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I need help in the following problem :

Assume that the function $f:[a,b]\to \mathbb{R}$ satisfies $f(a)=f(b)=0, \forall x\in (a,b): f(x)>0$, for all $x\in [a,b]$ and $f+f''>0$. Prove that $b-a\geq \pi$ .

Here is when I am stuck: let $h=f+f''$ , $h$ is a positive function in $[a,b]$ and $f$ is a solution of the differential equation $y+y"=h$

By standard method of resolution we can prove that there existe two constante $u,v\in \mathbb{R}$ such that $f(x)=u\cos(x)+v\sin(x)+\int_{a}^{x} h(t)\sin(x-t)dt$

We can notice that $f(x)+f(x+\pi)=\int_{0}^{\pi} h(u+x)\sin(u)du \geq 0 $ , but I don't see how to conclude that $b-a \geq \pi $ ,

any idea ? thank you .

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  • $\begingroup$ Added some text in answer below to clarify as promised $\endgroup$ – Rutger Moody Apr 4 '17 at 16:02
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The function : $g(x)=A\sin(x-a)$ has the properties : $g(a)=0$ , $g(\pi+a)=0$ , $\forall x \in (a,\pi+a) : g(x)>0$ and $\forall x \in [a,\pi +a] : g(x) + g''(x) =0$.

Now let $f(x)$ be any continuous differentiable function satisfying the conditions of the question that reaches zero before $x=\pi+a $ .
So : $\pi+a \gt \epsilon \gt 0 \land f(\pi+a-\epsilon)=0$. We can choose $A$ above such that : $A\sin(x-a) \ge f(x)$ in the interval $x \in [a,\pi+a-\epsilon]$ and that there is at least one $c \in (a,\pi+a-\epsilon)$ for which $f(c)=A\sin(c-a)$. Define $m$ to be the maximum of such points :
$ m=\max \left\{c\in (a,\pi+a-\epsilon) : f(c)=A\sin(c-a) \right\}$.

We see that $f(m)+f''(m)<0$. Therefore $b-a \ge \pi \enspace \square$

Point where f() more concave then g()

Update

Promised Oty and Paramanand Singh to give extra motivation as they were not yet convinced :

Proof outline :
A : Below I will show that the amplitude $A$ of the sine function $g(x)$ can be made large enough so that $g(x) > f(x) $ for $\forall x \in (a,\pi+a-\epsilon]$.
B : Next it's easy to see we can decrease the amplitude $A$ of $g(x)$ until it just 'touches' $f(x)$ in one or more points without ever having $g(x)<f(x)$.
C : We now see there is a point $m$ where $f(m)=g(m)$ and $f'(m)=g'(m)$ and $f''(m)<g''(m)$. But we have $g(x) + g''(x) =0$ so this means $f(x) + f''(x) <0$ which is not allowed.

Proof :
A : From the question we can conclude $f(x)$ must be continuous and twice differentiable. It follows that both $f'(x)$ and $f(x)$ have a finite maximum in $[a,\pi+a-\epsilon]$. We can take amplitude $A$ such that $g'(a)>f'(a)$. This means the function $g(x)-f(x)$ (see picture below) must be $\gt 0$ in some interval $(a,\lambda)$. Let $E$ be the minimum of $g(x)-f(x)$. And let $e$ be the value in which it is attained $ \implies g(e)-f(e)=E$ (with $E<0$). Now we replace function $g(x)$ with a new function $g_2(x)=A_2\sin(x-a)$ with $A_2>A$. $A_2$ can be made arbitrarily large, (see points $C$ and $D$ in picture below with : $g(C)-f(C)=g(D)-f(D)=0$) we choose $A_2$ such that : $g_2(C)-f(C)> -E \land g_2(D)-f(D)> -E $.
We now have : $\forall x \in (a,\pi+a-\epsilon] : A_2\sin(x-a) > f(x)$.

B : From the above it is clear we can also choose smaller $A_3$ such that $A_3\sin(x-a)$ touches $f(x)$ while maintaining : $A_3\sin(x-a) \ge f(x)$.

C : It follows by elementary calculus that there must be $m$ such that : $A_3\sin(m-a)=f(m) $ and $\frac{d}{dx}(A_3\sin(x-a))|_{x=m}=f'(m) $ and $\frac{d^2}{dx^2}(A_3\sin(x-a))|_{x=m}>f''(m) $ ($f()$ must be more convex to stay underneath curve of $A_3\sin(x-a)$). We know $A_3\sin(m-a) + \frac{d^2}{dx^2}(A_3\sin(x-a))|_{x=m}=0$ , it follows that $f(m)+f''(m)<0$. We conclude there can be no function $f(x)$ under the conditions of the question for which $f(\pi+a-\epsilon)=0$ , so $b-a \ge \pi$.

Feel free to comment but please provide good motivation.

I'll include another picture below not as proof but as illustration.

g(x)-f(x)

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  • $\begingroup$ how do u get the last equation $f(m) + f''(m) < 0$? $\endgroup$ – Paramanand Singh Apr 4 '17 at 7:46
  • $\begingroup$ and why do you assume $f''(a) > 0$ and where is it used in your proof? $\endgroup$ – Paramanand Singh Apr 4 '17 at 7:49
  • $\begingroup$ @ParamanandSingh I'll include picture to illustrate. The point is that $f(m)=A\sin(m-a)$ but $f''(m)<-A\cos(m-a)$ because $f()$ reaches zero first. $\endgroup$ – Rutger Moody Apr 4 '17 at 7:50
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    $\begingroup$ Thanks for all those illustrations. The only point which needs to be proven analytically is that $f''(m) < g''(m) $ where $g(x) =A\sin(x-a) $. This is done by Taylor's theorem using $F(x) =f(x) - g(x) $ and then $F(m) =F'(m) =0$ and $F''(m+h) = (f''(m) - g''(m)) h^{2}/2+o(h^{2})$. Now we want $f<g$ ie $F<0$ so we must have $f''(m) <g''(m) $. +1 for the otherwise nice proof. $\endgroup$ – Paramanand Singh Apr 4 '17 at 18:20
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    $\begingroup$ @RutgerMoody beautifful Proof congratulation :) $\endgroup$ – uvw Apr 5 '17 at 11:17

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