Nonconvexity can break the equation; nonsuperlinearity can limit the Lagrangian domain

Conjugate of the nonconvex Hamiltonian. For $H(p)=p^2-2\mathrm{\mid }p\mathrm{\mid }$ and $p\ge 0$ one has $pv-H(p)=p(v+2)-p^2$, whose supremum over $p\ge 0$ is $(v+2{)}_{+}^2\mathrm{/}4$; for $p\le 0$, writing $p=-q$ with $q\ge 0$ gives $pv-H(p)=-qv-q^2+2q=q(2-v)-q^2$, whose supremum is $(2-v{)}_{+}^2\mathrm{/}4$. Hence $L(v)=\max \{(v+2{)}_{+}^2\mathrm{/}4,(2-v{)}_{+}^2\mathrm{/}4\}$, where $r_{+}=\max (r,0)$; in particular $L(0)=1$, $L(v)=(2+\mathrm{\mid }v\mathrm{\mid }{)}^2\mathrm{/}4\ge 1$ for every $v$ and $L(v)\ge 4$ for $\mathrm{\mid }v\mathrm{\mid }\ge 2$, so ${\inf }_{v}L(v)=1$, attained at $v=0$.

Conjugate of the non-superlinear Hamiltonian. For $H(p)=\sqrt{1+p^2}$ and $\mathrm{\mid }v\mathrm{\mid }<1$, the function $p\mapsto pv-\sqrt{1+p^2}$ has derivative $v-p\mathrm{/}\sqrt{1+p^2}$, which vanishes exactly at $p=v\mathrm{/}\sqrt{1-v^2}$, where the value is $-\sqrt{1-v^2}$; at $v=\pm 1$ the supremum is $0$, approached along the infinite tail $p\to \pm \mathrm{\infty }$; and for $\mathrm{\mid }v\mathrm{\mid }>1$ the expression tends to $+\mathrm{\infty }$ along $p=\mathrm{sgn}(v)r$ as $r\to \mathrm{\infty }$. Hence $L(v)=-\sqrt{1-v^2}$ for $\mathrm{\mid }v\mathrm{\mid }\le 1$ and $L(v)=+\mathrm{\infty }$ for $\mathrm{\mid }v\mathrm{\mid }>1$: the Legendre transform is not real-valued on all of $\mathbb{R}$, finite action being available only when $\mathrm{\mid }x-y\mathrm{\mid }\mathrm{/}t\le 1$, that is $\mathrm{\mid }x-y\mathrm{\mid }\le t$. For $u_0\equiv 0$ the formula is $Q_t{u}_0=t{\inf }_{v}L(v)=-t$; the residual of $u(x,t)=-t$ is $u_t+H(u_x)=-1+H(0)=-1+1=0$, so as a formal expression it does satisfy the equation, and no failure of the solution property is claimed in this clause.

The formula is not a solution for the nonconvex Hamiltonian. With $u_0\equiv 0$, the Hopf--Lax formula of [F2] is $Q_t{u}_0(x)=t{\inf }_{v\in \mathbb{R}}L(v)=t$ for every $x$ and $t>0$ (the substitution $v=(x-y)\mathrm{/}t$ turns the infimum over $y$ into the infimum over $v$). The function $u(x,t)=t$ is smooth, and for a smooth function every point is both an upper and a lower contact with the test function itself; as an upper test, [F3] requires $u_t+H(u_x)\le 0$, but $u_t=1$ and $u_x=0$ give $1+H(0)=1>0$. Hence the Hopf--Lax output is not a viscosity subsolution, therefore not a viscosity solution, of $u_t+H(u_x)=0$.

Conclusion. Part (a) exhibits a finite superlinear nonconvex Hamiltonian whose Hopf--Lax output fails the subsolution inequality pointwise, and part (b) exhibits a finite convex non-superlinear Hamiltonian whose Legendre transform is finite only on a bounded velocity set, so superlinearity cannot be dropped from the full-domain finiteness conclusion. The two failures are of different kinds: convexity is needed for the equation to hold, superlinearity for the Lagrangian to be finite everywhere.