$L^2$ discrepancy bound for sequences in $[0,1)$ Given a sequence $x_1,x_2,\dots$, let $D_n$ be the $L^2$-norm of the function $f_n$ whose value at $t \in [0,1)$ is $nt$ minus the number of $1 \leq i \leq n$ with $x_i \leq t$.  What can be said about the rate at which $D_n$ must go to infinity, regardless of the choice of $x_1,x_2,\dots$?
That is, what theorem lower-bounds the $L^2$ norm of $f_n$ in analogy with the way Schmidt’s discrepancy theorem (see http://mathworld.wolfram.com/DiscrepancyTheorem.html) lower-bounds the $L^\infty$ norm of $f_n$?
 A: This problem has been addressed in work of Roth and Davenport.  Roth showed that for any sequence there must be $n$ with $D_n$ larger than a constant times $\sqrt{\log n}$, and Davenport constructed sequences for which $D_n$ grows like at most a constant times $\sqrt{\log n}$.  
More precisely, for any set ${\mathcal P}$of $N$ points in $[0,1)^2$, Roth showed that 
$$ 
\int_{\alpha,\beta=0}^{1} \Big| |{\mathcal P}\cap [0,\alpha)\times [0,\beta)| - N\alpha\beta \Big|^2 d\alpha d\beta \gg \log N. 
$$ 
Apply this to the points $(n/N,x_n)$ with $n=1$, $\ldots$, $N$.  It then follows that 
$$ 
\frac{1}{N} \sum_{n=1}^{N} D_n^2 \gg \log N,
$$ 
which proves the lower bound for $D_n$.  
As for the upper bound, Davenport showed that Roth's result above is best possible by looking at the set $(n/N,\{n \alpha\})$ where $\alpha$ is an irrational number with bounded partial quotients (e.g. $\alpha=\sqrt{2}$).  If you look at Davenport's argument (see page 133 of the paper), he really shows that for this sequence (i.e. $x_n=\{n\sqrt{2}\}$), the $L^2$ discrepancy is of size $\sqrt{\log n}$.  
Roth's paper is in Mathematika vol 1, 1954, and Davenport's in Mathematika vol 3, 1956.  
