I have encountered the following problem, when working trough the paper "The condition on the Asymptotic Elasticity of Utility Functions and Optimal Investment in Incomplete Markets" by D. Kramkov and W. Schachermayer (see link).
Let $U:(0,\infty)\to \mathbb{R}$ be a strictly increasing, strictly concave and continuously differentiable function, which satisfies the Inada conditions, i.e. $$ U'(0)=\lim_{x\to0}U'(x)=\infty, \quad U'(\infty)=\lim_{x\to\infty}U'(x)=0.$$ Further let $\mathcal{C}\subseteq L^0_+(\Omega,\mathcal{F},\mathbb{P})$ be convex, solid ( $f\in\mathcal{C}$ and $0\leq g\leq f$ imply $g\in\mathcal{C}$), bounded and closed in the topology of convergence in measure. Then for x>0 define the set $\mathcal{C}(x):=x\mathcal{C}$ as well as the value function $$ u(x)=\sup_{g\in\mathcal{C}(x)}\mathbb{E}[U(g)].$$ Further assume that $u(x)<\infty$ for some $x>0$, then it is remarked in the paper that $u(x)<\infty$ for all $x>0$, as well as that $u$ is concave.
I have some problems verifying those claims. My idea was to show that $u$ is concave and conclude that in this case the assumption implies $u(x)<\infty$ for all $x>0$. However I failed to do so, as the standard approach, i.e. showing $u(\lambda x+(1-\lambda)y)\geq \lambda u(x)+(1-\lambda) u(y)$, doesn't seem to work.
I would kindly appreciate some input and thoughts on the topic.
EDIT: Question is answered below.
Let $h_i \in C$. Note that
$\lambda E[U(x h_1)] + (1 - \lambda) E[U(y h_2)] = $
$E[\lambda U(x h_1) + (1-\lambda) U(y h_2)] \leq $
$E[U(\lambda x h_1 + (1 - \lambda) y h_2)]$
by concavity of $U$.
Now the function $h$, given by $\lambda x h_1 + (1 - \lambda) y h_2 = (\lambda x + (1 - \lambda) y) h$, is a convex combination of $h_1$ and $h_2$, and is therefore in $C$.
Thus we continue the estimate:
$... = E[U((\lambda x + (1-\lambda) y) h)] \leq u(\lambda x + (1 - \lambda) y)$.
Let $z$ be the point where $u(z)$ is finite. Let $0 < x < z = \lambda x + (1 - \lambda) y < y$.
Applying $\sup_{h_i \in C}[\cdots]$ to the above inequality shows that $u(x)$ and $u(y)$ are finite. Moreover, $u$ is convex.