both convex and superharmonic function on manifold concave?

Question

M is a non-compact Rimannian manifold without boundary.

$f\in W_{loc}^{1,2}(M)$ satisfies $\Delta f \leq c$ in the weak sense, i.e. $$ -\int_M \langle \nabla f,\nabla \psi \rangle dvol \leq c\int_M \psi dvol $$ for any nonnegative Lipschitz function $\psi$ with compact support, where c is a constant.

Suppose in addition f is a $\lambda$-convex ($\lambda$ is a constant) function on M, i.e. for any geodesic $\gamma(t)$, $f \circ\gamma(t)-\lambda t^2/2$ is a convex function.

Then is there a constant $\mu$ such that f is $\mu$-concave? (The definition is similar as above)

Answer 1

This is a partial answer.

The result is true if $f$ is smooth. In this case, $\lambda$-convex is equivalent to $Hess(f)\geq -\lambda$, and $\Delta f\leq c$ just means that $trace(Hess(f))\leq c$. In particular the eigenvalues of $Hess(f)$ cannot be greater than $\mu=c+(n-1)\lambda$, which implies that $f$ is $\mu$-concave.

I'll think about it and see if this can be adapted to the $W^{1,2}$ case.