Compact convergence applied to a sequence

Question

Suppose we have maps $f_n:X \to Y$, both $X$ and $Y$ are Hilbert spaces. We have that $f_n \to 0$ uniformly on compact subsets. This means (according to Wiki) that $$\lim_{n \to \infty}\sup_{x \in K} \lVert f_n(x) \rVert =0$$ whenever $K \subset X$ is compact.

Suppose we have a sequence $a_j \to a$ in $X$. Is there a chance that the compact convergence will imply $$\lim_{n \to \infty}f_n(a_j) \to 0$$ uniformly in $j$? In general $\{a_j\}$ is not compact since the limit point isn't inluded, so I don't think this holds. Is there any other way to ensure that this desired result holds?

Answer 1

$K'=\{a_j\}_{j\in\mathbb N}$ is not compact, but $K=\{a_j\}_{j\in\mathbb N}\cup\{a\}$ is, as you can easily see considering any open covering and noting that an open set, covering $a$, covers almost all points. Also supremum over $K'$ of some function is always not greater then the supremum of the same function over $K$, because $K'\subset K$. This means that you can actually replace $K$ by $K'$ in that $\lim\sup$, which implies uniform convergence.

Answer 2

The set $K:= \{a_j\}_j \cup \{a\}$ is compact, so now you may apply uniform convergence on compacts.