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Is it possible to construct an unbounded extension of bounded densely defined operator?

To be more concrete, let $\mathcal{H}$ be Hilbert space, $\mathcal{D}\subset\mathcal{H}$ - a dense subset, $A:\mathcal{D}\rightarrow\mathcal{H}$ an operator such that $\forall_{\psi\in\mathcal{D}}\|A\psi \|\leq const$. I wonder whether there can exist an operator $\tilde{A}:\mathcal{D}_1 \rightarrow \mathcal{H}$, $\mathcal{D}\subset\mathcal{D}_1$ such that $\tilde{A}|_\mathcal{D} = A$ and $\tilde{A}$ is unbounded.

$\tilde{A}$ cannot be closed since densely defined operator admits exactly one closed extension which is bounded.

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up vote 3 down vote accepted

Assuming the axiom of choice, yes. Let $A'$ be the unique bounded extension. Pick a Hamel basis $B$ for $\mathcal{D}$ and extend it to a Hamel basis $B'$ for $\mathcal{H}$. Now define $\tilde{A}$ on $B' \setminus B$ any way you like, as long as it's different from $A'$. The result will necessarily be an unbounded linear operator.

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