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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 $t$?

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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).

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  • $\begingroup$ What would be a good reference for the holomorphicity part? Also is it sufficient to have $p>\frac{\dim M}{2}$? $\endgroup$ – ougoah Nov 24 '17 at 5:39
  • $\begingroup$ $(D+t)^{-1} = ((D+t_0)(I+(t-t_0)(D+t_0)^{-1})^{-1} = \sum_{n=0}^\infty (-1)^n\left(D+t_0\right)^{-n}(t-t_0)^n (D+t_0)^{-1}$, so if $(D+t_0)^{-1}$ is a bounded operator bounded the series converges absolutely in a small disk $|t-t_0|<\epsilon$ showing that $t\mapsto (D+t)^{-1}$ is holomorphic in that disk. $\endgroup$ – Lior Silberman Nov 25 '17 at 0:47

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