# Equivariant band matrices and Fourier series

Recall that in the first post of this series we claimed that there exists an infinite matrix $$T$$ which is in the closure (in operator norm) of the band matrices with uniformly bounded entries, but for which we have $$\|T^{(R)}\| \to \infty$$. Here $T^{(R)}_{m,n} := \begin{cases} T_{m,n} & \text{ if } |m-n| \le R\\ 0 & \text{ otherwise }\end{cases}$

The goal of this post is to provide an example of such a matrix.

##### Equivariant band matrices

We will consider equivariant matrices, i.e., matrices $$T$$ such that for all $$m,n \in \mathbb{Z}$$ we have $$T_{m,n} = T_{m+k,n+k}$$ for all $$k \in \mathbb{Z}$$. Such matrices are completely determined by their entries $$T_{0,n}$$ for all $$n \in \mathbb{Z}$$.

Since an equivariant band matrix $$T$$ only has finitely many non-zero values $$T_{0,n}$$, we get a map $\{\text{equivariant band matrices}\} \to \mathbb{C}[\mathbb{Z}], \quad T \mapsto \sum_{n \in \mathbb{Z}} T_{0,n} \cdot n,$ where $$\mathbb{C}[\mathbb{Z}]$$ denotes the complex group ring of $$\mathbb{Z}$$. This is actually an isomorphism of $$\mathbb{C}$$-algebras.

We can use the above map to define a norm on $$\mathbb{C}[\mathbb{Z}]$$ by using the operator norm of the corresponding equivariant band matrix. The completion of $$\mathbb{C}[\mathbb{Z}]$$ under this norm is called the reduced group $$C^*$$-algebra of $$\mathbb{Z}$$ and denoted $$C_r^*(\mathbb{Z})$$. Taking the completion corresponds to taking the closure of the equivariant band matrices in the space of all infinite matrices with bounded operator norm, i.e., elements of $$C_r^*(\mathbb{Z})$$ can be written as equivariant infinite matrices.

##### Fourier series

Forming Fourier series can be thought of as the map $\ell^2(\mathbb{Z}) \to L^2(S^1), \quad (c_n)_{n \in \mathbb{Z}} \mapsto \sum_{n \in \mathbb{Z}} c_n \cdot e^{2\pi i t \cdot n},$ which is continuous.

On $$\ell^2(\mathbb{Z})$$ we can act with (equivariant) band matrices and on $$L^2(S^1)$$ we can act with $$C(S^1)$$, i.e., with continuous functions on the unit circle, by point-wise multiplication. These actions are intertwined with each other by the operation of forming Fourier series: the assignment $\sum_{n \in \mathbb{Z}} T_{0,n} \cdot n \mapsto \text{ point-wise multiplication by } \sum_{n \in \mathbb{Z}} T_{0,n} \cdot e^{2\pi i t \cdot n}$ defines a map $$\mathbb{C}[\mathbb{Z}] \to C(S^1)$$ and the operator norm on $$\mathbb{C}[\mathbb{Z}]$$ corresponds under it to the sup-norm on $$C(S^1)$$. The image of this map is dense in $$C(S^1)$$ and so we get an isomorphism $$C_r^*(\mathbb{Z}) \cong C(S^1)$$ by the Stone-Weierstrass theorem. Note that $$C_r^*(\mathbb{Z})$$ is viewed here as a subalgebra of $$B(\ell^2(\mathbb{Z}))$$ and $$C(S^1)$$ is viewed as a subalgebra of $$B(L^2(S^1))$$, where $$B(-)$$ denotes the bounded, linear operators.

##### The counter-example

What does all the above help us in our goal of providing an infinite matrix $$T$$ which is in the closure (in operator norm) of the band matrices with uniformly bounded entries, but for which we have $$\|T^{(R)}\| \to \infty$$?

We assume that $$T$$ is from $$C_r^*(\mathbb{Z})$$. Then it is in the closure (in operator norm) of the band matrices with uniformly bounded entries and can be represented as $$\sum_{n \in \mathbb{Z}} T_{0,n} \cdot n$$. The operators $$T^{(R)}$$ correspond then to $$\sum_{-R \le n \le R} T_{0,n} \cdot n$$.

Now we apply the isomorphism $$C_r^*(\mathbb{Z}) \cong C(S^1)$$ that we discussed above: the operator $$T$$ becomes the function $$\sum_{n \in \mathbb{Z}} T_{0,n} \cdot e^{2\pi i t \cdot n}$$, the operators $$T^{(R)}$$ the functions $$\sum_{-R \le n \le R} T_{0,n} \cdot e^{2\pi i t \cdot n}$$, and the operator norm becomes the sup-norm.

This means that if we can find a continuous function $$f$$ on $$S^1$$ with Fourier series $$\sum_{n \in \mathbb{Z}} c_n \cdot e^{2\pi i t \cdot n}$$ such that the continuous functions $$f^{(R)}$$ defined by $$\sum_{-R \le n \le R} c_n \cdot e^{2\pi i t \cdot n}$$ satisfy $$\|f^{(R)}\|_\infty \to \infty$$, then we have our counter-example: we just have to transform $$f$$ back to an operator from $$C_r^*(\mathbb{Z})$$ via the isomorphism $$C_r^*(\mathbb{Z}) \cong C(S^1)$$.

An example of such a continuous function can be found in Section 3.2.2 of the book “Fourier Analysis” by Stein and Shakarchi.

I thank Rufus Willett for providing me the reference for the construction of such a continuous function with divergent Fourier series.

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