Given a compact convex set $C\subset \mathbb R^d$ (possibly satisfying some TBD regularity conditions), can we show that, for $X$ distributed uniformly (Lebesgue measure) over $C$, we have $\operatorname{Var} X = \mathbb E \lVert X - \mathbb E X\rVert^2 = \Theta((\operatorname{diam} C)^2)$?
This is trivial for $d=1$ and I think I have a good proof sketch for $d=2$ that should generalize to arbitrary dimension but I'm having trouble formalizing it.
EDIT: This proof sketch is wrong. I'm still interested in any proof or counterexample for the original statement.
My proof sketch (probably overcomplicated):
WLOG take $\mathbb E X = 0$. Let $R = \max_{X\in C} \lVert X\rVert$. Observe that $R = \Theta(\operatorname{diam} C)$ since $\lVert x-y\rVert \le \lVert x\rVert + \lVert y\rVert \le 2R$ for all $x, y\in C$. So we we want to show $\mathbb E \lVert X\rVert^2 = \Omega(R^2)$. We can scale so that $R=1$. Then our problem is to show $\mathbb E\lVert X\rVert^2$ is $\Omega(1)$ (ie, bounded away from $0$).
My approach is to consider a small ball around $0$ of radius $\delta$, along with a point $x$ at distance $R$ from $0$ and note that most of the measure of the cone formed by $B_\delta(0)\cap C$ and $x$ is outside $B_\delta(0)$.
Specializing to the two-dimensional case, consider rotating two lines $\ell_1, \ell_2$, one counterclockwise and the other clockwise, about $x$ till they first hit any point in $B_\delta(0)\cap C$. Then the extent of $B_\delta(0)\cap C$ within $B_\delta(0)$ is obviously bounded entirely by $\ell_1$ and $\ell_2$. Further the arc $P$ of $\partial B_\delta(0)$ subtending the angle $\angle\ell_1 x\ell_2$ forms a roughly triangular figure when joined with $\ell_1$ and $\ell_2$. This is easier to understand with a diagram:
Let $\theta$ be the measure of the angle $\angle \ell_1 x\ell_2$. Note that as $\delta$ approaches $0$, $\theta$ also approaches $0$.
The area of $B_\delta(0)\cap C$ is bounded by the area of the figure $P\ell_1Q\ell_2$ (highlighted in red above). For sufficiently small $\delta$, this is $\Theta(\theta\delta^2)$. In that limit the area of figure $\ell_1P\ell_2$ (highlighted in green) is $\Theta(\theta\delta)$. Further it is not hard to see that the expected value of $\lVert X\rVert^2$ in $\ell_1P\ell_2$ is $\Omega(1)$ independent of $\delta$ and $\theta$.
So over all of $C$ must have $\mathbb E\lVert X\rVert^2 \ge \frac{|\ell_1P\ell_2|}{|\ell_1P\ell_2|+|P\ell_1Q\ell_2|}\Omega(1) = \frac1{1+\Theta(\delta)}\Omega(1) = \Omega(1)$ (EDIT: This estimate does not actually follow from the previous results. I forgot to account for measure from $C$ that's outside $B_\delta(0)$ and $\ell_1P\ell_2$ but still not distance $\Omega(1)$ from $0$.), which is what we wanted to prove.