# Continuity of argmax (moving domain and function), under uniqueness hypothesis

Let $f:[0,1]\times\mathbb{R}\to\mathbb{R}$ continuous and $c:\mathbb{R}\to[0,1]$ continuous. Consider $$F:\mathbb{R}\to\mathbb{R},\ \ F(x)=\max_{t\in[0,c(x)]}f(t,x)$$ Here Continuity of max (moving the domain and the function) it was proved that $F$ is continuous.

Now assume also that for every $x\in\mathbb{R}$ $\textrm{argmax}_{t\in[0,c(x)]}\ f(t,x)$ is unique (that is there exists a unique point $t\in[0,c(x)]$ s.t. $f(t,x)=F(x)$ ). And consider $$T:\mathbb{R}\to\mathbb{[0,1]},\ \ T(x)=\textrm{argmax}_{t\in[0,c(x)]}\ f(t,x)$$ Is $T$ continuous too?

Yes. This follows from Berge's maximum theorem. That theorem shows that the set-valued map to maximizers is upper-hemicontinuous. A set-valued map that is singleton-valued is upper-hemicontinous if and only if the corresponding function is continuous.

Here is a direct argument. Let $(x_n)$ be a convergent sequence in $\mathbb{R}$ with limit $x$. We want to show that $T(x_n)\to T(x)$. This means just that for every convergent subsequence of $(T(x_n))$, the limit maximizes $u(\cdot, x)$, since the maximizer is unique. So let $(y_n)$ be such a subsequence with $T(y_n)\to t$. Let $t^*$ be a maximizer of $u(\cdot,x)$. Then the sequence $u(t^*,y_n)$ converges to $u(t^*,x)$. Since $u(t^*,y_n)\leq u(t(y_n),y_n)$, this means by continuity that $u(t^*,x)\leq u(t,x)$. By the uniqueness of the maximizer, $t=t*$.

• Thank you. I saw that theorem on wikipedia, but is there a simple way to positively answer my question without working with set-valued maps? Jun 21 '13 at 15:15
• Some authors use upper-semicontinuous, also a closed map (some slight differences, but same idea). Jun 21 '13 at 15:29
• @qwertyuio One can give an elementary proof by backwards-engeneering the proof. I will do so later. Jun 21 '13 at 15:36
• @Michael Greinecker Ok thank you, I wait for you proof Jun 21 '13 at 16:06

It is straightforward, if unsatisfying, to show continuity using sequences:

Suppose $x_i \to x$. Since $c$ is continuous, it is bounded on bounded sets, so we may assume $c(x_i) \le K$ for some $K$. In particular, if $t_i \in [0,c(x_i)]$ then $t_i \le K$.

Let $t_i$ be a minimizer corresponding to $x_i$, that is, $F(x_i) = f(x_i, t_i)$. Suppose $t$ is an accumulation point of $t_i$, that is, $t_{n_i} \to t$ over some subsequence. Since $F,f$ are continuous, we have $F(x) = f(x,t)$. In particular, any accumulation point of $t_i$ is a maximizer. Since the maximizer is unique, the only accumulation point of the sequence $t_i$ is the maximizer $t$. It follows that $t_i \to t$ (otherwise the bounded sequence $t_i$ would have another accumulation point). Hence $T(x_i) = t_i \to t = T(x)$, and so $T$ is continuous.

Clarification: Suppose $t_i \in K$, where $K$ is compact, and $t \in K$. Further, suppose $t_i$ has the property that any subsequence has a further subsequence that converges to $t$. Then $t_i \to t$.

To see this, suppose $U$ is a neighborhood of $t$ such that $t_i \not\in U$ an infinite number of times. Hence this subsequence has a further subsequence which converges to $t$, which is a contradiction. Hence $U$ must contain all except a finite number of the $t_i$. It follows that $t_i \to t$.

• Thank you! Minor observation: under my hypothesis $t_i$ is trivially bounded in $[0,1]$. A question: at the end you use the fact that if every subsequence of $(t_n)_n$ admits a further subsequence convergent to $t$, then $t_n\to t$ ? Jun 21 '13 at 17:14
• @qwertyuio: The statement is true much more generally as Michael pointed out. Continuity of $F$ is more straighforward to establish if we note that $F(x) = \max_{\tau \in [0,1]} f(x,\tau c(x))$ (that is, the domain of maximization is a fixed compact set). I added some clarification about regarding my characterization of convergence. Jun 21 '13 at 18:54