Subspaces of $\ell^2$

Question

Let $\ell^2$ be the usual space of sequences $(x_n)_{n=1}^\infty$ of real/complex numbers such that $$\sum_{n=1}^\infty |x_n|^2<\infty$$ taken as a Hilbert space with the usual inner product and with canonical basis $\{U_n\}_{n=1}^\infty$.

I have some questions regarding its subspaces, mainly how big is $\ell^1$ inside $\ell^2$? I know it is dense but only because $c_{00}$, the space of "eventually zero" sequences is dense (in every $\ell^p$) so this doesn't seem very "satisfying".

Let $V\le\ell^2$ be an infinite dimensional subspace, is $V\cap\ell^1$ always non-trivial? This would mean $\ell^1$ is in fact very big inside $\ell^2$ as it is present in every "important" subspace. I have tried many proofs/counterexamples of this but failed every time.

The map $(x_n)_{n=1}^\infty\mapsto (x_n-x_{n+1})_{n=1}^\infty$ from the space $c_0$ of "converging to $0$" sequences isn't even well defined as the image need not be in $\ell^2$ (not even in $\ell^1$). Taking some sequence $\mathbf{x}=(x_n)_{n=1}^\infty\in\ell^2\setminus\ell^1$ and taking the space generated by the vectors $\{\mathbf{x}-x_nU_n\}_{n=1}^\infty$, that is $\mathbf{x}$ but with its $n$-th coordinate set equal to $0$, doesn't work. If $\{b_1,b_2,\dots\}\subseteq V$ is a linearly independent (infinite subset) then we can make all sorts of assumptions like $\Vert b_i\Vert_2=1$ and $b_i \perp b_j$ for every $i,j\ge 1$ and $i\ne j$ but I can't make a proof out of this.

So, if $V\le\ell^2$ is a subspace and $V\cap\ell^1=\{\mathbf{0}\}$ then must $V$ be finite-dimensional? You are free to make the assumption that $V$ is closed.

Also, can you give examples (maybe in a different answer) of some open and closed proper subspaces of $\ell^2$

Thanks!

Answer 1

Of course there's a correct example of the $V$ you ask for in @Paul 's comment. Another example, possibly interesting, and possibly such that it's more obvious that it has the required property:

Let $H^2$ denote the space of functions $f$ holomorphic in the unit disk such that $$||f||_{H^2}=\sup_{0\le r<1}\left(\frac1{2\pi}\int_0^{2\pi}|f(re^{it}|^2\,dt\right)^{1/2}<\infty.$$ It's well known and not hard to show that if $$T(x_n)_{n\ge0}=\sum_{n=0}^\infty x_nz^n$$then $T$ is a bijective linear isometry from $\ell_2$ onto $H^2$.

If $|a|=1$ and we define $$f_a(z)=\frac1{(z-a)^{1/4}}$$then $f_a\in H^2$. If $V$ is the span of the $f_a$ then it's clear that $$V\cap T(\ell_1)=\{0\},$$just because any element of $T(\ell_1)$ must be bounded in the unit disk.

Bonus: This is a convenient place to insert details for the other example in a comment above.

For $\alpha\in(1/2,1)$ let $e_\alpha$ be the sequence $$(e_\alpha)_n=n^{-\alpha}.$$Note that $e_\alpha\in\ell_2$.

Suppose $$x=\sum_{k=1}^na_ke_{\alpha_k},$$where $a_k\ne0$ for every $k$. Suppose $\alpha_1<\alpha_k$ for every $k>1$.

Then the first term in the sum dies asymptotically slower than all the other terms. So there exist $c>0$ and $N$ such that$$|x_n|\ge cn^{-\alpha_1}\quad(n>N);$$hence $x\notin\ell_1$. (And $x\notin\ell_1$ implies $x\ne0$, proving independence of the $e_\alpha$ at the same time.)

Answer 2

There are no subspaces "both open and closed" in $\ell_2$ except $\ell_2$ itself, because $\ell_2$ is a connected topological space, among many other things.

As far as topological category is concerned, $\ell_1$ is rather small inside $\ell_2$. Actually, it is of the first category in $\ell_2$, which means that it can be represented as a countable union of nowhere dense subsets of $\ell_2$.