Both analytic normal directions at a flat support boundary

Multiply $u$ by a compact smooth cutoff equal to one near $0$ and supported where the one-sided hypothesis holds. It remains supported in $y_n\ge 0$, and the wavefront hypothesis is unchanged by exponential Gaussian tail estimates. Denote this compact distribution again by $u$ and put $U(z,\lambda )=⟨u(y),e^{-\lambda (z-y{)}^2\mathrm{/}2}⟩,z\in {\mathbb{C}}^n\mathrm{.}$ This is entire by [F3]. A finite distributional derivative order gives the following bound on every fixed bounded complex set: $\mathrm{\mid }U(z,\lambda )\mathrm{\mid }\le C{\lambda }^M{e}^{\lambda \mathrm{\Phi }(z)},\mathrm{\Phi }(z)=\frac{\mathrm{\mid }\mathrm{Im}z{\mathrm{\mid }}^2-(\mathrm{Re}{z}_n{)}_{-}^2}{2},$ where $s_{-}:=\max (-s,0)$. To justify the sharp half-space bound for a distribution, insert a smooth factor $\vartheta (\lambda y_n)$ that is one for $\lambda y_n>-1\mathrm{/}2$ and zero for $\lambda y_n<-1$. It equals one on a neighborhood of the support, so does not change the pairing. Its derivatives of each fixed order cost only a power of $\lambda$. On its support, $y_n\ge -1\mathrm{/}\lambda$, and $\mathrm{Re}(z-y{)}^2=\mathrm{\mid }\mathrm{Re}z-y{\mathrm{\mid }}^2-\mathrm{\mid }\mathrm{Im}z{\mathrm{\mid }}^2$ gives the bound with $(s+1\mathrm{/}\lambda {)}_{-}^2$ in place of $s_{-}^2$. Their difference is $O(1\mathrm{/}\lambda )$ uniformly for bounded $s$, so it changes the exponential by a bounded factor. Gaussian derivatives contribute another fixed power of $\lambda$. This proves the displayed bound using the ordinary fixed-support distributional seminorm, without assuming an estimate on the exact support set.

We record the elementary quantitative continuation used in the left half-plane. If a holomorphic $g$ is bounded by $1$ on a disk of radius $R$ and by $e^{-d\lambda }$ on the concentric disk of radius $r<R$, then on the disk of radius $r<\rho <R$, $\mathrm{\mid }g\mathrm{\mid }\le \exp \text{\hspace{0.17em}⁣}(-d\lambda \frac{\log (R\mathrm{/}\rho )}{\log (R\mathrm{/}r)})\mathrm{.}$ For the annulus compare $\log \mathrm{\mid }g\mathrm{\mid }$ with the harmonic function that is $-d\lambda$ on the inner circle and $0$ on the outer circle. This comparison follows from [F2]: near any hypothetical positive interior maximum of their difference, subtract a local harmonic conjugate and apply the maximum modulus principle; a small strict harmonic perturbation excludes a non-strict maximum, then let that perturbation decrease to zero. Zeros of $g$ cannot be positive maxima. Chaining finitely many overlapping disks contained in the connected left half-plane propagates a strictly positive exponent from the initial disk to any chosen compact subset there. To make the finite chain explicit, cover a polygonal path from the initial disk to the compact set by equal small disks whose doubled disks lie in the half-plane, take successive center distances less than one quarter of the radius, and iterate the displayed estimate. The exponent is multiplied by finitely many positive radius ratios.

Write $z=(z^{\mathrm{\prime }},s+it)$. The identity $e^{-\lambda \mathrm{\mid }p{\mathrm{\mid }}^2\mathrm{/}2}U(x-ip,\lambda )=⟨u,e^{i\lambda (x-y)\cdot p-\lambda \mathrm{\mid }x-y{\mathrm{\mid }}^2\mathrm{/}2}⟩$ and [F1] imply that absence of either normal gives $\mathrm{\mid }U(z,\lambda )\mathrm{\mid }\le C{e}^{\lambda (\mathrm{\mid }\mathrm{Im}z{\mathrm{\mid }}^2\mathrm{/}2-c)}$ in a fixed complex neighborhood of $(0,\mp i)$ for some $c>0$. The use of [F1] here is only to let $p$ range over an open ball about $\pm e_n$, rather than the unit sphere. Fix a small neighborhood of $z^{\mathrm{\prime }}=0$ and treat $z^{\mathrm{\prime }}$ as a parameter. For each such $z^{\mathrm{\prime }}$ the holomorphic function of one variable $G(z_n,\lambda )=e^{-\lambda \mathrm{\mid }\mathrm{Im}{z}^{\mathrm{\prime }}{\mathrm{\mid }}^2\mathrm{/}2}{e}^{\lambda z_n^2\mathrm{/}2}U(z^{\mathrm{\prime }},z_n,\lambda )$ satisfies $\mathrm{\mid }G\mathrm{\mid }\le C{\lambda }^M{e}^{\lambda s_{+}^2\mathrm{/}2}$ on bounded rectangles. On one small disk lying in $s<0$ near $\mp i$ it satisfies $\mathrm{\mid }G\mathrm{\mid }\le C{e}^{-c_1\lambda }$: in this regularity estimate the additional exponent is $s^2\mathrm{/}2$, which is less than $c\mathrm{/}2$ after shrinking that disk. Constants are uniform in $z^{\mathrm{\prime }}$.

