TITLE: Differential inequality implies inequality for points at distance pi.
QUESTION [6 upvotes]: Given a function $f$ with $f+f''\ge 0$, show that $f(x)+f(x+\pi) \ge 0$ for all $x$.
Note that for sine and cosine both inequalities become equations. It seems reasonable to look at $f+f''=g$, but the resulting expressions seem inconclusive.

REPLY [6 votes]: \begin{align}
f(x)+f(x+\pi)&=\int_0^{\pi}\left(\sin(t)f(x+t)-\cos(t)f'(x+t)\right)dt\\
  &=\int_0^{\pi}\sin(t)f(x+t)dt-\int_0^{\pi}\cos(t)f'(x+t)dt\\
  &=\int_0^{\pi}\sin(t)f(x+t)dt-\left(-\int_0^{\pi}\sin(t)f''(x+t)dt\right)\\
  &=\int_0^{\pi}\left(\sin(t)f(x+t)+\sin(t)f''(x+t)\right)dt\ge0
\end{align}