I have attached [pg. 275](https://i.stack.imgur.com/nZRSh.png) and [pg. 276](https://i.stack.imgur.com/YY7nI.png) of [BS91]. My concern is with the claim (2.7) on pg. 276. To prove this claim, I require the following additional assumption, which is not made by the authors:


**Assumption**: $u^{\rho}$ is upper semicontinuous for each $\rho$.


*Remark*: In the supersolution case, the analogue of the above assumption is that $u^\rho$ is lower semicontinuous for each $\rho$.


*My proof of (2.7) under the additional assumption above*: Let $(\rho\_n,x\_n)\_n$ be a sequence with $\rho\_n\rightarrow 0$, $x\_n\rightarrow x\_0$, and $u^{\rho\_n}(x\_n)\rightarrow \overline{u}(x\_0)$. Then, by the assumption above, for each $n$, there exists $y\_n$ such that $$(u^{\rho\_n}-\varphi)(y\_n) = \sup\_{ \overline{B(x\_0;r)} } \{ u^{\rho\_n} - \varphi \}.$$ Extract a subsequence of $(\rho\_n,x\_n,y\_n)\_n$ converging to some $(0,x\_0,\hat{y})$ and relabel it, with a slight abuse of notation, $(\rho\_n,x\_n,y\_n)\_n$. It follows that 


\begin{multline\*}
0=(\bar{u}-\varphi)(x\_{0})=\lim\_{n\rightarrow\infty}(u^{\rho\_{n}}-\varphi)(x\_{n})\leq\limsup\_{n\rightarrow\infty} (u^{\rho\_{n}}-\varphi)(y\_{n}) \\
\leq\limsup\_{\substack{\rho\rightarrow0\\
y\rightarrow \hat{y}
}
}(u^{\rho}-\varphi)(y)=(\overline{u}-\varphi)(\hat{y}).
\end{multline\*}
Because it was assumed that $x\_0$ is a strict local maximum of $\overline{u}-\varphi$, we get $\hat{y}=x\_0$. This proves (2.7), with $(\rho\_n,y\_n)$ being the desired sequence.


**My problem**: I cannot see how to remove the assumption. While the assumption can be relaxed, as far as I can tell, we need to impose *something* to ensure that the maximum is attained in the proof below. Am I mistaken?


[BS91] Barles, Guy, and Panagiotis E. Souganidis. "Convergence of approximation schemes for fully nonlinear second order equations." *Asymptotic analysis* 4.3 (1991): 271-283.



Discontinuity of solutions to approximation schemes in the Barles-Souganidis framework
After checking the arguments with @Jeff, I am willing to conclude that the issue in the question above is an extremely minor error in the original paper [BS91]. Both Jeff and I came up with "fixes" for the case of $u^{\rho}$ not continuous. Both are detailed below.


**Update**: I had the chance to speak to Professor Barles about this,
who was kind enough to suggest a way to improve my fix. The result is below.


Parsiad and Professor Barles' fix
---------------------------------


Redefine the notion of an approximation scheme as a pair of functions
$$\underline{S},\overline{S}:(0,\infty)\times\overline{\Omega}\times\mathbb{R}\times B(\overline{\Omega})\rightarrow\mathbb{R}.$$
For fixed $\rho$, a solution of this approximation scheme is a bounded
function $u^{\rho}:\overline{\Omega}\rightarrow\mathbb{R}$ such that
$$
\underline{S}(\rho,x,(u^{\rho})^{\*}(x),(u^{\rho})^{\*})\leq0\leq \overline{S}(\rho,x,(u^{\rho})\_{\*}(x),(u^{\rho})\_{\*})\text{ for all }x\in\overline{\Omega}.
$$


*Remark*: Intuitively, $(u^{\rho})^\*$ and $(u^{\rho})\_\*$ are simply upper and lower semicontinuous envelopes of the solution $u^{\rho}$ of an approximation scheme in the sense of [BS91].


Monotonicity is extended in the obvious way: an approximation scheme
$(\underline{S},\overline{S})$ is *monotone* if for each $S\in\{\underline{S},\overline{S}\}$,
\begin{gathered}
S(\rho,x,t,u)\leq S(\rho,x,t,v)\text{ if }u\geq v\\
\text{ for all }\rho>0\text{, }x\in\overline{\Omega}\text{, }t\in\mathbb{R}\text{, and }u,v\in B(\overline{\Omega}).
\end{gathered}
Stability is also extended in the obvious way: an approximation scheme $(\underline{S},\overline{S})$ is *stable*
if
\begin{gathered}
\text{for all }\rho>0\text{, there exists a solution }u^{\rho}\in B(\overline{\Omega})\text{,}\\
\text{with }u^{\rho}\text{ bounded independent of }\rho.
\end{gathered}
The consistency requirement is modified slightly. In particular, we
require that $\overline{S}$ (resp. $\underline{S}$) only satisfy (2.4a) (resp.
(2.4b)) of [BS91]. That is, an approximation scheme $(\underline{S},\overline{S})$
is *consistent* if for all $x\in\overline{\Omega}$ and $\varphi\in C\_{b}^{\infty}(\overline{\Omega})$,
\begin{align\*}
\limsup\_{\substack{\rho\rightarrow0\\
y\rightarrow x\\
\xi\rightarrow0
}
}\frac{\overline{S}(\rho,y,\varphi(y)+\xi,\varphi+\xi)}{\rho} & \leq F^{\*}(D^{2}\varphi(x),D\varphi(x),\varphi(x),x)\\
\text{and }\liminf\_{\substack{\rho\rightarrow0\\
y\rightarrow x\\
\xi\rightarrow0
}
}\frac{\underline{S}(\rho,y,\varphi(y)+\xi,\varphi+\xi)}{\rho} & \geq F\_{\*}(D^{2}\varphi(x),D\varphi(x),\varphi(x),x).
\end{align\*}
**Proposition**: *Let $(\underline{S},\overline{S})$ be an approximation scheme that
is monotone, stable, and consistent (in the sense above). Assume $F=0$
satisfies a strong comparison principle (i.e., (2.5) of [BS91]).
Then, as $\rho\rightarrow0$, the solution $u^{\rho}$ of the scheme
converges locally uniformly to the unique viscosity solution of $F=0$ (in the sense of [BS91]).*


