Meeting 7: Michael Francis

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Cesaro averages for $\mathbb{B}(H)$

Let $[a_{ij}]_{i,j \geq 0}$ be the matrix of bounded operator $T$ on $\ell^2= \ell^2(\mathbb{N})$. For each $n \in \mathbb{Z}$, let $\Phi_n(T) \in \mathbb{B}(\ell^2)$ be the restriction of $T$ to the $n$th diagonal. Specifially, the $(i,j)$ entry of the matrix of $\Phi_n(T)$ is $a_{ij}$ wherever $i-j=n$ and $0$ everywhere else. Thus, for example: $$\begin{matrix} \Phi_0(T) = \begin{bmatrix} a_{00} & 0 & 0 & 0 & 0 & \ldots \\ 0 & a_{11} & 0 & 0 & 0 & \ldots \\ 0 & 0 & a_{22} & 0 & 0 & \ldots \\ 0 & 0 & 0 & a_{33} & 0 & \ldots \\ 0 & 0 & 0 & 0 & a_{44} & \ldots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\ \end{bmatrix} &&& \Phi_2(T) = \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & \ldots \\ 0 & 0 & 0 & 0 & 0 & \ldots \\ a_{20} & 0 & 0 & 0 & 0 & \ldots \\ 0 & a_{31} & 0 & 0 & 0 & \ldots \\ 0 & 0 & a_{42} & 0 & 0 & \ldots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\ \end{bmatrix}. &&& \Phi_{-1}(T)= \begin{bmatrix} 0 & a_{01} & 0 & 0 & 0 & \ldots \\ 0 & 0 & a_{12} & 0 & 0 & \ldots \\ 0 & 0 & 0 & a_{23} & 0 & \ldots \\ 0 & 0 & 0 & 0 & a_{34} & \ldots \\ 0 & 0 & 0 & 0 & 0 & \ldots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\ \end{bmatrix}. \end{matrix}$$ Notice that $\|\Phi_n(T)\| \leq \|T\|$ for all $n \in \mathbb{Z}$. Indeed, $\|\Phi_n(T)\| = \sup_{i-j=n} |a_{ij}|$. This is obvious if $n=0$, in which case $\Phi_0(T)$ is an honest diagonal matrix. For nonzero $n$, the same formula for the norm follows from expressing $\Phi_n(T)$ as the product of (a power of) the unilateral shift (or its adjoint) and an honest diagonal matrix.

In a naive sense, any $T \in \mathbb{B}(\ell^2)$ can be expressed as a formal summation $$T = \sum_{n \in \mathbb{Z}} \Phi_n(T)$$ where the right hand side at least makes sense as a formula for the matrix of $T$. It is natural to wonder: "Does this formal summation converge to $T$?". Generally, the answer is: "No, not really". However, it turns out that, as in classical Fourier analysis, by performing a Cesaro averaging process, we do get the convergence.

Defintion 7.1: Given $T \in \mathbb{B}(\ell^2)$ and $N \geq 0$, we define:

• The $N$th partial sum: $S_N(T)=\sum_{n=-N}^N \Phi_n(T)$
• The $N$th Cesaro sum: $\Sigma_N(T)=\tfrac{1}{N+1}\sum_{n=0}^N S_n(T)=\sum_{n=-N}^N \left(1-\tfrac{|n|}{N+1} \right) \Phi_n(T)$

In general, the partial sums $S_N(T)$ behave poorly. They do not converge to $T$ in any standard topology and the operator norms $\|S_N(T)\|$ can grow without bound. However, the Cesaro sums behave perfectly well. This is described in the following proposition:

Proposition 7.2: Let $T \in B(\ell^2(\mathbb{N}))$ be arbitrary. Then, the Cesaro sums $\Sigma_N(T)$ (defined above) satisfy:

1. $\Sigma_N(T) \to T$ in s.o.t. as $N \to \infty$
2. $\|\Sigma_N(T)\| \leq \|T\|$ for all $N$.

This is a special case of Lemma 1.1 in the paper Invariant subspaces and hyper-reflexivity for free semigroup algebras by Davidson and Pitts. However, the proof is not given there. For the proof, we can follow the outline of Theorem VIII.2.2 in Davidson's book C*-algebras by Example.

