Take-it-or-leave-it PBE I've found an interesting question looking at perfect-bayesian-equilibrium. I haven't seen a question where beliefs are not discrete.

There is a single potential buyer of an object which has zero value to
the seller. This buyer’s valuation v is uniformly distributed on [0, 1] and is
private information. The seller names a price $p_1$ which the buyer accepts or
rejects. 

If he accepts, the object is traded at the agreed price and the buyer’s
payoff is $v − p_1$ and the seller’s is $p_1$. 

If he rejects then the seller makes another
price offer, p2. If the buyer accepts this, his payoff is $\delta_(v − p_2)$ and the seller’s
is $\delta p_2$, where $\delta = 0.5$. 

If he rejects, both players get zero (there are no further
￼￼￼offers).

Find a Perfect Bayesian Equilibrium.

My usual approach is to fix beliefs, but I don't quite know how to do this with continuous beliefs. Any advice?
After posting a bad solution yesterday I believe I got a better one:

The strategy of the buyer consists of two functions, $(f_1(v,p_1),f_2(v,p_1,p_2))$ where both functions map to $\left\{A,R\right\}$ (where $A$ stands for Accept, $R$ for Reject).
The strategy of the seller is $(p_1,p_2(f_1(v,p_1)))$. You get the solution via backward induction. In PBE $f_2(v,p_1,p_2)$ maps to $A$ if and only if $v \geq p_2$. (There is inconsequential leeway at equality.) In PBE the seller believes that there is a set $H$ of types for which the buyer refused her offer $p_1$. Then
$$
p_2^* = \arg\max_{p_2} p_2 \cdot Prob(f_2(v,p_1,p_2) = A | f_1(v,p_1) = R).
$$
The buyer will accept offer $p_1$ if and only if
$$
v - p_1 \geq \delta \cdot (v - p_2). 
$$
From this you get
$$
v \cdot (1 - \delta) \geq p_1 - \delta \cdot p_2. 
$$
The left hand side of this equation is increasing in $v$, so types with high valuation will Accept. This means that in PBE the set $H$ is such that 
$$
H = [0, \bar{v}). 
$$
From this we get the optimal $p_2$ given $\bar{v}$:
$$
p_2^* = \arg\max_{p_2} p_2 \cdot Prob(v \geq p_2 | v \in [0, \bar{v})) = \frac{\bar{v}}{2}.
$$
In PBE $\bar{v}$ is a function of $p_1$: 
$$
\bar{v} \cdot (1 - \delta) = p_1 - \delta \cdot \frac{\bar{v}}{2}, 
$$
so
$$
\bar{v} = \frac{p_1}{1 - \frac{\delta}{2}}.
$$
We have determined all the PBE strategies but $p_1$. 
The expected payoff of the seller is
$$
p_1 \cdot \left( 1 - \frac{p_1 - \delta \cdot p_2(\bar{v}(p_1))}{1 - \delta} \right) + \frac{1}{2} \cdot p_2(\bar{v}(p_1)) \cdot \left( \frac{p_1 - \delta \cdot p_2(\bar{v}(p_1))}{1 - \delta} - p_2(\bar{v}(p_1)) \right),
$$
where
$$
p_2(\bar{v}(p_1)) = \frac{\bar{v}(p_1)}{2} = \frac{\frac{p_1}{1 - \frac{\delta}{2}}}{2} = \frac{p_1}{2 - \delta}.
$$
Substituting this we get
$$
p_1 \cdot \left( 1 - \frac{p_1 - \delta \cdot \frac{p_1}{2 - \delta}}{1 - \delta} \right) + \frac{1}{2} \cdot \frac{p_1}{2 - \delta} \cdot \left( \frac{p_1 - \delta \cdot \frac{p_1}{2 - \delta}}{1 - \delta} - \frac{p_1}{2 - \delta} \right),
$$

You have to maximize this w.r.t. $p_1$. With $\delta = 0.5$ I got
$$
p_1^* = \frac{9}{20}, \hskip 20pt \bar{v} = \frac{3}{5}, \hskip 20pt p_2^* = \frac{3}{10}.
$$