Suppose that $K$ is a field. Then for all $n$, define a bilinear operation $*$ (or $*_{n,K}$ in case there may be ambiguity) on $K^{2^n}$ along with a conjugation operation $^*$ on $K^{2^n}$ by setting $x*y=xy$ and $x^*=x$ for $x,y\in K^1$ (i.e. if $n=0$, then this is just standard multiplication) and where $*$ is defined recursively by setting $(x,y)^*=(x^*,-y)$ and $$(w,x)*(y,z)=(wy-z^*x,zw+xy^*).$$
This is the Cayley-Dickson construction that produces the complex numbers ($n=1$), quaternions ($n=2$), and the octonions $(n=3)$ from the real numbers (and when $n=4$ we call the algebra the sedenions but these are not as well-known as the octonions).
Suppose that $U,V,W$ are Banach spaces. Then suppose that $*:U\times V\rightarrow W$ is a bilinear operation. Then define the operator norm of $*$ by setting $\|*\|=\sup\{u*v:\|u\|\leq 1,\|v\|\leq 1\}$. We say that $*$ is bounded if $\|*\|$ is finite. The operation $*$ is bounded if and only if it is continuous.
What are the values of the operator norms $\|*_{n,\mathbb{R}}\|,\|*_{n,\mathbb{C}}\|$ for $n\geq 3$?
Proposition:
$\|*_{n,\mathbb{R}}\|\leq 2^{(n-3)/2}$ for all $n\geq 3$.
$\|*_{n,\mathbb{C}}\|\leq 2^{(n-2)/2}$ for all $n\geq 3$.
Proof: The base case for $1$ is a well-known basic fact about the octonions. For the induction step for $1$, suppose that $w,x,y,z\in\mathbb{R}^{2^n}$ and $\|(w,x)\|\leq 1,\|(y,z)\|\leq 1$. Then $\|(w,x)(y,z)\|^2\leq\|(wy-z^*x,zw+xy^*)\|^2=\|wy-z^*x\|^2+||zw+xy^*||^2$ $$\leq 2(\|wy\|^2+\|z^*x\|^2+\|zw\|^2+\|xy^*\|^2)$$ $$\leq 2(\|w\|\cdot\|y\|^2+\|z^*\|\cdot\|x\|^2+\|z\|\cdot\|w\|^2+\|x\|^2\cdot\|y^*\|^2)\cdot 2^{n-3}$$ $$=2(\|w\|\cdot\|y\|^2+\|z\|\cdot\|x\|^2+\|z\|\cdot\|w\|^2+\|x\|^2\cdot\|y\|^2)\cdot 2^{n-3}$$ $$=2\cdot (\|w\|^2+\|x\|^2)\cdot(\|y\|^2+\|z\|^2)\cdot 2^{n-3}$$ $$=2^{n-2}(\|(w,x)\|^2\cdot\|(y,z\|^2).$$ The base case and the induction step for the complex numbers are similar. Q.E.D.
I am wondering whether the reverse inequality in our proposition holds. Is $\|*_{n,\mathbb{R}}\|=2^{(n-3)/2}$ and $\|*_{n,\mathbb{C}}\|=2^{(n-2)/2}$ for all $n\geq 3$?
A motivation for this question should be clear. It is well-known that $\|x*y\|=\|x\|\cdot\|y\|$ whenever $x,y$ are octonions, so it is natural to ask how well the operation $*_{n,K}$ preserves the norm when $n\geq 3$ and $K$ is either the real or complex numbers since this sort of question helps us determine whether the sedenions are well-behaved or not. My computer calculations indicate that the reverse inequality holds whenever $n$ is small, so I expect the reverse inequality to hold for all $n$.
Analysis of inequalities
Set $$B_n=\{x,y\in\mathbb{R}^{2^n}:\|x\|=\|y\|=1,\|x*_ny\|=2^{(n-3)/2}\}$$ and $$C_n=\{x,y\in\mathbb{C}^{2^n}:\|x\|=\|y\|=1,\|x*_ny\|=2^{(n-2)/2}\}.$$ Set $$B_n^\sharp=B_n\cdot[0,\infty),C_n^\sharp=C_n\cdot[0,\infty).$$ Then we need to determine whether the sets $B_n,C_n$ are always non-empty. We can describe $B_n,C_n$ without explicitly referring to the norm. $B_3=S^7\times S^7.$
Two vectors $u,v$ are on a ray starting at $0$ precisely when $u=\alpha v$ or $v=\alpha u$ for some $\alpha\geq 0$.
Furthermore, $((w,x),(y,z))\in B_{n+1}$ ($((w,x),(y,z))\in C_{n+1}$ respectively) if and only if
$wy,-z^*x$ are both on a ray starting at $0$,
$zw,xy^*$ are both on a ray starting at $0$, and
$(w,y),(z^*,x),(z,w),(x,y^*)\in B_n^\sharp$ ($C_n^\sharp$ respectively).
We also observe that if $a,b,c,d$ are real octonions, and $\|a+bi\|=\|c+di\|=1$, then $(a+bi,c+di)\in C_3$ precisely when $ac,-bd$ are on a ray starting from $0$ and $bc,ad$ are on a ray starting from $0$. We now need to determine whether $C_n,B_n$ are always non-empty.
