Estimation of greatest speed in a polyhedron in order to control velocities in a three dimensional volume, I look for  a proof or a proof idea for the following assumption:
Given a non-empty solid polyhedron in 3D, all points inside this set have a  speed than  is lower or equal than the maximum of the speed of the vertices of this polyhedron.
This seems to be self-evident, but my inspirations how to show that are ugly and complicated and real messy and therefore unrepresentable. And I need a proof.
The same problem arises with ball and sphere: are the speeds inside  the ball volume lower or equal than the speeds of the sphere?
 A: It seems OP is essentially asking about the speed of a convex combination
$$ {\bf r} ~=~ \sum_{i=1}^N \alpha_i {\bf r}_i,\qquad  \sum_{i=1}^N \alpha_i~=~1, \qquad \alpha_i~\geq~0, $$
in a rigid body. The velocity is then
$$ {\bf v} ~=~ \sum_{i=1}^N \alpha_i {\bf v}_i. $$
Finally the speed is given by the 2-norm
$$\begin{align} |{\bf v}|~\stackrel{\text{triangle ineq.}}{\leq}& \sum_{i=1}^N \alpha_i |{\bf v}_i|\cr~\stackrel{\text{Hölder ineq.}}{\leq}&
 \max(|{\bf v}_1|,\ldots, |{\bf v}_N|)\sum_{i=1}^N \alpha_i\cr
~=~~~~& \max(|{\bf v}_1|,\ldots, |{\bf v}_N|).\end{align} $$
A: In 3D there is an axis of rotation. The point with the highest speed, the point furthest away from this axis.
The general instantaneous motion of 3D body is a rotation about an axis with rotational velocity $\vec{\omega}$ and a parallel translation along the axis $\vec{v}_{\rm axis} = h\, \vec{\omega} $
An arbitrary point called A has translational velocity $$ \vec{v}_A = \vec{v}_{\rm axis} + \vec{\omega} \times (\vec{r}_A - \vec{r}_{\rm axis}) $$
where $\vec{r}_A$ is the location of point A and $\vec{r}_{\rm axis}$ any point along the rotation axis.
The speed of A is $$ v_A = \| \vec{v}_A \| = \sqrt{ \| \vec{v}_{\rm axis} \|^2 + \|  \vec{\omega} \times (\vec{r}_A - \vec{r}_{\rm axis}) \|^2 } $$
If you move A parallel to the axis, then $\vec{\omega} \times (\vec{r}_A - \vec{r}_{\rm axis})$ does not change. Only movement where $(\vec{r}_A - \vec{r}_{\rm axis})$ is perpendicular to $\vec{\omega}$ has an effect in speed.
So you want the $\vec{r}_A$ on the body that has the highest perpendicular distance to the rotation axis.
You can find the location of the rotation axis $\vec{r}_{\rm axis}$, if you know the velocity vector of any point on the body. Say a point located at $\vec{r}_B$ has velocity $\vec{v}_B$. The location on the rotation axis closest to B is
$$ \vec{r}_{\rm axis} = \vec{r}_B + \frac{ \vec{\omega} \times \vec{v}_B}{ \| \vec{\omega} \|^2} $$
Proof of the above comes from the relationship
$$ \vec{v}_B - \vec{v}_{\rm axis} = \vec{\omega} \times (\vec{r}_B - \vec{r}_{\rm axis}) = -\vec{\omega} \times \frac{ \vec{\omega} \times \vec{v}_B}{ \| \vec{\omega} \|^2}$$
and the vector triple product $a\times(b \times c) = b (c \cdot a) -c (b \cdot a)$
$$\vec{v}_B - \vec{v}_{\rm axis} = -\frac{ \vec{\omega} ( \vec{\omega} \cdot \vec{v}_B) - \vec{v}_B ( \vec{\omega}\cdot \vec{\omega}) }{ \| \vec{\omega} \|^2} = \vec{v}_B - \left( \frac{\vec{\omega} \cdot \vec{v}_B}{\| \vec{\omega}\|^2} \right) \vec{\omega} $$
This means the parallel velocity on the axis is
$$ \vec{v}_{\rm axis} = \left( \frac{\vec{\omega} \cdot \vec{v}_B}{\| \vec{\omega}\|^2} \right) \vec{\omega} $$ or the pitch $h =\frac{\vec{\omega} \cdot \vec{v}_B}{\| \vec{\omega}\|^2}$ describing the amount of linear velocity per rotation.
