Exercise: American put-call parity bounds

Tasks

1. Upper bound. Prove the upper bound $C^{\text{Am}} - P^{\text{Am}} \le S_0 - Ke^{-rT}$ by comparing with European equivalents and using the fact that an American option is worth at least as much as the corresponding European.
2. Lower bound. Prove the lower bound $C^{\text{Am}} - P^{\text{Am}} \ge S_0 - K$ by a direct arbitrage argument: if $C^{\text{Am}} - P^{\text{Am}} < S_0 - K$, buy the call, short the stock, invest cash at rate $r$ — show this is riskless.
3. Call never exercised early (no dividends). Argue using the bounds or directly that for a non-dividend-paying stock, early exercise of an American call is never optimal, so $C^{\text{Am}} = C^{\text{Eur}}$.
4. American put exercised early. Give a numerical example where $P^{\text{Am}} > P^{\text{Eur}}$: specifically, for a deeply in-the-money put, compare the exercise value $K - S$ with the hold value. For $S_0 = 10$, $K = 100$, $r = 0.05$, $T = 1$, $\sigma = 0.3$: verify numerically that $P^{\text{Am}} \approx K - S_0 = 90$ (exercise now) while $P^{\text{Eur}} < 90$.
5. With dividends. The American-call-never-exercised-early result fails when the stock pays dividends. Why?

Hint

For part 3, combine the bound $C^{\text{Am}} \ge S_0 - Ke^{-rT}$ with $C^{\text{Am}} \ge 0$ (option value non-negative) and $S_0 - K \le S_0 - Ke^{-rT}$ to show the exercise value $S_0 - K$ is dominated by the hold value.