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?
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: $$ \|e_1-x_n\|_{l^2}^2 = \frac1{n^2} + \sum_{k=2}^{n-1}( \frac{n-k}{n}- \frac{n-k+1}{n})^2 + \frac1{n^2}= \frac{n}{n^2}=\frac1n \to 0 \quad\text{ for } n\to\infty. $$