My question is whether the following argument is correct. Is there any defect on my proof?
Let $f_n$ and $f$ be real-valued functions defined on a metric space $(S,\rho)$ such that $f_n \to f$ point-wise. Suppose that $f_n$ and $f$ are minimized at $x_n^*$ and $x^*$, respectively.
I wish to derive a sufficient condition for the convergence \begin{eqnarray} f_n(x_n^*) \to f(x^*) \end{eqnarray} as $n\to \infty$. To this end, first observe that \begin{eqnarray} |f_n(x_n^*) - f(x^*)| &\leq& |f_n(x_n^*) - f(x_n^*)| +f(x_n^*) - f(x^*)\\ &\leq& 2 |f_n(x_n^*) - f(x_n^*)| +|f_n(x^*) - f(x^*)|. \end{eqnarray} By point-wise convergence, the second term on the right converges to $0$.
To show the convergence of the first term on the right, let $A := \{x_n^*: n\in \mathbb{N}\}$. Then, \begin{eqnarray} |f_n(x_n^*) - f(x_n^*)| \leq \sup_{x\in A} |f_n(x) - f(x)|. \end{eqnarray} Thus,by a standard argument for uniform convergence, I conclude that the following conditions are sufficient for the convergence $f_n(x_n^*) \to f(x^*)$:
(a) $A$ is relatively compact, (b) $f$ is uniformly continuous on $A$ and (c) the family $\{f_n\}$ is uniformly equicontinuous on $A$.
As Matt A Pelto suggested, the condition (b) can be deleted since uniform equi-continuity of $\{f_n\}$ and point-wise convergence $f_n(x) \to f(x)$ imply uniform continuity of $f$.