Dual set of the unit ball is part of the unit ball.

Question

Define the unit ball centered at the origin as $B=\{x\in\mathbb{R}^d\mid \|x\|\leq 1\}$.

Define the dual set of set $X$ as $X^*=\{y\in\mathbb{R}^d\mid\langle x,y \rangle\leq 1\ \forall x\in X\}$.

I'm attempting to prove that for any set $C\subseteq\mathbb{R}^d$ it holds that $C=B$ if and only if $C=C^*$. I've managed to prove the implication from the right to the left by the Cauchy-Schwarz inequality.

However, I'm having some trouble with the converse implication. In particular, how to prove that $C=B$ implies $C^*\subseteq C$?

Answer 1

Ok, I think I got it. Thanks to @AhmedHussein for the hint.

Take any $y\in B^*$. If $y=0$, then clearly $y\in B$, so assume otherwise. Notice that $\frac{1}{||y||}y\in B$, and thus by the definition of $B^*$ it holds that $\langle\frac{1}{||y||}y, y\rangle\leq 1$.

Extract the coefficient: $\frac{1}{||y||}\langle y, y\rangle\leq 1$. Multiply both sides by $||y||$: $\langle y, y\rangle\leq ||y||$

We can now substitute $||y||^2=\langle y,y \rangle$ and obtain inequality: $||y||^2\leq ||y||$. This is true only if $||y||\leq 1$. Hence, $y$ belongs to $B$.

Answer 2

Hint:

For $y \in B^*$, notice that if $y \neq 0$ then $\frac1{\|y\|} y \in B$.

Answer 3

Suppose $C=C^*$, you have $<x,x>=\|x\|\leq 1$ since $x\in C,x\in C^*$ thus $C\subset B$. This implies $B^*=B\subset C^*=C$ since it is easily shown that $X\subset Y$ implies $Y^*\subset X^*$. done.