10
$\begingroup$

Let $\| \cdot \|$ be a norm on $\mathbb{R}^n$. The associated dual norm, denoted $\| \cdot \|_*$ is defined as $\| z \|_* = \sup\{ z^{t} x : \| x \| < 1 \}$.

Does someone know how prove that the dual norm of the $\mathcal l_{p}$ norm is the $\mathcal l_{q}$ norm? It's not homework. I've been reading about norms and it was stated without proof in a book. Thanks.

  • 2
    Are you familiar with [Hölder's inequality?](https://en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality)2012-12-27
  • 1
    hi: yes, I know that it says that \sum_{i=1}^{n} x_i y_i <= (\sum_{i=1}^{n} (|x_{i}|^p ))^(1/p) + (\sum_{i=1}^{n} (|y_{i}|^q ))^(1/q) where 1/p + 1/q = 1. In fact, that was the hint in the statement of the theorem. But I still couldn't figure out how to use that to prove it. Thanks.2012-12-27
  • 2
    For other readers: $$\sum_{i=1}^{n} |x_i y_i| \leq \left(\sum_{i=1}^{n} |x_{i}|^p \right)^{1/p} \cdot \left(\sum_{i=1}^{n} |y_{i}|^q \right)^{1/q} \text{ where } \frac{1}{p} + \frac{1}{q} = 1$$2012-12-27
  • 0
    Nice answer by the way @nullUser.2012-12-27

1 Answers 1