0
$\begingroup$

I have a nonempty convex set $C \subset \mathbb{R}^n$ with $0 \notin C$. Moreover, the inner of C is nonempty. I know that there exists a unique $ x \in \overline{C}$ with $\Vert x \Vert = d(0,C)$. Now I am supposed to show that for all $\lambda \in (0,1)$ and $y \in C$ such that $y \neq x$ I have \begin{align*} \Vert x \Vert < \Vert (1-\lambda) x + \lambda y\Vert. \end{align*} Unfortunately, I don't know how to obtain this strict inequality.

$\endgroup$
0
$\begingroup$

You said it yourself. There is a unique $x \in \overline C$ with the property that $\|x\| = d(0,C)$. If $y \in C$ and $0 < \lambda < 1$ then $\lambda x + (1-\lambda) y \in \overline C$ too, so that $\| \lambda x + (1-\lambda)y\| \ge d(0,C).$ But $x \not= y$ implies $\lambda x + (1-\lambda)y \not= x$, so you must have $\|\lambda x + (1-\lambda)y\| \not= d(0,C)$.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Consider a function $f(x) =||x||$ on compact set $ A=B(0, R)\cap \overline{C} $ where $B(x,r)=\{y\in\mathbb{R}^n : ||y-x||\leq r\} $ and $R=d(0, C) +1 .$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Ok, but how does this solve my problem? Sorry, I don't see this yet. $\endgroup$ – user559991 Jun 15 '18 at 17:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy