Let $f:R^n \to R$ be convex (may not $C^1$), $$[D^2f]=[D^2f]_{ac}+[D^2f]_s=[h_{ij}] L^n+[D^2f]_s$$ is the Lebesgue decomposition of the Hessian matrix. Where $[h_{ij}]$ is the density w.r.t the Lebesgue measure $L^n$. Then we know that $[h_{ij}](x)$ is non-negatively definite matrix at any point where the absolutely part supports. And the singular part $[D^2f]_s \geqslant 0$.

If we assume in addition there is a constant $\lambda>0$ such that for any $\psi=(\psi_1,...,\psi_n)$ with $\psi_i \in C^2_c(R^n)$ $$ \Sigma_{i,j}\int_{R^n} f \frac{\partial^2(\psi_i\cdot \psi_j)}{\partial x_i \partial x_j} dx \leqslant \lambda \Sigma_i \int |\psi_i|^2 $$ Then we have $[h_{ij}]-\lambda \cdot I$ is non-positively definite matrix and $[D^2f]_s \leqslant 0$, where $I$ is the unit $n\times n$ matrix. Thus the singular part vanishes $[D^2f]_s=0$.

My question is "Is $f$ $\lambda$-concave?"

Or for short, equivalently "Is the Hessian measure of a convex function $f$ has vanished singular part, and at any second differential points $D^2f(x) \leqslant \lambda \cdot I$, is $f$ $\lambda$-concave?".

My calculation is as follows: Let $f_{\epsilon}$ be the smooth modifiller. If we can get that $D^2 f_{\epsilon} \leqslant \lambda-\delta(\epsilon)$, Then $f_{\epsilon}$ is $\lambda-\delta(\epsilon)$-concave. Then by approximation we can get $f$ is $\lambda$-concave.

$$ \int D^2 f_{\epsilon} \psi_i \psi_j =\int f_{\epsilon} \frac{\psi_i \psi_j}{\partial x_i \partial x_j} \to \int f \frac{\psi_i \psi_j}{\partial x_i \partial x_j}=\int \psi_i \psi_j h_{ij}dx \leqslant \lambda \Sigma_i \int |\psi_i|^2 $$ But it seems we can't get $f_{\epsilon}$ is $\lambda-\delta(\epsilon)$-concave from above. The convergence is for measure, it's weak not strong.