Mixing
Uniform convergence for the Fourier series of $alpha$-Hölder periodic functions and continuous bounded...
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In the answers to this question it is proved that if $f:mathbb{R}tomathbb{C}$ is a $alpha$-Hölder $2pi$-periodic function, then the Fourier series of $f$ converges uniformly to $f$.
In the answer to this question it is proved that if $f:mathbb{R}tomathbb{C}$ is a continuous bounded variation $2pi$-periodic function, then the Fourier series of $f$ converges uniformly to $f$.
Note that since the Weierstrass function is $alpha$-Hölder continuous for every $alpha<1$ and differentiable nowhere (hence not of bounded variation), and since $xmapstofrac{1}{log(x)}$ is absolutely continuous in $[-frac{1}{2},frac{1}{2}]$ but not $alpha$-Hölder continuous for any $alphain(0,1)$, the first result doesn't imply directly the second and vice versa.
Can be both theorems be viewed as particular results of a theorem for a class of functions that contains both continuous bounded variation periodic functions and $alpha$-Hölder periodic functions?
fourier-series
asked Jan 24 at 12:27
BobBob
1,7361725
$endgroup$
add a comment |
$begingroup$
In the answers to this question it is proved that if $f:mathbb{R}tomathbb{C}$ is a $alpha$-Hölder $2pi$-periodic function, then the Fourier series of $f$ converges uniformly to $f$.
In the answer to this question it is proved that if $f:mathbb{R}tomathbb{C}$ is a continuous bounded variation $2pi$-periodic function, then the Fourier series of $f$ converges uniformly to $f$.
Note that since the Weierstrass function is $alpha$-Hölder continuous for every $alpha<1$ and differentiable nowhere (hence not of bounded variation), and since $xmapstofrac{1}{log(x)}$ is absolutely continuous in $[-frac{1}{2},frac{1}{2}]$ but not $alpha$-Hölder continuous for any $alphain(0,1)$, the first result doesn't imply directly the second and vice versa.
Can be both theorems be viewed as particular results of a theorem for a class of functions that contains both continuous bounded variation periodic functions and $alpha$-Hölder periodic functions?
fourier-series
asked Jan 24 at 12:27
BobBob
1,7361725
$endgroup$
$begingroup$
Possible duplicate of How to show for $alphain (0,1)$, any $fin C^alpha([0,1]/{sim})$ has a Fourier series $S_nf$ uniformly converging to $f$
$endgroup$
– Calvin Khor
Jan 24 at 12:39
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Thanks a lot. I'll modify the question accordingly citing your question and leaving only the second part
$endgroup$
– Bob
Jan 24 at 12:41
1
$begingroup$
Of course there's no such thing as a proof that the answer is no, but I tend to think the answer's no, just because if there were such a result I suspect I would have heard about it - people would prove the two results you mention as special cases of the general result, which they don't, as far as I've ever seen.
$endgroup$
– David C. Ullrich
Jan 24 at 13:38
add a comment |
$begingroup$
In the answers to this question it is proved that if $f:mathbb{R}tomathbb{C}$ is a $alpha$-Hölder $2pi$-periodic function, then the Fourier series of $f$ converges uniformly to $f$.
In the answer to this question it is proved that if $f:mathbb{R}tomathbb{C}$ is a continuous bounded variation $2pi$-periodic function, then the Fourier series of $f$ converges uniformly to $f$.
Note that since the Weierstrass function is $alpha$-Hölder continuous for every $alpha<1$ and differentiable nowhere (hence not of bounded variation), and since $xmapstofrac{1}{log(x)}$ is absolutely continuous in $[-frac{1}{2},frac{1}{2}]$ but not $alpha$-Hölder continuous for any $alphain(0,1)$, the first result doesn't imply directly the second and vice versa.
Can be both theorems be viewed as particular results of a theorem for a class of functions that contains both continuous bounded variation periodic functions and $alpha$-Hölder periodic functions?
fourier-series
asked Jan 24 at 12:27
BobBob
1,7361725
$endgroup$
In the answers to this question it is proved that if $f:mathbb{R}tomathbb{C}$ is a $alpha$-Hölder $2pi$-periodic function, then the Fourier series of $f$ converges uniformly to $f$.
In the answer to this question it is proved that if $f:mathbb{R}tomathbb{C}$ is a continuous bounded variation $2pi$-periodic function, then the Fourier series of $f$ converges uniformly to $f$.
