Toeplitz operator

Remarks

$T_\varphi$ is a bounded linear operator, and in fact $\norm{T_\varphi} = \norm{\varphi}_\infty$. With respect to the orthonormal basis $e_n(t) = e^{int}$, $n \ge 0$, its matrix entries are $\ip{T_\varphi e_k}{e_j} = \hat\varphi(j - k)$, so the matrix is a Toeplitz matrix. If $\varphi$ is continuous and has no zeros on $\mathbb{T}$, then $T_\varphi$ is a Fredholm operator whose index is minus the winding number of $\varphi$ around $0$.

Example: $\varphi(e^{it}) = e^{it}$. For $f \in H^2$ the product $\varphi f$ has no negative Fourier coefficients, so $P(\varphi f) = \varphi f$ and $T_\varphi$ is the shift operator $f \mapsto z f$.

Non-example: for $\varphi(e^{it}) = e^{-it}$, plain multiplication by $\varphi$ does not map $H^2$ into itself, since it sends $1$ to $e^{-it} \notin H^2$. So the projection $P$ in the definition cannot be omitted; the Toeplitz operator $T_\varphi$ is the backward shift.