Relative compact sequence

Question

I am reading a paper and there are some things I am struggling to understand. I will explain the setup below:

Let $X=C[0,1]$, with sup norm, and $(f_n)\subset X$. Suppose $(g_n)\subset X$ is relatively compact and $\|f_n-g_n\|_\infty\to 0$. The paper claims that this implies relative compactness of $(f_n)$.

Question 1) What exactly does relative compactness of $(f_n)$ mean? Since $X$ is a metric space, relative compactness of $Y\subset X$ is equivalent to every sequence in $Y$ having a convergent subsequence. But what exactly is a sequence in $(f_n)$? Is this just a subsequence of $(f_n)$?

Question 2) When thinking of compactness and $C[0,1]$, Arzela-Ascoli is usually the first thought. But in the event my last sentence of Q1 is false, is AA needed for such a deduction?

Thanks

Answer 1

Relative compactness of the sequence $(f_n)$ really means relative compactness of the set $S = \{f_n: n \in \mathbb{N}\}$. This means that the sequential characterisation of relative compactness for $(f_n)$ requires more than just that $(f_n)$ has a convergent subsequence.

The problem is that the sequence you choose in the set $S$ may not be in the same order as in the sequence $(f_n)$. However, any sequence in $S$ has a subsequence that is also a subsequence of $(f_n)$ and so $S$ is relatively compact iff every subsequence of $(f_n)$ has a further subsequence that converges.

This makes the proposition you want straightforward, since given a subsequence $(f_{n_k})$ of $(f_n)$, $(g_{n_k})$ has a convergent subsequence $(g_{n_{k_j}})$ and then it is easy to check that $(f_{n_{k_j}})$ converges to the same limit.

Answer 2

1) yes, a sequence in $(f_n)$ is a subsequence of $(f_n)$ (or possibly a sequence with values in $\{f_n,\,n \geq 1\}$, which works in the same way).

2) no: just take a subsequence $f_{k_n}$, there is a subsequence $g_{k_{l_n}}$ of $g_{k_n}$ that converges, and since $\|f_{k_{l_n}}-g_{k_{l_n}}\| \rightarrow 0$, $f_{k_{l_n}}$ converges to the same limit.