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Let $\mathbf{H}$ be a Hilbert space, and $D$ an unbounded densely-defined operator on $\mathbf{H}$. As is well-known, every such operator admits an adjoint, with domain possibly different from that of $D$.

Is it possible to define an adjoint for a non-densely defined operator?

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  • $\begingroup$ Are you asking about how the usual definition breaks down in this case, or are you asking about possible alternate definitions that might work instead? $\endgroup$
    – Nate Eldredge
    Commented Mar 4, 2019 at 21:10
  • $\begingroup$ I'm asking about alternate definitions $\endgroup$
    – Max Schattman
    Commented Mar 4, 2019 at 22:05
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    $\begingroup$ I guess one sort of trivial approach is to consider the operator $DP$ where $P$ is orthogonal projection onto the closure of $\operatorname{dom D}$. In other words, extend your operator by zero on $(\operatorname{dom} D)^\perp$. Now $DP$ is densely defined and you may consider its adjoint in the usual sense, which formally should equal $PD^*$. $\endgroup$
    – Nate Eldredge
    Commented Mar 5, 2019 at 1:51
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    $\begingroup$ The issue when $D$ is not densely defined is that, since $D^* x$ is only defined via its inner product with elements of $\operatorname{dom} D$, it is only well defined up to adding something from $(\operatorname{dom} D)^\perp$. But there is a "natural" choice of what to add, namely 0. $\endgroup$
    – Nate Eldredge
    Commented Mar 5, 2019 at 1:58

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