# Second order differential inequality and comparison theorem. If x:[0,1] \rightarrow \mathbb R

Second order differential inequality and comparison theorem.
If $x:\left[0,1\right]⇒\mathbb{R},x\in {C}^{\mathrm{\infty }}$ that satisfies
$\left\{\begin{array}{l}\stackrel{¨}{x}\le -2x\\ x\left(0\right)=x\left(1\right)=0\end{array}$
is it true that ?
I observed that if $x\ge 0$, then $\stackrel{¨}{x}\le 0$, so x is concave. However I could not find out any other properties. I think if there exists a solution of the following differential equation
$\left\{\begin{array}{l}\stackrel{¨}{x}=-2x\\ x\left(0\right)=x\left(1\right)=0\end{array}$
then some kind of comparative theorem can be used. However, there is no non-trivial solution for this.
If the above proposition is incorrect, could you give me a counterexample?
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Kendal Kelley
$x\left(t\right)=x\left(\left(1-t\right)\cdot 0+t\cdot 1\right)\ge \left(1-t\right)\cdot x\left(0\right)+t\cdot x\left(1\right)=0$.
Alternatively, by Rolle's theorem, ${x}^{\prime }\left({t}_{0}\right)=0$ for some ${t}_{0}\in \left(0,1\right)$, ans as , so x not decreases on $\left[0,{t}_{0}\right]$-thus is non-negative there, and x not increases on $\left[{t}_{0},1\right]$ - thus again is non-negative there.