Prove an integral inequality ($e^{\alpha x}u(x) \le e^{\alpha y}u(y) + \int_y^x e^{\alpha \xi} f(\xi) d\xi$) under certain hypotheses Let $A\in \mathbb{R}$ and $A>0$, $\alpha \in \mathbb{R}$, and  $u,f \in C([0,A])$.
Suppose that for every $g \in C^\infty((0,A))$ we have 
$$g'(x_0) + \alpha u(x_0) \le f(x_0)$$ if $x_0$ is a local maximum of $u-g$. 
How can I prove that 
$$e^{\alpha x}u(x) \le e^{\alpha y}u(y) + \int_y^x e^{\alpha \xi} f(\xi) d\xi$$ for any $x,y \in [0,A]?$

My failed attempts were mostly manipulations that included integrations by parts and several attempts at reasoning by contradiction.
 A: The function $u$ is a subsolution of the differential equation $v^{\prime
}+\alpha v=f$ in the viscosity sense. I'll use standard tricks for viscosity
solutions. Assume that $\alpha>0$. Fix $y\in\lbrack0,A]$. A solution of this
differential equation is given by
$$
g(x)=e^{-\alpha(x-y)}u(y)+e^{-\alpha x}\int_{y}^{x}e^{\alpha t}f(t)\,dt,\quad
x\in\lbrack0,A].
$$
We claim that $u(x)\leq g(x)$ for every $x\in(y,A)$. If not, then there is
$x_{1}\in(y,A)$ such that $u(x_{1})>g(x_{1})$. Let $x_{0}\in(y,A]$ be the
point of absolute maximum of $u-g$. We consider two cases:


*

*If $y<x_{0}<A$, then by hypothesis
\begin{align*}
& g^{\prime}(x_{0})+\alpha u(x_{0}) = (f(x_0) - \alpha \, g(x_0)) + \alpha \, u(x_0)\leq f(x_{0}),
\end{align*}
which implies that
$$
\alpha u(x_{0})\leq\alpha g(x_0).
$$
Since $\alpha>0$, we get that $u(x_{0})\leq g(x_{0})$, which is a
contradiction. 

*If $x_{0}=A$, then we can assume that $u-g$ has no local
maximum in $(y,A)$, which implies that $x_{0}=A$ is a point of strict maximum.
Consider the function
$$
g_{\varepsilon}(x)=g(x)+\frac{\varepsilon}{A-x} =: g(x) + h_{\varepsilon}(x).
$$
Then $g_{\varepsilon}\in C^{\infty}((0,A))$ and $(u-g_{\varepsilon
})(x)\rightarrow-\infty$ as $x\rightarrow A^{+}$, so $u-g_{\varepsilon}$ has a
maximum in $[y,A)$. Since $u(x_{1})-g(x_{1})=\delta>0$ and $u(y)-g(y)=0$, by
taking $\varepsilon$ sufficiently small we can assume that $(u-g_{\varepsilon
})(y)<(u-g_{\varepsilon})(x_{1})$, and so $u-g_{\varepsilon}$ has a maximum in
$[y,A)$ at some point $x_{\varepsilon}\in(y,A)$. Reasoning as before we get
$g_{\varepsilon}^{\prime}(x_{\varepsilon})+\alpha u(x_{\varepsilon})\leq
f(x_{\varepsilon})$, which gives
$$
\alpha u(x_{\varepsilon})\leq \alpha\,g(x_\varepsilon) - h_{\varepsilon}'(x_{\varepsilon}) = \alpha\,g(x_{\varepsilon}) - \frac{\varepsilon}{(A-x_\varepsilon)^2} < \alpha\,g(x_{\varepsilon}).
$$
We claim that $x_{\varepsilon}\rightarrow A$. Assuming that this is the case,
letting $\varepsilon\rightarrow0$ we get $\alpha u(A)\leq g(A)$, which is
again a contradiction since $u(x_{1})-g(x_{1})>0$.


To see that $x_{\varepsilon}\rightarrow A$, assume by contradiction that for a
subsequence (not relabelled) $x_{\varepsilon}\rightarrow x_{0}$ as
$\varepsilon\rightarrow0^{+}$ with $x_{0}\neq A$ (note that $[y,A]$ is
compact). But for every $x\in\lbrack y,A]$ since $x_{\varepsilon}$ is a point
of maximum,
$$
u(x)-g_{\varepsilon}(x)\leq u(x)-g_{\varepsilon}(x_{\varepsilon}%
)=u(x)-g(x_{\varepsilon})-\frac{\varepsilon}{A-x_{\varepsilon}}\leq
u(x)-g(x_{\varepsilon}).
$$
Letting $\varepsilon\rightarrow0$ it follows that%
$$
u(x)-g(x)\leq u(x_{0})-g(x_{0}).
$$
But since $A$ is a point of strict maximum, necessarily $x_{0}=A$. Thus, we
have shown that when $\alpha>0$, $u(x)\leq g(x)$ for every $x\in(y,A)$. With a
similar proof you can show that $u(x)\leq g(x)$ for every $x\in(0,y)$.
Not sure what to do for $\alpha<0$. Similar tricks, I assume.
