Comass norm of a differential form

Question

Let's consider the m-exterior algebra of $\mathbb{R}^n$ $\Lambda_m (\mathbb{R}^n)$ and its dual $\Lambda^m (\mathbb{R}^n) = (\Lambda_m (\mathbb{R}^n))^*$; a basis of $\Lambda^m (\mathbb{R}^n)$ is given by $$\{dx_{i_1} \wedge \dots \wedge dx_{i_m} \, : \, 1 \le i_1 < i_2 < \dots < i_m \le n \}. $$ I set now $m=n-1$. Let's consider then an $(n-1)-$form with compact support, id est $$\omega = \sum_{i=1}^n f_i(x) \, dx_1 \wedge \dots \wedge \widehat{dx_i} \wedge \dots \wedge dx_n$$ where $\widehat{dx_i}$ means "$dx_i$ is missing" and $f_i(x) \in \mathscr{C}^\infty _c(\mathbb{R}^n)$. The comass norm of $\omega$ is defined by $$\|\omega\|_C = \sup_{x \in \mathbb{R}^n} \sup_{|\tau| \le 1, \\ \, \tau \text{ simple}} |\omega (x)(\tau)|,$$ where a element $\tau \in \Lambda_{n-1}(\mathbb{R}^n)$ is called simple if it can be written $\tau = v_1 \wedge \dots \wedge v_{n-1}$ with $v_i \in \mathbb{R}^n$. Notice also that I indicate with $$\omega_1 \wedge \dots \wedge \omega_{n-1} (v_1 \wedge \dots \wedge v_{n-1}) = \text{det} (\omega_i (v_j))$$ the "duality product", where $\omega_i \in (\mathbb{R}^n)^*$ and $v_i \in \mathbb{R}^n$.

I would like to link $\|\omega\|_C$ with/to $\max_i \|f_i\|_{L^\infty (\mathbb{R}^n)}$ but I am stuck; I wonder if it is true that there exist two positive constants $C_1$ and $C_2$ such that $$\max_i \|f_i\|_{L^\infty (\mathbb{R}^n)} \le C_1 \Longleftrightarrow \|\omega\|_C \le C_2.$$I think I have a proof for $\Longrightarrow$ but $\Longleftarrow$ still confuses me.

Might anyone enlighten me? Thanks in advance!

Answer 1

Actually there's something more. An $(n-1)$-vector in $\mathbb{R}^n$ is $\tau=\sum_{i=1}^n \alpha_i \widehat{e_i}$ where $\alpha_i \in \mathbb{R}^n$ and we choose $\{e_1,\dots,e_n\}$ the canonical basis in $\mathbb{R^n}$ and indicate $\widehat{e_i}=e_1 \wedge \dots\wedge e_{i-1}\wedge e_{i+1} \wedge \dots \wedge e_n$. Moreover $|\tau|=|\alpha|$, where $\alpha=(\alpha_1,\dots,\alpha_n) \in \mathbb{R}^n$. It's easy to see that now $\omega(x)(\tau)=(f(x),\alpha)$, where $f=(f_1,\dots,f_n)$ and $(\;\;,\;\;)$ is the standard scalar product in $\mathbb{R}^n$. So: $$ \|\omega\|_C = \sup_{x \in \mathbb{R}^n} \sup_{|\tau| \le 1} |\omega (x)(\tau)| = \sup_{x \in \mathbb{R}^n} \sup_{|\alpha| \le 1} |(f(x),\alpha)| = \sup_{x \in \mathbb{R}^n} |f(x)| = \vert\vert f\vert\vert_{\infty}. $$ As a matter of fact there's an isometrical identification between $\Lambda_{n-1}(\mathbb{R}^n)$ and $\mathbb{R}^n$ itself, once you choose the canonical (or also any) basis, and the duality with $\Lambda^{n-1} (\mathbb{R}^n)$ becomes just the scalar product, and an $(n-1)$-form "is" just a vector field, and the comass coherently becomes the sup-norm.

Answer 2

Since you have problems with $\Longleftarrow$, I'll show only this part. If you need also the converse I'll add it.
For $i\in\{1,..,n\}$ choose $\tau_i$ of the form $$e_1\wedge \dots \wedge \widehat{e_i} \wedge \dots \wedge e_n,$$ where $\{e_i,\ \ i=1,..,n\}$ is the standard basis of $\mathbb{R}^n$. Now, by definition, the comass of $\omega$ will be larger than $$\sup_{x\in\mathbb{R}^n}\vert\omega(\tau_i)\vert=\sup_{x\in\mathbb{R}^n}\vert f_i(x)\vert.$$ Since you can repeat the game for every $i$ in $\{1,..,n\}$ you can conclude $$C_2\geq\sup_{x\in\mathbb{R}^n}\vert f_i(x)\vert=\vert\vert f_i\vert\vert_{L^{\infty}(\mathbb{R}^n)} \ ,\ \ \ \ \forall i\in\{1,...,n\}.$$ So you have the result with $C_2$. In general depending on the norm you fix on $\mathbb{R}^n$ you can get a different constant.