I have read that the adjoint of a linear operator $T$ on a Hilbert space $H$ is uniquely defined if and only if the domain of $T$, $\mathscr{D}(T)$ is dense in $H$. The "if" direction is a matter of simple computation, take any point $a$ of $H$ and any converging sequence $\{v_n\}_n$ to it and conclude that then $\left<a, T^*y - S^*y\right> = 0$ if $y\in\mathscr{D}(T^*)$ (set of $y$s s.t. $x\mapsto \left<x,y\right>$ is continuous in $\mathscr{D}(T)$), which in turn forces $T^*y = S^*y$ if $S^*y, T^*y$ are such that $\forall x\in\mathscr{D}(T):\left<Tx,y\right> = \left<x,T^*y\right> = \left<x,S^*y\right>$.
But how could one show the converse: That if $T^*$ is unique then necessarily $\mathscr{D}(T)$ is dense in $H$? One idea was to show the contrapositive, that is $\mathscr{D}(T)$ is not dense in $H$ then $T^*$ is not unique, by taking such $a\in H$ that there does not exist a sequence $\{v_n\}_n\subset\mathscr{D}(T)$ such that $\lim_{n\to\infty}v_n = a$. But then, what would be an explicit construction of some $S^*$ that for some $y\in\mathscr{D}(T^*)$ and for all $x\in\mathscr{D}(T)$ we have $\left<x,T^*y\right> = \left<x,S^*y\right>$ but $T^*y\neq S^*y$?