Can Analytic Functions on the Circle be Characterized by Sobolev Norms?
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In Exercise I.4.4 of Katznelson's book An Introduction to Harmonic Analysis we learn that the Fourier series $$f(z) = sum_{k=-infty}^{+infty} f_k z^k$$ defines an analytic function in the neighborhood of the unit circle if and only if there are constants $A_f , alpha_f>0$ so that $$|f_k| leq A_f e^{ - alpha_f |k|}.$$ In Chapter 7, Katznelson mentions that the compactness of the circle makes this characterization much simpler than the non-compact case of real line and the Paley-Wiener theorem.
To the question at hand: the exponential decay above implies that the $L^2$ Sobolev norms
$$||f||_s = sum_{k=0}^{infty} k^{2s} |f_k|^2 leq (2s)! A_f B_f^{2s}$$
of regularity $s=0,tfrac{1}{2}, 1, tfrac{3}{2}, 2, ldots$ can all be bounded as above using some constants $A_f, B_f$. For the implication, one plugs the first inequality into the expression for the Sobolev norm and uses geometric series to get $B_f = (1 - e^{-2alpha_f})^{-1}$. Notice that the radius $B_f$ goes to infinity as $alpha_f rightarrow 0$ as expected.
Question: is the converse true? That is, can we characterize the analytic $f$ among all smooth $f$ as being exactly those for which the sequence of Sobolev norms admits a bound $||f||_{s} leq (2s)! A_f B_f^{2s}$ for all $s$?
complex-analysis fourier-analysis sobolev-spaces harmonic-analysis analyticity
asked Dec 10 '18 at 21:14
Swallow TailSwallow Tail
512
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add a comment |
$begingroup$
In Exercise I.4.4 of Katznelson's book An Introduction to Harmonic Analysis we learn that the Fourier series $$f(z) = sum_{k=-infty}^{+infty} f_k z^k$$ defines an analytic function in the neighborhood of the unit circle if and only if there are constants $A_f , alpha_f>0$ so that $$|f_k| leq A_f e^{ - alpha_f |k|}.$$ In Chapter 7, Katznelson mentions that the compactness of the circle makes this characterization much simpler than the non-compact case of real line and the Paley-Wiener theorem.
To the question at hand: the exponential decay above implies that the $L^2$ Sobolev norms
$$||f||_s = sum_{k=0}^{infty} k^{2s} |f_k|^2 leq (2s)! A_f B_f^{2s}$$
of regularity $s=0,tfrac{1}{2}, 1, tfrac{3}{2}, 2, ldots$ can all be bounded as above using some constants $A_f, B_f$. For the implication, one plugs the first inequality into the expression for the Sobolev norm and uses geometric series to get $B_f = (1 - e^{-2alpha_f})^{-1}$. Notice that the radius $B_f$ goes to infinity as $alpha_f rightarrow 0$ as expected.
Question: is the converse true? That is, can we characterize the analytic $f$ among all smooth $f$ as being exactly those for which the sequence of Sobolev norms admits a bound $||f||_{s} leq (2s)! A_f B_f^{2s}$ for all $s$?
complex-analysis fourier-analysis sobolev-spaces harmonic-analysis analyticity
asked Dec 10 '18 at 21:14
Swallow TailSwallow Tail
512
$endgroup$
1
$begingroup$
No. For the converse you need to look at the $B_h^{2s}(c)$ you obtain when $h_k(c) = e^{-c |k|}$ and ask that $forall s, B_f^{2s} < A B_h^{2s}(c)$ for some $c,A$
$endgroup$
– reuns
Dec 10 '18 at 21:23
1
$begingroup$
Since the circle is compact that $f$ is analytic on some neighborhood of each point of the circle implies it is analytic on some annulus. For the real line it is different as $frac{1}{1-e^{2i pi z} e^{1/(z+1)^2}}$ is analytic on some neighborhood of the real line but not on $Im(z) in (-r,r)$ for any $r$
$endgroup$
– reuns
Dec 10 '18 at 21:27
$begingroup$
Thank you for your message I agree with your second comment about circle vs. real line case. I would like to know more about your first comment. Are you saying that yes there is some way of modifying what I originally wrote to get a converse, I just had to include a multiplicative constant A?
