Understanding Analytic Signals I know that a function $f_a\in \text{L}^2(\mathrm{R})$ is an analytic signal if its Fourier transform is 0 for negative frequencies, i.e.

$$ \hat{f_a}(\omega)=0$ if $\omega&lt;0 $$

We can characterize an analytic function by its real part $f = \text{Re}[f_a]$, and the Fourier transform of this function is

$$ \hat{f_a}(\omega)=\left\{\begin{matrix}
2\hat{f}(\omega) &amp; \text{if }\omega\geq0 \\ 
0 &amp; \text{if }\omega&lt;0
\end{matrix}\right. $$ 

Now if we have a discrete signal $f[n]$ of size $N$, we can also compute $f_a[n]$ by setting the negative frequency components of its discrete Fourier transform to 0.

Now why is it that if we want $\text{Re}[f_a]=f$, then we must have:

$$ \hat{f_a}[k]=\left\{\begin{matrix}
\hat{f}[k] &amp;&amp; \text{if }k=0,N/2\\   
2\hat{f}[k] &amp;&amp; \text{if }0&lt;k&lt;N/2\\ 
0 &amp;&amp; \text{if } N/2&lt;k&lt;N
\end{matrix}\right. $$

Could that be explained?
It's not clear what you're asking, but I think you want to know why the DFT of a signal has extra conditions for it to real. 

For the continuous case: the 2 is there because we need a function to be symmetric to have a real Fourier transform. 

In the discrete case, the DFT obey the symmtery: 

$$
X_{N-k} = X_{-k} = X_k^* 
$$
i.e. conjugate symmetry (half the values are just complex conjugates of the other half). Thus $X_0$ and $X_{N/2}$ are real numbers whilst the remainder of the DFT is specified by $N/2-1$ complex numbers (hence the 2 in the second line). 

Also, the answers to this question.

offer counterexamples to your first statement - that analytic functions have vanishing negative frequencies.