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No: uniform convergence of utilities on bounded sets does not imply closed convergence of preferences. It does imply the upper condition. Indeed, suppose $x_j\to x$, $y_j\to y$ and $u_{n_j}(x_j)\ge u_{n_j}(y_j)$. These bundles all lie in one bounded ball. Uniform convergence there, together with continuity of $u$, givesso $x⪰y$.

For a strict limiting comparison $u(x)>u(y)$, the fixed pair itself satisfies $u_n(x)>u_n(y)$ eventually. The obstruction is the lower condition at an indifference relation, where the strict approximating rankings need not disappear.

For a counterexample valid in every dimension $L\ge 1$, takeOn $\Vert x\Vert \le k$, $|u_n(x)-u(x)|\le k/n\to 0$, exactly as required. Every $u_n$ induces $x{⪰}_{n}y$ if and only if $x_1\ge y_1$, while $u$ makes every pair indifferent. In particular, $0⪰e_1$ holds in the limit. Any sequences $x_n\to 0$, $y_n\to e_1$ eventually have $(x_n{)}_1<(y_n{)}_1$, so they cannot satisfy $x_n{⪰}_n{y}_n$. The lower graph-limit condition fails.

Equivalently, an open neighborhood of $(0,e_1)$ sufficiently small to keep the first coordinate of its first bundle below that of its second bundle meets the limiting graph and none of the approximating graphs. The corresponding hit-open neighborhood in the Fell topology proves failure of convergence directly. HenceThe uniform utility convergence need not preserve closed preference convergence phenomenon reflects the loss of ranking information when the limiting utility has a flat indifference region.