I am a bit surprised by the following and would like to share it with you. I expect I am mistaken somewhere and will be happy to be corrected.
I have searched StackExchange not only in Mathematics but also in DSP, and found some confirmation about the need of the factor of two that you see below, but not in connection with energy conservation/variation.
Given a real sequence $x_k$ of length $N$, I am reducing it to even length $M=2m$ by means of DFT technique, I am evaluating how much energy I am losing, directly in the (discrete) frequency domain.
I.e. working on the positive frequency samples, I obtain the downsampled spectrum $X_{ds}$ by just truncating the original spectrum $X$ [square brackets denote vector subscripts] (the negative frequency samples being the complex conjugates): $$X_{ds}[i]=X[i], \quad \forall i \quad|\quad 0\le i \le m-1 $$ $$X_{ds}[m]=2 Re(X[m])$$
The reason for the factor of two is to keep the whole amplitude of any component of the original signal that is oscillating exactly at the Nyquist frequency of the new sampling, and in phase. (My intent is not to lose what need not be dropped: the imaginary part is to be dropped as it is in quadrature).
(E.g, if you resample e.g. $x_k=cos(2\pi*4*k/16)$ in the points with even index, you need to have a $1$ at the Nyquist spectrum sample..., where in the original spectrum you had $0.5$ on both sides).
Considering only the Nyquist term, assuming that its amplitude before downsampling was $a_m+i \,b_m$, we have after downsampling $(2a_m)^2=4a_m^2$, where we had $2(a_m^2+b_m^2)$ before, therefore the energy variation is $+2a_m^2-2b_m^2$, which I don't like and I am really not very comfortable with, because the positive term implies unexpected power generation.
Given this result I would rather have a good reason to drop the factor of two or maybe replace it with a factor of $\sqrt2$, so that the energy variation is exactly $-2b_m^2$, but at the moment I don't know how to justify it. I expect that somebody has already worked on this or there is some conventional / accepted decision how to do this.
Of course I understand that, if by previous filtering I zero the Nyquist sample, this problem disappears. But I am not looking for a workaround, but rather for some more fundamental solution/explanation.