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completed a mathematical correction and added some references
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Yemon Choi
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If $0 < \alpha < 1/2$ then a continuous function on the circle is $\operatorname{Lip}_\alpha$ only if the Fourier coefficients satisfy $a_n = {\rm O}( n^{-\alpha})$; this is in Katznelson's book (Chapter I, Corollary 4.6) for instance. 

[EDIT (2013-07-10): at the time I had thought this was "iff" but a comment points out that I may have misremembered; I will check in Katznelson later.] Thisany case, for lacunary series such as the one in the question, a lot more is known than in Katznelson's bookthe general case; see e.g. Katznelson Chapter V for instancethe basics.]

So the function you defined above isn't going to be Hölder continuous for any positive exponent, even though it's clearly continuous (absolutely convergent Fourier series).

Off the top of my head, I don't know of any particularly good source for the higher-dimensional stuff. There might be something in Katznelson's book, but I don't recall.

If $0 < \alpha < 1/2$ then a continuous function on the circle is $\operatorname{Lip}_\alpha$ if the Fourier coefficients satisfy $a_n = {\rm O}( n^{-\alpha})$. [EDIT: I had thought this was "iff" but a comment points out I may have misremembered; I will check in Katznelson later.] This is in Katznelson's book for instance. So the function you defined above isn't going to be Hölder continuous for any positive exponent, even though it's clearly continuous (absolutely convergent Fourier series).

Off the top of my head, I don't know of any particularly good source for the higher-dimensional stuff. There might be something in Katznelson's book, but I don't recall.

If $0 < \alpha < 1/2$ then a continuous function on the circle is $\operatorname{Lip}_\alpha$ only if the Fourier coefficients satisfy $a_n = {\rm O}( n^{-\alpha})$; this is in Katznelson's book (Chapter I, Corollary 4.6) for instance. 

[EDIT (2013-07-10): at the time I thought this was "iff" but a comment points out that I misremembered; in any case, for lacunary series such as the one in the question, a lot more is known than in the general case; see e.g. Katznelson Chapter V for the basics.]

So the function you defined above isn't going to be Hölder continuous for any positive exponent, even though it's clearly continuous (absolutely convergent Fourier series).

Off the top of my head, I don't know of any particularly good source for the higher-dimensional stuff.

Joined in the LaTeX fun, added mathematical caveat/correction
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Yemon Choi
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If $0 < \alpha < 1/2$ then a continuous function on the circle is $Lip_\alpha$ if and only$\operatorname{Lip}_\alpha$ if the Fourier coefficients satisfy $a_n = O( n^{-\alpha})$$a_n = {\rm O}( n^{-\alpha})$. [EDIT: I had thought this was "iff" but a comment points out I may have misremembered; I will check in Katznelson later.] This is in Katznelson's book for instance. So the function you defined above isn't going to be Hölder continuous for any positive exponent, even though it's clearly continuous (absolutely convergent Fourier series).

Off the top of my head, I don't know of any particularly good source for the higher-dimensional stuff. There might be something in Katznelson's book, but I don't recall.

If $0 < \alpha < 1/2$ then a continuous function on the circle is $Lip_\alpha$ if and only if the Fourier coefficients satisfy $a_n = O( n^{-\alpha})$. This is in Katznelson's book for instance. So the function you defined above isn't going to be Hölder continuous for any positive exponent, even though it's clearly continuous (absolutely convergent Fourier series).

Off the top of my head, I don't know of any particularly good source for the higher-dimensional stuff. There might be something in Katznelson's book, but I don't recall.

If $0 < \alpha < 1/2$ then a continuous function on the circle is $\operatorname{Lip}_\alpha$ if the Fourier coefficients satisfy $a_n = {\rm O}( n^{-\alpha})$. [EDIT: I had thought this was "iff" but a comment points out I may have misremembered; I will check in Katznelson later.] This is in Katznelson's book for instance. So the function you defined above isn't going to be Hölder continuous for any positive exponent, even though it's clearly continuous (absolutely convergent Fourier series).

Off the top of my head, I don't know of any particularly good source for the higher-dimensional stuff. There might be something in Katznelson's book, but I don't recall.

Fixed maths rendering (since I was already fixing the question here)
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Andrew Stacey
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If 0 < α < 1/2$0 < \alpha < 1/2$ then a continuous function on the circle is Lipα$Lip_\alpha$ if and only if the Fourier coefficients satisfy a_n = O( n )$a_n = O( n^{-\alpha})$. This is in Katznelson's book for instance. So the function you defined above isn't going to be Hölder continuous for any positive exponent, even though it's clearly continuous (absolutely convergent Fourier series).

Off the top of my head, I don't know of any particularly good source for the higher-dimensional stuff. There might be something in Katznelson's book, but I don't recall.

If 0 < α < 1/2 then a continuous function on the circle is Lipα if and only if the Fourier coefficients satisfy a_n = O( n ). This is in Katznelson's book for instance. So the function you defined above isn't going to be Hölder continuous for any positive exponent, even though it's clearly continuous (absolutely convergent Fourier series).

Off the top of my head, I don't know of any particularly good source for the higher-dimensional stuff. There might be something in Katznelson's book, but I don't recall.

If $0 < \alpha < 1/2$ then a continuous function on the circle is $Lip_\alpha$ if and only if the Fourier coefficients satisfy $a_n = O( n^{-\alpha})$. This is in Katznelson's book for instance. So the function you defined above isn't going to be Hölder continuous for any positive exponent, even though it's clearly continuous (absolutely convergent Fourier series).

Off the top of my head, I don't know of any particularly good source for the higher-dimensional stuff. There might be something in Katznelson's book, but I don't recall.

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Yemon Choi
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