Let $(E,\|\cdot\|)$ be a separable Hilbert space on $\mathbb{R}$. We consider a sequence of strongly continuous semigroup $\{T^{(n)}\}_{n=1}^\infty$ on $E$ [of course, for $n \in \mathbb{N}$ and $t>0$, $T^{(n)}_t$ is a bounded linear operator].
For $n \in \mathbb{N}$, we write $(A_n,\text{Dom}(A_n))$ for the generator of $T^{(n)}=\{T^{(n)}_t\}_{t>0}$. That is, \begin{align} A_nf&:=\lim_{t\to 0}\frac{T^{(n)}_tf-f}{t},\quad \text{Dom}(A_n):=\left\{f \in E : \lim_{t\to 0}\frac{T^{(n)}_tf-f}{t} \text{exists in }E\right\}. \end{align}
Question.
Let $T=\{T_t\}_{t>0}$ be a strongly continuous contraction semigroup on $E$. Then, the corresponding generator $(A,\text{Dom}(A))$ is a densely defined closed linear operator on $E$.
We now assume the following conditions.
- $T$ is a symmetric semigroup on $E$. In other words, $(A,\text{Dom}(A))$ is self-adjoint on $E$.
- For every $f \in \text{Dom}(A)$ and $n \in \mathbb{N}$, we have $f \in \text{Dom}(A_n)$ and $\lim_{n \to \infty}\|A_nf-Af\|=0$.
Under these conditions, can we show that $\lim_{n \to \infty} \|T_t^{(n)} f-T_{t}f\|=0$ for any $t>0$ and $f \in E$ ?
Note that each $T^{(n)}_t$ is not necessarily contractive. In my impression, such a result generally does not hold. However, I have no counterexamples.
