I've read that:
$$ var(r) = \frac{1}{T}\sum_{t=0}^{T-1}(r_{t}-\bar{r})^2$$
is equal to:
$$ var(r) = \frac{1}{T^2} \sum_{k=1}^{T-1} R_{k}R_{k}^*$$
with $R_k$ the $T$-point DFT coefficients of $r_t$:
$$ R_{k}= \sum_{t=0}^{T-1}{r_t}e^{-j\frac{2\pi kt}{T}}$$
I tried to play with the formulas to match the equations, but without success. Any hint on the solution?
(let us assume that we work on a centered function (mean = $0$).
Nothing astonishing : the continuous Fourier Transform is an isometry :
$$\|{\frak FT}(r)\|=\|r\|$$
This property is known to be preserved for the (DFT) Discrete Fourier Transform (see Parseval/Plancherel theorems there; morever, if you observe the variance, it is defined as a sum of squares, thus is by itself a norm (in the hermitian sense) ; from there your formula follows.
(don't pay too much attention to coefficient $1/N$ ; it is possible to eliminate it into by tuning the definition of the DFT).