I'm trying to find an efficient way to compute maximum of a signal using its DFT. More formally:
$$\max\left\{ \mathcal F^{-1}\left(X_k\right)\right\}, X_k\text{ is the DFT of the signal and } \mathcal F^{-1} \text{ is the IDFT}$$
The original signal $x(n)$ is real, and it has some noise when sometimes there are wide peaks.
I'm looking for a solution that is quicker than the actual IFFT since the signal is very long (but we have its DFT).
This sounds like a good candidate for the Sparse Fast Fourier Transform (sFFT). Have a look at H. Hassanieh, P. Indyk, D. Katabi, and E. Price, sFFT: Sparse Fast Fourier Transform, 2012. I don't know details of the algorithm. You have the domains swapped, which should be fine as the discrete Fourier transform (DFT) and its inverse (IDFT) are nearly identical.
sFFT has been discussed also here on dsp.stackexchange.com.
There is no way to compute the (exact) maximum of the time domain signal directly from its DFT (without doing an IDFT first). Without any further knowledge, the best you can do is
$$|x[n]|=\left|\frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{j2\pi nk/N}\right|\le\frac{1}{N}\sum_{k=0}^{N-1}|X[k]| $$
which gives you an upper bound. However, this upper bound is usually not very tight, especially for large values of the DFT length $N$.
Generally:
There can be no shortcut here – every single element of the (inverse or forward, doesn't matter) DFT needs every element of the input. So, the FFT is already pretty much as fast as you can go.
However, you say:
The original signal $x(n)$ is real, and it has some noise when sometimes there are wide peaks.
Aha! That means that your $X$ is (hermitian) symmetric. Which means you might be able to pick an FFT algorithm that benefits from that knowledge and can omit quite a few calculation – but I don't think that'll change the general $\mathcal O(N \log N)$ complexity of the IFFT, nor the $\mathcal O(N)$ of finding the maximum.
The maximum of $x[n]$, where $X[k] = \texttt{DFT}\{x\}$, is
$$ \begin{align} & \texttt{max}\{x\} = \frac{1}{N}\underset{s\in[0, N-1]}{\texttt{max}}\left\{ \sum_{l=0}^{N-1} (X[k] \cdot e^{-j2\pi s k/N})[l] \right\} \tag{1} \\ & k = [0, 1, ..., N - 1] \end{align} $$
or in code-like syntax,
$$ \texttt{max}\{x\} = \frac{1}{N}\texttt{max}\big\{ \big[\texttt{sum}\{\texttt{FFT}\{x\} \cdot e^{-j2\pi s [0:N-1] / N}\} \ \text{for}\ s=[0:N-1]\big]\big\} \tag{2} $$
The DC bin $ = \texttt{sum}\{x\}$. Per duality, $x[0]$ is the DC bin for the DFT: $x[0] = \texttt{sum}\{X\} / N$, for any $x$, where $1/N$ is the only difference between the domains. Hence, if we shift $\texttt{max}\{x\}$ to index 0, it's retrievable by summing $X$. We don't know where max is ahead of time, so we try all possible shifts $s$, and by definition the max will be the max of all sums (sum at shift $-s$ equals $x[s]$).
$\texttt{min}\{x\}$ is supported, and so are real and complex-valued inputs. In fact any $\texttt{func}\{x\}$ is supported, since we're quite literally retrieving $x$.
This is a very slow algorithm, and I don't know if it is reducible to a satisfactory speed. If we seek to fetch the maximum by working purely in frequency, or understanding the relationship, then it's one way. I found it useful for proving that "time aliasing" cannot cause peaks in convolution.
import numpy as np
for func in (max, min, lambda x: max(abs(x))):
# N==1 works but requires more pure coding related stuff
for N in range(2, 129):
x = np.random.randn(N) + 1j*np.random.randn(N)
xf = fft(x)
exparg = -1j*2*np.pi*np.arange(N)/N
shifts = np.array([sum(xf * np.exp(exparg * s)) for s in range(N)])
out = func(shifts) / N
assert np.allclose(out, func(x))
