Question: $2.22$ The purpose of this exercise is to construct an unbounded operator $A: D(A) \subset$ $E \rightarrow E$ that is densely defined, closed, and such that $\overline{D\left(A^*\right)} \neq E^* .$
Let $E=\ell^{1}$, so that $E^*=\ell^{\infty}$. Consider the operator $A: D(A) \subset E \rightarrow E$ defined by
$$
D(A)=\left\{u=\left(u_{n}\right) \in \ell^{1} ;\left(n u_{n}\right) \in \ell^{1}\right\} \text { and } A u=\left(n u_{n}\right)
$$
- Check that $A$ is densely defined and closed
- Determine $D\left(A^*\right), A^*$, and $\overline{D\left(A^*\right)}$.
Brezis solution claims that:
$$
\begin{aligned}
D\left(A^*\right) &=\left\{v=\left(v_{n}\right) \in \ell^{\infty} ;\left(n v_{n}\right) \in \ell^{\infty}\right\}, \\
A^* v &=\left(n v_{n}\right) \text { and } \overline{D\left(A^*\right)}=c_{0} .
\end{aligned}
$$
My attempt
- First doubt: I know that the definition of
$$\ell^{1}=\left\{(x_n)_{\mathbb{N}}: \|x\|= \displaystyle\sum_{k=1}^{\infty}|x_k|<+\infty \right\}$$
so, I can accept intuitively that $D(A)$ contains all finite sequences. However, I'm struggling to formalize this idea. I've tried to use this result from metric spaces:
$F\subset M$ is dense $\iff \forall a\in M, \; \exists \; x_n \in F: x_n \rightarrow a$
I do not know if I have to use an specific norm, do I have to use uniform convergence?
- Second doubt: How can I infer from the general definition of $D(A^*)$ given in Brezis' book, i.e.
$D(A^*)=\left\{v \in F^*; \exists c\geq 0 \; \text{such that}\; |\langle v, Au \rangle \leq c\|u\| \; \forall u \in D(A)\right\}.$
That $D(A^*)$ in this case is going to be $D\left(A^*\right) =\left\{v=\left(v_{n}\right) \in \ell^{\infty} ;\left(n v_{n}\right) \in \ell^{\infty}\right\}$?
- Finally, third and last doubt: I could not see that $c_0=\{(a_j)_{\mathbb{N}}: \lim a_j=0\}$ is equal to the closure of $D(A^*)$, because for me is not intuitively that all sequences in $D(A^*)$ goes to zero, or maybe I misunderstood (I believe more in this).
Thanks in advance.
Best Answer
$\ell^1$ and $\ell^\infty$ each have a specific norm. You have not shown denseness in $\ell^1$ or in $\ell^\infty$ unless you use that norm (or show that whatever norm you did use is equivalent to that norm).
The point here is that the set of finite sequences (those that end in tails of $0,0, \ldots$) is dense in $\ell^1$. This is easy enough to prove. If $u \in \ell^1$, then because $\sum_n |u_n|$ converges to a finite value, for any $\epsilon > 0$ there has to be some $N$ so that $\sum_{n >N} |u_n| < \epsilon$. And therefore the sequence $$\tilde u = \begin{cases}u_n& n \le N\\0& n > N\end{cases}$$ has the property $$\|u - \tilde u\| = \sum_n |u_n - \tilde u_n| = \sum_{n >N} |u_n| < \epsilon$$ and hence there is a finite sequence in every neighborhood of any point $\ell^1$.
Now if you can show that $D(A)$ contains all finite sequences, then $D(A)$ must be dense as well.
For $u \in D(A), Au =(nu_n)_n$ and for $v \in \ell^\infty, \langle v, Au\rangle = \sum_n v_n(nu_n) = \sum_n (nv_n)u_n$. If $(nv_n)_n \in \ell^\infty$, then there is some $c$ with $|nv_n|< c$ for all $n$. So $$|\langle v, Au\rangle| \le \sum_n |nv_n||u_n| \le \sum_n c|u_n| = c\|u\|$$ Conversely, if $|\langle v, Au\rangle| \le c\|u\|$ holds for every $u \in D(A)$, it holds in particular for $u = (\delta_{nk})_n$ for each $k\in\Bbb N$.
$v \in D(A^*)$ if $(nv_n)_n \in \ell^\infty$. That is, if there is some $M$ such that $|nv_n| < M$ for all $n$. And that is, if $$|v_n| < \frac Mn$$ for all $n$. Do you see it now?