1
$\begingroup$

I have Cauchy-Schwarz inequality in the form

$$\lvert x_1y_1+\dots+x_ny_n\rvert \leq \sqrt{x_1^2+...+x_n^2}\sqrt{y_1^2+...+y_n^2}.\tag{1}$$

But I have also seen the following form:

$$\lvert x_1y_1\rvert +...+ \lvert x_ny_n\rvert \leq \sqrt{x_1^2+...+x_n^2}\sqrt{y_1^2+...+y_n^2}\tag{2}$$

But I don't see why these are equivalent. Isn't it true that in general $$\lvert x_1y_1+\dots+x_ny_n\rvert \neq \lvert x_1y_1\rvert +...+ \lvert x_ny_n\rvert?$$

$\endgroup$
1
$\begingroup$

One can see by triangle inequality that the (2) implies (1).

Now suppose (1) is true. In particular, it is true for any $(y_1,\cdots,y_n)$. Thus, one can change the sign of $y_i$, so that $$ |x_1y_1+\cdots+x_ny_n|=|x_1y_1|+\cdots+|x_ny_n| $$ Since the right hand side of (1) stays the same, one has (2).

The equivalency between (1) and (2) does not imply that in general, $$ |x_1y_1+\cdots+x_ny_n|=|x_1y_1|+\cdots+|x_ny_n|. $$ and thus there is no contradiction.


To see what I mean by "change the sign", it is instructive to look at the case when $n=2$. Suppose for all $x,y\in{\bf R}^2$, $$|x_1y_1+x_2y_2|\leq\sqrt{x_1^2+x_2^2}\sqrt{y_1^2+y_2^2}. $$ Given $a=(a_1,a_2)$ and $b=(b_1,b_2)$, one wants to prove that $$ |a_1b_1|+|a_2b_2|\leq |a|\cdot|b|. $$ If $(a_1b_1)(a_2b_2)\geq 0$, then $$ |a_1b_1|+|a_2b_2|=|a_1b_1+a_2b_2|\leq |a|\cdot|b|. $$ If $(a_1b_1)(a_2b_2)<0$, then $$ |a_1b_1|+|a_2b_2|=|a_1b_1-a_2b_2|\leq |a|\cdot|b|. $$ where we use (1) and the fact that $|(b_1,b_2)|=|(b_1,-b_2)|$.

$\endgroup$
0
$\begingroup$

Yes, they are not equal but mod of sums is less than or equal to sum of mods and hence it follows. So, second implies the first one for sure.

$\endgroup$
  • $\begingroup$ The direction follows from the first sentence should be (2) implies (1). $\endgroup$ – Jack Jun 29 '17 at 13:43
  • $\begingroup$ @Jack Corrected! $\endgroup$ – user251050 Jun 29 '17 at 13:47
0
$\begingroup$

By C-S $$\sqrt{(x_1^2+x_2^2+...+x_n^2)(y_1^2+y_2^2+...+y_n^2)}=$$ $$=\sqrt{(|x_1|^2+|x_2|^2+...+|x_n|^2)(|y_1|^2+|y_2|^2+...+|y_n|^2)}\geq$$ $$\geq|x_1|\cdot|y_1|+|x_2|\cdot|y_2|+...+|x_n|\cdot|y_n|=$$ $$=|x_1y_1|+|x_2y_2|+...+|x_ny_n|.$$ The equality occurs for $$\left(|x_1|,|x_2|,..., |x_n|\right)||\left(|y_1|,|y_2|,..., |y_n|\right)$$. In another hand, by C-S $$\sqrt{(x_1^2+x_2^2+...+x_n^2)(y_1^2+y_2^2+...+y_n^2)}\geq|x_1y_1+x_2y_2+...+x_ny_n|.$$ the equality occurs for $(x_1,x_2,...,x_n)||(y_1,y_2,...,y_n)$.

$\endgroup$

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.