Let $f : R \to R$ be twice differentiable, $f (x+\pi$) = $f(x)$, $f''(x) + f(x) \geq 0$ $∀$ $x\in {R}$. Prove that $f(x)\geq0$ $∀$ $x\in {R}$

Question

I came across this question:

Let $f : R \to R$ be a twice differentiable function such that $f (x+\pi$) = $f(x)$ and $f''(x) + f(x) \geq 0$ $∀$ $x\in {R}$. Prove that $f(x)\geq0$ $∀$ $x\in {R}$

My teacher suggested me to define a function $g: R\to R$ for a real number $a\in R$, where $g(x) = f'(x+a) \sin x - f(x+a)\cos x$

$\therefore g'(x) = \sin x(f(x+a) + f''(x+a))\geq 0$ $∀$ $x\in [0,\pi]$

Hence for every $a\in R$

$g(\pi) - g(0)\geq 0$

$\because$ $g(\pi) -g(0) = f(a+\pi)+f(a) = 2f(a)$

$\therefore$ $2f(a) \geq0$

$\therefore$ $f(a) \geq0$

My Problem:

Everything seems okay with the method he suggested, but how do I know how to define $g(x)$ the way he did. Is there any other way to do this question? If not, can someone please help me in identifying why my teacher went for that particular definition of $g(x)$?

Thanks for comments/suggestions.

Answer 1

I have seen this before, but can't properly remember the thinking to get you started off.

Having said that, you have clues in that $f(x) = f(x + \pi)$, which shows you the function is periodic, and the relationship between $f''(x)$ and $f(x)$ should make you immediately start to look for a function that looks like a combination of $sin(x)$ and $cos(x)$.

I know this isn't a full answer, but hopefully it gets you started. If I suddenly recall the thinking I'll come back!