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I am not sure about the following. Let us consider the space of positive bounded sequences and a functional $(x_i)\rightarrow \sum_{i=1}^{\infty}\beta(i) x_i$, where $\beta$ is a function $\beta: N \rightarrow (0,1]$ on the natural numbers strictly decreasing and s.t. $\sum_{i=1}^\infty \beta(i)=B<+\infty$. Is $$(x_i)\rightarrow\sum_{i=1}^{\infty}\beta(i) x_i$$ continuous w.r.t. the product topology on the set of positive bounded sequences?

My intuition is yes and my idea is the following. Consider a net $x^{\alpha}\rightarrow x$ $\forall \delta>0$, $\forall k\in N$, $\exists \alpha_k$ s.t. for all $i=1,\dots,k$, $|x_i^{{\alpha}_k}-x_i|<\delta$. Therefore $\forall k\in N$, $\sum_{i=0}^k \beta(i)|x_i^{{\alpha}_k}-x_i|<\sum_{i=0}^k \beta(i)\delta$. So $$\lim_k \sum_{i=0}^k \beta(i)|x_i^{{\alpha}_k}-x_i|<B\delta$$ Is this true or there is a problem when I take the $lim$ for $\alpha_k$?

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You want to show that for each $\delta$, there is a fixed $\alpha_k$ that makes the difference small. What you have shown is that you can make the sum small on arbitrarily large finite subsets of indices, but you still must provide an argument on how to bound the rest of the sum if you want to show this is continuous. But you have no control over the rest of the $x_i$, other than the bound you get from the fact they form a bounded sequence. But you only get that bound for the point, not points near it. So you have points near the given point that take on arbitrarily large values for large N.

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I think it is not continuos. Take the net of sequences $x^n=(0,\dots,\frac{1}{b(n)},0\dots)$ where the term $\frac{1}{b(n)}$ appears in the $n$-th position of the sequence $x^n$. We have that $x^n\rightarrow x:=(0,0,\dots)$ in the product topology. But the functional above maps $x$ to $0$, whereas $\forall n\in N$, $$ \sum_{i=1}^{\infty} \beta(i) x^n_i=\beta(n)\frac{1}{\beta(n)}=1. $$ Therefore, calling the map above $f$ we will have $x^n\rightarrow_n x$, but $f(x^n)\nrightarrow_n f(x)$.

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