Note that since the Weierstrass function is $alpha$-Hölder continuous for every $alpha<1$ and differentiable nowhere (hence not of bounded variation), and since $xmapstofrac{1}{log(x)}$ is absolutely continuous in $[-frac{1}{2},frac{1}{2}]$ but not $alpha$-Hölder continuous for any $alphain(0,1)$, the first result doesn't imply directly the second and vice versa.
Can be both theorems be viewed as particular results of a theorem for a class of functions that contains both continuous bounded variation periodic functions and $alpha$-Hölder periodic functions?
fourier-series
fourier-series
asked Jan 24 at 12:27
BobBob
1,7361725
asked Jan 24 at 12:27
BobBob
1,7361725
edited Jan 24 at 15:11
Bob
asked Jan 24 at 12:27
BobBob
1,7361725
asked Jan 24 at 12:27
BobBob
1,7361725
asked Jan 24 at 12:27
BobBob
1,7361725
1,7361725
$begingroup$
Possible duplicate of How to show for $alphain (0,1)$, any $fin C^alpha([0,1]/{sim})$ has a Fourier series $S_nf$ uniformly converging to $f$
$endgroup$
– Calvin Khor
Jan 24 at 12:39
$begingroup$
Thanks a lot. I'll modify the question accordingly citing your question and leaving only the second part
$endgroup$
– Bob
Jan 24 at 12:41
1
$begingroup$
Of course there's no such thing as a proof that the answer is no, but I tend to think the answer's no, just because if there were such a result I suspect I would have heard about it - people would prove the two results you mention as special cases of the general result, which they don't, as far as I've ever seen.
$endgroup$
– David C. Ullrich
Jan 24 at 13:38
add a comment |
$begingroup$
Possible duplicate of How to show for $alphain (0,1)$, any $fin C^alpha([0,1]/{sim})$ has a Fourier series $S_nf$ uniformly converging to $f$
$endgroup$
– Calvin Khor
Jan 24 at 12:39
$begingroup$
Thanks a lot. I'll modify the question accordingly citing your question and leaving only the second part
$endgroup$
– Bob
Jan 24 at 12:41
1
$begingroup$
Of course there's no such thing as a proof that the answer is no, but I tend to think the answer's no, just because if there were such a result I suspect I would have heard about it - people would prove the two results you mention as special cases of the general result, which they don't, as far as I've ever seen.
$endgroup$
– David C. Ullrich
Jan 24 at 13:38
$begingroup$
Possible duplicate of How to show for $alphain (0,1)$, any $fin C^alpha([0,1]/{sim})$ has a Fourier series $S_nf$ uniformly converging to $f$
$endgroup$
– Calvin Khor
Jan 24 at 12:39
$begingroup$
Possible duplicate of How to show for $alphain (0,1)$, any $fin C^alpha([0,1]/{sim})$ has a Fourier series $S_nf$ uniformly converging to $f$
$endgroup$
– Calvin Khor
Jan 24 at 12:39
$begingroup$
Thanks a lot. I'll modify the question accordingly citing your question and leaving only the second part
$endgroup$
– Bob
Jan 24 at 12:41
$begingroup$
Thanks a lot. I'll modify the question accordingly citing your question and leaving only the second part
$endgroup$
– Bob
Jan 24 at 12:41
1
1
$begingroup$
Of course there's no such thing as a proof that the answer is no, but I tend to think the answer's no, just because if there were such a result I suspect I would have heard about it - people would prove the two results you mention as special cases of the general result, which they don't, as far as I've ever seen.
$endgroup$
– David C. Ullrich
Jan 24 at 13:38
$begingroup$
Of course there's no such thing as a proof that the answer is no, but I tend to think the answer's no, just because if there were such a result I suspect I would have heard about it - people would prove the two results you mention as special cases of the general result, which they don't, as far as I've ever seen.
$endgroup$
– David C. Ullrich
Jan 24 at 13:38
add a comment |
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$begingroup$
Possible duplicate of How to show for $alphain (0,1)$, any $fin C^alpha([0,1]/{sim})$ has a Fourier series $S_nf$ uniformly converging to $f$
$endgroup$
– Calvin Khor
Jan 24 at 12:39
$begingroup$
Thanks a lot. I'll modify the question accordingly citing your question and leaving only the second part
$endgroup$
– Bob
Jan 24 at 12:41
1
$begingroup$
Of course there's no such thing as a proof that the answer is no, but I tend to think the answer's no, just because if there were such a result I suspect I would have heard about it - people would prove the two results you mention as special cases of the general result, which they don't, as far as I've ever seen.
$endgroup$
– David C. Ullrich
Jan 24 at 13:38