Is entropy being maximized and the norm being minimized by the same (unique) probability vector somehow related?

Consider the set of probability vectors

$$\mathcal P_n=\Big\lbrace x\in[0,1]^n\,\Big|\,\sum_{i=1}^n x_i=1\Big\rbrace\subset\mathbb C^n$$

for any $n\in\mathbb N$ where $x_i$ is the $i$-th component of $x$. Since $\mathcal P_n$ is a non-empty convex and closed set as is readily verified, by the Hilbert projection theorem (or minimizing vector theorem) there exists a unique vector of minimal norm which nonsurprisingly turns out to be the equilibrium $\frac1n(1,\ldots,1)$ as a simple consequence of Cauchy-Schwarz.

On the other hand for any $x\in \mathcal P_n$ we can define the entropy as in quantum information via

$$S:\mathcal P_n\to\mathbb R_0^+\qquad x\mapsto-\sum_{i=1}^n x_i\log(x_i).$$

It is well known that the entropy is maximal if and only if $x=\frac1n(1,\ldots,1)$.

Question: I wondered if the entropy being maximized and the norm being minimized by the same unique vector $\frac1n(1,\ldots,1)$ is somehow related? I don't see a direct connection since the logarithm obviously is not linear but I still feel like there might be some kind of link between those two results.

Thanks in advance for any answer or comment!

• mathoverflow.net/questions/118169/… – jkabrg Oct 25 '17 at 9:46
• Another viewpoint on this is related to schur-convex and schur-concave functions from the majorization theory. Essentially norm is a schur-convex and entropy is a schur-concave function. In most scenarios, the solution to minimizing ( maximizing) a schur-convex function (concave function) is the symmetrical solution where every co-ordinate is the same. – dineshdileep Oct 25 '17 at 12:40