Suppose $ f, g $ are two smooth functions and that for all $ h: \mathbb{R} \rightarrow \mathbb{R} $: \begin{align*} \int f(h(x)) + g(h(x)) \frac{d^2 h}{dx^2} dx = 0 \end{align*} Can I conclude that $ f $ and $ g $ are separately 0? Assume $ h$ is smooth and vanishing at infinity.
EDIT: And what about: $$ \int f(h(x)) \left(\frac{d h}{dx}\right)^2 + g(h(x)) \frac{d^2 h}{dx^2} dx = 0 $$ Here I think it is not possible to conclude $ f $ and $ g $ are separately zero. What is the general statement for this sort of thing?
For the first equation, first consider $h(x) \equiv 0$. Then you have $$ \int f(0) \mathrm{d}x = 0 \iff f(0) = 0 $$ Next, let $\chi$ be a smooth monotonically increasing function so that $\chi \equiv 0$ on $(-\infty,-1)$ and $\chi = 1$ on $(1,\infty)$. For $\lambda,\mu$ let $$ h_{\lambda,\mu}(x) = \chi(x - \lambda) \chi(\mu - x) $$ Observe if $\mu < \lambda$ then $h_{\lambda,mu} = 0$, and if $\mu - \lambda > 2$ we have that $h_{\lambda,\mu} = 1$ on $[\lambda + 1, \mu - 1]$.
It is easy to see that by definition $$ \int g(h_{\lambda,\mu}(x)) h_{\lambda,\mu}''(x) \mathrm{d}x $$ is independent of $\lambda$ and $\mu$. Whereas for $\mu - \lambda > 2$ we have $$ \int f(h_{\lambda,\mu}(x)) \mathrm{d}x = \int f(h_{-1,1}(x)) \mathrm{d}x + f(1) (\mu - \lambda - 2) $$ From this we easily conclude that $f(1) = 0$. The same argument applied to $\sigma h_{\lambda,\mu}$ for $\sigma \in \mathbb{R}$ gives that $f \equiv 0$. So you are down to $$ \int g(h(x)) h'' \mathrm{d}x = 0 $$ for every $h\in C^{\infty}_0$. This, of course, is insufficient for concluding that $g$ vanishes identically. For example, $g$ can be any constant function. On the other hand, integrating by parts you get $$ \int g'(h(x)) (h'(x))^2 \mathrm{d}x = 0 $$ Now let $\eta$ be the function $\eta = 0$ if $|x| \geq 1$, $\eta = 1 - |x|$ if $|x| \leq 1$. Consider $h_{\sigma,\tau}$ to be a smooth approximation to $\sigma \eta(x/\tau)$. This shows $$ 2 \frac{\sigma}{\tau} \int_0^\sigma g'(y) \mathrm{d}y = 0 $$ which by the fundamental theorem of calculus tells us that $g$ must be constant.
The second equation is somewhat easier to see. Let $f = g'$ and you have that
$$ [ g(h(x)) h'(x) ]' = f(h(x)) (h'(x))^2 + g(h(x)) h''(x) $$
so from the fundamental theorem of calculus its integral vanishes. So the solution set is much much larger than just $\{ f\equiv 0, g\equiv 0\}$.