This question stems out of: which sequences $(a_n)_{n\in\mathbb{Z}}$ of complex numbers have the property that if there exists a continuous function $f$ on the circle with Fourier coefficients $b_n$, then there also exists a continuous function with Fourier coefficients $a_nb_n$? First we note that $(a_n)$ must be bounded, or else we could take a subsequence $a_{m_n}$ with $|a_{m_n}|\geq 2^n$, and define $f(x)=\sum_{n\geq 0}\frac{1}{a_{m_n}} e^{im_n x}$ and then the hypothesis would give us a continuous function whose Fourier coefficients do not tend to zero, which cannot happen.

The natural question arises: then does $(a_n)$ being bounded suffice? What can be said about the $(a_n)$ that do work? One thing we can note is that any such sequence $a=(a_n)$ determines a continuous linear operator $T_a:C(S^1)\to C(S^1)$ taking $f$ to the function $T_a(f)$ with $n$th Fourier coefficient $a_n\hat{f}(n)$. Thus, we get a commutative Banach subalgebra $A\subseteq \mathcal{L}(C(S^1),C(S^1))$ whose elements are the $T_a$ for sequences $a$ such that the hypothesis holds. There is a Banach algebra homomorphism $F:A\to \ell^{\infty}(\mathbb{Z})$ taking $T_a$ to $a$, which is contractive.

In many of the basic cases, we even have $\|T_a\|=\|a\|$, so this prompts the questions: is $F$ an isomorphism? If $F$ isometric? Is $A$ even a $C^*$-algebra?

There are (non-continuous) $S^1$-actions by isometries on $A$ and $\ell^{\infty}(\mathbb(Z))$ underlying everything here, where $e^{iz}$ acts by taking a sequence $(x_n)$ to $(e^{inz}x_n)$, so perhaps there’s some nice way to use some $S^1$-equivariant generators of $\ell^{\infty}(\mathbb{Z})$ to see that $F$ is surjective, if it is at least. Since $\ell^{\infty}(\mathbb{Z})\simeq C(\beta \mathbb{Z})$, perhaps there’s some way to utilize Stone-Weierstrass here. This would let us show the image of $A$ is dense so long as for any infinite set $S$, the sequence $(\delta_{n\in S})$ is in the image of $A$.


1 Answer 1


$a_n=\operatorname{sign}(n)$ is a bounded sequence that is not a multiplier on $C(S)$. The corresponding operator $L^2(S)\to L^2(S)$ is the Hilbert transform $$ (Hf)(x) = \lim_{h\to 0+} \int_{|y-x|>h} f(y)\cot (x-y)\, dy , $$ and $Hf$ need not be continuous (or even bounded) when $f\in C(S)$. For example when $f$ is odd and $|f(x)|=1/\log |x|$ near $x=0$, then $Hf$ is unbounded there.

  • $\begingroup$ Nice example! So F can’t be surjective, in fact your example shows that $(\delta_{n>0})$ isn’t in the image. At the same time, for any $m,i$, an averaging trick will show that $(\delta_{n\equiv i\mod(m)})$ is in the image. So the condition for an infinite subset S of the integers such that the indicator sequence for $S$ is in the image of $F$ lies somewhere between these… $\endgroup$ Commented Feb 23 at 22:28
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    $\begingroup$ @LoganHyslop: I think the answer is the same as for multipliers on $L^{\infty}$: they are exactly the Fourier coefficients of measures. Clearly, these sequences work (since convolution with a measure is a bounded operator on $C(S)$), and conversely, if convolution with a distribution is bounded on $C(S)$, it seems "clear" that its order must be $0$, so it's a measure by the Riesz representation theorem. $\endgroup$ Commented Feb 23 at 22:36
  • $\begingroup$ indeed, that should be true. My only questions left above are about whether or not $A$ is a $C^*$-algebra/if $F$ is isometric. $\endgroup$ Commented Feb 23 at 22:51
  • $\begingroup$ @LoganHyslop: I don't think $F$ can be isometric. The lazy version is to say that in my example $\|a\|=1$, $\|T_a\|=\infty$, and it should not be too difficult to make a real argument out of this by considering a cut off version of $a$ that still has large (but finite) norm on $C$. $\endgroup$ Commented Feb 23 at 23:24

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