Closures of the following subsets of $\ell^1(\mathbb{N})$:

Question

I'm trying to work out how to compute the closures of the following sets:

1. $$\left\{a ∈ ℓ^1(\mathbb{N}) \,\middle|\, \sum |a_j|^2 \le 1 \right\}$$
2. $$\left\{a ∈ ℓ^1(\mathbb{N}) \,\middle|\, \sum |a_j|^2 < \infty \right\}$$

How do I go about doing this?

Answer 1

Define $T$ on $\ell^1(\mathbb{N})$ by $ T(a) = \sum_{j} |a_j|^2 $. Then

$$ |T(b) - T(a)| \leq \sum_j |a_j^2 - b_j^2| \leq \|a - b\|_{1} (\|a\|_{1} + \|b\|_{1}), $$

and so, $T$ is a well-defined function $\ell^1(\mathbb{N}) \to [0, \infty)$ that is locally Lipschitz and hence continuous. In particular, the inverse image of a closed set under $T$ is also closed. Now,

$$ \left\{a \in \ell^1(\mathbb{N}) \,\middle|\, \sum |a_j|^2 \leq 1 \right\} = T^{-1}([0, 1]) $$

and

$$ \left\{a \in \ell^1(\mathbb{N}) \,\middle|\, \sum |a_j|^2 < \infty \right\} = T^{-1}([0, \infty)) = \ell^1(\mathbb{N}). $$

Answer 2

The second set is the whole space $\ell^1(\mathbb{N})$; this is easily seen by Hölder's inequality: $$ \|a\|_2^2 \leq \|a\|_1 \|a\|_\infty < \infty, $$ since if $a \in \ell^1(\mathbb{N})$ then necessarily $\|a\|_\infty < \infty$.

For the first set, say $a^{(j)} \to a$ in $\ell^1(\mathbb{N})$ with $\|a^{(j)}\|_2 \leq 1$ for all $j$. To show $\|a\|_2 \leq 1$ it suffices to show that $a^{(j)} \to a$ in $\ell^2(\mathbb{N})$. For this, one can use $$ \|a^{(j)} - a\|_2^2 \leq \|a^{(j)} - a\|_1 (\|a^{(j)}\|_1 + \|a\|_1), $$ where we use $(u-v)^2 \leq |u - v|(|u| + |v|)$. Evidently the righthand side vanishes as $j \to \infty$, as needed.