Which of these subsets are dense in $l^2$?

Question

Let $l^2$ be the Hilbert Space of Square summable sequence and $e_{k}$ denote the coordinate vector in which the $k$th position is $1$ and rest $0$.Then which of the following subspaces is not dense in $l^2$?

1.$\operatorname{span}\{e_{1}-e_{2},e_{2}-e_{3},e_{3}-e_{4}, \ldots\}$

2.$\operatorname{span}\{2e_{1}-e_{2},2e_{2}-e_{3},2e_{3}-e_{4},\ldots\}$

3.$\operatorname{span}\{e_{1}-2e_{2},e_{2}-2e_{3},e_{3}-2e_{4},\ldots\}$

4.$\operatorname{span}\{e_{2},e_{3},\ldots\}$

I think option (4) cannot be dense in $l^2$ as we cannot find any sequence from the set in option 4 which converges to $(1,0,0,....)$.

And observing option (3) (simply adding the vectors)leads to $(1,-1,1,-1,1,-1,...)$ which is not square summable and hence cannot be dense in $l^2$, please correct me if I am wrong?

How to observe other options?

Answer 1

A subspace $M$ of $\ell^2$ is dense if and only if $M^\perp = \{0\}$. This is evident from the Riesz projection theorem: $\overline{M} \oplus M^\perp = \ell^2$.

1. Assume $x \perp e_1-e_2, e_2-e_3, \ldots$. We get $\langle x, e_k\rangle = \langle x, e_{k+1}\rangle, \forall k \in \mathbb{N}$ so $\langle x, e_k\rangle = \langle x, e_1\rangle, \forall k \in \mathbb{N}$. By Parseval we have $$\|x\|^2 = \sum_{n=1}^\infty \left|\langle x, e_n\rangle\right|^2 = \left|\langle x, e_1\rangle\right|^2 \cdot \sum_{n=1}^\infty 1$$ For this to converge it must be $\langle x, e_1\rangle = 0$ so $x = 0$.
2. Assume $x \perp 2e_1-e_2, 2e_2-e_3, \ldots$. We get $2\langle x, e_k\rangle = \langle x, e_{k+1}\rangle, \forall k \in \mathbb{N}$ so $\langle x, e_k\rangle = 2^{k-1}\langle x, e_1\rangle, \forall k \in \mathbb{N}$. By Parseval we have $$\|x\|^2 = \sum_{n=1}^\infty \left|\langle x, e_n\rangle\right|^2 = \left|\langle x, e_1\rangle\right|^2 \cdot \sum_{n=1}^\infty 2^{k-1}$$ For this to converge it must be $\langle x, e_1\rangle = 0$ so $x = 0$.

3. Assume $x \perp e_1-2e_2, e_2-2e_3, \ldots$. We get $\langle x, e_k\rangle = 2\langle x, e_{k+1}\rangle, \forall k \in \mathbb{N}$ so $\langle x, e_k\rangle = \frac1{2^{k-1}}\langle x, e_1\rangle, \forall k \in \mathbb{N}$. By Parseval we have $$\|x\|^2 = \sum_{n=1}^\infty \left|\langle x, e_n\rangle\right|^2 = \left|\langle x, e_1\rangle\right|^2 \cdot \sum_{n=1}^\infty \frac1{2^{n-1}} = \left|\langle x, e_1\rangle\right|^2$$ We see that $\langle x, e_1\rangle$ can be anything, so the vector of the form $$x = \langle x, e_1\rangle\cdot \sum_{n=1}^\infty \frac1{2^{n-1}}e_n = \langle x, e_1\rangle\left(1, \frac12,\frac14, \ldots\right)$$ will be orthogonal to all $e_1-2e_2, e_2-2e_3, \ldots$.

4. Clearly $e_1$ is orthogonal to all $e_2, e_3, \ldots$.

We conclude that $1$ and $2$ are dense in $\ell^2$ while $3$ and $4$ aren't.