The key to proving Proposition 7.2 is to express $T \mapsto \Phi_n(T)$, introduced above as a rather brutal operation on the matrix of $T$, in a more useful way in order that we may connect things to classical Fourier analysis. Let $X$ the unbounded diagonal matrix $\mathrm{diag}(0,1,2,3,\ldots)$. Correspondingly, we have the s.o.t.-continuous 1-parameter unitary group $$e^{itX}=\begin{bmatrix} 1 & 0 & 0 & 0 & \\ 0 &e^{it}& 0 & 0 & \\ 0 & 0 &e^{i2t} & 0 & \\ 0 & 0 &0 &e^{i3t} & \\ & & & & \ddots \\ \end{bmatrix}$$ One may check that $$\Phi_0(T) = \tfrac{1}{2\pi}\int_{-\pi}^\pi e^{-itX} T e^{itX} \ dt$$ where the integral is intended to be s.o.t. convergent. Roughly speaking, the point is that the matrix of $e^{-itX}T e^{itX}$ looks the same as the matrix of $T$, except that the entries on the $n$th diagonal are multiplied by $e^{-int}$. Thus, averaging over $t \in [-\pi,\pi]$, all entries become zero, except those on the main diagonal.

Similarly, we may achieve any $\Phi_n(t)$ via the following integral: $$\Phi_n(T)=\tfrac{1}{2 \pi} \int_{-\pi}^\pi e^{int} e^{-itX} T e^{itX} \ dt$$ since the entries of $e^{int} e^{-itX} T e^{itX}$ will have nontrivial phase on every diagonal except for the $n$th one.

Now we can make contact with classical Fourier analysis. Quite generally, any $\mathbb{C}$-linear combination $\sum_{n=-N}^N \lambda_n \Phi_n$ of the operations $\Phi_n$ is going to be given by $$\sum_{n=-N}^N \lambda_n \Phi_n(T) = \frac{1}{2\pi} \int_{-\pi}^\pi f(t) e^{-itX}T e^{itX} \ dt$$ where $$f(t)=\sum_{n=-N}^N \lambda_n e^{int}.$$ In particular, taking $\lambda_n=1-\frac{|n|}{N+1}$, we get that $$\Sigma_N(T) = \frac{1}{2\pi} \int_{-\pi}^\pi \sigma_N(t) e^{-itX} T e^{itX} \ dt$$ where $$\sigma_N(t)=\sum_{n=-N}^N \left(1-\tfrac{|n|}{N+1} \right)e^{int}$$ are the classical Fejér kernel functions whose properties include:

1. $\int_{-\pi}^\pi \sigma_N(t) \ dt = 1$ (obvious)
2. $\sigma_N(t) \geq 0$ everywhere (not obvious)
3. $\sigma_N(t) \to 0$ uniformly on any deleted neighbourhood $[-\pi,-\epsilon) \cup (\epsilon,\pi]$.
From the positivity and the fact that $\frac{1}{2\pi} \int \sigma_N = 1$, we have that $\|\sigma_N\|_{L^1}=1$ whence $\|\Sigma_N(T)\| \leq \|T\|$ by an easy estimate. Thus, we have the second part of Proposition 7.2. For the first part, i.e. the fact that $\Sigma_N(T) \to T$ in s.o.t., one uses the Dirac delta-like properties of the Fejér kernels.

Relevance to the Kaplansky density theorem

As a corollary of Proposition 7.2, we obtain ways of concretely realizing the approximations in the Kaplansky density theorem in certain situations. Suppose that $A \subseteq \mathbb{B}(\ell^2)$ is some algebra whose s.o.t. closure is $M$. In some circumstances, it may by the case that $\Phi_n(T) \in A$ whenever $T \in M$. If this occurs, then the Cesaro averages $\Sigma_N(T)$ faciliate a sequence of operators in $A$, norm-bounded by $\|T\|$ and converging s.o.t. to $T \in M$. This happens occasionally, even for non-self-adjoint algebras $A$. For example, we can consider $A$ to be the (non-self-adjoint) algebra consisting of all finite linear combinations of the unilateral shift. Pretty clearly, the s.o.t. closure of $A$ is going to be contained in the bounded operators having lower triangular matrices which are constant along diagonals. Evidently, applying $\Phi_n$ to such an operator will map us back into $A$.

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