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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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@Arturo: $\{h_{n}\}$ is any sequence of functions in $H$. So to make the definition clear, $S=\{\,\{s_{n}\}: s_{n}=\sup_{\mathbb R}|h_{n}(x)|, \{h_{n}\}\subset H, s_{n}\to 0\}$. –  Paul Jun 12 '12 at 4:35
    
It just came to my mind: if there is a sequence $\{s_{n}\}\in S$ which converges to 0 then there will be many such sequences in $S$ by taking subsequences of that $\{s_{n}\}$, which all will convereg to 0. –  Paul Jun 12 '12 at 4:54
    
So, any help!!? –  Paul Jun 12 '12 at 5:06
    
Patience, patience. –  Gerry Myerson Jun 12 '12 at 6:02
    
@Paul Isn't $S$ just the set of all positive sequences that converge to $0$? Given any such sequence $\{\sigma_n\}$ I can consider $h_n=\sigma_n f$ where $f\in H$ is a function with $\sup |f|=1$. This choice of $h_n$ gives me $s_n=\sigma_n$. –  user31373 Jun 12 '12 at 23:44

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