Are deterministic context-free languages closed under outfix (or other erasing operations)
A proof that uses closure properties:

DCF languages are not closed under union, so take,  $L_1, L_2 \in DCF$ s.t. $L = L_1 \cup L_2 \notin DCF$

Add three new symbols $\{\alpha, \beta, \#\}$ to the original alphabet $\Sigma$ and build the languages:

$L'_1 = \{ \alpha \# w \mid w \in L_1\}$
$L'_2 = \{ \beta \# w \mid w \in L_1\}$

We have $L'_1, L'_2$, but also $L' = L'_1 \cup L'_2 \in DCF$ (it is enough to add a starting state that leads to the recognition of $\#L_1$ after reading a leading symbol $\alpha$ or the recognition of $\#L_2$ after reading a leading symbol $\beta$).

Now suppose that $Outf(L') \in DCF$; we know that DCFs are closed under intersection with regular languages, so: 

$$Outf(L') \cap \{ \#w \mid w\in \Sigma^*\} =  \{ \#w \mid w\in L_1 \lor w \in L_2 \} \in DCF$$

too.

But given a DPDA for $Outf(L') \cap \{ \#w \mid w\in \Sigma^*\}$ it is immediate to build a DPDA for $\{ w \mid w \in L_1 \lor w \in L_2\} = L$ (just skip the recognition of the first leading symbol $\#$), so $L \in DCF$ contradicting the hypothesis.