# Rescaling Fourier coefficients of a continuous function by a bounded sequence

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 $$A$$.

$$a_n=\operatorname{sign}(n)$$ 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.
• 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… Commented Feb 23 at 22:28
• @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. Commented Feb 23 at 22:36
• indeed, that should be true. My only questions left above are about whether or not $A$ is a $C^*$-algebra/if $F$ is isometric. Commented Feb 23 at 22:51
• @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 $C$. Commented Feb 23 at 23:24