If $C$ and $K$ are symmetric convex bodies with $C\subset K$. Is it true that there is another symmetric convex body $B$ with $C \subset B \subset K$?

Question

Let $C$ and $K$ be symmetric convex bodies(i.e. unit balls of some norm in $\mathbb{R}^n$), with $C\subset K$ (strictly). I was wondering if is true that there is a different convex body $B$ such that

$$ C \subset B \subset K. $$

I can visually imagine this kind of picture but I couldn't prove it, I was thinking maybe using that $C$ and $K$ both come from norms, say $\lVert\cdot\rVert_{C}$ and $\lVert\cdot\rVert_{K}$ and try to construct a new norm $\lVert\cdot\rVert_{B}$, but couldn't think how. Can anyone help me?

Answer 1

What about the unit ball connected with $\Vert \cdot\Vert:=\frac12\left(\Vert \cdot\Vert_C+\Vert \cdot\Vert_K\right)$?

Answer 2

I'll give you two choices for $B$.

Let $K^\circ:=\{y\in\mathbb{R}^n: y^\top x\leq 1 \;\forall x\in K\}$ ("polar set of $K$"). For $\|\cdot\|$ a norm, let $\|y\|^\circ:=\sup\{y^\top x:\|x\|\leq 1\}$ ("dual norm").

1. Take $B=\frac{K+C}{2}:=\{\frac{x+y}{2}:x\in K,y\in C\}$. This corresponds to $\|\cdot\|_B=(\frac{1}{2}\|\cdot\|_K^\circ+\frac{1}{2}\|\cdot\|_C^\circ)^\circ$.

2. Take $B=(\frac{K^\circ + C^\circ}{2})^\circ$. This corresponds to $\|\cdot\|_B=\frac{1}{2}\|\cdot\|_K+\frac{1}{2}\|\cdot\|_C$.