Let $$u'(x) + \alpha u(x) - f(x) = 0,$$ with $x \in [0,\infty)$ and $\alpha \in \mathbb{R}$. Suppose $f \in C(\mathbb{R})$.


If 


* $u\_1$ is a viscosity **supersolution** (or a viscosity **solution**, or a **$C^1(\mathbb{R})$ solution**) of $u'(x) + \alpha u(x) - f(x) = 0$,
* $u\_2$ is a viscosity **subsolution** of $u'(x) + \alpha u(x) - f(x) = 0$,
* $u\_1(0)=u\_2(0)$,


how do I prove that $$u\_1(x) \ge u\_2(x)$$ for all $x \in [0,\infty)$? 



Viscosity solutions for $u'(x) + \alpha u(x) - f(x) = 0$: supersolutions dominate subsolutions
If you set $v(x)=e^{\alpha x}u(x)$ then 


$$v'(x) - e^{\alpha x}f(x) = 0. \ \ \ \ \ \ \ \ \ \ (\*)$$


This equation has no zeroth order term, so the maximum principle applies with no restrictions. This is basically the trick @PaulBryan mentioned in a comment. 


This works in the viscosity sense as well. Indeed, suppose $u$ is a viscosity subsolution of


$$u' + \alpha u - f = 0.$$


Let $x\_0 \in \mathbb{R}$ and let $\phi$ be a smooth function for which $\phi(x\_0)=v(x\_0)$ and $v(x) \leq \phi(x)$ for $x$ near $x\_0.$ Then $\psi(x\_0) = u(x\_0)$ and $\psi(x) \leq u(x)$ for $x$ near $x\_0$, where $\psi(x) = e^{-\alpha x}\phi(x)$. Therefore $u-\psi$ has a local max at $x\_0$ and hence


$$\psi'(x\_0) + \alpha \psi(x\_0) - f(x\_0) \leq 0,$$


and so


$$\phi'(x\_0) - e^{\alpha x\_0}f(x\_0) \leq 0.$$


This verifies that $v$ is a viscosity subsolution of (\*). The supersolution verification is similar.


EDIT: Let me add, after removing the zeroth order term, the comparison principle argument is quite standard in the theory of viscosity solutions. You can see the book by Bardi and Capuzzo-Dolcetta for instance. You should pose the problem on the bounded domain $[0,M]$ as @PaulBryan did for the classical argument.


