A calculus problem with functions such that $f''(x) = g(x)$ and $g''(x) = f(x)$ Let:


*

*$f(x)$ and $g(x)$ be twice differentiable, non-decreasing functions. 

*$f''(x) = g(x)$ and $g''(x) = f(x)$.

*$f(x) \cdot g(x)$ is a linear function. 


Then we have to show that $f(x) = g(x) = 0$.
I am really not able to do anything useful, any help would be appreciated :)

Let $f(x)g(x)=ax+b$ for some reals $a,b$. Differentiate w.r.t. x gives $f^\prime(x)g(x)+f(x)g^\prime(x)=a$, and differentiate again gives 
$$
f^{\prime\prime}(x)g(x)+2f^\prime(x)g^\prime(x)+f(x)g^{\prime\prime}(x)=0.
$$
So,
$$
g^2(x)+2f^\prime(x)g^\prime(x)+f^2(x)=0
$$
Since $f$ and $g$ are non-decreasing functions, then $f^\prime(x)g^\prime(x)\ge 0$. It follows that
$$
0=g^2(x)+2f^\prime(x)g^\prime(x)+f^2(x)\ge f^2(x)+g^2(x)
$$
which of course implies $f(x)=g(x)=0$.
