($x(t)\to\bar{x}$ and $f_{t}\to f$) implies $f_{t}(x(t))\to f(\bar{x})$

Question

For each $t\geq0$, let $f_{t}:\mathbb{R}^{n}\to\mathbb{R}$ be continuously differentiable. Suppose, in addition, that $x:[0,\infty)\to\mathbb{R}^{n}$ is continuously differentiable. Let $f:\mathbb{R}^{n}\to\mathbb{R}$ and $\bar{x}\in\mathbb{R}^{n}$. I would like to understand under which conditions (on $f_{t}$ and $f$) and notion of convergence the following is true: \begin{equation} (x(t)\to\bar{x}\quad \text{and}\quad f_{t}{\color{red}\to}f)\quad\text{implies}\quad f_{t}(x(t))\to f(\bar{x})\quad \text{as $t\to\infty$}. \end{equation} Is there a common expression for this kind of convergence?

Answer 1

It is enough to assume that $f_t \overset{U}{\rightrightarrows} f$ where $U$ is a neighbourhood of $\bar x$ (w.l.o.g. $U= B(\bar x, r)$ for some $r > 0$). Note that the traditional definition of uniform convergence extends naturally to allow $t\geq 0$ as an index. Like @kabel abel mentioned, we won’t need differentiability.

Let $\epsilon > 0$. Because of uniform convergence we have $$ \exists t_1 > 0 : \forall t\geq t_1 : \forall x_0 \in U : |f(x_0)-f_t(x_0)| < \frac{\epsilon}2. $$ Similarly, because of continuity we have $$ \exists t_2 > 0 : \forall t\geq t_2 : |f(x(t))-f(\bar x)| < \frac{\epsilon}2. $$ Since $\lim_{t\rightarrow \infty} x(t) = \bar x$, we have a $t_3 > 0$ such that for all $t \geq t_3$ we have $x(t) \in U$. Now let $t_0 := \max(t_1, t_2, t_3)$ and note that this implies $\forall t\geq t_0 : |f(x(t))-f_t(x(t))| < \epsilon/2$. Adding the two inequalities we get $$ \forall t \geq t_0 : |f(\bar x) - f_t(x(t))| \leq |f(x(t))-f(\bar x)|+ |f(x(t))-f_t(x(t))| < \epsilon. $$ Thus, $\lim_{t\rightarrow \infty} f_t(x(t)) = f(\bar x)$. I’d also like to note that point-wise convergence won’t be enough.