Question: 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).

Answer: 
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](http://groups.csail.mit.edu/netmit/sFFT/)**, 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](https://dsp.stackexchange.com/questions/2317/what-is-the-sparse-fourier-transform) 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 $N$.