Showing a subspace is not closed

Question

I have been asked the following problem which I have so far failed to solve rigourosly.

Let $M=\{ (a_n)_{n=1}^\infty\in l^2\vert \sum_{n=1}^\infty \frac{a_n}{\sqrt{n}}=0 \}$. Determine whether $M$ is a closed subspace of $l^2$.

My attempt:

I think that the answer is no because this seems like the kernel of a functional

$$ \phi(a):= \Big \langle a,\big(\frac{1}{\sqrt{n}}\big) \Big \rangle $$.

And since $(l^2)^*=l^2$, and $\Big( n^{-1/2} \Big)_{n=1}^\infty\notin l^2$, we should know that $\phi$ is not continuous and its kernel is dense, i.e., not closed.

However it is not clear that $\phi$ is defined and hence not even necessarily a functional.

My question is whether this argument is salvagable, and if not whether someone knows of a solution to this question?

Answer 1

No, $\phi$ is not well-defined. It follows from this nice result:

Let $(b_n)_n$ be a sequence of scalars such that $\sum_{n=1}^\infty a_nb_n$ converges for every $(a_n)_n \in \ell^2$. Then $(b_n)_n \in \ell^2$.

Also, an indication that $\phi$ is not well-defined is the fact that we cannot explicitly construct an unbounded linear functional on a Banach space. This is because it is consistent with the axiom of choice that all linear functionals on a Banach space are bounded.

There is a better way to show that $M$ is not closed. Prove that $M^\perp = \{0\}$. Then, if $M$ were closed, The Riesz projection theorem would imply that $\ell^2 = M \oplus M^\perp = M$, which is false since e.g. $e_1 \notin M$.

Let $x = (x_n)_n \in M^\perp$ be arbitrary; we claim that $x = 0$. For every $n \ge 2$ we have that $\sqrt{n-1}e_{n-1} - \sqrt{n}e_n \in M$ and hence $$x \perp \left(\sqrt{n-1}e_{n-1} - \sqrt{n}e_n\right) \implies x_n = \sqrt{\frac{n-1}{n}}x_{n-1}$$ so iterating this gives $x = x_1\left(\frac1{\sqrt{n}}\right)_n \in M$ which implies $x_1 = 0$ and hence $x = 0$. Therefore $M^\perp = \{0\}$.

Answer 2

For an explicit example, consider the sequence $a^{(n)}$ of $\ell^2$ defined as follows.

• For $k \in \{2, \dots, n-1 \}$, $a^{(n)}_k = \frac{1}{\sqrt{k}\log k}$.
• For $k \in \{n, \dots, r_n\}$, $a^{(n)}_k = - \frac{x_n}{\sqrt{k}\log k}$.
• Otherwise, $a^{(n)}_k = 0$.

We denote here $r_n = \lfloor e^{(\log n)^2} \rfloor$, and $$ x_n = \frac{\sum_{k=2}^{n-1} \frac{1}{k\log k}}{\sum_{k=n}^{r_n} \frac{1}{k\log k}}. $$

By comparison with integrals, we see that $$ \sum_{k=2}^{n-1} \frac{1}{k \log k} = \log \log n + O(1), $$ and thus a short computation shows that, by the very choice of $(r_n)$, we have $x_n \to 1$. Finally the sequence $(a^{(n)})$ has the following properties.

• By construction and definition of $x_n$, $a^{(n)} \in M$.
• If $(b_k)$ is the sequence defined by $b_k = \frac{1}{\sqrt{k}\log k}$ for $k \geq 1$, then $$ \|a^{(n)}-b\|_{\ell^2} \leq x_n^2 \sum_{k \geq n} \frac{1}{k\log^2 k}, $$ and this tends to $0$ since $(x_n)$ is bounded and the series $\sum \frac{1}{k \log^2 k}$ converges.
• Yet, $(b_k)$ is obviously not in $M$.