$f''(x) = g(x)$ and $g''(x) = f(x).$ Suppose also that $f(x)g(x)$ is linear in $x$ on $(a,b).$ Show that $f(x) = g(x) = 0$ for all $x ∈ (a,b).$ 
QUESTION: Let $f$ and $g$ be two non-decreasing twice differentiable functions defined on an interval $(a,b)$ such that for each $x ∈ (a,b), f''(x) = g(x)$ and $g''(x) = f(x).$ Suppose also that $f(x)g(x)$ is linear in $x$ on $(a,b).$ Show that we must have $f(x) = g(x) = 0$ for all $x ∈ (a,b).$


MY ANSWER: I have done the proof, but it's not a rigorous one.. this is what I did-
Let $f(x)=x^k$, $(k>0)$ which is increasing on $(a,b)$. Now, $$f'(x)=kx^{k-1}$$$$f''(x)=k(k-1)x^{k-2}$$ According to the question, $g(x)=k(k-1)x^{k-2}$, therefore, $$g'(x)=k(k-1)(k-2)x^{k-3}$$$$g''(x)=k(k-1)(k-2)(k-3)x^{k-4}$$ Now according to the question again, $f(x)=g''(x)$, but $f(x)=x^k$, therefore our statement implies that, $$x^k=k(k-1)(k-2)(k-3)x^{k-4}$$
Also, it is said that $f(x)g(x)$ must be linear in $x$. Therefore, we observe that,
$$k(k-1)(k-2)(k-3)x^kx^{k-4}$$ must be linear in $x$. Which clearly states that, $$k+(k-4)=1$$$$\therefore 2k-4=1$$$$\implies k=\frac{5}2$$
Putting $k=\frac{5}2$ in the previous equation, we get, $$x^4=\frac{5}2(\frac{5}2-1)(\frac{5}2-2)(\frac{5}2-3)$$$$\implies x^4=-\frac{15}{16}$$ which is clearly impossible for any $x$ in $\mathbb{R}$. Therefore, we may conclude that $$k\neq\frac{5}2$$ and the only way to satisfy both the above statements is to make $x=0$.
Therefore, we can conclude that $f(x)=0$ and consequently $g(x)=0$
Note 1: we observe, if $k<4$ then the the value of the derivatives becomes zero somewhere in between and our proof works.
Note 2: if we had assumed the function more generally as $f(x)=x^k+c$ then too, it would have worked, only $c$ would have become zero at last (it is forced to become)..
Now, there are a hell lot of non-decreasing functions out there (even trigonometric functions if defined in suitable intervals) and obviously this proof is not rigorous. Without assuming any function, how do I proceed to do this?
Any help will be much appreciated. Thank you.
 A: $x=0$ is not a logically correct way to solve these types of functional equations, since $x$ is a variable which is supposed to vary from $a$ to $b$.You cannot fix its value any way whatsoever.

Better to proceed like this:
Assume $f(x)g(x) = kx$ where $k$  is some constant (because of the linearity condition).
Therefore, you get
$$f'(x)g(x) + f(x)g'(x) = k$$
$$\Rightarrow f''(x)g(x) + 2f'(x)g'(x) + f(x)g''(x)= 0$$
$$\Rightarrow g^2(x) + 2f'(x)g'(x) + f^2(x)= 0 \tag1$$
Differentiating again gives,
$$\Rightarrow 2g(x)g'(x) + 2f''(x)g'(x) + 2f'(x)g''(x) + 2f(x)f'(x)= 0$$
$$\Rightarrow 4g(x)g'(x) + 4f(x)f'(x)= 0$$
$$\Rightarrow g(x)g'(x) + f(x)f'(x)= 0 \tag{2}$$
If you differentiate two more times, you will get
$$\Rightarrow f(x)g'(x) + g(x)f'(x)= 0 \tag3$$
Now can you solve $(2)$ and $(3)$ for $f(x)$ and $g(x)$?
A: Since $f(x)g(x)$ is linear then its second derivative is 0.So,$f''(x)g(x)+2f'(x)g'(x)+g''(x)f(x) =0$ that is $g^{2}(x)+f^2(x) =-2f'(x)g'(x)$ and as $f,g$ are twice differentiable and non-deacreasing,so $f'(x)$ and $g'(x)$ can not be negative.So $f(x)=g(x)=0$.
