I've to show that the function $\displaystyle Tv$ is concave in $\displaystyle x$.

$\displaystyle Tv: \mathbb{R}_+ \to \mathbb{R}$ defined by
$\displaystyle Tv(x):=\sup\limits_{a \in [0,1]}\mathbb{E}[v(e^{rh}(1+a(R-1)))]$
$\displaystyle v:\mathbb{R}_+ \to \mathbb{R}$ is a continous function with the follwing properties
$\displaystyle v(x)\leq c(1+x)$ for a constant $\displaystyle c \geq 0$,
strictly concave and strictly increasing.

$\displaystyle \mathbb{E} $ expectation,
$\displaystyle R$ lognormal distributed random variable,
$\displaystyle r,h>0$ are constants

Let $\displaystyle \lambda \in (0,1)$ and $\displaystyle x_1, x_2 \in \mathbb{R}_+.$ $\displaystyle v$ is concave, so it holds
$\displaystyle Tv(\lambda x_1+ (1-\lambda)x_2)\geq \sup\limits_{a \in [0,1]}{(\lambda \mathbb{E}v(e^{rh}x_1(1+a( R-1))+(1-\lambda)\mathbb{E}v(e^{rh}x_2(1+a( R-1))) $

But why is
$\displaystyle Tv(\lambda x_1+ (1-\lambda)x_2)\geq \lambda Tv(x_1)+(1-\lambda) Tv(x_2)$?

Can anybody help me?
Thanks in advance!