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Need a way to identify the anonymous map shown in pmap
pmap -d shows the 1.1GB mapped against an anonymous map.
debian:~# pmap -d 1598 | more
1598:   /usr/bin/X :0 -br -nocursor -auth /tmp/serverauth.O1gWpWvWuP
Address   Kbytes Mode  Offset           Device    Mapping
08048000    1644 r-x-- 0000000000000000 008:00001 Xorg
081e3000      48 rw--- 000000000019b000 008:00001 Xorg
081ef000      44 rw--- 0000000000000000 000:00000   [ anon ]
08231000 1164236 rw--- 0000000000000000 000:00000   [ anon ]
b5422000    3752 rw--- 0000000000000000 000:00000   [ anon ]
b59a1000     384 rw-s- 0000000000000000 000:00004   [ shmid=0x520000 ]

I need a way to identify the owner of address 08231000 which is occupying 1.1GB. Kindly help.

It is not an anonymous process, it is an anonymous memory region. 
All memory regions in the output are "owned" by the process whose pid you've specified on the command line (i.e. 1598). Looks like it is Xorg in this case. 
The fact that this line mentions [ anon ] is because this memory block is anonymously mapped, it is just local memory to this process, in contrast to e.g. the shmid=... block which is shared memory.
To find the process with this pid, just do
 # ps -fq 1598

but you'll likely find it is the process Xorg

Yes, it is an anonymous mem region. I realized it after I posted it and continued my reading on pmap. So, it is Xorg which is owning it locally. But, my original question was why is Xorg needing 1.1GB. My RAM is only 512MB. And I run only one GTK application under IceWM. Actually, my original question is posted in link . Please help me if you can spare some time.

Wow your question at the link was actually way better ;-) And more on-topic over there. So I will answer it there if you don't mind.
