Ishii's proof of uniqueness for Hamilton-Jacobi equations of eikonal type I am reading Ishii's paper "A simple, direct proof of uniqueness for Hamilton-Jacobi equations of eikonal type" (Proc. Am. Math. Soc. 100, 247–251 (1987)) and I am stuck on a step of the proof which is probably trivial, but I cannot seem to be able to work out.
The setting is as follows. Let $\Omega\subset\mathbb{R}^n$ be a bounded domain in $\mathbb{R}^n$ and let $H:\mathbb{R}^n\times\mathbb{R}^n\rightarrow\mathbb{R}$ satisfy the following conditions:

*

*There is a modulus of continuity $\omega$ such that $$H(x,p)-H(y,p)\leq\omega(|x-y|(1+|p|))$$ for all $x,y\in\Omega$ and $p\in\mathbb{R}^n$

*The map $p\mapsto H(x,p)$ is convex for all $x\in\Omega$

*There exist $\varphi\in C^\infty(\overline{\Omega})$ and $\gamma>0$ such that $$H(x, D\varphi(x))+\gamma \geq0$$ for all $x\in\Omega$
Suppose that an upper semicontinuous $u\in\text{USC}(\overline{\Omega})$ satisfies $H\leq 0$ in the viscosity sense, i.e. for every $x\in\Omega$ and every $\psi\in C^2(\Omega)$ such that $u-\psi$ has a local maximum at $x$ with respect to $\Omega$, we have $H(x,D\psi(x))\leq 0$.
We now define for $\theta\in(0,1)$, $$u_\theta=\theta\varphi+(1-\theta)u$$ and we claim that $u_\theta$ satisfy $H\leq-\theta\gamma$ in the viscosity sense.
The intuition is quite clear: if $u$ were $C^1$, we would have $$H(x,Du_\theta(x))=H(x,\theta D\varphi(x)+(1-\theta)Du(x))\leq \theta H(x, D\varphi(x))+(1-\theta)H(x,Du(x))\leq \theta H(x, D\varphi(x))\leq-\theta \gamma$$
However, I haven't been able to turn this into the corresponding statement in the viscosity sense.
Let $\psi\in C^2(\Omega)$ be such that $u_\theta-\psi$ has a local maximum at $x\in\Omega$, then I want to show that $H(x,D\psi(x))\leq -\theta \gamma$, using the fact that $H$ is convex in the second variable.
Now, if $u_\theta-\psi$ has a local maximum at $x\in\Omega$, then $(1-\theta)u-(\psi-\theta\varphi)$ has a local maximum at $x$ and $$H(x, D\psi(x)-\theta D\varphi(x))\leq 0$$ However, what I would like to show is that $H(x,D\psi(x))\leq\theta H(x,D\varphi(x))$ and convexity does not seem to help me. What am I missing?
 A: As I suspected, the answer is straightforward, so let me record it here, in case it might be of use to someone else.
First, recall that the definition of viscosity subsolution can be recast as follows: an upper semicontinuous $u\in\text{USC}(\overline{\Omega})$ satisfies $H\leq 0$ in the viscosity sense if, whenever a $C^2$ function $\tilde{\psi}$ touches $u$ from above at $x\in\Omega$ (i.e. $u\leq\tilde{\psi}$ in $\Omega$ and $u(x)=\tilde{\psi}(x)$), we have $H(x, D\tilde{\psi}(x))\leq 0$. This is relatively easy to check directly.
So, now consider the setting in the question and say $\tilde{\psi}\in C^2(\Omega)$ touches $u_\theta$ from above at $x\in\Omega$.
Note that if we set $\psi=\frac{1}{1-\theta}(\tilde{\psi}-\theta\varphi)$, then $\psi\in C^2(\Omega)$ and $\tilde{\psi}=\theta\varphi+(1-\theta)\psi$.
Then it's easy to check that our newly defined $\psi$ touches $u$ from above at $x$, so $H(x,D\psi(x))\leq 0$ since $u$ satisfies $H\leq 0$ in the viscosity sense.
Now we can use convexity:
$$\begin{equation}\begin{aligned}
H(x,D\tilde{\psi}(x)) &= H(x,\theta D\varphi(x)+(1-\theta)D\psi(x))\\
  &\leq \theta H(x,D\varphi(x))+(1-\theta)H(x,D\psi(x))\\
  &\leq \theta H(x,D\varphi(x))\\
  &\leq -\theta\gamma\\
\end{aligned}\end{equation}$$
which is what we wanted to show.
