This question comes from the Limiting Absorption Principle (LAP). I want to obtain the most general statement possible, so I proceed as follows.
Let $M$ be a topological space, $(H, \langle \cdot, \cdot \rangle)$ a Hilbert space, and $\mathcal L(D, H)$ the space of linear operators from the dense domain $D \subset H$ into $H$ (not necessarily bounded). Suppose $P : M \to \mathcal L(D, H)$ is continuous in the sense that $P(x) - P(y)$ is a bounded operator and its norm is continuous over $M \times M$. Assume further that $P(x)$ has a bounded inverse for all $x$ in a subset $\Omega \subset M$. This inverse may be locally bounded, but it may blow up close to the boundary $\partial \Omega$. To remedy this, we look for a Banach subspace $Y \subset H$ with a stronger norm $||\cdot ||_Y$ than that of $H$. By the canonical identification of $H = H ^*$, we may regard $H$ as a subset of $Y^*$, and, if we picked the right subspace, we will have that $\sup_{x \in \Omega} ||P(x)^{-1}||_{Y \to Y ^*} < \infty$.
Now, for any $u \in Y$, the sequence $u_n = P(x_n)^{-1} u$ is bounded in $Y^*$, so the Banach-Alaoglu theorem furnishes a weak* limit $u_\infty \in Y^*$ s.t. $u_n(y) \to u_\infty(y)$ for all $y \in Y$. Since $u_\infty$ is a bounded linear functional on the subspace $Y$, the Hahn-Banach theorem furnishes us an extension $\tilde u_\infty$ of $u_\infty$ to all of $H$. By the Riesz Representation theorem, $\tilde u_\infty(y) = \langle g, y\rangle$ for some $g \in H$. If we show that this limit is unique, this $g$ would be a reasonable extension of $P^{-1}$ to a point on the boundary of $\Omega$.
In the classical LAP, $M = \Bbb C$, $H = L^2(\Bbb R)$, $P(x)$ would be something like $-\Delta + x$, with $\Omega = \Bbb C \setminus [0, \infty)$, and $Y$ an appropriately weighted $L^2$ space. The problem is that in this case, the extension of the resolvent $P(x)^{-1}$ to the real line is a map into the dual of $Y$ (which is just another weighted $L^2$ space with in inverse weight). This evinces a flaw in my previous argument because according to the above paragraph, the extension of the resolvent should take values in the same Hilbert space $H = L^2(\Bbb R)$. What am I doing wrong?