1
$\begingroup$

More specifically, let $x_1\geq x_2\geq x_3\geq x_4>0$, and $y_1\geq y_2\geq y_3\geq y_4>0$. Suppose $x_1\leq y_1$, $x_1+x_2\leq y_1+y_2$, $x_1+x_2+x_3\leq y_1+y_2+y_3$, $x_1+x_2+x_3+x_4=y_1+y_2+y_3+y_4$; also, $x_1x_2\leq y_1y_2$, $x_1x_2x_3\leq y_1y_2y_3$, $x_1x_2x_3x_4\leq y_1y_2y_3y_4$. Is it true that $x_i=y_i$ for all $i=1,2,3,4$? In general? Thanks

$\endgroup$
1
  • $\begingroup$ in fact, all that you need is the final product inequality $ \prod x_i \leq \prod y_i$ $\endgroup$ – Calvin Lin Jan 4 at 16:34
1
$\begingroup$

(Fill in the details as needed. If you're stuck, show your work and explain what you've tried.)

Claim: If $ y $ majorizes $x$, then $ \prod x_i \geq \prod y_i$, with equality iff $x_i = y_i$.

Proof:

  • Since $y$ majorizes $x$, so $x$ is in the convex hull of $y$ and it's permutations. So we have $ x = \sum_i \alpha_i y^{(i)}$, where $ y^{(i)}$ are permutations of $y$ and $ 0 \leq \alpha \leq 1$.
  • Specifically, for each coordinate $j$, $x_j = \sum_i \alpha_i y_j ^{(i)}$.
  • Since $ \log x$ is concave, so $ \log x_j \geq \sum_i \alpha_i \log y_j ^{(i)}$.
  • Hence, $ \sum_j \log x_j \geq \sum_j \sum_i \alpha_i \log y_j^{(i)} = \sum_j \log y_j$. Thus, $ \prod x_i \geq \prod y_j$.
  • Check the equality cases to show that equality holds iff $x_i = y_i$. Sight care has to be taken, esp when some coordinates of $y$ are equal.

Corollary: If $ \prod x_i \leq \prod y_i$, then we're in the equality case, and the desired result follows.

$\endgroup$
4
  • $\begingroup$ Since $\log$ is strictly concave, we can obtain $\alpha_i=1$ for some $i$, or all coordinates of $y$ are equal? Okay, got it, thanks! $\endgroup$ – Green Jan 4 at 20:12
  • $\begingroup$ @Green Not quite. We could have $ \alpha_1 = \alpha_2 = 0.5%$, leading to "some coordinates are equal", E.g. $y = (4, 2, 2, 1) $. In which case, because of the strictly concave, we know that "If $x_j$ has support on some $y$ coordinates, then those must all be the same value. $\endgroup$ – Calvin Lin Jan 4 at 20:19
  • $\begingroup$ Right. Could we say that the equality $\log \sum\alpha_iy_{j}^{i}=\sum\alpha_i\log y_{j}^{i}$ leads to $\sum\alpha_iy_{j}^{i}= y_{j}^{i} $ some $i$? For $j=4$, $x_4= y_{4}^{i} \geq y_4=x_k\geq x_4$, hence, $x_4=y_4$, and then by induction? $\endgroup$ – Green Jan 5 at 13:39
  • $\begingroup$ @Green Yes, essentially. As to how much detail one needs to add, that's dependent on the context. To me, that seems obvious after thinking about it (esp with the phrasing of "support"), but there are still some details to check if one is reasonably pedantic. $\endgroup$ – Calvin Lin Jan 5 at 17:00

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.