convex function with distributional Hessian $D^2 f \leqslant \lambda$, $\lambda$-concave?

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.

I assume $\lambda>0$. $\lambda$ concave means that the function must be concave. A function that is both convex and concave must be linear. Take $f(x) = \lambda/2 x^2$, clear not a linear function. Then $f''(x) = (D^2 f) = \lambda \le \lambda$. However, $f$ is not $\lambda$-concave.
• I assume the OP refers to the following notion of $\lambda$-concavity: for all $t\in[0,1]$ and all $x$, $y$, $f((1-t)x+ty)\ge (1-t)f(x)+tf(y)+\frac\lambda 2 t(1-t)\|x-y\|^2$. Then a convex function can be $\lambda$-concave if $\lambda<0$. – Sebastian Goette Nov 4 '15 at 7:28
• Sorry, in the above comment, replace $\lambda$ by $-\lambda$, so that the right hand side looks as in the definition of $\lambda$-convexity, but with the direction of the inequality sign reversed. So a convex function can be $\lambda$-concave for $\lambda>0$. – Sebastian Goette Nov 4 '15 at 7:38
• Interesting. I would only flip the sign of $\lambda$ in the inequality you have because then this would look like $\lambda$ smoothness, see e.g.,Nesterov: Introductory Lectures on Convex Optimization, Theorem 2.1.5. However, rather than guessing what @mafan meant, I prefer to wait for his feedback, because based on the definition, we can flip the answer from yes to no and back. In fact, smoothness would make a lot of sense, see Theorem 2.1.6 in the same book. But I am truly puzzled about this convex/concave confusion. – Csaba Jan 7 '17 at 9:07