Suppose $K \subset \mathbb{R}^{n}$ is a convex body i.e. a compact convex set with a non-empty interior. Additionally assume that $\mathbf{0} \in K$, where $\mathbf{0}$ is the zero-vector in $\mathbb{R}^{n}$.
Now, a well known concept in convex body theory is the Brightness Function of a convex body $K$ (see cited paper below). In summary, we quote the following from the paper:
The brightness function of K $$ b_{K}(u)=V\left(K \mid u^{\perp}\right) $$ for $u \in S^{n-1}$. Cauchy's projection formula (the case $i=n-1$ of $[10,(\mathrm{A.} 45), \mathrm{p} .408]$ ) states that $$ b_{K}(u)=\frac{1}{2} \int_{S^{n-1}}|u \cdot v| d S(K, v) $$ for $u \in S^{n-1}$
Here $K \mid u^{\perp}$ denotes the (orthogonal) projection of $K$ onto the orthogonal subspace of $\{ u \}$, where $u \in S^{n-1}$ (i.e. a hyperplane in $\mathbb{R}^{n -1}$ passing through the origin).
Based on this I have a few related questions:
- I wanted to check if $K \mid u^{\perp}$ is a convex body in $\mathbb{R}^{n -1}$ i.e. a compact convex set with non-empty interior? I expect it to be, but wasn't sure how to go about proving it (particularly that the non-empty interior is preserved under orthogonal image).
- If we scale $K$ i.e. $K^{'} = \alpha K$, for some $\alpha > 0$, then does $b_{K^{'}}(u) = \alpha^{n - 1} b_{K}(u)$. This is a conjecture, but can it be shown formally using the Cauchy projection formula?
Campi, Stefano; Gardner, Richard J.; Gronchi, Paolo; Kiderlen, Markus, Lightness functions, Adv. Math. 231, No. 6, 3118-3146 (2012). ZBL1261.52002..
This is a partial answer in response to a comment.
First, I want to deal with the question of why the intersection of a ball in $\Bbb{R}^n$ with $u^\perp$ is a ball in $u^\perp$. Basically, the answer is "by definition".
Remember that $u^\perp$ is a subspace of $\Bbb{R}^n$, and it inherits its norm from $\Bbb{R}^n$. Of course, it's isometrically isomorphic to $\Bbb{R}^{n-1}$, but in order for that statement to mean anything, it needs to be a normed linear space in its own right. A subspace is always endowed with norm of the superspace, in that we simply restrict the norm function's domain to the subspace. So, we have $$x \in B_{u^\perp}[y; r] \iff x \in u^\perp \text{ and }\|x - y\| \le r \iff x \in B_{\Bbb{R}^n}[y; r] \cap u^\perp.$$ This holds similarly for open balls, of course!
Now, let's say that we have a convex body $K$. I'm assuming you're happy with projection being both linear and continuous, hence $K \mid u^\perp$ is both convex and compact. Since $K$ has non-empty interior, some there exists some subset of $K$ of the form $B_{\Bbb{R}^n}(y; s)$, where $s > 0$.
Let $K' = K - \langle y, u\rangle u = \{k - \langle y, u \rangle u : k \in K\}$. Then $K'$ is simply a translation of $K$, and hence still a convex body.
Recall that the projection function is linear (again) and that it maps $u$ (and hence all its multiples) to $0$. So, the projection of $k - \langle y, u \rangle$ onto $u^\perp$ is the same as the projection of $k$ onto $u^\perp$, and so $$K' \mid u^\perp = K \mid u^\perp.$$
Now, since $B_{\Bbb{R}^n}(y; s) \subseteq K$, we have $$B_{\Bbb{R}^n}(y - \langle y, u \rangle u; s) = B_{\Bbb{R}^n}(y; s) - \langle y, u \rangle u \subseteq K'.$$ The centre, $y - \langle y, u \rangle u$, lies in $u^\perp$ (you can calculate this for yourself). As discussed before, $$B_{\Bbb{R}^n}(y - \langle y, u \rangle u; s) \cap u^\perp = B_{u^\perp}(y - \langle y, u \rangle u; s),$$ and since the projection onto $u^\perp$ fixes every vector in $u^\perp$, \begin{align*} B_{u^\perp}(y - \langle y, u \rangle u; s) &= B_{u^\perp}(y - \langle y, u \rangle u; s) \mid u^\perp \\ &\subseteq B_{\Bbb{R}^n}(y - \langle y, u \rangle u; s) \mid u^\perp \\ &\subseteq K' \mid u^\perp \\ &= K \mid u^\perp. \end{align*}
I hope that helps.