What is the range of the real discrete inverse fourier transform? And would the range be the same for both 1D and 2D version of the transform?
To state it in a different manner if we know the {min, max, length} of $X$, do we know the range of $\mathbf{F}^{-1}(X)$?
Let's denote the input vector by $(X_n)_{n=0}^{N-1}$. I assume that the inverse DFT is defined as something like $$\frac{1}{N}\sum_{n=0}^{N-1}X_n e^{2\pi ink/N}$$ By the triangle inequality, we have $$\begin{aligned} \left|\frac{1}{N}\sum_{n=0}^{N-1}X_n e^{2\pi ink/N}\right| &\leq \frac{1}{N}\sum_{n=0}^{N-1} |X_n e^{2\pi ink/N}| \\ &= \frac{1}{N}\sum_{n=0}^{N-1} |X_n| \\ \end{aligned}$$ which shows that the inverse DFT samples are bounded in magnitude by the average of the magnitudes of the input samples. This bound holds for any input vector, and it's the best we can say in general, since equality is achieved when, for example, $$X_n = \begin{cases} 1 & n = 0 \\ 0 & n \neq 0 \\ \end{cases}$$