How to apply the Cauchy-Schwartz inequality on the following inner product?











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Reading a research article, I came across the following statement:



The following function (where $x_i$ and $p_j$ are two vectors in $mathbb{R}^n$ and $mu_{i,j}$ is a constant) - it doesn't matter where it came from, is just an inner product actually:



$sum_{i} mu_{i,j} - left< sum_{i} mu_{i,j} frac{x_i}{||x_i||}, frac{p_j}{||p_j||} right>$



is minimized by using the Cauchy-Schwartz inequality iff



$p_j propto sum_{i} mu_{i,j} frac{x_i}{||x_i||}$.



I want to apply the Cauchy-Schwartz inequality on $left< sum_{i} mu_{i,j} frac{x_i}{||x_i||}, frac{p_j}{||p_j||} right>$ in order to understand where the proportionality thing came from.



So basically the Cauchy-Schwartz inequality states that:
$left< x, yright> le ||x|| cdot ||y||$. But how can I apply this formula on that monstrous term?










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  • 1




    begin{align*}leftlangle sum_imu_{i,j}frac{x_i}{|x_i|},frac{p_j}{|p_j|}rightrangle &= sum_imu_{i,j}leftlanglefrac{x_i}{|x_i|},frac{p_j}{|p_j|}rightrangle \&= sum_imu_{i,j}frac{langle x_i,p_jrangle}{|x_i||p_j|}\&leqsum_imu_{i,j} end{align*} Does that help?
    – user3482749
    Nov 24 at 18:02










  • @user3482749 no, this is actually where the formula I stated came from (like going backwards), and it doesn't use the Cauchy-Schwartz inequality, but thanks for your contribution anyway
    – Hello Lili
    Nov 24 at 18:12










  • It does use the Cauchy-Schwartz inequality: that's precisely what that inequality is.
    – user3482749
    Nov 24 at 18:18










  • @user3482749 I don't think this is the correct answer because as I said, in the article the equations you stated are used to generate the formula in my question. So it's like going backwards.
    – Hello Lili
    Nov 24 at 18:22










  • @user3482749 I can give you a link if this helps: google.com/…
    – Hello Lili
    Nov 24 at 18:24















up vote
0
down vote

favorite












Reading a research article, I came across the following statement:



The following function (where $x_i$ and $p_j$ are two vectors in $mathbb{R}^n$ and $mu_{i,j}$ is a constant) - it doesn't matter where it came from, is just an inner product actually:



$sum_{i} mu_{i,j} - left< sum_{i} mu_{i,j} frac{x_i}{||x_i||}, frac{p_j}{||p_j||} right>$



is minimized by using the Cauchy-Schwartz inequality iff



$p_j propto sum_{i} mu_{i,j} frac{x_i}{||x_i||}$.



I want to apply the Cauchy-Schwartz inequality on $left< sum_{i} mu_{i,j} frac{x_i}{||x_i||}, frac{p_j}{||p_j||} right>$ in order to understand where the proportionality thing came from.



So basically the Cauchy-Schwartz inequality states that:
$left< x, yright> le ||x|| cdot ||y||$. But how can I apply this formula on that monstrous term?










share|cite|improve this question


















  • 1




    begin{align*}leftlangle sum_imu_{i,j}frac{x_i}{|x_i|},frac{p_j}{|p_j|}rightrangle &= sum_imu_{i,j}leftlanglefrac{x_i}{|x_i|},frac{p_j}{|p_j|}rightrangle \&= sum_imu_{i,j}frac{langle x_i,p_jrangle}{|x_i||p_j|}\&leqsum_imu_{i,j} end{align*} Does that help?
    – user3482749
    Nov 24 at 18:02










  • @user3482749 no, this is actually where the formula I stated came from (like going backwards), and it doesn't use the Cauchy-Schwartz inequality, but thanks for your contribution anyway
    – Hello Lili
    Nov 24 at 18:12










  • It does use the Cauchy-Schwartz inequality: that's precisely what that inequality is.
    – user3482749
    Nov 24 at 18:18










  • @user3482749 I don't think this is the correct answer because as I said, in the article the equations you stated are used to generate the formula in my question. So it's like going backwards.
    – Hello Lili
    Nov 24 at 18:22










  • @user3482749 I can give you a link if this helps: google.com/…
    – Hello Lili
    Nov 24 at 18:24













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Reading a research article, I came across the following statement:



The following function (where $x_i$ and $p_j$ are two vectors in $mathbb{R}^n$ and $mu_{i,j}$ is a constant) - it doesn't matter where it came from, is just an inner product actually:



$sum_{i} mu_{i,j} - left< sum_{i} mu_{i,j} frac{x_i}{||x_i||}, frac{p_j}{||p_j||} right>$



is minimized by using the Cauchy-Schwartz inequality iff



$p_j propto sum_{i} mu_{i,j} frac{x_i}{||x_i||}$.



I want to apply the Cauchy-Schwartz inequality on $left< sum_{i} mu_{i,j} frac{x_i}{||x_i||}, frac{p_j}{||p_j||} right>$ in order to understand where the proportionality thing came from.



So basically the Cauchy-Schwartz inequality states that:
$left< x, yright> le ||x|| cdot ||y||$. But how can I apply this formula on that monstrous term?










share|cite|improve this question













Reading a research article, I came across the following statement:



The following function (where $x_i$ and $p_j$ are two vectors in $mathbb{R}^n$ and $mu_{i,j}$ is a constant) - it doesn't matter where it came from, is just an inner product actually:



$sum_{i} mu_{i,j} - left< sum_{i} mu_{i,j} frac{x_i}{||x_i||}, frac{p_j}{||p_j||} right>$



is minimized by using the Cauchy-Schwartz inequality iff



$p_j propto sum_{i} mu_{i,j} frac{x_i}{||x_i||}$.



I want to apply the Cauchy-Schwartz inequality on $left< sum_{i} mu_{i,j} frac{x_i}{||x_i||}, frac{p_j}{||p_j||} right>$ in order to understand where the proportionality thing came from.



So basically the Cauchy-Schwartz inequality states that:
$left< x, yright> le ||x|| cdot ||y||$. But how can I apply this formula on that monstrous term?







vectors cauchy-schwarz-inequality






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share|cite|improve this question











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asked Nov 24 at 17:57









Hello Lili

1104




1104








  • 1




    begin{align*}leftlangle sum_imu_{i,j}frac{x_i}{|x_i|},frac{p_j}{|p_j|}rightrangle &= sum_imu_{i,j}leftlanglefrac{x_i}{|x_i|},frac{p_j}{|p_j|}rightrangle \&= sum_imu_{i,j}frac{langle x_i,p_jrangle}{|x_i||p_j|}\&leqsum_imu_{i,j} end{align*} Does that help?
    – user3482749
    Nov 24 at 18:02










  • @user3482749 no, this is actually where the formula I stated came from (like going backwards), and it doesn't use the Cauchy-Schwartz inequality, but thanks for your contribution anyway
    – Hello Lili
    Nov 24 at 18:12










  • It does use the Cauchy-Schwartz inequality: that's precisely what that inequality is.
    – user3482749
    Nov 24 at 18:18










  • @user3482749 I don't think this is the correct answer because as I said, in the article the equations you stated are used to generate the formula in my question. So it's like going backwards.
    – Hello Lili
    Nov 24 at 18:22










  • @user3482749 I can give you a link if this helps: google.com/…
    – Hello Lili
    Nov 24 at 18:24














  • 1




    begin{align*}leftlangle sum_imu_{i,j}frac{x_i}{|x_i|},frac{p_j}{|p_j|}rightrangle &= sum_imu_{i,j}leftlanglefrac{x_i}{|x_i|},frac{p_j}{|p_j|}rightrangle \&= sum_imu_{i,j}frac{langle x_i,p_jrangle}{|x_i||p_j|}\&leqsum_imu_{i,j} end{align*} Does that help?
    – user3482749
    Nov 24 at 18:02










  • @user3482749 no, this is actually where the formula I stated came from (like going backwards), and it doesn't use the Cauchy-Schwartz inequality, but thanks for your contribution anyway
    – Hello Lili
    Nov 24 at 18:12










  • It does use the Cauchy-Schwartz inequality: that's precisely what that inequality is.
    – user3482749
    Nov 24 at 18:18










  • @user3482749 I don't think this is the correct answer because as I said, in the article the equations you stated are used to generate the formula in my question. So it's like going backwards.
    – Hello Lili
    Nov 24 at 18:22










  • @user3482749 I can give you a link if this helps: google.com/…
    – Hello Lili
    Nov 24 at 18:24








1




1




begin{align*}leftlangle sum_imu_{i,j}frac{x_i}{|x_i|},frac{p_j}{|p_j|}rightrangle &= sum_imu_{i,j}leftlanglefrac{x_i}{|x_i|},frac{p_j}{|p_j|}rightrangle \&= sum_imu_{i,j}frac{langle x_i,p_jrangle}{|x_i||p_j|}\&leqsum_imu_{i,j} end{align*} Does that help?
– user3482749
Nov 24 at 18:02




begin{align*}leftlangle sum_imu_{i,j}frac{x_i}{|x_i|},frac{p_j}{|p_j|}rightrangle &= sum_imu_{i,j}leftlanglefrac{x_i}{|x_i|},frac{p_j}{|p_j|}rightrangle \&= sum_imu_{i,j}frac{langle x_i,p_jrangle}{|x_i||p_j|}\&leqsum_imu_{i,j} end{align*} Does that help?
– user3482749
Nov 24 at 18:02












