Question
Suppose a stock pays 2 discrete dividends $d_1, d_2$ at times $t_1, t_2$ respectively, where $ t < t_1 < t_2 < T.$ Assume the risk-free rate, $r$, is a positive constant. Given that
The lower and upper bounds for an American call, $C_t$, is trivially determined as $$\max{ \left( S_t - D_t - Ke^{-r(T-t)}, S_t - K, 0 \right)} \leq C_t \leq S_t.$$
Consider a strategy that exercises the option at the instant before time $t_1$, we have $$S_t - Ke^{-r(t_1 -t)} \leq C_t.$$
Consider a strategy that exercises the option at the instant before time $t_2$, we have $$S_t - d_1e^{-r(t_1 -t)} - Ke^{-r(t_2 -t)} \leq C_t.$$
Combining the three inequalities above, what is the improved lower bound for an American call in this situation?
My attempt
I understand that American call with discrete dividends should at most be exercised at the instant before the ex-dividend dates for optimal payoff, so I obtained the lower bound as
$$\max{\left(S_t - Ke^{-r (t_1 - t)}, S_t - d_1e^{-r (t_1 - t)} - Ke^{-r (t_2 - t)}, S_t - D_t- Ke^{-r (t_1 - t)}, 0\right)} \leq C_t $$ where $D_t = d_1e^{-r (t_1 - t)} + d_2e^{-r (t_2 - t)}.$
However, I am not sure if I should include $S_t - K$ into the max function which was given in the original question.
Any intuitive explanation will be highly appreciated!
You can avoid to include it simply because it can never be the maximum value since
$$ S_t - K e^{-r(t_+t)}>S_t - K $$ is verified for every $t<t_1$ which is always the case by contruction. Hence including it gives no information since it can never be reached