One-sided support forces its analytic conormal wavefront

Work in an analytic ambient chart centered at $x_0$. The hypersurface remains $C^2$ in that chart, since changing from its given smooth chart is a smooth diffeomorphism. A real linear coordinate change makes its tangent plane nonvertical. The $C^2$ implicit graph theorem then writes it as $x_n=\psi (x^{\mathrm{\prime }})$, with $\psi$ of class $C^2$: the inverse of a $C^2$ map with invertible derivative is $C^2$, by differentiating the identity for its inverse twice. Translate to have $\psi (0)=0$. Shrink to a convex $x^{\mathrm{\prime }}$ ball on which $\mathrm{\parallel }{D}^2\psi \mathrm{\parallel }\le 2C$. Taylor's integral formula gives $q_{-}(x^{\mathrm{\prime }}):=D\psi (0)x^{\mathrm{\prime }}-C\mathrm{\mid }{x}^{\mathrm{\prime }}{\mathrm{\mid }}^2\le \psi (x^{\mathrm{\prime }})\le D\psi (0)x^{\mathrm{\prime }}+C\mathrm{\mid }{x}^{\mathrm{\prime }}{\mathrm{\mid }}^2=:q_{+}(x^{\mathrm{\prime }})\mathrm{.}$ Both supporting quadratics are real analytic and have the same differential as $\psi$ at $0$.

Suppose first that $f$ vanishes on the strict subgraph of $\psi$. Its local support lies in $x_n\ge \psi (x^{\mathrm{\prime }})$, hence in $x_n\ge q_{-}(x^{\mathrm{\prime }})$. The analytic shear $(x^{\mathrm{\prime }},x_n)\mapsto (x^{\mathrm{\prime }},x_n-q_{-}(x^{\mathrm{\prime }}))$ has analytic inverse $(y^{\mathrm{\prime }},y_n)\mapsto (y^{\mathrm{\prime }},y_n+q_{-}(y^{\mathrm{\prime }}))$, determinant $1$, and sends this latter closed side into $y_n\ge 0$. If instead $f$ vanishes on the strict epigraph, its support lies in $x_n\le \psi (x^{\mathrm{\prime }})\le q_{+}(x^{\mathrm{\prime }})$ and use the corresponding analytic shear with $q_{+}$. In either case the pulled-back normal $d{y}_n$ at $0$ is $d{x}_n-D\psi (0)d{x}^{\mathrm{\prime }}$, a nonzero scalar multiple of the supplied conormal ${\nu }_0$. This flattens an analytic supporting quadratic; it does not flatten the original $C^2$ graph analytically.

If $f$ already vanishes near $x_0$, there is nothing to prove. Otherwise $x_0\in \mathrm{supp}f$ by [F2]. Localize inside the chart. Under the appropriate shear, [F3] then puts both $d{y}_n$ and $-d{y}_n$ in the analytic wavefront set at the origin. By [F4] these are exactly both conormal orientations at $x_0$, up to positive frequency scaling. Either absence hypothesis in the statement contradicts this conclusion. Thus $f=0$ near $x_0$. Every construction used finite-dimensional coordinates and explicit quadratics; it introduces no choice assumption.