Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is probably very obvious, I was looking at this.

It looks so much like a Cauchy-Schwarz though, and I would say it is very obvious from the definition if it wasn't for the condition that $\|x\| \le 1$ in the definition:

$$\| z \| := \sup_x\{ z^T x : \|x\| \le 1\}.$$

How is that restriction still ok?

share|cite|improve this question
In a word, scaling. – Robert Israel Sep 24 '12 at 2:56

1 Answer 1

Using Robert Israel's hint, let $y = \dfrac{x}{\|x\|}$. Note that $\|y\| = 1$ and $x = \|x\|y$, so we have

$$z^Tx = z^T(\|x\|y) = \|x\|(z^Ty) \leq \|x\|\sup_y\{z^Ty \mid \|y\| \leq 1\} = \|x\|\|z\|_*.$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.