Let $I$ be a finite nonempty set, and $D$ a nonvoid compact subset of an N-dimensional Euclidean space. The functions $V:D\rightarrow\mathbb{R}$ and $u_i:D\rightarrow\mathbb{R}$ are given. They are all continuous, and, moreover, each $u_i$ satisfies $u_i(\lambda p + (1-\lambda) q) = \lambda u_i( p) + (1-\lambda) u_i(q)$ when $\lambda \in (0,1)$ (I don't believe this last piece of information is important, though).
$V$ and $u_i$ also satisfy: (P) if $u_i(p) = u_i(q)$ for all $i\in I$, then $V(p) = V(q)$. This condition allows us to construct a mapping $W:L\rightarrow\mathbb{R}$, where $L =$ {$ (u_1(p),...,u_I(p)): p \in D$}, and $W$ is defined as $W(u) = V(p)$ for any $p \in D$ such that $u = (u_1(p),...,u_I(p))$. Condition (P) ensures W is well defined.
Question: Is $W$ continuous?
Motivation: I am sketching a "proof" from a (published) paper in "applied math" I found online, but that I cannot verify. Let $(u^n)$ be a sequence in $L$ converging to $u \in L$. Let $(p_n)$ be a sequence in $D$ such that $u^n = (u_1(p_n),...,u_I(p_n))$ for all $n$, and $p \in D$ satisfy $u = (u_1(p),...,u_I(p))$. Because $D$ is compact, there exists a convergent subsequence $p_{n_{k}}\rightarrow p^\prime$. Because $V$ is continuous, we obtain $V(p_{n_{k}})\rightarrow V(p^\prime)$. Moreover, since each $u_i$ is also continuous, we have $\lim u^n = \lim u^{n_{k}} $, and hence $(u_1(p),...,u_I(p)) = (u_1(p^\prime),...,u_I(p^\prime))$. By property (P), we have $V(p) = V(p^\prime)$. Finally, the article goes on and concludes that $W(u^n) = V(p_n) \rightarrow V(p^\prime) = V(p) = W(u)$.
I don't think the conclusion follows from the previous arguments. In particular, nothing guarantees that $V(p_n)$ converges to $p$ in the first place, right? Nevertheless, I cannot think of an easy counter-example...