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While I was studying functional analysis I found in the script the following counterexample:

Let $X = l^1$ and consider the linear operator $$ (Ax)_n\left\{ \begin{array}{ll} n x_{n-1} & \text{if $n$ is even} \\ 0 & \text{if $n$ is odd} \end{array} \right.$$ and let $D(A) := \{x ∈ l^1: Ax ∈ l^1\}.$ It is easy to see that $A$ is closed.

However, $B := A + (−A) = 0$ with $D(B) = D(A)$ and $C := AA = 0$ with $D(C) = D(A)$ are not closed.

My question is how one can check this. It is said that $D(A)$ is dense in $l^1$ (and therefore if $B$ or $C$ are closed, $D(A)$ must be also closed which is a contradiction), but I don't know how $D(A)$ looks like or how can I prove its density in $l^1$.

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$D(A)=\{(x_n) \in \ell^{1}:\sum |2nx_{2n-1}| <\infty\}$. Every sequence $(y_n)$ with the property that only finite number of $y_n$'s are non zero is on $D(A)$ do $D(A)$ is dense.

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  • $\begingroup$ I get it now. Thank you for your help! $\endgroup$ – gmirsan May 14 at 13:34

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