# Inequalities in special function cones

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 \end{array}\right.$$

and $$(i i)$$ $$v(t) \leq\left\{\begin{array}{ll} \frac{t}{p} v(p) & t>p \\ \frac{1-t}{1-p} v(p) & t Moreover, for all $$0 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 $$u|_{\partial\Omega}=0$$?.

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