Let $A:=[a,b]$ be a closed interval in $\mathbb{R}$. Let $F(x,p,q,r)$ be a function from $[a,b]\times \mathbb{R} \times \mathbb{R} \times\mathbb{R} \rightarrow \mathbb{R}$ describing a second order ODE. We assume that $F$ is continuous on $[a,b]$ and satisfies all the nice properties we need for viscosity solutions.

Let our DE be -

$F(x,v(x), v'(x), v''(x)) = 0$ (1)

Suppose $v$ is a viscosity solution to $F(x,v(x),v'(x),v''(x))=0$ on $(a,b]$ i.e. it is both a subsolution and supersolution. Moreover, I know that $v$ is continuous on $[c,d] \supset [a,b]$ Then, can I say that $v$ is also a viscosity solution to Equation (1) on $[a,b]$?

The reason why I think this should be true is the following. Suppose my claim is false. In particular, let's assume that $v$ is not a supersolution of $(1)$. Then, there exists a $C^2$ function $h$ such that $v - h$ is minimized at $a$ and $F(a, v(a), h'(a),h''(a)) < 0$.

But then, since $v$ is continuous and $h$ is $C^2$, I should get a function that is "very similar" to $h$, say $\bar{h}$, such that $v-\bar{h}$ is minimized at some $a + \epsilon$ and $F(a+\epsilon, v(a+\epsilon), \bar{h}'(a+\epsilon),\bar{h}''(a+\epsilon)) < 0$ Therefore, $v$ could not have been a supersolution on $(a,b)$ in the first place.

Is it absolutely bizarre to expect such a thing? And if I can get it, what are the nice conditions I would need on $v$ besides continuity?