Trembling hand perfection and weakly dominated strategies

It is well known that players cannot use weakly dominated strategies in a trembling hand perfect equilibrium. My question, however, is a little different: does iterated deletion of weakly dominated strategies preserve trembling hand perfect equilibria? In other words, if $$\sigma$$ is a trembling hand perfect equilibrium of a finite game $$G$$, is $$\sigma$$ also a trembling hand perfect equilibrium of the game $$G'$$ obtained by iteratively deleting weakly dominated strategies from $$G$$?

A trembling hand perfect (THP) equilibrium may not survive iterated deletion of weakly dominated strategies, as is shown by the following example. Let game $$G$$ be

$$\begin{array}{|c|c|c|c|}\hline &A&B&C\\\hline A&0,0&0,0&0,-2\\\hline B&0,0&1,1&-1,-2\\\hline C&-2,0&-2,-1&-2,-2\\\hline \end{array}$$

There are two pure strategy NEs: $$(A,A)$$ and $$(B,B)$$, with the former involving iteratively weakly dominated strategies. And yet $$(A,A)$$ is a THP equilibrium. To see this, consider the following sequence of totally mixed strategies $$$$\sigma_i^k=\left(1-\epsilon^{-2k}-\epsilon^{-k},\epsilon^{-2k},\epsilon^{-k}\right),\qquad i\in\{1,2\} \text{ and }\epsilon>2,$$$$ which converges to $$(1,0,0)=A$$ as $$k\to\infty$$. It is easy to verify that, for any $$k$$, the best response of player $$j\ne i$$ to $$\sigma_i^k$$ is $$A$$: \begin{align} u_j(A,\sigma_i^k)&=0 \\ u_j(B,\sigma_i^k)&=\epsilon^{-2k}-\epsilon^{-k}<0 \\ u_j(C,\sigma_i^k)&=-2<0. \end{align} Therefore, $$(A,A)$$ is THP.

However, the game $$G'$$ that survives iterated deletion of weakly dominated strategies does not contain the strategy $$A$$ for either player. Hence, the THP equilibrium $$(A,A)$$ of the original game $$G$$ is not preserved in game $$G'$$.