Black-Scholes American Put Option

Here is my question: This is a question about Black-Scholes model, but it may be applicable to more complicated models. Throughout the discussion, the strike price $K$, interest rate $r$ and volatility $\sigma$ will be assumed to be constant. The thing we are interested in is the time decay of the option.

First consider a perpetual American put problem. This has an optimal exercising level which I will call $b$ and this is given by $b=\frac{K}{1+\sigma^2/2r}$

Now consider any $x>b$, this $x$ is a level, for which it would not be optimal to exercise in a perpetual option. In this case, does this necessarily mean, there exist a $T$ such that $t\in (0,T)$, the equation below holds

$\mathbb{E}_x e^{-r(t\hat{}\tau_b)}(K-X_{t\hat{}\tau_b})^+ > (K-x)^+$

here $\tau_b$ denotes the first hitting time of level $b$.

I think I have shown this is the case for some $t$. Here is my proof:

Denote $V(x)=\mathbb{E}_x e^{-r\tau_b}(K-X_{\tau_b})^+$. It is widely known that $V(x)>(K-x)^+$. Assume the statement above is false, then

$\mathbb{E}_x e^{-r(t\hat{}\tau_b)}(K-X_{t\hat{}\tau_b})^+ \leq (K-x)^+$ for all $t>0$.

Then we can take the limit as $t\uparrow\infty$ on the left-hand side and apply the dominated convergence theorem, we see that

$V(x)\leq (K-x)^+$

Contradicting the inequality $V(x)>(K-x)^+$

However, I have struggled to show $\mathbb{E}_x e^{-r(t\hat{}\tau_b)}(K-X_{t\hat{}\tau_b})^+ > (K-x)^+$ for small value of $t$. I believe this conjecture is true for all $t>0$ but struggled to prove it. Anyone has any ideas?

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Consider a finite horizon problem with the time to maturity $t$, then there exists an optimal stopping boundary $B(t)$, such that if the stock price is at this level, we should exercise the option. Notice, this $B(t)>b$ and it is an increasing function of $t$.
If we start our process, at $B(t)$ with amount of time $t$ left to run, then since we started at the optimal stopping boundary so $(K-x)^+ > \mathbb{E}_{B(t)} (K-X_\tau)^+$ for any stopping time $\tau<t$.
In fact, the result I asked for is totally ridiculous, because as $t$ goes to $0$, $B(t) \uparrow K$. For small times, it is a lot better to stay where you are than let the process run on.