Generally, weak convergence of a sequence of operators $T_n$ on a dense subset $\mathcal D \subset \mathcal F$ does not imply weak convergence on the full set, $$ \lim_{n\to \infty} T_n(f) = T(f) ~~\forall f \in \mathcal D \not \Rightarrow \lim_{n\to \infty} T_n(f) = T(f) ~~\forall f \in \mathcal F, $$ see e.g. Is it sufficient to check weak convergence on a (weak* or strongly) dense subset of the dual?. But is it true in more specific cases, where more is known of $T_n$?
I am considering an operator which is the boundary value of an analytic function $T(z) : \mathbb R + i\mathbb R_{>0} \to \mathbb C$ on Schwartz functions,
$$
T_\epsilon(f) := \int dx~ f(x) T(x + i\epsilon),~ \lim_{\epsilon \to 0} T_\epsilon(f) = T(f) ~~\forall f \in \mathscr S.
$$
It satisfies typical bounds for boundary values of tempered distributions of the form
$$
|T_\epsilon(f)| \leq \frac{C}{\epsilon^k} ||f||_2.
$$
Additionally, $T(f)$ satisfies $L^2$ bounds, $|T(f)| \leq C ||f||_2$, such that by Riesz lemma $T(x) \in L^2(\mathbb R)$.
Does this imply
$$
\lim_{\epsilon\to 0} T_\epsilon(f) = T(f) ~~\forall f \in L^2(\mathbb R), \text{ and/or } \lim_{\epsilon\to 0} ||T_\epsilon - T||_2 = 0?
$$