$endgroup$
– Swallow Tail
Dec 10 '18 at 21:38
add a comment |
$begingroup$
In Exercise I.4.4 of Katznelson's book An Introduction to Harmonic Analysis we learn that the Fourier series $$f(z) = sum_{k=-infty}^{+infty} f_k z^k$$ defines an analytic function in the neighborhood of the unit circle if and only if there are constants $A_f , alpha_f>0$ so that $$|f_k| leq A_f e^{ - alpha_f |k|}.$$ In Chapter 7, Katznelson mentions that the compactness of the circle makes this characterization much simpler than the non-compact case of real line and the Paley-Wiener theorem.
To the question at hand: the exponential decay above implies that the $L^2$ Sobolev norms
$$||f||_s = sum_{k=0}^{infty} k^{2s} |f_k|^2 leq (2s)! A_f B_f^{2s}$$
of regularity $s=0,tfrac{1}{2}, 1, tfrac{3}{2}, 2, ldots$ can all be bounded as above using some constants $A_f, B_f$. For the implication, one plugs the first inequality into the expression for the Sobolev norm and uses geometric series to get $B_f = (1 - e^{-2alpha_f})^{-1}$. Notice that the radius $B_f$ goes to infinity as $alpha_f rightarrow 0$ as expected.
Question: is the converse true? That is, can we characterize the analytic $f$ among all smooth $f$ as being exactly those for which the sequence of Sobolev norms admits a bound $||f||_{s} leq (2s)! A_f B_f^{2s}$ for all $s$?
complex-analysis fourier-analysis sobolev-spaces harmonic-analysis analyticity
asked Dec 10 '18 at 21:14
Swallow TailSwallow Tail
512
$endgroup$
In Exercise I.4.4 of Katznelson's book An Introduction to Harmonic Analysis we learn that the Fourier series $$f(z) = sum_{k=-infty}^{+infty} f_k z^k$$ defines an analytic function in the neighborhood of the unit circle if and only if there are constants $A_f , alpha_f>0$ so that $$|f_k| leq A_f e^{ - alpha_f |k|}.$$ In Chapter 7, Katznelson mentions that the compactness of the circle makes this characterization much simpler than the non-compact case of real line and the Paley-Wiener theorem.
To the question at hand: the exponential decay above implies that the $L^2$ Sobolev norms
$$||f||_s = sum_{k=0}^{infty} k^{2s} |f_k|^2 leq (2s)! A_f B_f^{2s}$$
of regularity $s=0,tfrac{1}{2}, 1, tfrac{3}{2}, 2, ldots$ can all be bounded as above using some constants $A_f, B_f$. For the implication, one plugs the first inequality into the expression for the Sobolev norm and uses geometric series to get $B_f = (1 - e^{-2alpha_f})^{-1}$. Notice that the radius $B_f$ goes to infinity as $alpha_f rightarrow 0$ as expected.
Question: is the converse true? That is, can we characterize the analytic $f$ among all smooth $f$ as being exactly those for which the sequence of Sobolev norms admits a bound $||f||_{s} leq (2s)! A_f B_f^{2s}$ for all $s$?
complex-analysis fourier-analysis sobolev-spaces harmonic-analysis analyticity
complex-analysis fourier-analysis sobolev-spaces harmonic-analysis analyticity
asked Dec 10 '18 at 21:14
Swallow TailSwallow Tail
512
asked Dec 10 '18 at 21:14
Swallow TailSwallow Tail
512
edited Dec 11 '18 at 19:24
Swallow Tail
asked Dec 10 '18 at 21:14
Swallow TailSwallow Tail
512
asked Dec 10 '18 at 21:14
Swallow TailSwallow Tail
512
asked Dec 10 '18 at 21:14
Swallow TailSwallow Tail
512
512
1
$begingroup$
No. For the converse you need to look at the $B_h^{2s}(c)$ you obtain when $h_k(c) = e^{-c |k|}$ and ask that $forall s, B_f^{2s} < A B_h^{2s}(c)$ for some $c,A$
$endgroup$
– reuns
Dec 10 '18 at 21:23
1
$begingroup$
Since the circle is compact that $f$ is analytic on some neighborhood of each point of the circle implies it is analytic on some annulus. For the real line it is different as $frac{1}{1-e^{2i pi z} e^{1/(z+1)^2}}$ is analytic on some neighborhood of the real line but not on $Im(z) in (-r,r)$ for any $r$
$endgroup$
– reuns
Dec 10 '18 at 21:27
$begingroup$
Thank you for your message I agree with your second comment about circle vs. real line case. I would like to know more about your first comment. Are you saying that yes there is some way of modifying what I originally wrote to get a converse, I just had to include a multiplicative constant A?