*Proof*: With the setup above, the proof is essentially identical to that of [BS91] (and avoids the issue of being unable to take extrema).


Jeff's fix
----------


**Proposition**: *Theorem 2.1 of [BS91] holds if we replace (2.4) of [BS91] by the following modified consistency requirement*
\begin{align\*}
\limsup\_{\substack{\rho\rightarrow0\\
y\rightarrow x\\
\xi\rightarrow0
}
}\frac{S(\rho,y,\varphi(y)+\xi+e^{-1/\rho},\varphi+\xi)}{\rho} & \leq F^{\*}(D^{2}\varphi(x),D\varphi(x),\varphi(x),x) & \text{(2.4a}\text{)}\phantom{.}\\
\text{and }\liminf\_{\substack{\rho\rightarrow0\\
y\rightarrow x\\
\xi\rightarrow0
}
}\frac{S(\rho,y,\varphi(y)+\xi-e^{-1/\rho},\varphi+\xi)}{\rho} & \geq F\_{\*}(D^{2}\varphi(x),D\varphi(x),\varphi(x),x) & \text{(2.4b}\text{)}.
\end{align\*}
*Proof*: While the claim (2.7) cannot be proved, we can instead establish the existence of a sequence $(\rho\_{n},y\_{n})$ such that
$$
\rho\_{n}\rightarrow0\text{, }y\_{n}\rightarrow x\_{0}\text{, and }(u^{\rho\_{n}}-\varphi)(y\_{n})+e^{-1/\rho\_{n}}\geq(u^{\rho\_{n}}-\varphi)(\cdot).
$$


To see this, let $(\rho\_{n},x\_{n})\_{n}$ be a sequence with $\rho\_{n}\rightarrow0$, $x\_{n}\rightarrow x\_{0}$, and $u^{\rho\_{n}}(x\_{n})\rightarrow\overline{u}(x\_{0})$. Then, for each $n$, there exists $y\_{n}$ such that
$$
(u^{\rho\_{n}}-\varphi)(y\_{n})+e^{-1/\rho\_{n}}\geq\sup\_{\overline{B(x\_{0};r)}}\left\{ u^{\rho\_{n}}-\varphi\right\} .
$$
Extract a subsequence of $(\rho\_{n},x\_{n},y\_{n})\_{n}$ converging
to some $(0,x\_{0},\hat{y})$ and relabel it, with a slight abuse of
notation, $(\rho\_{n},x\_{n},y\_{n})\_{n}$. It follows that
\begin{multline\*}
0=(\overline{u}-\varphi)(x\_{0})=\lim\_{n\rightarrow\infty}(u^{\rho\_{n}}-\varphi)(x\_{n})\leq\limsup\_{n\rightarrow\infty}\left\{ (u^{\rho\_{n}}-\varphi)(y\_{n})+e^{-1/\rho\_{n}}\right\} \\
\leq\limsup\_{\substack{\rho\rightarrow0\\
y\rightarrow\hat{y}
}
}\left\{ (u^{\rho}-\varphi)(y)+e^{-1/\rho}\right\} =\limsup\_{\substack{\rho\rightarrow0\\
y\rightarrow\hat{y}
}
}(u^{\rho}-\varphi)(y)=(\overline{u}-\varphi)(\hat{y}).
\end{multline\*}
Because it was assumed that $x\_{0}$ is a strict local maximum of
$\overline{u}-\varphi$, we get $\hat{y}=x\_{0}$, as desired.


Now, the remainder of the proof in [BS91] is modified as follows. Let $\xi\_{n}=(u^{\rho\_{n}}-\varphi)(y\_{n})+e^{-1/\rho\_{n}}$.
Then, $u^{\rho\_{n}}\leq\varphi+\xi\_{n}$ and hence
$$
0=S(\rho\_{n},y\_{n},u^{\rho\_{n}}(y\_{n}),u^{\rho\_{n}})\geq S(\rho\_{n},y\_{n},\varphi(y\_{n})+\xi\_{n}-e^{-1/\rho\_{n}},\varphi+\xi\_{n}).
$$
The desired result now follows by dividing by $\rho$, taking limit inferiors, and applying the consistency condition. $\square$


*Remark*: the exponential decay is chosen as it does not cause any problems in
creating approximation schemes, practically speaking. For example, consider the approximation $$\frac{u^\rho(x+\rho) - u^\rho(x)}{\rho}.$$ If we substitute the corresponding entities in the revised consistency requirement, we get $$\frac{\varphi(x+\rho) - \varphi(x) \pm e^{-1/\rho}}{\rho} = \varphi^\prime(x) + \mathcal{O}(\rho)$$ by a Taylor approximation.


