Quadratic sup convolution and transfer of upper jets

The maximum exists by [F1], is finite above by ${\max }_{K}v$, and is finite below by evaluating at any fixed point of nonempty $K$. Expanding the square gives $\widehat{v}(\xi )+\lambda \mathrm{\mid }\xi {\mathrm{\mid }}^2\mathrm{/}2={\sup }_{x\in K}\{v(x)-\lambda \mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}2+\lambda x\cdot \xi \}$. Every expression in the supremum is affine in $\xi$, so the supremum is convex; finiteness and [F2] give continuity and the asserted semiconvexity.

Fix a maximizer $y$ at $\eta$. Evaluation at $y$ and the upper-jet inequality [F3] give, for small $h$, $\lambda (y-\eta )\cdot h-\lambda \mathrm{\mid }h{\mathrm{\mid }}^2\mathrm{/}2\le \widehat{v}(\eta +h)-\widehat{v}(\eta )\le q\cdot h+h^{\mathsf{T}}Yh\mathrm{/}2+o(\mathrm{\mid }h{\mathrm{\mid }}^2)$. Put $h=te$ for every unit $e$ and both signs of $t$, divide by $\mathrm{\mid }t\mathrm{\mid }$, and let $t\to 0$; then $\lambda (y-\eta )=q$, so the maximizer is unique and the value identity follows. For $x$ near $y$, set $\xi =\eta +x-y$, so the penalty is unchanged; the same jet inequality gives $v(x)-v(y)\le \widehat{v}(\eta +x-y)-\widehat{v}(\eta )\le q\cdot (x-y)+(x-y{)}^{\mathsf{T}}Y(x-y)\mathrm{/}2+o(\mathrm{\mid }x-y{\mathrm{\mid }}^2)$. This is the transferred upper jet, and for an interior $y$ the extension and original function agree locally.