How to show that the set $S$ is closed in $(c_{0},\|\cdot\|_{\infty})$ 
Suppose $S$ is the set of all $x\in (c_{0},\|\cdot\|_{\infty}$) (infinitely small sequences), such that there exists $f\in\mathbb{L}^{2}[0,1],\;\|f\|_{\mathbb{L}^{2}}\leq 1$ and $\forall k\in\mathbb{N}$ $$x^{k}=(\text{k-th component of x}) = \displaystyle \int\limits_{[2^{-k},\;2^{1-k}]}f(t)dt$$
Is $S$ closed in $(c_{0},\|\cdot\|_{\infty})$?

I solved another version of this problem with $f\in\mathbb{L}^{1}[0,1],\;\|f\|_{\mathbb{L}^{1}}\leq 1$, which seems much more straightforward, because it all boils down to showing that $S$ is exactly those sequences, for which $\displaystyle \sum\limits_{j=1}^{\infty}|x^{j}|\leq 1$. For this problem we can obtain $$|x^{j}| \leq \displaystyle \int\limits_{[2^{-j},\;2^{1-j}]}|f(t)|dt\stackrel{\text{Hölder ineq.}}{\leq}\sqrt{\int\limits_{[2^{-j},\;2^{1-j}]}|f(t)|^2dt}\sqrt{2^{-j}}\leq\sqrt{2^{-j}}$$ So squaring both sides and summing over $\mathbb{N}$ gives $\displaystyle\sum\limits_{j=1}^{\infty}|x^{j}|^2\leq 1$. Is it a sufficient condition? I tried many things including trying to prove that the set $f$ of functions that correspond to at least one sequence is totally bounded in $\mathbb{L}^2[0,1]$ using Kolmogorov-Riesz criterion (and then trying to extract a Cauchy sequence), which doesn't seem to be the case, extracting subsequence, which can be bounded by some integrable function and then using Dominated Convergence Theorem, however it is not guaranteed that this subsequence converges pointwise almost everywhere.
It is clear that the strategy here is to somehow construct function $f$ for a sequence $x$ from the closure of $S$, which would satisfy that stuff with components and then prove that $\|f\|_{\mathbb{L}^{1}[0,1]}\leq 1$. But what should I consider in order to construct such $f$?
 A: Well, that turned to be very easy. I claim, that $S = \left\{x\in c_{0}:\displaystyle \sum\limits_{j=1}^{\infty}2^{j}x^{2}(j)\leq 1\right\}$.
Necessity: from what I derived from the start we get $2^{j}x^{2}(j)\leq \displaystyle \displaystyle \int\limits_{[2^{-j},\;2^{1-j}]}|f(t)|^2dt$, so we need just to sum over $\mathbb{N}$ to obtain the result.
Sufficiency: if $x\in c_{0}$ satisfies the inequality then set $f(t) = 2^{j}x(j)$ for $t\in [2^{-j},\;2^{1-j})$. So $$\displaystyle\int\limits_{[2^{-j},2^{1-j}]}f(t)dt = 2^{j}x(j)\cdot 2^{-j} = x(j)$$ and $$\|f\|_{\mathbb{L}^{2}} = \displaystyle \int\limits_{[0,1]}f^{2}(t)dt = \sum\limits_{j=1}^{\infty}\int\limits_{[2^{-j},\;2^{1-j}]}f^{2}(t)dt = \sum\limits_{j=1}^{\infty}4^{j}x^{2}(j)2^{-j} = \sum\limits_{j=1}^{\infty}2^{j}x^{2}(j)\leq 1$$
Suppose $x\in [S]$ ($[S]$ -- the closure of $S$). Then consider the sequence $x_{n}\stackrel{\|\cdot\|_{\infty}}{\to} x$. It follows that $x_{n}$ converges to $x$ pointwise (or coordinate-wise), so $$\displaystyle\sum\limits_{j=1}^{N}2^{j}x^2(j) = \sum\limits_{j=1}^{N}\lim\limits_{n\to\infty}2^{j}x^{2}_{n}(j) = \lim\limits_{n\to\infty}\sum\limits_{j=1}^{N}2^{j}x^{2}_{n}(j)\leq 1$$ So taking limit $N\to\infty$ we obtain the desired.
I hope I didn't fail to notice anything.
