## 6.1 Regularity of Operator

We will fix a manifold $N$, which will be thought of as $\partial M$, and consider the manifold $N \times \mathbb {R}_{\geq 0}$. We will write $y \in N$ and $u \in \mathbb {R}_{\geq 0}$. We let $E$ be a Hermitian vector bundle on $N$.

Let $A$ be a first-order self-adjoint elliptic operator on $E$, and consider the differential operator

$D = \frac{\partial }{\partial u} + A.$This is the differential operator of interest on $N \times \mathbb {R}_{\geq 0}$. While the definition resembles that of a heat operator, it may not be the best idea to think of it as a heat operator. We will later introduce a further time variable $t$ and consider $e^{-t D^*D}$, and confusion might arise.

The basic problem we want to solve is the equation

$Df = g.$For now, we assume $g \in \Gamma _ c(E \times \mathbb {R}_{\geq 0})$, and later extend to more general functions after we establish the right bounds.

As before, since $A$ is self-adjoint and elliptic, we can find an eigenbasis $\{ \psi _\lambda \}$ of $A$. We write

$f(y, u) = \sum _\lambda f_\lambda (u) \psi _\lambda (y),\quad g(y, t) = \sum _\lambda g_\lambda (u) \psi _\lambda (y).$The we see that for the equation to hold, we must have

$\left(\frac{\mathrm{d}}{\mathrm{d}u} + \lambda \right) f_\lambda (u) = g_\lambda (u).$We have previously decided that the boundary condition we want is $f_\lambda (0) = 0$ for $\lambda \geq 0$. There is another way we can think about this boundary condition. Observe that the solution to this differential equation is well defined up to adding $C e^{-\lambda u}$. If $\lambda < 0$, then there is at most one solution with reasonable growth as $u \to \infty$. However, if $\lambda \geq 0$, then we can add any multiples of $e^{-\lambda u}$ and have a sensible solution ($\lambda = 0$ is somewhat of an edge case). Thus, fixing $f_\lambda (0) = 0$ gives us a unique “sensible” solution.

For convenience, we write $P$ for the composition

$H^1(E \times \mathbb {R}_{\geq 0}) \overset {r}{\to } H^0(E \times \{ 0\} ) \to H^0(E \times \{ 0\} ),$where $r$ is restriction and the last map is the spectral projection onto the subspace spanned by the eigenvectors with non-negative eigenvalues. Then our boundary condition is $Pf = 0$. By an abuse of notation, we write $1 - P$ for $r - P$, and the boundary condition for $D^*$ will then by $(1 - P)f = 0$. We shall write $\Gamma (E\times \mathbb {R}_{\geq 0}; P)$ for the space of smooth sections $f$ such that $Pf = 0$.

Imposing this boundary condition, we can write down explicit solutions for $f_\lambda$:

$f_\lambda (u) = \begin{cases} \displaystyle \int _0^ u e^{\lambda (v - u)} g_\lambda (v) \; \mathrm{d}v & \lambda \geq 0\\ \displaystyle -\int _ u^\infty e^{\lambda (v - u)} g_\lambda (v) \; \mathrm{d}v & \lambda < 0 \end{cases}.$We check that for large $u$, the function $f_\lambda (u)$ is either exponentially decreasing (if $\lambda > 0$), constant (if $\lambda = 0$) or identically zero (if $\lambda < 0$). The constant term wrecks our hope that this takes values in $H^ s$, but we can still hope it takes values in $H^ s_{\mathrm{loc}}$.

Note also that for $\lambda < 0$, the value of $f_\lambda (u)$ depends on the *future* values of $g_\lambda$, which ties in with our previous point — at large $u$, all future values of $g_\lambda$ are zero, hence $f_\lambda (u) = 0$.

We let $Q\colon \Gamma _ c(E \times \mathbb {R}_{\geq 0}) \to \Gamma (E \times \mathbb {R}_{\geq 0}; P)$ be the function that sends $g_\lambda$ to $f_\lambda$ above, which is clearly linear. The next proposition will, in particular, show that this map is well-defined, i.e. $Qf$ is actually smooth:

$Q$ extends to a continuous map $H^ s \to H^{s + 1}_{\mathrm{loc}}$.

Recall that since $A$ is elliptic, we have

$\| f\| _{H^{s + 1}} \leq C (\| f\| _{H^ s} + \| Af\| _{H^ s}).$So we can equivalently define the $H^1$ norm by

$\| f\| _{H^1}^2 = \| f\| ^2_{L^2} + \| \partial _ u f\| ^2_{L^2} + \| Af\| ^2_{L^2}.$Using that the eigenspace decomposition is orthogonal in $L^2$ and the defining equations

$A \psi _\lambda = \lambda \psi _\lambda ,\quad \frac{\partial f_\lambda }{\partial u} = g_\lambda - \lambda f_\lambda ,$we get an inequality

$\| f\| _{H^1}^2 \leq C \cdot \sum _\lambda (1 + \lambda ^2) \| f_\lambda \| ^2_{L^2} + \| g_\lambda \| ^2_{L^2}.$To bound $\| f_\lambda \| _{L^2}$ in terms of $\| g_\lambda \| _{L^2}$, we use the (rotated) Laplace transform

$\hat{f}_\lambda (\xi ) = \int _0^\infty e^{-iu\xi } g_\lambda (u) \; \mathrm{d}u.$One checks easily that up to a constant which we omit, we have Parseval's identity:

$\| \hat{f}_\lambda \| _{L^2} = \| f_\lambda \| _{L^2}.$We can explicitly compute

$\hat{f}_\lambda (\xi ) = \frac{\hat{g}_\lambda (\xi ) + f_\lambda (0)}{\lambda + i \xi }.$Using Parseval's identity, and the fact that

$f_\lambda (0) = -\int _0^\infty e^{\lambda u} g_\lambda (u) \; \mathrm{d}u\text{ if } \lambda < 0,$we obtain bounds

$|\lambda | \| f_\lambda \| _{L^2} \leq 2 \| g_\lambda \| _{L^2}.$So if we ignore the $\lambda = 0$ term, $Q$ would send $H^0$ into $H^1$. The $f_0$ term is eventually constant, given by the integral of the compactly supported function $\int _0^\infty g_0(v)\; \mathrm{d}v$. So $Q$ maps $H^0$ into $H^1_{\mathrm{loc}}$.

Using the equation

$\frac{\mathrm{d}^ s}{\mathrm{d}u^ s} \left(\frac{\mathrm{d}}{\mathrm{d}u} + \lambda \right) f_\lambda = \frac{\mathrm{d}^ s g_\lambda }{\mathrm{d}u^ s}$and calculating as above, we see that $H^ s$ gets mapped into $H^{s + 1}_{\mathrm{loc}}$.

□The main takeaway from the section is then the following theorem:

There is a linear operator

$Q\colon \Gamma _ c(E \times \mathbb {R}_{\geq 0}) \to \Gamma (E \times \mathbb {R}_{\geq 0}; P)$such that

$\begin{aligned} DQg & = g\text{ for all }g \in \Gamma _ c(E \times \mathbb {R}_{\geq 0}),\\ QDf & = f\text{ for all }f \in \Gamma _ c(E \times \mathbb {R}_{\geq 0}; P). \end{aligned}$Moreover, $Q$ extends to a continuous map $H^ s \to H^{s + 1}_{\mathrm{loc}}$ for all integral $s$. It is given by convolution with a kernel $Q_ u(y, z)$, where $u$ is now allowed to take negative values. The kernel is smooth away from $u = 0$.□