essentially reduced to proving the following:

Let $n \geq 1$ be an integer and let $X$ be a random vector with an isotropic, log-concave density in $\RR^n$. Then for all $0 \leq \eps \leq 1$, $$Prob \left \{ \left| \, \frac{|X|}{\sqrt{n}} -1 \, \right| \geq \eps
\right \} \leq C n^{-c \eps^2},$$ where $c, C > 0$ are universal constants. \[cor\_202\]

A significantly superior estimate to that of Theorem \[cor\_202\], for the case where $\eps$ is a certain universal constant greater than one, is given by Paouris [@Pa2], [@Pa3]. It would be interesting to try and improve the bound in Theorem \[cor\_202\] also for smaller values of $\eps$.

Returning to the sketch of the proof, suppose that we are given a random vector $X$ in $\RR^n$ with an isotropic, log-concave density. We need to show that most of its marginals are almost-gaussian. Select a random $k$-dimensional subspace $E \subset \RR^n$, for a certain integer $k$. We use a concentration of measure inequality – in a way similar to Milman’s proof of Dvoretzky’s theorem – to show that with large probability of choosing the subspace $E$, the distribution of the random vector $Proj_E(X)$ is approximately spherically-symmetric. This step is carried out 