In variational problems, sometimes we want to show that minimizers of a sequence of functionals $F_n$ converge to the minimizers of some $F$. To do this, we usually show 1) $\Gamma$-convergence of $F_n$ to $F$ and 2) compactness.
$\Gamma$-convergence is a weak way to describe the convergence $F_n \rightarrow F$. It involves showing liminf and limsup inequalities. In addition to $\Gamma$-convergence, a compactness property is needed: namely that if $F_n(u_n) < \infty$ for all $n$, then there is a subsequence $u_{n_k}$ converging to some limit $u$.
However, what if we show strong/uniform convergence of $F_n \rightarrow F$? Do we still need a compactness property? If so, can we use a weaker version? By strong convergence, I mean that $F_n(u_n) \rightarrow F(u)$ for all $n$, and all $u_n \rightarrow u$ (wrt whatever norm the domain uses).
We cannot substitute continuous convergence for the compactness condition. Consider the following example.
Let $f : \mathbb{R} \to \mathbb{R}$ be defined by $f(x)=\exp(-|x|)$. For each $n = 1, 2, \dots$, let $f_n : \mathbb{R} \to \mathbb{R}$ be defined by \begin{equation} f_n(x) = \begin{cases} f(x) & \text{if $|x| \leq n$,} \\ [f(n) + f'(n)(|x|-n)]_+ & \text{otherwise,} \end{cases} \end{equation} where $[c]_+ := \max\{0, c\}$. In English, $f_n$ agrees with $f$ on the interval $[-n,n]$, but then linearly interpolates with the line $y=0$. A Desmos graph is worth a thousand words.
Let's show that $f_n \to f$ continuously. Let $x_n \to x$. Then there exists $M > 0$ such that $x,x_n \in [-M, M]$ for all $n$. Letting $n_0 \geq M$, it follows that $f_n(x_n) = f(x_n)$ for all $n \geq n_0$; hence, $f_n(x_n) \to f(x)$.
On the other hand, $\DeclareMathOperator{\argmin}{\mathrm{argmin}}$
where $c = n - f(n)/f'(n)$.
I am not aware of any general condition that can be traded for the compactness criterion, except for a more stringent compactness criterion (e.g., $f, f_n$ share the same compact support).