# Example of an unbounded operator whose adjoint is not densely defined

In his book "Quantum Theory for Mathematicians", B. C. Hall mentions that there are some pathological examples of unbounded operators on separable Hilbert spaces whose adjoint is not densely defined (p. 171). What is such an example?

## 1 Answer

The adjoint $$A^*$$ of an unbounded operator $$A$$ on a Hilbert space $$\mathcal H$$ is densely defined if and only if $$A$$ is closable, i.e. the closure of the graph $$\Gamma(A) = \{(x, Ax) \in \mathcal H^2: x \in \mathcal D(A)\}$$ is the graph of an operator. So for a counterexample all we need is a sequence $$x_n \in \mathcal D(A)$$ such that $$x_n \to 0$$ while $$A x_n$$ converges to some $$y \in \mathcal H$$. For example, if $$e_n$$ is an orthonormal sequence in $$\mathcal H$$ you could take $$A e_n = n e_1$$.

• Why is the adjoint $A^*$ densely defined iff $A$ is closable? – Rodrigo May 22 at 17:59
• See e.g. Rudin, "Functional Analysis", theorem 13.12. – Robert Israel May 22 at 23:50
• "If $T$ is a densely defined closed operator in $H$, then $\mathcal D(T^*)$ is dense and $T^{**}=T$" ? – Rodrigo May 24 at 1:42