# Computations of a counterexample in order to check that the sum and product of closed operators are not always closed

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 $$l^1$$.

## 1 Answer

$$D(A)=\{(x_n) \in \ell^{1}:\sum |2nx_{2n-1}| <\infty\}$$. 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.

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