Prove the function is nonpositive for every $x$ in the given intervals Let function $f: [0,1] \mapsto \mathbb{R}$ has a second derivative on $(0,1)$. If $f(0)=f(1)=1$ and $f''+2f'+f\leq 0$ on $(0,1)$, prove that $f(x) \geq 0$ for every $x \in [0,1]$
Please help me. We have discussed this in class, but found no way. We are stuck on finding the first step to on. 
 A: Well, the reason why you guys can't prove it is because the claim is wrong.
Consider the solution to
\begin{align}
y''+2y'+y = 10, \ \ y(0) = 1,\  y(1) = 1
\end{align}
which is given by
\begin{align}
y(t) = e^{-t}(-9et+9t+10e^t-9). 
\end{align}
However, $y$ is negative for some $t \in (0, 1)$. 
Additional: If we are working with $f''+2f'+f=g(t)  \leq 0$, then the claim is true since
\begin{align}
f(t) = e^{-t}+(e-1-e\alpha)te^{-t}+\int^t_0 (t-s)e^{-(t-s)}g(s)\ ds
\end{align}
where
\begin{align}
-e\alpha = -\int^1_0 (1-s)e^{s}g(s)\ ds \geq 0. 
\end{align}
Hence it follows $f(t) \geq 0$. 
Additional 2: Assuming $f''+2f'+f\leq 0$, then using the idea of @dezdichado, we see that
\begin{align}
(e^tf)''= e^t g \leq 0
\end{align}
meaning $u(t) = e^t f(t)$ is concave down for all $t \in (0, 1)$. But since $u(0) = 1$ and $u(1) = e$, we see that $u(t)\geq 1$ which means $f(t) \geq e^{-t} \geq 0$. 
A: Assuming that $f$ is continuous at $0$ and at $1.$
Let $g(x)=e^xf(x).$ Then $g''(x)\leq 0$ for $x\in (0,1).$   
Suppose   $x_0\in (0,1)$ and $f(x_0)<0.$ Take $x_1\in (0,x_0)$ and $x_2\in (x_0,1)$ where $x_1$ is close enough to $0$ that $f(x_1)>0,$ and $x_2$ is close enough to $1$ that $f(x_2)>0.$ 
Then $0> \frac {g(x_0)-g(x_1)}{x_0-x_1}=g'(y_1)$ for some $y_1\in (x_1,x_0).$
And $0< \frac {g(x_2)-g(x_0)}{x_2-x_0}=g'(y_2)$ for some $y_2\in (x_0,x_2).$ 
Then for some $x_3\in (y_1,y_2)$ we have $g''(x_3)=\frac {g'(y_2)-g'(y_1)}{y_2-y_1}>0,$. But $x_3\in (0,1)$ so  $g''(x_3)\leq 0,$ a contradiction.
