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No. Let $f_n=x^2/n$ for $n$ odd and $(x_1)^2/n$ for $n$ even. Then $x_n^*$ is an alternating sequence of $1$s and $0$s, which does not converge to anything. But $f_n$ converges pointwise to $f=0$.
We can modify this to make the convergence uniform, by using an absolute value instead of a square, or to make $f$ nonconstant, by adding $\max(|x-1/2|,1)$ to $f_n$.
If $f$ has only a single zerounique minimum, the statement is true. Let $a$ be the $\lim\inf$ of $x_n^*$ and $b$ be the $\lim\sup$. Let $y$ be the unique minimum of $f$. Assume $ay$, a< y$. Then$f_n(a)$converges to$f(a)$, and$f_n(y)$converges to$f(y)$, and since$f(a)>f(y)$,$f_n(a)\leq f_n(y)$only finitely many times. But every time$x_n^*\leq a$we have$f_n(a)\leq f_n(y)$since$a$is closer to the minimum$x_n^*$then$y$. This Since that occurs infinitely many times, this is a contradiction. Therefore$a\geq y$. Similarly$y\geq b$, and by properties of$\lim\sup$and$\lim\inf$we have$b\geq a$, so$a=b=y$and the limit is$y$. 1 No. Let$f_n=x^2/n$for$n$odd and$(x_1)^2/n$for$n$even. Then$x_n^*$is an alternating sequence of$1$s and$0$s, which does not converge to anything. But$f_n$converges pointwise to$f=0$. We can modify this to make the convergence uniform, by using an absolute value instead of a square, or to make$f$nonconstant, by adding$\max(|x-1/2|,1)$to$f_n$. If$f$has only a single zero, the statement is true. Let$a$be the$\lim\inf$of$x_n^*$and$b$be the$\lim\sup$. Let$y$be the unique minimum of$f$. Assume$ay$,$f_n(a)\leq f_n(y)$only finitely many times. But every time$x_n^*\leq a$we have$f_n(a)\leq f_n(y)$since$a$is closer to the minimum$x_n^*$then$y$. This is a contradiction. Therefore$a\geq y\geq b\geq a$, so$a=b=y$and the limit is$y\$.