Proving of regular language
L is not a regular language.
We can contradict by pumping lemma for regular languages.
Let assume L is regular, so there exists an integer $ k $ of pummping lemma.
$ w = a^kb $
$ w^R = ba^k $
$ ww^R = a^kbba^k \in L $
$ w' = a^{k-r}bba^k \notin L , r&gt;0 $
Contradiction to $ w' \in L $.