Hallo!
I've to show that the function Tv is concave in x.

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

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


Let \lambda \in (0,1) and x_1, x_2 \in \mathbb{R}_+. v is concave, so it holds
 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
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!