Let $\{a_i\}$ be $n$ real numbers. It can be shown that there is $k\ge 0$ such that for all $x_i,y_i\in\mathbb{R}$ satisfying $\sum_{i=1}^n x_i^2\le 1$ and $\sum_{i=1}^n y_i^2\le 1$, $$ (\sum_{i=1}^n a_i(x_i^2-y_i^2))^2\le k^2 \sum_{i=1}^n (x_i-y_i)^2. $$ I proved the above with $k=2\sqrt{\sum_{i=1}^n {a_i^2}}$, but I suspect that $k=2\max\{|a_i|\}$ is the optimal value for $k$. I struggle to produce a proof.
$\begingroup$
$\endgroup$
6
-
$\begingroup$ Try first for $n=1.$ Then extend to any $n,$ using CS. And you are not asked for "an optimal value". $\endgroup$– Anne BauvalDec 3, 2022 at 6:16
-
$\begingroup$ @AnneBauval I actually managed to prove the existence of $k$, but now I would like to find the optimal $k$. Changed the question a bit. $\endgroup$– durianiceDec 3, 2022 at 6:29
-
$\begingroup$ Again: instead (or before) proving $k=2\max\{a_i\}$ is "an optimal value", prove it is a value. Try when all $a_i$'s are equal. $\endgroup$– Anne BauvalDec 3, 2022 at 6:39
-
$\begingroup$ @AnneBauval It doesn't seem that the equality is always attainable for $k=2\max\{|a_i|\}$. I'm considering the quotient $(\sum a(x^2-y^2))^2/(\sum (x-y)^2)$, but I don't know how to maximize this quotient. $\endgroup$– durianiceDec 3, 2022 at 7:09
-
2$\begingroup$ by using C-S, $k^2\geq\sum a_i^2(x_i+y_i)^2$ is sufficient for above inequality to hold. And from it you can show $k^2\geq 4\max a_i^2$ would be sufficient. It remains to check the choice $x_i=1, y_i=1-\epsilon$ where $i$ is the index for maximal $a_i$ to justify your optimal argument. $\endgroup$– Oolong MilkteaDec 3, 2022 at 7:51
|
Show 1 more comment
1 Answer
$\begingroup$
$\endgroup$
Since Oolong Milktea does not seem to want to post a full-fledged solution here, I would answer the question myself following their idea.
If all $a_i$ are $0$, then clearly $k=0$ is the optimal value. Otherwise, there is $a_j$ such that $|a_j|=\max\{|a_i|\}>0$. Choose $k:=2|a_j|$. Then $$ \begin{align*} \left( \sum_{i=1}^{n} a_i(x_i^2-y_i^2) \right)^2 &= \left( \sum_{i=1}^{n} a_i(x_i+y_i)(x_i-y_i) \right)^2 \\ &\le \sum_{i=1}^{n} a_i^2(x_i+y_i)^2 \sum_{i=1}^{n} (x_i-y_i)^2 & \text{Cauchy's inequality} \\ &= a_j^2 \sum_{i=1}^{n} \frac{a_i^2}{a_j^2} (x_i+y_i)^2 \sum_{i=1}^{n} (x_i-y_i)^2 & a_j\neq 0\\ &\le a_j^2 \sum_{i=1}^{n} (x_i+y_i)^2 \sum_{i=1}^{n} (x_i-y_i)^2 & \forall i:a_j^2\ge a_i^2\\ &\le a_j^2 \left( \sqrt{\sum_{i=1}^{n} x_i^2} + \sqrt{\sum_{i=1}^{n} y_i^2} \right)^2 \sum_{i=1}^{n} (x_i-y_i)^2 & \text{Triangle inequality} \\ &\le 4a_j^2 \sum_{i=1}^{n} (x_i-y_i)^2 & \sum_{i=1}^{n} x_i^2 \le 1,\sum_{i=1}^{n} y_i^2\le 1 \\ &= k^2 \sum_{i=1}^{n} (x_i-y_i)^2. \end{align*} $$ We now prove that $k$ is optimal. Let $0<r<1$. Choose $x_j:=1$ and $y_j:=1-r$, and all other $x_i,y_i:=0$. Then $$ \begin{align*} \lim_{r\to 0^+} \frac{(\sum_{i=1}^{n} a_i(x_i^2-y_i^2))^2}{\sum_{i=1}^{n} (x_i-y_i)^2} &= \lim_{r\to 0^+} \frac{a_j^2(1-(1-r)^2)^2}{r^2} \\ &= 4a_j^2 \\ &= k^2. \end{align*} $$ This then shows that $$ k^2=\sup\left\{\frac{(\sum_{i=1}^{n} a_i(x_i^2-y_i^2))^2}{\sum_{i=1}^{n} (x_i-y_i)^2}:x_i,y_i\in\mathbb{R};\sum_{i=1}^{n}x_i^2\le 1,\sum_{i=1}^{n}y_i^2\le 1 ;\sum_{i=1}^{n} (x_i-y_i)^2\neq 0\right\}, $$ so $k$ is optimal.