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I come across the following problem in my study.

Let $x_i, y_i\in \mathbb{R}, i=1,2\cdots,n$ with $\sum\limits_{i=1}^nx_i^2=\sum\limits_{i=1}^ny_i^2=1$, and $a_1\ge a_2\cdots \ge a_n>0 $. Is it true $$\left(\frac{\sum\limits_{i=1}^na_i(x_i^2-y_i^2)}{a_1-a_n}\right)^2\le 1-\left(\sum\limits_{i=1}^nx_iy_i\right)^2~~?$$

Has anyone seen this inequality before, or can you give a counterexample?

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3  
The LHS is invariant under affine transformations of the $(a_i)$, so you might as well assume $a_1=1$ and $a_n=0$. –  Douglas Zare Apr 2 '10 at 18:54
    
You accepted an answer that is, according to the author, incorrect... –  Mariano Suárez-Alvarez Apr 3 '10 at 7:45
    
@Mariano: Maybe the outer square in the LHS should not be there after all? I'm sure I did not see it yesterday –  Sergei Ivanov Apr 3 '10 at 9:16
    
@Sergei: if it makes you feel better, I didn't see the square either. So after failing to prove the inequality, I "verified" your counterexample was correct, and up-voted it. It seems that the original post didn't use \left( and \right) for the brackets on the LHS... I blame that and my new glasses. –  Willie Wong Apr 3 '10 at 9:34

5 Answers 5

up vote 9 down vote accepted

[Wrong ounter-example deleted]

This it true for all $n$. The case $n=2$ is handled by Hailong Dao, let's reduce the general case to $n=2$.

First, we may assume that $a_1=1$ and $a_n=0$ as others mentioned. So remove the denominator in LHS. Then forget the condition that $a_i$ are monotone, let's only assume that they are in $[0,1]$ (we can rearrange the indices anyway). Also we may assume that the sum under the square in the LHS is nonnegative - otherwise swap $x$ and $y$.

Then, for every $i$ such that $x_i^2-y_i^2>0$, set $a_i=1$, otherwise $a_i=0$. The LHS grows, the RHS stays. So it suffices to prove the inequality for $a_i\in\{0,1\}$. Rearrange indices so that the first $k$ of $a_i$'s are 1. We arrive to $$ \left(\sum_{i=1}^k x_i^2-\sum_{i=1}^ky_i^2 \right)^2 \le 1 - \left(\sum_{i=1}^n x_iy_i\right)^2 . $$ Define $X_1,X_2,Y_1,Y_2\ge 0$ by $$ X_1^2 = \sum_{i=1}^k x_i^2, \ \ \ X_2^2 = \sum_{i=k+1}^n x_i^2, \ \ \ Y_1^2 = \sum_{i=1}^k y_i^2, \ \ \ Y_2^2 = \sum_{i=k+1}^n y_i^2. $$ Then the LHS equals $(X_1^2-Y_1^2)^2$ and $X_1^2+X_2^2=Y_1^2+Y_2^2=1$. By Cauchy-Schwarz, $$ \left|\sum x_iy_i\right| \le X_1Y_1+X_2Y_2 , $$ so the RHS is greater or equal to $1-(X_1Y_1+X_2Y_2)^2$. Now the inequality follows from $$ (X_1^2-Y_1^2)^2 \le 1-(X_1Y_1+X_2Y_2)^2 $$ which is the same inequality for $n=2$.

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5  
Erm... And how exactly is this a counterexample? If $x^2+y^2=1$, then $(x^2-y^2)^2=1-(2xy)^2$. –  fedja Apr 3 '10 at 1:36
    
I think fedja is right! –  Hailong Dao Apr 3 '10 at 3:35
    
Yup, I overlooked the square. In fact, inequality is correct modulo the case n=2. Please unaccept or read the new answer. –  Sergei Ivanov Apr 3 '10 at 7:59
    
Darnit, you beat me by ten minutes! What are the odds?? –  zeb Apr 3 '10 at 8:08

The inequality is true for all $n$.

First of all, we can simplify it a little - from Douglas Zare's comment, we can assume $a_0 = 1$, $a_n = -1$, and try to maximize the LHS by varying the $a_i$s. Since the set of values for the $a_i$s under these conditions is compact, there is a maximum value, and the LHS is a convex function of each $a_i$ so we must have $a_i = \pm 1$ for all $i$. Then, we see that the LHS is clearly maximized when we take $a_i = \frac{|x^2-y^2|}{x^2-y^2}$, so we just have to prove that:

$(\frac{\sum |x_i^2-y_i^2|}{2})^2 \le 1 - (\sum x_iy_i)^2$

whenever $\sum x_i^2 = \sum y_i^2 = 1$. This follows from plugging $\alpha_i = \mbox{max}(x_i,y_i)$, $\beta_i = \mbox{min}(x_i,y_i)$ into the inequality

$(\frac{\sum \alpha_i^2 - \sum \beta_i^2}{2})^2 \le (\frac{\sum \alpha_i^2 + \sum \beta_i^2}{2})^2 - (\sum \alpha_i\beta_i)^2$,

which is just Cauchy-Schwartz in disguise.

