Suppose a convex compact subset $X\subset \mathbb R^n$ is the intersection of all closed halfspaces that contain it, then with the inner product and the standard norm $\|.\|_2$,

$$X=\bigcap_{f\in \mathbb R^n, \|f\|_2\leq 1} \{ x\in \mathbb R^n : \langle f, x\rangle\leq \sup_{x\in X} \langle f,x\rangle\}$$

Since $f$ is used as a normal vector and only direction matters, can I replace $\|f\|_2\leq 1$ above equivalently by

$$\|f\|_{\infty}\leq 1$$

where $\|f\|_{\infty}=\max_{1\leq i\leq n} |f_i|$?



Yes. It is important to note that the set on the right-hand side (within the intersection) does not depend on the length of the vector $f$. And for every vector $f \in \mathbb{R}^n \setminus \{0\}$, you have $$ \Bigg\| \frac{f}{\|f\|_2} \Bigg\|_2 = 1$$ and $$ \Bigg\| \frac{f}{\|f\|_\infty} \Bigg\|_\infty = 1.$$


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.