Let $f \in C^{2}([a,b]) \ $, $f(a)= f(b) = 0 \ $, $f(x) > 0 \ \forall x \in (a,b) \ $, f(x) + f(x)''>0 \ . Then $b-a \ge \pi $. Any hint?
Functional disequality
3
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analysis
1 Answers
5
Put $g(t):=f(t-a)$ (so we can assume that $a=0$). We put $C:=\int_0^bf(t)\sin\left(\frac tb\pi\right)dt$, then $C>0$. Now, since -f''(t)\leq f(t) we have \begin{align*} C&=\left[-f(t)\cos\left(\frac tb\pi\right)\frac b{\pi}\right]_0^b+\int_0^bf'(t)\cos\left(\frac tb\pi\right)\frac b{\pi}dt\\ &=\frac b{\pi}\left(\left[f'(t)\sin\left(\frac tb\pi\right)\frac b{\pi}\right]_0^b-\int_0^bf''(t)\frac b{\pi}\sin\left(\frac tb\pi\right)dt\right)\\ &=-\frac{b^2}{\pi^2}\int_0^bf''(t)\sin\left(\frac tb\pi\right)dt\\ &\leq \frac{b^2}{\pi^2}\int_0^bf(t)\sin\left(\frac tb\pi\right)dt\\ &\leq \frac{b^2}{\pi^2}C, \end{align*} and the result follows since $b>0$.