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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...

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You say "nothing guarantees that V(pn) converges to p in the first place, right?" Do you mean $V(p_n)$ converges to $V(p)$ or $p_n$ converges to $p$. –  alext87 Sep 19 '10 at 14:41
Sorry for the typo, I actually meant $V(p_n)$ converges to $V(p)$. But I am also not able to see why $p_n$ should converge to $p$ as well. –  student Sep 19 '10 at 14:52
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1 Answer

up vote 1 down vote accepted

You can think of your $u_i$ as the components of a continuous map $U:D\to\mathbb{R}^n$ for some $n$. So you have the condition

  • if $U(p)=U(q)$ then $V(p)=V(q)$

and so $V=W\circ U$ for some map from $L=U(D)$ to $\mathbb{R}$. To show $W$ is continuous, it suffices to prove that $U:D\to L$ is a quotient map, that is $X\subseteq L$ is closed in $L$ iff $U^{-1}(X)$ is closed in $D$.

By continuity if $X$ is closed in $L$ then $U^{-1}(X)$ is closed in $D$, so suppose that $Y=U^{-1}(X)$ is closed in $D$. As $D$ is compact, then $Y$ is compact (Heine-Borel), and as $U$ is continuous, $U(Y)$ is a compact subspace of $\mathbb{R}^n$. Thus $U(Y)$ is a closed subset of $\mathbb{R}^n$ and a fortiori of $L$. But $U(Y)=X$, as $U$ maps $D$ onto $L$.

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Thanks, I believe one can also provide a sequential argument to show that it must be the case that $V(p_n)$ converges to $V(p)$. –  student Sep 19 '10 at 18:21
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