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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.

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    It would improve your Question to indicate what approaches you have already considered or actually tried, together with the difficulty encountered. This will help Readers focus on the obstacle that is most germane to your attempt. – 2017-01-24
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    @hardmath Unfortunately, my attempts didn't really amount to anything worthwhile. – 2017-01-24
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    Please show the attempts that led nowhere. – 2017-01-24
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    @Did Unfortunately, nothing useful came out of them, nor were they based on any substantial idea. They were mostly manipulations that included several attempts at reasoning by contradiction; and integrations by parts. – 2017-01-24
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    Quote: "Please show the attempts that led nowhere." – 2017-01-24
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    Maybe helpful : [Example ODE 0](http://www.alternatievewiskunde.nl/jaar2015/example0.htm) . – 2017-03-05

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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 $y0$, 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.

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    How do you know that $g \in C^\infty((0,A))$. From your definition, it is clearly $C^1$, but how do you show additional regularity? – 2017-03-08
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    I think there is a mistake: as $\varepsilon \to 0$, the denominator of the fraction tends to $0$ as well, so you can't conclude that this term vanishes in the limit. I'll award the bounty after we figure out the details. – 2017-03-08
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    I only let $\varepsilon$ tend to zero in the inequalities $u(x)-g_\varepsilon(x)\le u(x)-g(x_\varepsilon)$ and $\alpha u(x_\varepsilon)$g_\varepsilon(x)$ does not tend to zero. As for regularity, yes you are right. This works for $f\in C^\infty$. I guess one needs to regularize $f$ first and then do my proof – 2017-03-08
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    Oh, you are right, my mistake. The proof is pretty neat, apart from the $C^\infty$ issue. But you mean $\alpha u(x_\varepsilon) < \alpha \, g(x_\varepsilon)$, right? (with $\alpha$ in the right-hand side as well). – 2017-03-08
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    Anyway, bounty awarded. – 2017-03-08
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    Thanks. yes, sorry. I forgot the $\alpha$. For the $C^\infty$ I guess one could find a sequence $f_n\in C^\infty ([0,A])$ such that $f_n\to f$ in $C^0([0,A])$ and then take $n$ large enough that $\Vert f_n-f\Vert_\infty\le \delta/2$. Then define $g_n$ with $f_n$ in place of $f$ – 2017-03-08
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    And as I said these are standard tricks for viscosity solutions, especially the part with the $\varepsilon$. So I cannot take the credit for it. But I agree, it is a neat trick (just not mine!!!), – 2017-03-08
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    Well, anyway, thanks for sharing! I think you're right about the regularity; taking a sequence $f_n$ converging to $f$ in $L^\infty$ would probably work. – 2017-03-08
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    Did you get any ideas for the $\alpha < 0$ case? – 2017-03-10
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    No, I did not. Maybe fix $x$ and work with the function $$ g(y)=e^{-\alpha(x-y)}u(x)-e^{-\alpha y}\int_{y}^{x}e^{\alpha t}f(t)\,dt,\quad x\in\lbrack0,A]? $$ – 2017-03-12