$endgroup$
– Swallow Tail
Dec 10 '18 at 21:38
add a comment |
1
$begingroup$
No. For the converse you need to look at the $B_h^{2s}(c)$ you obtain when $h_k(c) = e^{-c |k|}$ and ask that $forall s, B_f^{2s} < A B_h^{2s}(c)$ for some $c,A$
$endgroup$
– reuns
Dec 10 '18 at 21:23
1
$begingroup$
Since the circle is compact that $f$ is analytic on some neighborhood of each point of the circle implies it is analytic on some annulus. For the real line it is different as $frac{1}{1-e^{2i pi z} e^{1/(z+1)^2}}$ is analytic on some neighborhood of the real line but not on $Im(z) in (-r,r)$ for any $r$
$endgroup$
– reuns
Dec 10 '18 at 21:27
$begingroup$
Thank you for your message I agree with your second comment about circle vs. real line case. I would like to know more about your first comment. Are you saying that yes there is some way of modifying what I originally wrote to get a converse, I just had to include a multiplicative constant A?
$endgroup$
– Swallow Tail
Dec 10 '18 at 21:38
1
1
$begingroup$
No. For the converse you need to look at the $B_h^{2s}(c)$ you obtain when $h_k(c) = e^{-c |k|}$ and ask that $forall s, B_f^{2s} < A B_h^{2s}(c)$ for some $c,A$
$endgroup$
– reuns
Dec 10 '18 at 21:23
$begingroup$
No. For the converse you need to look at the $B_h^{2s}(c)$ you obtain when $h_k(c) = e^{-c |k|}$ and ask that $forall s, B_f^{2s} < A B_h^{2s}(c)$ for some $c,A$
$endgroup$
– reuns
Dec 10 '18 at 21:23
1
1
$begingroup$
Since the circle is compact that $f$ is analytic on some neighborhood of each point of the circle implies it is analytic on some annulus. For the real line it is different as $frac{1}{1-e^{2i pi z} e^{1/(z+1)^2}}$ is analytic on some neighborhood of the real line but not on $Im(z) in (-r,r)$ for any $r$
$endgroup$
– reuns
Dec 10 '18 at 21:27
$begingroup$
Since the circle is compact that $f$ is analytic on some neighborhood of each point of the circle implies it is analytic on some annulus. For the real line it is different as $frac{1}{1-e^{2i pi z} e^{1/(z+1)^2}}$ is analytic on some neighborhood of the real line but not on $Im(z) in (-r,r)$ for any $r$
$endgroup$
– reuns
Dec 10 '18 at 21:27
$begingroup$
Thank you for your message I agree with your second comment about circle vs. real line case. I would like to know more about your first comment. Are you saying that yes there is some way of modifying what I originally wrote to get a converse, I just had to include a multiplicative constant A?
$endgroup$
– Swallow Tail
Dec 10 '18 at 21:38
$begingroup$
Thank you for your message I agree with your second comment about circle vs. real line case. I would like to know more about your first comment. Are you saying that yes there is some way of modifying what I originally wrote to get a converse, I just had to include a multiplicative constant A?
$endgroup$
– Swallow Tail
Dec 10 '18 at 21:38
add a comment |
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No. For the converse you need to look at the $B_h^{2s}(c)$ you obtain when $h_k(c) = e^{-c |k|}$ and ask that $forall s, B_f^{2s} < A B_h^{2s}(c)$ for some $c,A$
$endgroup$
– reuns
Dec 10 '18 at 21:23
1
$begingroup$
Since the circle is compact that $f$ is analytic on some neighborhood of each point of the circle implies it is analytic on some annulus. For the real line it is different as $frac{1}{1-e^{2i pi z} e^{1/(z+1)^2}}$ is analytic on some neighborhood of the real line but not on $Im(z) in (-r,r)$ for any $r$
$endgroup$
– reuns
Dec 10 '18 at 21:27
$begingroup$
Thank you for your message I agree with your second comment about circle vs. real line case. I would like to know more about your first comment. Are you saying that yes there is some way of modifying what I originally wrote to get a converse, I just had to include a multiplicative constant A?
$endgroup$
– Swallow Tail
Dec 10 '18 at 21:38