It's easy to bound a volume of a half. For example, the points $(0,0,0),(0,0,1),(0,1,0),(3,0,0)$ can do it. The problem is harder if no two points can be further than 1 apart. Bound a volume of 1/2 with a diameter $\le 1$ point set.
With infinite points at distance 1/2 from the origin, a volume of $\pi/6 = 0.523599...$ can be bound. But we want a finite point set. What is the minimal number of points?
(A 99 point set used to be here. See Answers for a much better 82 point set)
Here's a picture of the hull. Each vertex is numbered. Green vertices have one or more corresponding blue faces with vertices at distance 1. Each blue face has a brown number giving the opposing green vertex. Red vertices and yellow faces lack a face/vertex pairing.
Some may think that Thomson problem solutions might give a better answer. The first diameter 1 Thomson solution with a volume of 1/2 is 121 points with volume .500069.
These points will not fit in a diameter 1 sphere, but the maximal distance between points is less than 1. Similarly, a unit equilateral triangle will not fit in a diameter 1 circle.
Is 99 points minimal for bounding a volume of 1/2 using a point set with diameter 1? Or, to phrase it as a hypothesis:
99 Point Hypothesis
99 points of diameter 1 in Euclidean space.
99 points with a volume of a 1/2.
Take one off, move them around (without increasing diameter)
You can't get a volume of 1/2 any more.
(Update)
My current result is $82$ points:
consider this point set:
Then (Mathematica code)
is $\approx 9.00744\times10^{15}$.
And Mathematica sketch:
Another picture. If all vertices of a face are at distance one from another vertex, the face is colored blue.
Since all point coordinates are integer, then one can write it directly (with arbitrary small computational errors):
$$Diameter = \sqrt{68\;719\;348\;253} \approx 262\;143.\;754\;938;$$ $$Volume = \dfrac{54\;044\;635\;971\;533\;362}{6} \approx 9\;007\;439\;328\;588\;893.\;666\;667.$$
If multiply all coordinates by $\dfrac{1}{2^{18}}$, then we'll get:
$$Diameter = \frac{\sqrt{68\;719\;348\;253}}{262\;144} \approx 0.999\;999\;065;$$ $$Volume = \dfrac{54\;044\;635\;971\;533\;362}{2^{54}\times 6} \approx 0.\;500\;013\;326.$$
Note: when add any point (with real coordinates) rather close to (the center of) any face, one will get the set of $83, 84, ...$ points with described property.