Divide $G$ by its uniform polynomial bound $C{\lambda }^M$. In the left half-plane this quotient has modulus at most $1$. Step 1.2 gives $r,b,d>0$ such that on the segment $s=-r$, $\mathrm{\mid }t\mathrm{\mid }\le b$, it is at most $e^{-d\lambda }$, after decreasing $d$ and taking $\lambda$ large. The same construction works whether the initial disk is near $i$ or $-i$; both lie in the same connected left half-plane neighborhood joined by a bounded path to this segment. All paths, rectangles and constants can be fixed uniformly for the chosen $z^{\mathrm{\prime }}$ parameters.

Extend to the rectangle $-r\le s\le a$, $\mathrm{\mid }t\mathrm{\mid }\le b$, where $a>0$ will be small. For $k=\pi \mathrm{/}(2b)$ define $w(s,t)=\frac{\sinh (k(a-s))}{\sinh (k(a+r))}\cos (kt)\mathrm{.}$ This is harmonic, nonnegative in the rectangle, zero on the top, bottom and right sides, and between zero and one on its left side. For $H=G\mathrm{/}(C{\lambda }^M)$, the global bound in step 2.1 and the left-side improvement give, by [F2] with a harmonic conjugate of $w$, $\log \mathrm{\mid }H(s+it)\mathrm{\mid }\le \lambda (a^2\mathrm{/}2-dw(s,t))\mathrm{.}$ On the left the proposed upper bound is at least $-d\lambda$, and on the other three sides it is $\lambda a^2\mathrm{/}2$, so all boundary inequalities hold. At $(s,t)=(0,0)$, $w(0,0)=\frac{\sinh (ka)}{\sinh (k(a+r))}\ge c_{2}a$ for sufficiently small $a$, with fixed $c_2>0$. Choose $a$ so $a^2\mathrm{/}2<d{c}_{2}a\mathrm{/}2$. Continuity then gives a fixed disk about $z_n=0$ on which $\mathrm{\mid }H\mathrm{\mid }\le e^{-d_2\lambda }$ for some $d_2>0$. This is the needed strict improvement across the boundary.

For real $z=x$ in a sufficiently small neighborhood of $0$, the exponential factors defining $G$ give $\mathrm{\mid }U(x,\lambda )\mathrm{\mid }\le C{\lambda }^M{e}^{-d_2\lambda }$. On this real neighborhood the normalized Gaussian convolutions $(\lambda \mathrm{/}(2\pi ){)}^{n\mathrm{/}2}U(x,\lambda )$ converge distributionally to $u$. For completeness, after pairing with a compact test $\vartheta$, Fubini for a compact distribution and a rapidly decreasing Gaussian writes the pairing as $⟨u,{\rho }_{\lambda }\ast \vartheta ⟩$, where ${\rho }_{\lambda }$ has mass one. Every fixed derivative of ${\rho }_{\lambda }\ast \vartheta -\vartheta$ tends uniformly to zero on the support compact: split the integral into $\mathrm{\mid }y\mathrm{\mid }<\delta$, where uniform continuity controls the translated derivative, and $\mathrm{\mid }y\mathrm{\mid }\ge \delta$, where the Gaussian tail tends to zero; the same argument bounds the derivatives outside the test support. Thus [F3] proves the asserted distributional convergence. The exponential bound makes its limit zero for tests in the smaller neighborhood, so $u=0$ there. Reflecting $x_n$ interchanges the half-spaces and interchanges the two normal orientations. Both claims follow without a reality assumption on $u$.