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Let $H$ be a separable Hilbert space of functions which are nonzero continuous and bounded on $\mathbb R$. Let $s_{n}=\sup_{x\in \mathbb R}|h_{n}(x)|$, for $h_{n}\in H$, and define the set $S$ to be the set of all sequences $\{s_{n}\}$ which converges to zero (so we pick only the sequences of functions from $H$ where the limit of the sup's is zero for these sequences, and of course we can assume without lose of generality that these sequences are all decreasing)

Now, assuming that $S$ is nonempty, can we find a sequence, call it $\{s'_{n}\}$ in $S$ such that $\lim_{n\to\infty}\frac{s'_{n}}{t_{n}}\, x_{n}=0$ for any $\{t_{n}\}\in S$? (where $\{x_{n}\}\subset \mathbb R$ could be any sequence of our choice-not necessary from $S$, which converges to 0, to support our argument; for example if $S$ contains only one sequence, hence $t_{n}=s'_{n}$, then we can choose $x_{n}=1/n$)

I think the problem here is looking for a sequence which converges to zero faster than any sequence in $S$?

Edit: Maybe a good choice for $s'_{n}$ to be the product (I'm confused about the definition of the product of sequences) of all sequences in $S$? I don't know!! Please advice!

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    If the above holds, then there is no $s'_n$ with the desired property: one can always take $t_n=s'_n x_n$.2012-06-12

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