I have been working through the exercises for one of my upcoming exams and came across the following problem:

Let $B:D(B)\to\mathcal{H}$ be a bounded self-adjoint operator in a Hilbert space $\mathcal{H}$. Show that $D(B)=\mathcal{H}$.

Now, in my attempt to show this I have shown the implication: bounded $\implies$ continuous, before showing that: continuous $\implies$ closure of the operator. However, I am now unsure how to use the self-adjointness of the operator to show that the domain $D(B)$ is indeed $\mathcal{H}$.

Any help would be greatly appreciated.

  • $\begingroup$ This might be a silly question, but... Is it assumed that the domain of $B$ is maximal? I believe that given an operator $B:D(B)\rightarrow \mathcal{H}$, the domain of the adjoint is defined to be the maximal domain. The fact that $B$ is self-adjoint would guarantee that the domain of $B$ is maximal. $\endgroup$ – Peter Apr 26 '17 at 9:20
  • $\begingroup$ The only information given in lectures was that $D(B)=D(B^*)$, with the domain of the adjoint operator given by: $D(B^*)=\{f\in\mathcal{H}:\exists \;g\in D(B)\text{ such that } f=Bg\}$. There was no mention of the domain of $B$ being maximal, unfortunately. Unless this was missed out by accident? $\endgroup$ – David Apr 26 '17 at 9:29

Assuming that $B\colon D(B) \rightarrow \mathcal{H}$ is densely defined implies that its adjoint exists and it is closed. Since $B$ equals to $B^*$ that it has to be closed itself. We know that the closure of $D(B)$ equals to $\mathcal{H}$ so it is enough to show that $D(B)$ has to closed in $\mathcal{H}$.

Since $B$ it is closed then for any sequence $(x_n)$ from $D(B)$ that converges to some $x$ in $\mathcal{H}$, if $Bx_n$ converges to some $y \in \mathcal{H}$ then $x \in D(B)$ and $Bx = y$.

Take a sequence $(x_n)$ from $D(B)$ that converges to some $x$ in $\mathcal{H}$, we show that $Bx_n$ converges. It is enough to show that $(Bx_n)$ is a Cauchy sequence (since $\mathcal{H}$ is complete), this can be obtain from the boundedness of $B$ and the fact that $(x_n)$ is Cauchy, namely $$ \left\| Bx_n - Bx_m \right\| \leq \| B\| \| x_n - x_m \|.$$


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