0

$ \forall x, y \in \mathbb{R^n}$ this fact is true:

$$\sum\limits_{i=1}^n |x_iy_i| \le \left(\sum\limits_{i=1}^nx_i^2 \right)^{\frac{1}{2}} \left(\sum\limits_{i=1}^ny_i^2 \right)^{\frac{1}{2}}$$

This inequality is very similar to C.S. but is not C.S.,in particular is easy to see that it implies C.S. so it is a stronger inequality. I asked it beacuse some professors called it Cauchy-Schwarz inequality.

Wikipedia says, given an inner product space $(V, <,>)$ -in our space we consider $\mathbb{R^n}$ with usual inner product- is true that

$$|<x,y>| \le \sqrt{<x,x>}\sqrt{<y,y>}$$

i.e

$$|\sum\limits_{i=1}^n x_iy_i| \le \left(\sum\limits_{i=1}^nx_i^2 \right)^{\frac{1}{2}} \left(\sum\limits_{i=1}^ny_i^2 \right)^{\frac{1}{2}}$$

which is different from the first inequality. In the first i take the sum of absolute values, in C.S. I take absolute value of sum.

ictibones
  • 1,914
  • Why do you think this is not Cauchy-Schwarz? – Did Sep 02 '17 at 08:19
  • 1
    @MANMAID The link shows this is C-S... Sorry to say so but I seem to have met an awful lot of misleading or offtopic or squarely wrong comments from you on the site these last days. Maybe it is time to reconsider the mode of your interventions on mse? – Did Sep 02 '17 at 08:22
  • @Did I missed the word very similar, so I did that comment. – MAN-MADE Sep 02 '17 at 08:30
  • Also if you see the title which says, "Is this Cauchy-Schwarz inequality...", which mislead me actually. Certainly the expression OP gave @Did not match with the one I read, so I made that comment. – MAN-MADE Sep 02 '17 at 08:36

2 Answers2

2

Note that by Cauchy-Schwarz inequality, $\langle u,v\rangle \le \|u\|\|v\|$ for all $u,v\in\mathbb{R}^n$. By taking $u=(|x_1|,\dots,|x_n|)$ and $v=(|y_1|,\dots,|y_n|)$, we obtain $$\sum\limits_{i=1}^n |x_i||y_i| \le \left(\sum\limits_{i=1}^n|x_i|^2 \right)^{\frac{1}{2}} \left(\sum\limits_{i=1}^n|y_i|^2 \right)^{\frac{1}{2}}$$ which is equivalent to your inequality. Hence Cauchy-Schwarz inequality implies your inequality.

Moreover, let $u=(x_1,\dots,x_n)$ and $v=(y_1,\dots,y_n)$, then by your inequality $$\langle u,v\rangle=\sum\limits_{i=1}^n x_iy_i\leq \left|\sum\limits_{i=1}^n x_iy_i\right|\leq \sum\limits_{i=1}^n |x_i||y_i| \leq \left(\sum\limits_{i=1}^nx_i^2 \right)^{\frac{1}{2}} \left(\sum\limits_{i=1}^ny_i^2 \right)^{\frac{1}{2}}=\|u\|\|v\|.$$ Hence your inequality implies Cauchy-Schwarz inequality.

We may conclude that your inequality and Cauchy-Schwarz inequality are equivalent.

Robert Z
  • 145,942
  • This is equivalent to my inequality, but is C.S. defined in this way? – ictibones Sep 02 '17 at 08:31
  • @Leonardo Vannini It is equivalent to CS because $\sum\limits_{i=1}^n |x_i||y_i|\geq |\sum\limits_{i=1}^n x_iy_i|$. – Robert Z Sep 02 '17 at 08:42
  • I can't understand why $||y||*||x|| \ge |x_1y_1+x_2y_2+...+x_ny_n|$ (which is C.S.) implies the first inequality.

    $||y||*||x|| \ge |x_1y_1+x_2y_2+...+x_ny_n| \not\ge |x_1y_1|+|x_2y_2|+...+|x_ny_n|$

     – ictibones Sep 02 '17 at 08:48
  • @Leonardo Vannini You should take $u=(|x_1|,\dots |x_n|)$ and not $u=(x_1,\dots x_n)$. Similarly for $v$. Then use $\langle u,v\rangle \le |u||v|$. – Robert Z Sep 02 '17 at 09:00
  • I see that if I take $u=(|x_1|,...,|x_n)$ it works. But in C.S. there is not this limitation, I can take every n-tuple, or not? – ictibones Sep 02 '17 at 09:04
  • For this reason seems to me that first ineq. is stronger than CS in real case. – ictibones Sep 02 '17 at 09:05
  • @Leonardo Vannini In the proof, CS implies your inequality, we use that restriction. What's the problem? – Robert Z Sep 02 '17 at 09:09
  • Take x=(1,1) and y=(1,-1). Applying CS I find $2 \ge 0$, applying the first inequality I find $2 \ge 2$. Now these results are obvious but in general first inequality is more accurate that CS: you can minore ||x||*||y|| with a greater lower bound so tells me something more than CS. – ictibones Sep 02 '17 at 09:22
  • 1
    I am not saying that CS and your inequality are equivalent for a given $x$ and $y$. I say that the statement $$\forall x,y\in\mathbb{R}^n,\sum_{i=1}^n |x_iy_i|\leq|x||y|$$ is equivalent to the statement $$\forall x,y\in\mathbb{R}^n,\sum_{i=1}^n x_iy_i\leq|x||y|.$$ – Robert Z Sep 02 '17 at 09:38
  • @Leonardo Vannini Any further doubt? – Robert Z Sep 02 '17 at 15:37
  • Now I see that a inequality implies the other, the fact is that I obtain two different bound of the product of the norm. I don't know how to explain in a rigorous way: in the question with stronger I meant that the first give an "optimal" or "more accurate" lower bounding for that product that CS can't give me. – ictibones Sep 02 '17 at 15:49
  • @Leonardo Vannini I see your point but my answer shows that the two statements are formally equivalent. This is why $CS$ sometimes is stated in different ways. – Robert Z Sep 02 '17 at 15:57
2

It is a common variant of the standard C.S. $\langle u,v\rangle \le \|u\|\|v\|$.

The standard C.S. is, for any $u,v\in\Bbb R^n$: $$\sum_{i=1}^n u_iv_i \le \left(\sum_{i=1}^n u_i^2\right)^{1/2}\left(\sum_{i=1}^n v_i^2\right)^{1/2}$$ To obtain your version, just add the restriction $u,v\in\Bbb [0,\infty)^n$.

Vim
  • 13,640