Recently in my problem solving class we had the question: If $f: [0,\infty) \to \Bbb R$ is a twice differentiable function such that f''(x) + e^x f(x) = 0, then prove that $f$ is bounded.
The solution I found was to create a function g(x)=f(x)^2 + e^{-x}f'(x)^2. Looking at g', we have g'=2ff' -e^{-x}f'^2 + 2e^{-x}f'f'' g'=-e^{-x}f'^2+2ff'+2e^{-x}f'f'' g' = -e^{-x}f'^2 +2e^{-x}f' \left(e^xf + f'' \right) = -e^{-x}f'^2 \leq 0 Thus we have that $g$ is a decreasing function. This means that for $x>0$, we have $g(x) \leq g(0)$ Since $g(0)=R$ for some constant $R$, we have R \geq g(x)=f(x)^2+e^{-x}f'(x)^2 > f(x)^2 thus $f^2$ is bounded, so $f$ itself must be bounded.
It is not so hard to show that this works if we replace $e^x$ with any differentiable function $k(x)$ that is positive and increasing on $[0,\infty)$. Thus we have the more general result that if $f: [0,\infty) \to \Bbb R$ is a twice differentiable function such that f''(x) + k(x) f(x) = 0 on $[a,\infty)$ for some $a \geq 0$ and $k(x)$ is positive, increasing, and differentiable on $[a,\infty)$, then $f$ is bounded.
However, my teacher also gave us a similar problem: if $f: [0,\infty) \to \Bbb R$ is a twice differentiable function such that f''(x) + \frac{1}{x} f(x) = 0 on $[a,\infty)$ for some $a > 0$, then $f$ is bounded above.
Unfortunately for us, our above result does not apply, since $\frac{1}{x}$ is decreasing, rather than increasing on $[a,\infty)$. In fact, defining $g(x)$ analogously as g(x)=f(x)^2+xf'(x)^2 gives us that g' \geq 0, so the technique fails. All similarly defined functions in terms of $f(x)$ and f'(x) that I've tried have also failed to tell me anything about the boundedness of $f$. However, since there is a degree of intuition required for competition type problems like this, it's certainly very very likely that I missed one that does work.
I will mention that defining h(x) = \frac{1}{x}f(x)^2 + f'(x)^2 does tell us that f' must be bounded, but I couldn't get anything about the boundedness of $f$ out of this.
I've been a member of stackexchange for a couple of months now and I really wanted my first posted question to be something really interesting and I believe that I found one! Since the problem involving $e^x$ generalizes very nicely to all increasing, positive, differentiable functions, I imagine that a solution of the problem involving $\frac{1}{x}$ would also generalize to all decreasing, positive, differentiable functions, so a proof of this could possibly provide a very interesting result concerning boundedness of functions that satisfy relatively simple differential equations. I've hit a bit of an impasse in solving this problem and I would like to know if anyone has any suggestions on how to prove this.