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For the Hardy space $H^2$, every $\phi \in L^\infty (\mathbb T)$ induces a bounded Toeplitz operator $T_\phi$ on the Hardy space and $\lVert T_\phi \rVert = \lVert \phi \rVert _{\infty}$. Consequently, a sequence of Toeplitz operators $(T_{\phi_n})$ converges to a Toeplitz operator $T_\phi$ in the operator norm iff $\phi_n$ converges to $\phi$ in $L^{\infty}$.

Is there something that can be said about convergence of the symbols in some sense when $(T_{\phi_n})$ converges to a Toeplitz operator $T_\phi$ in the strong operator topology and the weak operator topology? References will be appreciated.

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  • $\begingroup$ For $\|\phi_n\|$ bounded (which is necessary for WOT/SOT convergence), isn't that just equivalent to the pointwise convergence of the matrix entries, which are the Fourier coefficients of the symbols? $\endgroup$ Commented Jul 25 at 22:42
  • $\begingroup$ I believe pointwise convergence of the matrix is just a necessary condition because operator norm convergence implies WOT convergence as well and WOT convergence implies pointwise convergence of the matrix. I am not sure if it is sufficient. $\endgroup$
    – ash
    Commented Jul 26 at 2:45
  • $\begingroup$ A modest suggestion (as a conjecture). SOT convergence of sequences corresponds to norm boundedness and convergence in $L^1$, WOT to norm boundedness and weak (i.e., $\sigma(L^\infty,L^1)$) convergence. $\endgroup$
    – terceira
    Commented Jul 26 at 6:08

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As essentially discussed in the comments already, we have $T_{\phi_n}\to T_{\phi}$ in WOT if and only if $\|\phi_n\|_{\infty}\le C$ and $\widehat{\phi_n}(k)\to\widehat{\phi}(k)$.

Moreover, $T_{\phi_n}\to T_{\phi}$ in SOT if and only if $\|\phi_n\|_{\infty}\le C$ and $Pz^k\phi_n\to Pz^k\phi$ in $L^2$ for all $k\in\mathbb Z$, with $P$ denoting the projection onto $H^2$. (In particular, this will hold if $\phi_n\to\phi$ in $L^2$.)

To see this, recall first of all that $T_{\psi}$ may be represented as the Toeplitz matrix $T_{jn}$, $j,n\ge 0$ on $\ell^2(\mathbb N_0 )$, with $T_{jn}=\widehat{\psi}(n-j)$.

Let's focus on the second part. If $T_{\phi_n}\to T_{\phi}$ in SOT, then $\|T_{\phi_n}\|=\|\phi_n\|_{\infty}$ is bounded, and the truncated versions of the Fourier coefficients $\widehat{\phi_n}$, $\widehat{\phi}$ converge in $\ell^2$ because these are the columns ($=Te_k$) of our matrices.

Conversely, if these conditions holds, then $T_{\phi_n}e_k\to T_{\phi}e_k$ for each fixed $k$, and since these vectors span $\ell^2$ and our operators are uniformly bounded, this shows that $T_{\phi_n}\to T_{\phi}$ in SOT.

The argument for the first part is similar.

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