Define the unit ball centered at the origin with radius r as $B(0,r)=\{x \in \mathbb{R}^d:||x||\leq 1\}$
Define the dual set of a set X as $ X^*=\{y \in \mathbb{R}^d: x^Ty \leq 1,\; \forall x \in X \} $
I'm trying to prove that the dual set of $B(0,r)$ is $B(0,\frac{1}{r})$.
What I attempted:
By definition the dual set of $B(0,r)$ is $X^*_B$={ y\in\mathbb{R}^d: x^Ty\leq 1,\; \forall x\in B(0,r) }
First I showed that $B(0,\frac{1}{r})\subset X^*_{B}$ using the Cauchy-Schwarz inequality.
Now, I'm trying to prove that $ X^*_{B}\subset B(0,\frac{1}{r})$
I attempted use the Cauchy Schwarz inequality again as follows: $x^Ty\leq ||x|| ||y||$, then $\frac{x^Ty}{||x||}\leq ||y||$, but this inequality is not useful for me. If someone has a sugestion for prove this part, I will be grateful.
Try proving the contrapositive. Suppose $y \notin B(0, \frac{1}{r})$, and consider $x = r\frac{y}{\|y\|} \in B(x; r)$ (i.e. the point on the sphere which points parallel to $y$). See if you can finish from there.