About Cauchy–Schwarz inequality

Question

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.

Answer 1

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.

Answer 2

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.