Consider $n-$dimensional Euclidean ball centred at 0 with radius $\sqrt{n}$. We want to show that the uniform distribution $X$ in this ball is sub-gaussian and $||X||\_{\psi\_2}<C$ where $C$ is absolute constant. 


Clarify: $X$ is subgaussian if $\langle X,x \rangle$ is subgaussian for any $x \in \mathbb{R}^n$ and $||X||\_{\psi\_2}=||\sup\_x\langle X,x\rangle||\_{\psi\_2}$ where sup is over all unit vector $x$.


Attempt: 
Uniform distribution on ball can be represented by $R,\varphi\_1,..,\varphi\_{n-1}$ jointly where $R$ is a uniform distribution on $[0, \sqrt{n}]$ representing radius, $\varphi\_i$ representing the angles in spherical coordinates and they are uniform on $[0,\pi]$. All these variables are independent. 


By symmetry, I only need to show $||\langle X, (1,0,0,0,...)\rangle||\_{\psi\_2}=||X\_1||\_{\psi\_2}=||R\cos\varphi\_1||\_{\psi\_2}<C$. Then it is not clear to me how to proceed



Uniform distribution in Euclidean ball is sub-gaussian
$\newcommand\Ga\Gamma$
For each unit vector $x$, the random variable (r.v.) $\langle X,x\rangle$ equals
$$V:=\sqrt n\,W\_nR$$
in distribution, where
$$W\_n:=\frac{Z\_1}{\sqrt{Z\_1^2+\dots+Z\_n^2}},$$
$Z\_1,\dots,Z\_n$ are iid $N(0,1)$ r.v.'s, and $R$ is a r.v. (independent of $V$ and) such that $0\le R\le1$.


So, it suffices to show that for some real $c>0$
$$\sup\_{n\ge2}Ee^{cnW\_n^2}<\infty. \tag{1}$$
Note that $W\_n^2$ has the beta distribution with parameters $1/2,(n-1)/2$. So, for any $c\in(0,1/2)$ and $n\ge3$
\begin{align}
Ee^{cnW\_n^2}
&=\frac{\Ga(n/2)}{\sqrt\pi\,\Ga((n-1)/2)}\int\_0^1 e^{cnw^2}w^{-1/2}(1-w)^{(n-3)/2}\,dw \\
&\le\frac{\Ga(n/2)}{\sqrt\pi\,\Ga((n-1)/2)}\int\_0^1 e^{cnw}w^{-1/2}e^{-(n-3)w/2}\,dw \\
&=O(\sqrt n)O(1/\sqrt n)=O(1). 
\end{align}
Also, clearly $Ee^{cnW\_n^2}<\infty$ for $n=2$. Thus, (1) holds, as desired.


