1. ## Example of operators

Do you know the example of next operators:
$\displaystyle B:H_1\to H_2$ is a closed operator, $\displaystyle A:H_2\to H_3$ is a bounded linear operator such that
$\displaystyle AB$ is not a closable operator?
$\displaystyle H_1,H_2,H_3$ are Hilbert spaces.

2. Originally Posted by karkusha
Do you know the example of next operators:
$\displaystyle B:H_1\to H_2$ is a closed operator, $\displaystyle A:H_2\to H_3$ is a bounded linear operator such that
$\displaystyle AB$ is not a closable operator?
$\displaystyle H_1,H_2,H_3$ are Hilbert spaces.
Let $\displaystyle \{e_n\}_{n=1}^\infty$ be an orthonormal basis for the Hilbert space $\displaystyle H$. Define A, B on H by $\displaystyle Ae_n = \frac1ne_1$ and $\displaystyle Be_n=n^2e_n$ (for all n). Then B is closable (it is densely defined and contained in its adjoint), A is bounded (use Cauchy–Schwarz to show that), but AB is not closable (if $\displaystyle x_n = \frac1ne_n$ then $\displaystyle x_n\to0$ but $\displaystyle ABx_n\to e_1$).

3. Thank you for the reply, but Is A linear operator?

4. Originally Posted by karkusha
Is A linear operator?
I should have said that you need to extend both A and B by linearity to the space of all finite linear combinations of the basis vectors, and then take their closures. So if $\displaystyle x = \sum\lambda_ne_n$ (only finitely many $\displaystyle \lambda$s nonzero) then $\displaystyle Ax = \sum\frac{\lambda_n}ne_1$, and $\displaystyle Bx = \sum n^2\lambda_ne_n$. That way, A and B are both linear.