Negative resolvent bounds and the retarded Fourier argument

By the self-adjoint resolvent identity and real multiplication, $K_z^{\ast }=K_{\bar{z}}$ on $L^2$. The conjugate of $(\kappa +i\epsilon {)}^2$ is $(\kappa -i\epsilon {)}^2$, and conjugation preserves $\mathrm{Re}z<0$. The pivot-adjoint construction in [F4] applies the $H_D^1\to H_D^2$ bound at $\bar{z}$ to obtain the $H_D^{-2}\to H_D^{-1}$ bound at $z$. The negative-real-part bound has the decay stated in [F4].

Put $X=H_D^s$ and extend $g$ by zero outside $(0,T)$. By Holder, $g\in L^1(0,T;X)$ with norm at most $B=\sqrt{T}\mathrm{\parallel }g{\mathrm{\parallel }}_{L^2(0,T;X)}$. Apply the continuous-convolution supplier to the realification of the complex Hilbert space $X$, the group $W$, group bound $M=1$, and base time zero. Its simple-integral construction is the same Bochner integral over the complex space. Thus $u(t)={\int }_0^{t}W(t-r)g(r)dr$ exists and is continuous on $[0,T]$, with $\mathrm{\parallel }u(t){\mathrm{\parallel }}_X\le B$. For $t>T$ put $u(t)=W(t-T)u(T)$, which agrees with the same retarded formula by the group law and commutation with integration; put $u(t)=0$ for $t<0$. Strong continuity gives continuity at $T$ and zero. For any $\delta >0$, set $h(t)=e^{-\delta t}g(t)$ and $v(t)=e^{-\delta t}u(t)$. The zero extension of $g$ makes $h\in L^1(\mathbb{R};X)\cap L^2(\mathbb{R};X)$, while $\mathrm{\parallel }v(t){\mathrm{\parallel }}_X\le B{e}^{-\delta t}$ for $t\ge 0$ makes $v$ belong to both spaces as well.

Let $J_s:X\to L^2(\mathrm{\Omega })$ be the onto unitary scale identification $J_s=A^{s\mathrm{/}2}$ supplied by the spectral-scale contract, and let $D_s=J_s^{-1}{\mathrm{\Delta }}_D{J}_s$ with domain $J_s^{-1}D({\mathrm{\Delta }}_D)$. It is self-adjoint on $X$ by unitary transport; the scale commutation gives $W(a)=J_s^{-1}{e}^{ia{\mathrm{\Delta }}_D}{J}_s$. Thus, for each real $\tau$, $U_{\tau }(a)=e^{-ia\tau }W(a)$ is the spectral group of the self-adjoint $D_s-\tau I$. The flow-calculus supplier identifies its generator as $G_{\tau }=i(D_s-\tau I)$ on exactly $D(D_s)$. Apply the positive-damping Laplace lemma on $X$: $Q_{\tau }x={\int }_0^{\mathrm{\infty }}{e}^{-\delta a}{e}^{-ia\tau }W(a)xda=(\delta -G_{\tau }{)}^{-1}x=-i(-D_s+\tau -i\delta {)}^{-1}x$. Indeed $\delta -G_{\tau }=i(-D_s+\tau -i\delta )$, which fixes the sign. By unitary transport and agreement on the common spectral core, the last inverse is the consistent $H_D^s$ realization of $(-{\mathrm{\Delta }}_D+\tau -i\delta {)}^{-1}$. The assumed AC supplies both choice assumptions of the Laplace lemma.

