This particular language:$$L = \{ u u^R v \,:\, u, v \in \{0, 1\}^+\}$$is giving me a lot of trouble.I highly suspect that its non-regular, considering that $\{ u u^R : u \in \{0, 1\}^+\}$ is non-regular, and I can't seem to reason out a DFA for it.However, trying to achieve a contradiction with the pumping lemma is giving me a lot of trouble. because of the arbitrary $v$ in the construction.In particular, if the string starts with $00$ or $11$ it is automatically in the language because we can pick $v$ as the remainder and the first two characters are trivially the reverse of each other.Issues like this seem to thwart my every attempt at applying the pumping lemma. With $p$ as the pumping length, I tried something like $(01)^p (10)^p 1$ but you can simply pump up the starting character to obtain a string that starts with $00$ (and pumping down also works), so this string doesn't work for contradicting the pumping lemma.I'm pretty stuck on ideas for strings that will contradict the pumping lemma, so I could appreciate a hint on the problem. (Perhaps it is regular? That's still in the back of my mind too)
Proving non-regularity of $u u^R v$?
formal languages;pumping lemma
You can also use another approach: suppose that your language $L$  is regular; then you can intersect it with another regular language: $L_R =  (01)^* (10)^* 1$  and - by the closure properties of regular languages - you get another regular language:$L' = L \cap L_R = \{ (01)^n\,(10)^m1 \mid m \geq n > 0 \}$.Now it is easier to prove that $L'$ is not regular using the pumping lemma (just pick $n = p$, the pumping length), leading to a contradiction.