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For the vectors $x$ and $y$, the Cauchy–Schwarz inequality reads $$ |x\cdot y|\leq||x||\cdot||y|| $$ Does this inequality only hold for 2-norm? Or for any norms?

Thanks in advance.

Rodrigo de Azevedo
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user143763
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  • In any inner product space $|\langle x, y \rangle| \leqslant |x||y|$. – luka5z Nov 13 '15 at 09:34
  • @luka5z -- Different inner-product spaces will yield different inner products. The question is this: Suppose a norm in $\mathbb{R}^n$ satisfies $\sum_i x_i y_i\leq ||x||\cdot ||y||$ for all $x,y$. Must it be the Euclidean norm? – uniquesolution Nov 13 '15 at 09:39
  • @uniquesolution -- perhaps I am missing something but since \mathcal{R}^n is finite dimensional, norms are equivalent. Now, since the inequality is satisfied by Euclidean norm, it should also hold for other norms? – ewcz Nov 13 '15 at 09:57
  • I suppose you are missing something, because if you take the max-norm for instance and $x=y=(1,1,...,1)$ in $\mathbb{R}^n$ then the inner product is $n$, but the product of the norms is $1$. – uniquesolution Nov 13 '15 at 10:04
  • @uniquesolution Max-norm doesnt come from any inner product...... Your example is irrelevant. A norm induced by inner product is defined as $|x|:=\sqrt{<x,x>}$. The inequality $|\langle x, y \rangle| \leqslant |x||y|$ always holds in any inner product (and thus normed) space. – luka5z Nov 13 '15 at 10:12
  • @luka5z -- I believe we are not reading the same question, and that's the source of confusion. I understand the question as asking whether one can replace the Euclidean norm by any other norm and still have the inequality. How do you understand the question? Also, you are wrong in saying that the inequality holds in any normed space. – uniquesolution Nov 13 '15 at 10:25
  • Will the user please clarify their question, as there seems to be some confusion going on. – uniquesolution Nov 13 '15 at 10:28
  • @uniquesolution I said any inner space, not normed. Any inner space induces normed space (but not the other way around). I agree, user could have formulated his/her question more precisely. There is serious notational problem in his question too. What $\cdot$ means on LHS and what on RHS? – luka5z Nov 13 '15 at 10:37
  • You said "and thus normed", did you not? The l.h.s is $|x\cdot y|$. – uniquesolution Nov 13 '15 at 10:41
  • I did. But it does mean that the CS holds for any norm that is induced by inner product (in which case $|x|:=\sqrt{<x,x>}$). It does not mean it holds for any normed space, since there are norms which are not induced by any inner product, ok? You are confusing things. Please read https://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality – luka5z Nov 13 '15 at 10:42
  • @luka5z Thank you for the wikipedia reference to CS inequality. – uniquesolution Nov 13 '15 at 11:12

2 Answers2

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With finite sequences $x_i$ and $y_i$, $1\le i\le N$ (assumed positive, or add absolute values to the $x_i$s and $y_i$s) you have a generalized Rogers-Hölder's inequality: for $u,v, w>0$ and $1/u+1/v\le 1/w$:

$$\Big(\sum_1^N (x_i y_i)^w\Big)^\frac{1}{w} \le \Big(\sum_1^N (x_i )^u\Big)^\frac{1}{u} \Big(\sum_1^N (y_i )^v\Big)^\frac{1}{v} $$

See for instance P. S. Bullen, Handbook of Means and Their Inequalities, 2003, p. 188. I called them Rogers-Hölder from L. Maligranda, “Why Hölder’s inequality should be called Rogers' inequality? ” Math. Inequal. Appl., vol. 1, no. 1, pp. 69–83, 1998.

You recover your case with $w=1$, and the inequality on $u,v,w$ offers you different options.

Laurent Duval
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The Cauchy-Schwarz inequality is a special case of Hölder's inequality, which reads as follows: $$\left \vert \vec{a} \cdot \vec{b} \right \vert \leq \left \Vert \vec{a} \right\Vert_p \left \Vert \vec{b} \right\Vert_q$$ where $\dfrac1p + \dfrac1q = 1$ and $\Vert \cdot \Vert_s$ is the $s$-norm of the vector.

Daniel Fischer
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Adhvaitha
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  • How does this answer the question? – uniquesolution Nov 13 '15 at 10:04
  • @uniquesolution Doesn't it answer the question? Instead of $2$ norm, we can choose the $p$ norm for one vector and $q$ norm for another vector such that $1/p+1/q = 1$. – Adhvaitha Nov 13 '15 at 10:07
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    No, you can't because the OP asks you whether there is a norm -- one norm -- that you can put on the r.h.s of the inequality while the l.h.s of the inequality remains as it is, with the natural inner product. – uniquesolution Nov 13 '15 at 10:08
  • @uniquesolution The OP says "Or for any norms?" – Adhvaitha Nov 13 '15 at 10:09
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    It is pretty obvious to me he or she do not mean different norms on the r.h.s, but maybe I'm wrong. – uniquesolution Nov 13 '15 at 10:26
  • @uniquesolution OK. In which case, could you remove your down vote? Thanks. – Adhvaitha Nov 13 '15 at 10:32
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    Until proven otherwise, I just don't think you are answering the question at all, sorry. – uniquesolution Nov 13 '15 at 10:36
  • @uniquesolution Your logic is flawed. Proven otherwise is decided by the OP not by someone else. – Adhvaitha Nov 13 '15 at 10:37
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    I believe the question is much more interesting than rendering it trivial by quoting Holder's inequality. It actually asks you to prove or disprove the following: "Let $ ||\cdot|| $ be a norm in $\mathbb{R}^n$ such that the following inequality (the OP's inequality) holds for every $x,y\in\mathbb{R}^n$. Must $||\cdot||$ be the Euclidean norm?" That is what I believe to be the question, and your answer is nowhere near. – uniquesolution Nov 13 '15 at 10:39