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The restriction doesn't change the inequality - read Douglas Zare's comment. Letting $a_n < 0$ slightly generalizes the inequality, and makes the proof look a lot nicer. –  zeb Apr 3 '10 at 20:59

Here is a proof for the case $n=2$:

Some general facts: As Douglas pointed out, one can assume $a_1=1, a_n=0$. Also, note that the RHS is:

$$(\sum x_i^2)(\sum y_i^2) -(\sum x_iy_i)^2 = \sum_{i<j}(x_iy_j-x_jy_i)^2$$

Now let $n=2$. The inequality becomes: $$(x_1^2 -y_1^2)^2 \leq (x_1y_2-x_2y_1)^2$$ or

$$|x_1^2 -y_1^2| \leq |x_1y_2-x_2y_1|$$

Let $x_1 = cos(\alpha), x_2=sin(\alpha), y_1=cos(\beta), y_2=sin(\beta)$. Then LHS is $$|\frac 12 (cos(2\alpha) -cos(2\beta))| = |sin(\alpha-\beta)sin(\alpha+\beta)|$$

while the RHS is $|sin(\alpha-\beta)|$

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The following proof is a bit heavy-handed; I'm sure you it can be simplified. Assume $a_1=1, a_n=0$ as suggested above and write:

Write $$F(x,y) = \sum_{i=1}^n a_i(x_i^2-y_i^2)$$, $$G(x,y)=F(x,y)^2+\langle x,y\rangle^2.$$ Let $(x,y)\in S^{n-1}\times S^{n-1}$ be a point where $G$ is maximized, where we may assume that for each $i$ at least one of $x_i,y_i$ is non-zero. It is clear that $-\sum_i y_i^2 \leq F(x,y) \leq \sum_i x_i^2$ so we may also assume $\langle x,y\rangle \neq 0$.

By the method of Lagrange multipliers there exist $\xi,\eta$ such that for all $i$ $$ 4a_i x_i F+2\langle x,y\rangle y_i=2\xi x_i$$ and $$ -4a_i y_i F+2\langle x,y\rangle x_i=2\eta y_i.$$

Multiplying the first equation by $x_i$, the second by $y_i$, adding the two and summing over $i$ gives $ 4G = 2(\xi+\eta)$. Multiplying the second equation by $y_i$, the first by $x_i$ and adding gives $$ \langle x,y\rangle (y_i^2+x_i^2) = (\xi+\eta)x_i y_i = 2G\cdot x_i y_i. $$ By assumption one of $x_i,y_i$ is non-zero. Dividing by the square of that number we see that the quadratic $$\langle x,y\rangle t^2 - 2G t + \langle x,y\rangle = 0$$ has a real root. Evaluating the discriminant it follows that $$ G^2 \leq \langle x,y\rangle^2 \leq 1.$$

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Looks like you got ninja'd as well... –  zeb Apr 3 '10 at 8:29

A similar (maybe slightly simpler) counterexample: $n=2$, $x=(2,0)$, $y=(0,2)$, arbitrary $a_1>a_2$.

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3  
There is a condition $\sum x_i^2=\sum y_i^2=1$. –  Sergei Ivanov Apr 2 '10 at 18:50
    
Oops, sorry. I didn't see the "=1" (even though I was suspecting that it should be there!), because the jsMath formula extended too far out on the right into the rest of the sentence so that it was overwritten by the word "and". This annoying problem with formulas happens from time to time, and quite unpredictably. For example, the "1" and the "S" in your comment are rendered on top of each other as I'm writing this, although it was shown correctly before I reloaded the page. Has anyone else experienced this too? –  Hans Lundmark Apr 3 '10 at 10:18
    
This annoys me too. My preferred browser (Konqueror) sometimes skips jmath symbols or renders them on top on one another. The only solution I found, however unpleasant, was to use another browser (Firefox). –  Sergei Ivanov Apr 3 '10 at 17:25
    
Well, I'm using Firefox already, so that's not an option for me! Maybe I should try Konqueror. ;) –  Hans Lundmark Apr 4 '10 at 16:15

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