If $f''(x)+f(x)>0$ and $f(x)>0$ $\forall x\in(a,b)$; $f(a)=f(b)=0$; prove that $b-a>\pi$.

Solution 1:

I think I have a solution when $f''$ is assumed to be continuous.

When $x\in (a,b)$, we have that $\sin\left(\frac{x-a}{b-a}\pi\right)>0$ so using the first assumption, $$0<\int_a^bf(x)\sin\left(\frac{x-a}{b-a}\pi\right)dx+\int_a^bf''(x)\sin\left(\frac{x-a}{b-a}\pi\right)dx=I_1+I_2\tag{1}.$$ We have, integrating by parts and using $f(a)=f(b)=0$ that $$I_1=-\frac{\pi}{b-a}\left(\int_a^b-f'(x)\cos\left(\frac{x-a}{b-a}\pi\right)dx.\right)$$ Doing the same for $I_2$, we finally get $$\small 0<\left(\frac{(b-a)^2}{\pi^2}-1\right)\int_a^bf'(x)\cos\left(\frac{x-a}{b-a}\pi\right)=\left(\frac{(b-a)^2}{\pi^2}-1\right)\int_a^bf(x)\sin\left(\frac{x-a}{b-a}\pi\right)dx.$$ As $\int_a^bf(x)\sin\left(\frac{x-a}{b-a}\pi\right)dx>0$, we are done.

Solution 2:

My answer is under the assumption that $f$ and $f'$ are continuous on $[a,b]$ and $f''$ exists on $(a,b)$.

Define $$g:[a,b]\to \mathbb{R}, \quad x\mapsto f'(x)\sin(x-a)-f(x)\cos(x-a).$$ By definition, $g$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Moreover, since $f(a)=f(b)=0$, $g(a)=0$ and $g(b)=f'(b)\sin(b-a)$. Then by mean value theorem, there exists $c\in(a,b)$, such that $$\frac{f'(b)\sin(b-a)}{b-a}=\frac{g(b)-g(a)}{b-a}=g'(c)=(f''(c)+f(c))\sin(c-a).\tag{1}$$ On the one hand, beacause $f(x)>0$ on $(a,b)$ and $f(b)=0$, $f'(b)\le 0$; on the other hand, $f''(c)+f(c)>0$. Combing these facts with $(1)$, we can conclude that $b-a>\pi$.

Edit: After reading Sanchez's answer, I realized that without assuming that $f$ is differentiable at $a,b$, the proof can be modified as follows.

Assume that $b-a\le \pi$. Then $g$ is differentiable on $(a,b)$ and $$g'(x)=(f''(x)+f(x))\sin(x-a)>0.$$ Therefore, $g$ is strictly increasing on $(a,b)$. As a result, both $$g(a^+):=\lim_{x\to a^+} g(x)=\lim_{x\to a^+}f'(x)\sin(x-a)$$ and $$g(b^-):=\lim_{x\to b^-} g(x)=\lim_{x\to b^-}f'(x)\sin(x-a)$$ exist, and $g(b^-)>g(a^+)$. However, since $f(x)>0$ on $(a,b)$ and $f(a)=f(b)=0$, by mean value theorem, $$\limsup_{x\to a^+}f'(x)\ge\liminf_{x\to a^+}\frac{f(x)-f(a)}{x-a}\ge 0 $$ and $$\liminf_{x\to b^-}f'(x)\le\limsup_{x\to b^-}\frac{f(x)-f(b)}{x-b}\le 0 .$$ It follows that $g(a^+)\ge 0\ge g(b^-)$, a contradiction.