We consider the Banach space $X=C([0,1])$ endowed with the norm $\|v\|_{\infty}=\max _{t \in[0,1]}|v(t)|$ and, we define the cone $\mathcal{C}=\{u \in X \mid u \mbox{ is concave, } u \geq 0, u(0)=u(1)=0\}$. We can prove the following result:

Given a function $v$ in the cone $\mathcal{C}$ and a point $p \in(0,1),$ the following estimates hold: $(i)$ $$ v(t) \geq\left\{\begin{array}{ll} \frac{t}{p} v(p) & t<p \\ \frac{1-t}{1-p} v(p) & t>p \end{array}\right. $$ and $(i i)$ $$ v(t) \leq\left\{\begin{array}{ll} \frac{t}{p} v(p) & t>p \\ \frac{1-t}{1-p} v(p) & t<p \end{array}\right. $$ Moreover, for all $0<t_{0}<t_{1}<1,$ we have $$ \min _{t \in\left[t_{0}, t_{1}\right]} v(t) \geq c_{t_{0}, t_{1}}\|v\|_{\infty} \,\,\,\,\,\,\,\,\,\, (1) $$ where $c_{t_{0}, t_{1}}:=\min \left\{t_{0}, 1-t_{1}\right\}$.

It is possible to obtain a generalization of (1) for some cone of functions $u$ defined in a subset $\Omega$ of $\mathbb{R}^n$ ($n\geq 2$) satisfying $u|_{\partial\Omega}=0$?.

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    $\begingroup$ What is concave down? concave? $\endgroup$ – Dieter Kadelka Jan 13 at 18:57
  • $\begingroup$ Sorry, is only concave!!! $\endgroup$ – Anderson de Araujo Jan 13 at 22:48

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