lfrac{l - |k|}{l}c_kz^k,$$
we get uniform convergence in $z$ by Fejér's theorem.
Note that $sigma_l(z,x)$ is still continuous in $x$.
In his book K-Theory, Atiyah claims that the $sigma_l$ converge uniformly in $x$, that is given $epsilon > 0$ there exists $l_0 in mathbb{N}$, such that for all $(z,x) in S^1 times X$ and $l geq l_0$ we have
$$|f(z,x) - sigma_l(z,x)| < epsilon .$$
I don't have any idea why this should be true, but my knowledge about Fourier series and analysis is pretty limited. Any help would be appreciated.
Here are some thoughts I had:
One obstacle is that the $sigma_l$ do not form a true series, because of the factor $frac 1 l$. So the classical Weierstrass test does not even apply here. Is there another kind of Weierstrass test that one could apply to this situation?
Because $X$ is compact, it would suffice to show uniform convergence locally.
analysis convergence fourier-analysis fourier-series uniform-convergence
analysis convergence fourier-analysis fourier-series uniform-convergence
edited Nov 27 at 18:10
asked Nov 23 at 22:26
red_trumpet
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