3
$\begingroup$

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.

$\endgroup$

1 Answer 1

-1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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 $\lambda<0$. $\endgroup$ Nov 4, 2015 at 7:28
  • 2
    $\begingroup$ 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 $\lambda>0$. $\endgroup$ Nov 4, 2015 at 7:38
  • $\begingroup$ 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. $\endgroup$
    – Csaba
    Jan 7, 2017 at 9:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.