5
$\begingroup$

Is there a generalization of Jensen's inequality for convex multivariate functions? By convex, let's say $f$ is a multivariate function defined on the convex set $A$, and for all $x,y \in A$ and $\lambda \in [0,1]$, $$f(\lambda x + (1-\lambda)y) \leq \lambda f(x) + (1-\lambda)f(y).$$ Then, letting $x_1,\ldots,x_n$ denote points in $A$, the result would be something to the effect of saying that for any $n$ points in $A$, $$\frac{\sum_{i=1}^n f(x_i)}{n} \geq f \left(\frac{\sum_{i=1}^n{x_i}}{n} \right). $$ I do see a few articles that may be related:

  1. Perlman, Michael D. "Jensen's inequality for a convex vector-valued function on an infinite-dimensional space." Journal of Multivariate Analysis 4.1 (1974): 52-65.
  2. Merkle, Milan. "Jensen's inequality for multivariate medians." Journal of Mathematical Analysis and Applications 370.1 (2010): 258-269.
  3. Aras-Gazic, G., et al. "GENERALIZATION OF JENSEN’S INEQUALITY BY HERMITE POLYNOMIALS AND RELATED RESULTS." Mathematical reports 17.2 (2015): 201-223.
  4. Agnew, Robert A. "Multivariate version of a Jensen-type inequality." J. Inequal. in Pure and Appl. Math 6.4 (2005).

I do not think the first is particularly related if I'm interested in finite dimensional spaces, and my function is not vector-valued in any case. The second may be more related, but it seems to be generalizing in slightly different directions. The third is beyond my comprehension and the fourth, again, seems to be working on a slightly different generalization.

Are there no less technical generalizations of Jensen's to multivariate functions out there?

$\endgroup$
  • $\begingroup$ @Did I'm not sure I follow. I'm thinking $f:\mathbb{R}^n \rightarrow \mathbb{R}$. $u$ and $v$ (or $x$ and $y$) are vectors, then, but where am I using any notion of their ordering? Or could you perhaps point me someplace to better understand your first sentence? $\endgroup$ – Shane Jul 4 '16 at 8:53
  • 1
    $\begingroup$ From that proof: "$\ldots \varphi$ be a convex function on the real numbers. Since $\varphi$ is convex, at each real number $x \ldots$". Isn't this univariate? $\endgroup$ – Shane Jul 4 '16 at 9:37
  • 2
    $\begingroup$ Just yell if the mystery does not dissolve by itself... $\endgroup$ – Did Jul 4 '16 at 10:16
  • 2
    $\begingroup$ Rereading your question and the comments, I must admit being a little lost. Is your goal to show that, if $f$ is a multivariate function defined on some convex set $A$, and if, for all $x$ and $y$ in $A$ and all $\lambda$ in $[0,1]$, $f(\lambda x + (1-\lambda)y) \leq \lambda f(x) + (1-\lambda)f(y)$, then, for all $n$ and all points $x_1$, $\ldots$, $x_n$ in $A$, one has $\frac1n\sum\limits_{i=1}^n f(x_i)\geqslant f \left(\frac1n\sum\limits_{i=1}^n{x_i} \right)$? Because this one has self-contained short proofs, for example, by induction on $n$. $\endgroup$ – Did Jul 10 '16 at 15:24
  • 1
    $\begingroup$ To deduce the $n+1$ case of the inequality from the $n$ case, use $\lambda=\frac1{n+1}$, $x=x_{n+1}$, $y=\frac1n\sum\limits_{k=1}^nx_k$, then $z=\frac1{n+1}\sum\limits_{k=1}^{n+1}x_k=\lambda x+(1-\lambda)y$ hence $f(z)\leqslant\lambda f(x)+(1-\lambda)f(y)$, now by the recurrence hypothesis, $f(y)\leqslant\frac1n\sum\limits_{k=1}^nf(x_k)$, hence $f(z)\leqslant\lambda f(x_{n+1})+(1-\lambda)\frac1n\sum\limits_{k=1}^nf(x_k)$, qed. $\endgroup$ – Did Jul 11 '16 at 19:44
1
$\begingroup$

Jensen's inequality always holds when you're dealing with a convex function whose domain is finite-dimensional. By restricting the domain of $f$ to $S := \text{span} \{x_1, \ldots, x_n\}$, we put the question into that setting and know that $f(\mathbb{E} X) \leq \mathbb{E} f(X)$ for any random vector $X$ taking values in $S$. In particular, it holds when the distribution of $X$ is uniform on $\{x_1, \ldots, x_n\}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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