$B^n_2 \subset B^n_K$ implies that $\|x\|_K \le \|x\|$ for all $x\in\mathbb{R}^n$

Question

I want to show that

$B^n_2$ is the Euclidean ball of the unit radius in $\mathbb{R}^n$, and $K$ is a centrally symmetric convex body. Prove that $B^n_2 \subset K$ implies that $\|x\|_K \le \|x\|$ for all $x\in\mathbb{R}^n$. Here, $\|x\|_K$ refers to the norm on which $K$ is the unit ball.

Note: In the title of this question, I have loosely used the notation $B^n_K$ to denote the unit ball under the norm $\|\cdot\|_K$ - this notation is by no means standard.

We know that every centrally symmetric convex body can be represented as the unit ball of some norm on $\mathbb{R}^n$, and $\|x\|_K$ is precisely that norm for $K$. A proof can be found here. $\|x\|$ refers to the usual Euclidean norm.

How do I show this? $\|x\|_K$ is basically $$p_K(x) = \inf\{t>0, x\in tK\}$$ i.e. the Minkowski functional, so I believe I need to show $$ \inf\{t>0, x\in tK\} \le \|x\| \text{ for all }x\in\mathbb{R}^n$$

I tried to start off a proof by contradiction, i.e. assume there is some $y\in\mathbb{R}^n$ such that $$ \inf\{t>0, y\in tK\} > \|y\|$$ but I'm not able to take it from here. I'd appreciate any help!

Answer 1

Let $||x||=a$. Then $x\in aB_2^{n}\subset aK$ (since containment is preserved under scaling), so we have $x\in aK$ and thus $||x||_k\le a=||x||$.

Answer 2

Just observe that, for any given $x$ $$ \{t: x\in tB\}\subset\{t: x\in tK\} $$ where $B$ is the Euclidean unit ball (because $tB\subset tK$). Then the inf of the right hand side is less than or equal to the one of the left one.