Extending the fundamental lemma of the calculus of variations so that the integral is proportional to the endpoint of the integrand

Question

Consider the analytic and bounded function $f:\mathbb{R}_{\leq x_0}\rightarrow \mathbb{R}$ with bounded first derivative such that $\alpha\in\mathbb{R}$ and \begin{equation} \alpha\varphi''(x_0)f(x_0)=\int_{-\infty}^{x_0}\varphi''(x)f(x)\mathop{\mathrm dx}, \end{equation} for every choice of sufficiently smooth function $\varphi:\mathbb{R}_{\leq x_0}\rightarrow\mathbb{R}$ where $\lim_{x\rightarrow -\infty}\varphi(x)=\varphi(x_0)=0$. It is known that $\varphi'(x_0)=\alpha\varphi''(x_0)$ and I have written the above condition in terms of $\varphi''(x_0)$ so that the statement applies $\forall\varphi$. If either $f(x_0)=0$ or $\lim_{x\rightarrow -\infty}f(x)=0$, then it is a direct consequence of the fundamental lemma of the calculus of variations that $f(x)=Ax+B$ for $A,B\in\mathbb{R}$ (the latter case requiring $\varphi$ to be continuously thrice-differentiable). If there were no such conditions on $f$, is a more general solution admissible for $f$?

Answer 1

The fundamental lemma of the calculus of variations can still be applied as long as the equation is true for all smooth functions $\varphi$ that are compactly supported on $(-\infty,x_0)$. However, any compactly supported function would have all of its derivatives be $0$ outside its support, which means $\varphi''(x_0)=0$ and so : $$ \int_{-\infty}^{x_0} \varphi''(x)f(x)dx = 0, \ \ \ \forall \varphi \in \mathcal{C}_c((-\infty,x_0)), $$ and so we still have $f(x)=Ax+B$ according to the lemma. There's still hope that it might be slightly more general since we didn't assume that $f(x_0)=0$, but notice that, for any smooth $\varphi$ with compact support in $(-\infty,x_0]$ and with $\varphi(x_0)=0$ : $$ \int_{-\infty}^{x_0} f(x)\varphi''(x)dx = f\varphi'\vert_{-\infty}^{x_0} - \int_{-\infty}^{x_0}\underbrace{f'(x)}_{=A}\varphi'(x)dx = f(x_0)\varphi'(x_0) - A\underbrace{\varphi(x_0)}_{=0}, $$ and so the original equation becomes : $$ -\alpha\varphi''(x_0)f(x_0) = f(x_0)\varphi'(x_0), \ \ \ \forall \varphi \in \mathcal{C}_c((-\infty,x_0]) \text{ such that } \varphi(x_0)=0. $$ Taking any $\varphi$ that's equal to $(x-x_0)$ near $x_0$ yields $f(x_0)=0$, and so the solution is of the exact same form.