For any complex Hilbert space $E$, the space $L^2(\mathbb{R};E)$ is complete: apply [F12] after translation on each $(-m,m)$ to a Cauchy sequence, use uniqueness of the restricted limits to glue them, and take the supremum over $m$ of the squared restricted-norm bounds to obtain a global $L^2$ limit and norm convergence. Finite sums ${\sum }_j{a}_j(t)e_j$ with scalar $a_j\in L^2(\mathbb{R})$ and fixed $e_j\in E$ are dense. Indeed, for a strongly measurable $f\in L^2(\mathbb{R};E)$ choose finite-range simple $S_n\to f$ almost everywhere and put $T_n={\mathbf{1}}_{[-n,n]}{\mathbf{1}}_{\{\mathrm{\parallel }{S}_n{\mathrm{\parallel }}_E\le 2\mathrm{\parallel }f{\mathrm{\parallel }}_E\}}{S}_n$. These are finite-range simple maps supported on finite-measure sets; $T_n\to f$ almost everywhere and $\mathrm{\parallel }{T}_n-f{\mathrm{\parallel }}_E^2\le 9\mathrm{\parallel }f{\mathrm{\parallel }}_E^2$, so [F12] gives convergence in $L^2$. On finite sums define ${\mathcal{F}}_E({\sum }_j{a}_j{e}_j)={\sum }_j({\mathcal{F}}_2{a}_j)e_j$, using the angular-frequency normalization of [F6]. Choosing an orthonormal basis of their finite-dimensional vector span and applying scalar Plancherel shows that this definition is independent of representation and isometric. Scalar inverse Plancherel defines an isometric inverse on the same finite-sum space. Density and completeness extend both maps to mutually inverse unitaries on $L^2(\mathbb{R};E)$. This construction requires no higher-scale regularity of $u$.

Fix real $\tau$ and a bounded complex-linear coordinate functional $\mathrm{\ell }$ on $X$. The scalar integrand obtained by pairing $e^{-it\tau }{e}^{-\delta t}{\mathbf{1}}_{0\le r\le t}W(t-r)g(r)$ with $\mathrm{\ell }$ is product-measurable: simple approximants to $g(r)$ and continuity of each group orbit give pointwise scalar limits outside the null set in the $r$ coordinate. Its absolute value is bounded by $\mathrm{\parallel }\mathrm{\ell }\mathrm{\parallel }{e}^{-\delta t}{\mathbf{1}}_{0\le r\le t}\mathrm{\parallel }g(r){\mathrm{\parallel }}_X$. Tonelli computes the integral of this bound as $\mathrm{\parallel }\mathrm{\ell }\mathrm{\parallel }{\delta }^{-1}{\int }_0^T{e}^{-\delta r}\mathrm{\parallel }g(r){\mathrm{\parallel }}_{X}dr<\mathrm{\infty }$. Scalar Fubini therefore interchanges $t,r$. In the inner integral put $a=t-r\ge 0$; commuting $\mathrm{\ell }$ with Bochner integrals and using step 1.3 gives $\widehat{v}(\tau )=Q_{\tau }\widehat{h}(\tau )=-i(-{\mathrm{\Delta }}_D+\tau -i\delta {)}^{-1}\widehat{h}(\tau )$ in $X$. These integral transforms exist since $h,v\in L^1$; equality follows from all coordinate functionals separating vectors. Scalar $L^1\cap L^2$ agreement in [F6], applied to orthonormal coordinates and then finite-coordinate $L^2$ limits, identifies them almost everywhere with the Hilbert-valued Plancherel transforms. This argument applies directly to the general strongly measurable input in the Statement: the simple approximations establish measurability, and the displayed absolute-integrability bound justifies Fubini without assuming smooth time dependence or spectral-core values.

On the squared-parameter family, [F2] gives the $H_D^{-1}\to H_D^0$ endpoint for $K_z$. For $\mathrm{Re}z<0$, choose a nonnegative smooth compactly supported $\rho$ equal to one on $\mathrm{supp}{\eta }_1$. The identity $K_z=M_{{\eta }_1}[\rho (-{\mathrm{\Delta }}_D-z{)}^{-1}\rho ]M_{{\eta }_1}$ holds on $L^2$; [F3] and the multiplier bounds extend it to $H_D^{-1}\to H_D^0$, with the same negative-real-part decay up to a cutoff-dependent constant. The negative endpoint from step 1.1 and this order-minus-one endpoint are consistent extensions of the same $L^2$ operator. Interpolation therefore gives the claimed uniform $H_D^s\to H_D^{s+1}$ bounds for $-2\le s\le -1$, with negative-real-part decay retained.

