Why do most seismic inversion methods ignore high frequencies? Different post-stack inversion methods such as model based, sparse spike, colored and recursive inversion methods use high cut frequency impedance logs (low frequency) and mid-frequency seismic trace in order to generate an inverted impedance model. Then why are high frequency components are not taken into account during inversion?
Because those frequencies are not present in the seismic data.

Here's an example with some typical numbers:


Seismic sample interval: 4 ms
Therefore Nyquist is 0.5 $\times$ 1/0.004 = 125 Hz
Usually there is a filter on the recorder at 0.8 $\times$ Nyquist = 100 Hz
So the seismic does not contain frequencies beyond 100 Hz


The maximum frequency $f_\text{max}$ of the data determines its resolving power. There are various ways to think about and calculate this; here's one formula (Kallweit &amp; Wood, 1982):

$$\tau_\text{min} = \frac{1}{1.4\,f_\text{max}}$$

So in our example, the thinnest resolvable bed is about 1 / (1.4 $\times$ 100) = 7.1 ms, and it will be thicker than that if there is a lot of noise or the bandwidth is not optimal (the earth attenuates high frequencies more than low ones, so this is an absolute minimum). 

If we assume a P-wave velocity of 2500 m/s, then 7.1 ms two-way time is about 2500 * (7.1/2) = 8.9 m (55 ft). Ordinary wireline logs have a resolution of 0.15 m or 6 in, so there's a substantial discrepancy. I call this 'the integration gap'.

Because the seismic contains no information about geology thinner than this 8.9 m (in this example), not only is it pointless trying to invert for it, but it would be dangerous: anything we thought we'd resolved would be erroneous. This is why it's necessary to filter the logs back to the resolution of the seismic.

Reference

Kallweit, R &amp; L Wood (1982). The limits of resolution of zero-phase wavelets. Geophysics 47 (7), p 1035–1046, DOI:10.1190/1.1441367.