How to Classify Image Filter as Low Pass Filter (LPF) or High Pass Filter (HPF)? I am currently reading on the topics of image filtering and convolution kernels. While the process of convolution is clear to me, I do not understand, how one arrives  at the statement, that a certain filter is a low or a high pass? 

How can one derive that? For example, if one averages the value of a pixel respecting its surrounding pixels, does that for all pixels, then each pixel would have a value between the minimum and the maximum value of the previous image. It treads low values and high values equally, however literate often states, that it is a low pass filter. 

For that specific case I found this question here. Reading it, however didn't help me, as I seem to lack something to see how these arguments are correct and intuitive. Thus I hope, that a more mathematical approach (can be a general one), would benefit me.
Welcome to DSP Community.

General

We assume 2 types of filters: LPF or HPF.

Classifying Filter Type

Usually, if it is a well planned LPF and well Planned HPF a simple test will do.
Calculate the sum of all coefficients.
The sum of the coefficients is the first element of the DFT of the signal.
It means it is the DC gain and well behaved LPF has gain of 1 and well behaved HPF have DC gain of 0.

So sum the coefficients and:


If the sum is $ 1 $ then this is an LPF.  
If the sum is $ 0 $ then this is an HPF.


Pay attention that those method can't differentiate between HPF and Band Pass or between LPF and Band Stop.

The more general method is to look on its DFT Spectrum to see its behavior.

Converting Filter Type

There is very intuitive formula - ALL_PASS_FILTER = LPF_FILTER + HPF_FILTER.
It's not accurate but this is the idea.

Be more accurate in the Frequency Domain All Pass Filter is a filter which has value of 1 for any frequency.
Assuming we have LPF Filter then HPF_FILTER = ALL_PASS_FILTER - LPF_FILTER.
In the spatial domain it means $ h \left[ n \right] = \delta \left[ n \right] - l \left[ n \right] $.

Where $ \delta \left[ n \right] $ is the filter with 1 in its origin and $ 0 $ anywhere else.

In your case $ \delta \left[ n \right] = [0, 0, 0, 0, 1, 0, 0, 0, 0] $.
So the HPF Filter becomes $ h \left[ n \right] = \delta \left[ n \right] - l \left[ n \right] = [-1/9, -1/9, -1/9, -1/9, 8/9, -1/9, -1/9, -1/9, -1/9] $.