If $B:E\to E^{\mathrm{\prime }}$ is bounded and complex-linear between Hilbert spaces, its pointwise action preserves strong measurability and has $L^2$ norm at most $\mathrm{\parallel }B\mathrm{\parallel }$. On finite sums, scalar Fourier linearity gives ${\mathcal{F}}_{E^{\mathrm{\prime }}}(Bf)=B{\mathcal{F}}_{E}f$; density from step 1.4 extends this identity to every $f\in L^2(\mathbb{R};E)$. Set $Y=H_D^{s+1}$ and retain $X=H_D^s$. By [F11] the canonical embedding $\iota :Y\to X$ is contractive and injective, so in particular ${\mathcal{F}}_X(\iota y)=\iota {\mathcal{F}}_{Y}y$ for all $y\in L^2(\mathbb{R};Y)$.

Choose $\delta >0$ sufficiently small that $\sqrt{\delta \mathrm{/}2}$ lies in the allowed $\epsilon$ range of step 2.2. For every real $\tau$, set $z=-\tau +i\delta$. If $\mathrm{Re}z<0$, the half-plane estimate applies. Otherwise write $z=(\kappa +i\epsilon {)}^2$ with $\kappa >0$ and $2\kappa \epsilon =\delta$; since ${\kappa }^2\ge {\epsilon }^2$, one has $\epsilon \le \sqrt{\delta \mathrm{/}2}$. Thus $K_{-\tau +i\delta }$ is uniformly bounded from $H_D^s$ to $H_D^{s+1}$ for all real $\tau$.

Put $w=e^{-\delta t}\eta u=M_{\eta }v\in L^2(\mathbb{R};X)$, as follows from step 1.2 and the multiplier bound. Since ${\eta }_{1}g=g$ and $\eta {\eta }_1=\eta$, step 2.1 and step 2.3 give, in $X$ almost everywhere, ${\mathcal{F}}_{X}w=\iota q$, where $q(\tau )=-i{M}_{\eta }{K}_{-\tau +i\delta }\widehat{h}(\tau )$ is a vector in $Y$ by step 3.1. The bound is $\mathrm{\parallel }q(\tau ){\mathrm{\parallel }}_Y\le C\mathrm{\parallel }\widehat{h}(\tau ){\mathrm{\parallel }}_X$. To establish strong measurability in $Y$, rather than assume it from the $X$ identity, apply the bounded maps $E_N:X\to Y$ of [F11] to a strongly measurable representative of ${\mathcal{F}}_{X}w$. Consistency gives $E_N{\mathcal{F}}_{X}w=E_{N}q$ in $Y$ almost everywhere, and $E_{N}q\to q$ in $Y$ by [F11]. Thus $q$ has a strongly measurable $Y$-valued representative and belongs to $L^2(\mathbb{R};Y)$, with $\mathrm{\parallel }q{\mathrm{\parallel }}_{L^2(Y)}\le C\mathrm{\parallel }h{\mathrm{\parallel }}_{L^2(X)}$ by step 1.4.

Use surjectivity in step 1.4 to define $y={\mathcal{F}}_Y^{-1}q\in L^2(\mathbb{R};Y)$. Step 2.3 yields ${\mathcal{F}}_X(\iota y)=\iota q={\mathcal{F}}_{X}w$; injectivity of the $X$-valued unitary transform gives $\iota y=w$ in $L^2(\mathbb{R};X)$. Hence $w$ has the $Y$-valued $L^2$ representative $y$, with $\mathrm{\parallel }y{\mathrm{\parallel }}_{L^2(Y)}=\mathrm{\parallel }q{\mathrm{\parallel }}_{L^2(Y)}\le C\mathrm{\parallel }h{\mathrm{\parallel }}_{L^2(X)}$. On $(0,T)$ removing the weight costs at most $e^{\delta T}$, and $\mathrm{\parallel }h{\mathrm{\parallel }}_{L^2(\mathbb{R};X)}\le \mathrm{\parallel }g{\mathrm{\parallel }}_{L^2(0,T;X)}$. This proves the retarded estimate with the claimed spatial regularity.