Dense sets in Hilbert space of square summable sequences? Let $l^2 = \{x = (x_n)| x_n \in \mathbb{R}, \sum_{n=1}^{\infty} x_n^2 < \infty\}$  be the Hilbert space of square summable sequences and let $e_k$ denote the $k^{th}$ co-ordinate vector (with $1$ in $k^{th}$ place, $0$ elsewhere). Which of the following subspaces is NOT dense in $l^2$?


*

*span$\{e_1-e_2, e_2-e_3, \ldots\}$

*span$\{2e_1-e_2, 2e_2-e_3, \ldots\}$

*span$\{e_1-2e_2, e_2-2e_3, \ldots\}$ 

*span$\{e_2, e_3, \ldots\}$


By definition, we have to prove that for any $x \in l^2$ and for any $\epsilon > 0$, there exists $y$ in the respective sets such that $\|x-y\| < \epsilon$. How to proceed further?
 A: It is much easier to try to analyze the orthogonal complement of these sets instead of constructing the approximating element explicitly.
For the example (1), trying to write $e_1$ as a linear combination of $e_k-e_{k+1}$ leads to the fruitless sum
$$
e_1 = (e_1-e_2) + (e_2-e_3) + \dots
$$
Truncating early does not deliver something that has small distance to $e_1$.
To solve (1) and (2), take an arbitrary vector in the orthogonal complement of the sets in question. It is easy to prove that this vector is constant (1)
or is a multiple of $(1,2,4,8,\dots)$ (2). Hence the vector is zero, and the sets are dense.
Trying to do the same should easily generate a non-zero vector in the orthogonal complement of the sets in (3) and (4).

Now that we know, that the set in (1) is dense, we can try to construct a convergent approximation of $e_1$.
It turns out that
$$
x_n = \sum_{k=1}^{n-1} \frac{n-k}n (e_k-e_{k+1})
$$
does the trick:
$$
\|e_1-x_n\|_{l^2}^2 = \frac1{n^2} + \sum_{k=2}^{n-1}( \frac{n-k}{n}- \frac{n-k+1}{n})^2 + \frac1{n^2}= \frac{n}{n^2}=\frac1n \to 0 \quad\text{ for } n\to\infty.
$$
