Suppose $D$ is a first-order differential operator on a manifold $M$ and that the inverse $(D+t)^{-1}:H^0(M)\rightarrow H^1(M)$ exists for all $t > 0$, where $H^i(M)$ is the $i^\text{th}$ Sobolev space. 


Let $\psi\in C\_c^\infty(M)\subseteq H^0(M)$. Then in particular $(D+t)^{-1}\psi$ is smooth. Also, for any $\phi\in C\_c^\infty(M)$, the value of the function $(D+t)\phi$ at $x\in M$ is simply 


$$((D+t)\phi)(x) = D\phi(x) + t\phi(x),$$ 


which in particular depends continuously on the parameter $t$. I'm interested in whether this is true for the inverse operator. More precisely:


**Question:** At a fixed point $x\in M$, does the value $((D+t)^{-1}\psi)(x)$ depend continuously on $t$?



Continuity of image of resolvent operator with respect to resolvent parameter
Yes, $((D + t)^{-1}\psi)(x)$ is continuous provided there is a $p > dim(M)$ such that
$$
 (D + t)^{-1} : H^p(M) \to H^{p+1}(M), 
$$
exists and is continuous as a map from $\mathbf R$ to $B(H^p, H^{p+1})$. This is true if $D$ is a first order differential operator with smooth coefficients and the resolvent is not just continuous but holomorphic in a neighbourhood of $\mathbf{R}^+$


We then have 
$$
\begin{align}
\mathbf{R} &\to B(H^p, H^{p+1}) &\to &H^{p+1}(M) &\to &C^0(M) &\to &\mathbf{R}
\\
t &\mapsto (D + t)^{-1} &\mapsto &(D + t)^{-1}(\psi) &\mapsto &(D + t)^{-1}(\psi) &\mapsto &((D + t)^{-1}(\psi))(x)
\end{align}
$$


Note that it suffices that the first map is continuous in the Ultraweak topology (i.e the map is continuous if all matrix elements $\langle\phi, (D + t)^{-1}\psi\rangle$ are continuous). 


