I am studying DFT and am having trouble with the notation system.
The frequency is from $0$ to $2B$ - in DFT the frequency domain does not have negative frequencies. But if this is the case, and we know the signal is real, won't summing up the signal in frequency domain lead to complex valued signal as there is no negative frequency in to make the signal in frequency domain real (via summation).
In order to clarify the notation used in DFT, note that the frequency operation is modulo the sampling frequency $f_s$. From the Nyquist sampling theorem, the maximum frequency component of the signal being sampled must be less than or equal to $f_s/2$.
For an $N$ point DFT, the frequency resolution is $\Delta f=\frac{f_s}{N}$, so the $N$ frequency bins can be labelled $0$,$1$,$\cdots$$,N-1$ in one notation system i.e. where all the frequencies are positive. The notation in the question goes from $0$ to (I am assuming) $2B-1$, where $B$ is the bin corresponding to half the sampling frequency $f_s/2$.
In another notation, the bins are labelled $-(N/2)+1$, $-(N/2)+2$,$\cdots$,$-1$,$0$,$1$,$\cdots$,$(N/2)$ - where you have both positive and negative frequencies.
Both notations are equally valid and have a one-to-one mapping with each other.
Sticking with the all-positive notation $0$,$1$, $N-1$, bins $0$ to $N/2$ correspond to the positive frequencies, where the frequency for the $k^{th}$ bin is $k\Delta f$.
Thus bin $0$ has frequency $0$, bin $1$ has frequency $\Delta f$, bin $2$ has frequency $2\Delta f$, $\cdots$, bin $(N/2)$ has frequency $(N/2)\Delta f$.
However, bins $(N/2)+1$ to $N-1$ correspond to the negative frequencies, where for bin index $k$ the frequency of is $(k-N)\Delta f$.
Thus bin $(N/2)+1$ corresponds to $(-(N/2)+1)\Delta f$, bin $(N/2)+2$ corresponds to $(-(N/2)+2)\Delta f$ and the last bin $N-1$ corresponds to $-\Delta f$ .
