If $a_n(t) \to a(t)$ and $t \mapsto a_n$ is continuous, are the $a_j(t)$ in a compact subset independent of $t$?

Question

I have a sequence $a_n(t) \to a(t)$ in a Hilbert space $H$. Furthermore, for each $n$, $a_n:[0,1] \to H$ is continuous. Is it possible to deduce that $$\{a_n(t)\}_{n \in \mathbb{N}}$$ is contained in a compact subset $K$ of $X$, where $K$ is independent of $t$ as well as $n$???

I think the asnwer is no, since we only get inclusion in compact sets that either depend on $t$ or on $n$. Does anyone know if it holds under additional assumptions on the sequence?

I assume you mean $\lim_{n\rightarrow\infty} a_n(t) = a(t)$ for all $t \in [0,1]$. You might consider a counter-example for real-valued functions $a_n(t)$ that converge to the all-zero function $a(t)=0$. — 2017-02-21

If $a_n(t) \to a(t)$ and $t \mapsto a_n$ is continuous, are the $a_j(t)$ in a compact subset independent of $t$?

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