Suppose $ U : \mathbb{R} \to \mathbb{R} $ is concave, and that the random variable $ \epsilon $ has zero mean. Assuming that the function $ \phi : \mathbb{R} \to \mathbb{R} $, defined by $ \phi(\lambda) = \mathbb{E} U(\mu + \lambda \epsilon) $ is everywhere finite-valued, prove that $ \phi $ is concave.
I've tried a few different things, including Jensen's inequality, but I can't get it to work. Any help would be greatly appreciated. Thanks
EDIT: I'll show some of my working
$ p \phi(\lambda_1) + (1-p)\phi(\lambda_2) = p \mathbb{E} U (\mu + \lambda_1 \epsilon) + (1-p) \mathbb{E} U(\mu + \lambda_2 \epsilon) $
$ \leq pU(\mathbb{E}(\mu+\lambda_1 \epsilon)) + (1-p)(U(\mathbb{E}(\mu + \lambda_2 \epsilon)) = pU(\mu) + (1-p)U(\mu) = U(\mu) $
The inequality comes from Jensen and the fact that $U$ is concave. But I'm not sure where to go from here. Am I right in thinking $ U(\mu) = \phi(0) $?