Say you are trying to sell a good at the highest price. You draw independently from $F(p)$. The history of observations is $P = \{p_1, p_2,\dots\}$. Observing the $i$th price costs $c_i$, where the observation cost weakly increases with every additional observation: $c_i \geq c_j$ whenever $i > j$.
In the end, you will only want to sell to the highest bidder. Hence, we don't care about the whole history of observed prices, but just about the highest observed price $\bar p = \max P$. After every observation, you can decide whether you stop and enjoy $U(\bar p)$.
Here's how I attacked the problem.
Denote $V_i(p, p^N)$ the value of observing $p^N$ as the $i$ths observation when the highest price so far is $p$. Denote by $W_i(p^*)$ the value of having $p^*$ as the highest observed price after $i$ observations. Then,
$$ W_i(p) = \max\{ U(p), -c_{i+1} + \int V_{i+1}(p, \tilde p)dF(\tilde p)\}\\ V_i(p, p^N) = \begin{cases} W_i(p) & \text{ if } p > p^N \\ W_i(p^n) & \text{else} \end{cases}$$
So far, so good. How to attack this? In these class of problems, there is typically a reservation price $R$ such that one stops only if the draw $p > R$. However here, the value is a draw is not stationary. Hence, I think
$$\forall i \exists R_i: W_i(R_i) = U(R_i) = -c_{i+1} + \int V_{i+1}(R_i, \tilde p)dF(\tilde p) $$
Then,
$$ \forall p > R_i: W_i(p) = U(p) \\ \forall p < R_i: W_i(p) = -c_{i+1} + \int V_{i+1}(p, \tilde p)dF(\tilde p) $$
I can distinguish the latter case as follows:
$$ \forall R_i > p > R_{i+1}: W_i(p) = -c_{i+1} + F(p)U(p) + \int_{\tilde p > p} U(\tilde p) dF(\tilde p)\\ \forall R_i > R_{i+1} > p: W_i(p) = -c_{i+1} + F(p)W_{i+1}(p) + \int_{\tilde p > p} W_{i+1}(\tilde p) dF(\tilde p)$$
However, I cannot see how I could use this recursive structure to solve for the set of $\{W_i(p)\}_i$. Even proving useful Lemmas is not easy. For example, it should be true (and I have already been using this) that $R_j \leq R_i$ whenever $j > i$ (as $c_i$ is a weakly increasing sequence). How should I proceed with this?