In their paper User's guide to viscosity solutions of second order differential partial equations, Crandall, Ishii and Lions define the superjet of an upper semicontinuous function $u$ at $\hat x \in \mathcal O$, $\mathcal O$ a locally compact subset of $\mathbb R^N$, to be the set of $(p, X) \in \mathbb R^N \times \mathcal S_N$ ($\mathcal S_N$ the set of symmetric matrices) such that
\begin{equation}
u(x) \leq u(\hat x) + \langle p, x – \hat x \rangle + \frac 12 \langle X(x – \hat x), x – \hat x \rangle + o(|x – \hat x|^2) \quad \quad (*)
\end{equation}
holds. The superjet is denoted $J_{\mathcal O}^{2, +}u(\hat x)$. It is left as an exercise that
$$
J_{\mathcal O}^{2, +}u(\hat x) = \{(D\varphi(\hat x), D^2 \varphi(\hat x)) \ : \ \varphi \in C^2 \text{ and } u – \varphi \text{ has a local maximum at } \hat x\}.
$$
My question is: given $(p, X) \in \mathbb R^N \times \mathcal S_N$ such that $(*)$ holds, how to find such a $\varphi$?
Thanks in advance.
Best Answer
Given a $(p, X)$ in this set, we can construct a $\phi$ as follows, by using these as data to construct $\phi$ with a specified first/second derivative. Some extra work has to be done to ensure this local maxima property. First, set $$ \tilde\phi(x) = u(\hat x) + \langle p, x - \hat x\rangle + \frac{1}{2}\langle X(x - \hat x), x - \hat x\rangle. $$ From the definition of $\tilde\phi$ we see that $\tilde\phi$ is $C^2$ with $D\tilde\phi(\hat x) = p$ and $D^2 \tilde\phi(\hat x) = X$. By the fact that $(p, X)$ was in the superjet, we have $$ u(x) \leq \tilde\phi(x) + o(|x - \hat x|^2). $$ This isn't quite what we want, since we want $u - \tilde\phi$ to have a local max at $\hat x$, i.e $u - \tilde \phi \leq 0$ (since $u(\hat x) = \tilde\phi(\hat x)$). To fix this, we use the definition of little-o notation. There exists a $\delta > 0$ such that for $|x- \hat x| < \delta$, $\frac{1}{|x - \hat x|^2}o(|x - \hat x|^2) \leq 1$, hence the little-o error term is $\leq |x - \hat x|^2$ on $B(\hat x, \delta)$. Define now $$ \phi(x) = \tilde\phi(x) + 2|x - \hat x|^2. $$ Then $\phi$ has $D\phi(\hat x) = D\tilde\phi(\hat x)$, $u(\hat x) = \phi(\hat x)$, and for all $x \in B(\hat x, \delta)$, we have $$ u(x) - \phi(x) = u(x) - \tilde\phi(x) - 2|x - \hat x|^2 \leq |x - \hat x|^2 - 2|x - \hat x|^2 = - |x - \hat x|^2 \leq 0. $$ Thus $u - \phi$ has a local maximum at $x = \hat x$. Note also that $$ D^2\phi(\hat x) = D^2\tilde\phi(\hat x) + 2I $$ and thus $$ u(x) \leq u(\hat x) + \langle D\phi(\hat x), (x - \hat x)\rangle + \langle D^2\phi(\hat x)(x - \hat x), x - \hat x \rangle + o(|x - \hat x|^2), $$ since $D^2\phi \geq D^2 \tilde \phi$ (the identity matrix is positive definite). Thus $(D\phi(\hat x), D^2\phi(\hat x))$ is in the superjet.