prove a subset of $l^2$ is closed? Let $\{f_i\}$ be a sequence of (nice) functions in $L^p[0,1],$ and $p>1, \frac{1}{p}+\frac{1}{q}=1.$
Define a subset $A$ of the space $l^2$ as
$$A=\left\{(a_1, a_2, \ldots)\in l^2: \text{ such that } a_i=\int_0^2g(x)f_i(x)dx, \text{ for }g\in L^q[0,1] \text{ and } \|g\|_q\leq1\right\}.$$
Here $\{f_i\}$ are good enough in order that for any $\|g\|_q\leq1,$ the resulting $(a_1, a_2, \ldots)\in l^2.$
Can we prove this subset $A$ is closed in $l^2$? 
Have been working this for a while. Any suggestion is greatly appreciated. (more assumptions may be imposed on $\{f_i\},$ for instance, smoothness)

I will assume the limits of the integral should be $0$ and $1$, not $0$ and $2$. I also won't really comment on the requirements of the sequence $f_i$; the set you've provided is well-defined in any case. If we can establish that this set is always closed, regardless of the condition that $(a_1, a_2, \ldots) \in l^2$ for any $\lVert g \rVert \le 1$, then surely this will be of some use to you!
I also plan to generalise the result, for no other reason than to simplify the notation. We replace $L^p[0, 1]$ with a reflexive Banach space $X$, and $L^q[0, 1]$ with its dual $X^*$. The functions $f_i$ shall now be a sequence $x_i \in X$. We are therefore looking at the set
$$A = \left\lbrace (f(x_1), f(x_2), \ldots) \in l^2 : f \in B_{X^*} \right\rbrace.$$
Suppose we have a sequence $(f_i(x_1), f_i(x_2), \ldots)_{i=1}^\infty \in A$ that converges in $l^2$ to some sequence $(a_1, a_2, \ldots)$. That is,
$$\sum\limits_{j = 1}^\infty ~ \lvert f_i(x_j) - a_j \rvert^2 \rightarrow 0$$
as $i \rightarrow \infty$.
In order to prove $(a_1, a_2, \ldots) \in A$, we must construct its corresponding $f \in X^*$. Since $X$ is reflexive, so is $X^*$, hence $B_X^*$ is weakly compact. By the Eberlein-Smulian theorem, it's also sequentially weakly compact. Hence the sequence $f_i$ has a weakly convergent subsequence $f_{n_i}$, weakly converging to some $f \in B_{X^*}$. That is, for any $x \in X$,
$$f_{n_i}(x) \rightarrow f(x)$$
as $i \rightarrow \infty$. Now we must prove that $f(x_j) = a_j$ for all $i$.
But this is now straightforward. If $\sum\limits_{j = 1}^\infty ~ \lvert f_i(x_j) - a_j \rvert^2 \rightarrow 0$, then $\lvert f_i(x_j) - a_j \rvert \rightarrow 0$ for any fixed $j$. That is, $f_i(x_j) \rightarrow a_j$ as $i \rightarrow \infty$. Considering the subsequence $f_{n_i}(x_j)$ subsequence, which converges to $f(x_j)$, we see by the uniqueness of limits that $a_j = f(x_j)$ as required. Hence, the set $A$ is closed.