@user3482749 no, this is actually where the formula I stated came from (like going backwards), and it doesn't use the Cauchy-Schwartz inequality, but thanks for your contribution anyway
– Hello Lili
Nov 24 at 18:12




@user3482749 no, this is actually where the formula I stated came from (like going backwards), and it doesn't use the Cauchy-Schwartz inequality, but thanks for your contribution anyway
– Hello Lili
Nov 24 at 18:12












It does use the Cauchy-Schwartz inequality: that's precisely what that inequality is.
– user3482749
Nov 24 at 18:18




It does use the Cauchy-Schwartz inequality: that's precisely what that inequality is.
– user3482749
Nov 24 at 18:18












@user3482749 I don't think this is the correct answer because as I said, in the article the equations you stated are used to generate the formula in my question. So it's like going backwards.
– Hello Lili
Nov 24 at 18:22




@user3482749 I don't think this is the correct answer because as I said, in the article the equations you stated are used to generate the formula in my question. So it's like going backwards.
– Hello Lili
Nov 24 at 18:22












@user3482749 I can give you a link if this helps: google.com/…
– Hello Lili
Nov 24 at 18:24




@user3482749 I can give you a link if this helps: google.com/…
– Hello Lili
Nov 24 at 18:24










1 Answer
1






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up vote
1
down vote



accepted










Beyond the inequality itself, the full Cauchy-Schwarz theorem states that $|langle u, vrangle|$ is maximized exactly when $u$ and $v$ are parallel or anti-parallel. I.e., when $v propto u$. For your inner product, this is



$$frac{p_j}{|p_j|} propto sum_{i} mu_{i,j} frac{x_i}{||x_i||}$$



Since $1/|p_j|$ is just a scalar multiplier, it can be absorbed into the constant of proportionality, leaving
$$p_j propto sum_{i} mu_{i,j} frac{x_i}{||x_i||}$$






share|cite|improve this answer





















  • Thank you very much!
    – Hello Lili
    Nov 25 at 9:20











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










Beyond the inequality itself, the full Cauchy-Schwarz theorem states that $|langle u, vrangle|$ is maximized exactly when $u$ and $v$ are parallel or anti-parallel. I.e., when $v propto u$. For your inner product, this is



$$frac{p_j}{|p_j|} propto sum_{i} mu_{i,j} frac{x_i}{||x_i||}$$



Since $1/|p_j|$ is just a scalar multiplier, it can be absorbed into the constant of proportionality, leaving
$$p_j propto sum_{i} mu_{i,j} frac{x_i}{||x_i||}$$






share|cite|improve this answer





















  • Thank you very much!
    – Hello Lili
    Nov 25 at 9:20















up vote
1
down vote



accepted










Beyond the inequality itself, the full Cauchy-Schwarz theorem states that $|langle u, vrangle|$ is maximized exactly when $u$ and $v$ are parallel or anti-parallel. I.e., when $v propto u$. For your inner product, this is



$$frac{p_j}{|p_j|} propto sum_{i} mu_{i,j} frac{x_i}{||x_i||}$$



Since $1/|p_j|$ is just a scalar multiplier, it can be absorbed into the constant of proportionality, leaving
$$p_j propto sum_{i} mu_{i,j} frac{x_i}{||x_i||}$$






share|cite|improve this answer





















  • Thank you very much!
    – Hello Lili
    Nov 25 at 9:20













up vote
1
down vote



accepted







up vote
1
down vote



accepted






Beyond the inequality itself, the full Cauchy-Schwarz theorem states that $|langle u, vrangle|$ is maximized exactly when $u$ and $v$ are parallel or anti-parallel. I.e., when $v propto u$. For your inner product, this is



$$frac{p_j}{|p_j|} propto sum_{i} mu_{i,j} frac{x_i}{||x_i||}$$



Since $1/|p_j|$ is just a scalar multiplier, it can be absorbed into the constant of proportionality, leaving
$$p_j propto sum_{i} mu_{i,j} frac{x_i}{||x_i||}$$






share|cite|improve this answer












Beyond the inequality itself, the full Cauchy-Schwarz theorem states that $|langle u, vrangle|$ is maximized exactly when $u$ and $v$ are parallel or anti-parallel. I.e., when $v propto u$. For your inner product, this is



$$frac{p_j}{|p_j|} propto sum_{i} mu_{i,j} frac{x_i}{||x_i||}$$



Since $1/|p_j|$ is just a scalar multiplier, it can be absorbed into the constant of proportionality, leaving
$$p_j propto sum_{i} mu_{i,j} frac{x_i}{||x_i||}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 25 at 4:22









Paul Sinclair

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  • Thank you very much!
    – Hello Lili
    Nov 25 at 9:20


















  • Thank you very much!
    – Hello Lili
    Nov 25 at 9:20
















Thank you very much!
– Hello Lili
Nov 25 at 9:20




Thank you very much!
– Hello Lili
Nov 25 at 9:20


















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