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?
 A: 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, $t=t*$. 
A: 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 $t_i \to t$.
