What is a Toeplitz operator? Define the Hardy space \(H^2(\mathbb{T})\), the Toeplitz operator \(T_\varphi\) associated with a symbol \(\varphi \in L^\infty(\mathbb{T})\), and describe its matrix representation.
Solution
Let \(\mathbb T=\{z\in\mathbb C:|z|=1\}\) with normalized Lebesgue measure. The Hardy space is \[ H^2(\mathbb T) = \left\{f\in L^2(\mathbb T):\widehat f(n)=0\text{ for }n<0\right\} = \overline{\operatorname{span}}\{1,z,z^2,\ldots\}. \] Let \[ P_+:L^2(\mathbb T)\to H^2(\mathbb T) \] be the orthogonal (Szegő) projection.
For \(\varphi\in L^\infty(\mathbb T)\), multiplication by \(\varphi\) is bounded on \(L^2\), and the Toeplitz operator with symbol \(\varphi\) is \[ T_\varphi=P_+M_\varphi|_{H^2}, \qquad T_\varphi f=P_+(\varphi f). \] In particular, \[ \|T_\varphi\|\le \|\varphi\|_\infty. \]
Write the Fourier coefficients of \(\varphi\) as \[ a_k=\widehat\varphi(k). \] For the orthonormal basis \(e_j(z)=z^j\), \(j\ge0\), \[ \langle T_\varphi e_j,e_i\rangle = \langle \varphi z^j,z^i\rangle =a_{i-j}. \] Thus the matrix is the semi-infinite Toeplitz matrix \[ [T_\varphi]=(a_{i-j})_{i,j\ge0} = \begin{pmatrix} a_0&a_{-1}&a_{-2}&\cdots\\ a_1&a_0&a_{-1}&\cdots\\ a_2&a_1&a_0&\cdots\\ \vdots&\vdots&\vdots&\ddots \end{pmatrix}, \] so its entries are constant along diagonals.
Also \(T_\varphi^*=T_{\overline\varphi}\), and for \(\varphi(z)=z\), \(T_z\) is the unilateral shift. The Brown–Halmos characterization says that a bounded operator \(T\) on \(H^2\) is Toeplitz exactly when \[ S^*TS=T, \] where \(S=T_z\).