Is the language $K=\{u \in\{0,1\}^n\mid n \geq 0, \forall_{v \in\{0,1\}^n} (u+v) \in L \}$ regular? For two words $w,v \in\{0,1\}^*$ of equal length, let $w+v \in\{0,1,2\}^*$ denote the word in which the $i$-th word is the sum of $i$-th position of $w$ and $v$, as follows: if $w=a_1 \ldots a_n$ and $v=b_1 \ldots b_n$, then $w+v=c_1 \ldots c_n$, where $c_i=a_i+b_i$ for each $i\in \{1, \ldots, n\}$.
Assume $L \subseteq \{0,1,2\}^*$ is a regular language. Decide whether the language $$K=\{u \in\{0,1\}^n\mid n \geq 0, \forall_{v \in\{0,1\}^n} (u+v) \in L \}$$ is regular.

I would appreciate any hint as I don't know how to start. I know that the every word u+v can be easily converted to be over the alphabet $\{0,1\}^*$ such that
$h(a_i) = 0$ if $a_i=0$ and $h(a_i) = 1$ if $a_i=1$ and
if $a_i = 2$, then $h(a_{i+1})=a_{i+1}+1$. Then $u+v$ is just binary addition.
It might be easier to consider the complement of your language:
$$
\overline{K} = \{ u \in \{0,1\}^* : \exists v \in \{0,1\}^{|u|} \text{ s.t. } u+v \in \overline{L} \}.
$$

You can now easily convert an NFA for $\overline{L}$ to one for $\overline{K}$ by replacing all edges labelled $1$ with edges labelled $0,1$ and all edges labelled $2$ with edges labelled $1$.

I'll let you figure out why this works.