After Reading again the answer to this question and the answer to this question,
I am wondering if the language $L=\{\langle M \rangle | M $is a Turing machine and $\exists$ input $x$ such that in $M(x)$running $M$ doesn't move left $\}$ can be decide as well.
Maybe we can use the same method as in the last question above, and somehow locate a path which will lead us to this conclusion, by examining the transitions and etc.
What do you think?
This is decidable in almost the same way as in the (second) linked question. The issue here is that the Turing machine is allowed to sit still for a while; but at worst on input of length $n$, if the machine has $k$ states, it must get past the initial input within $nk$ steps. Otherwise it would have to see the same symbol in the same cell twice, i.e. get into a loop.
After running past the input, we look again for a cycle in the transition graph, but now one which starts from a blank symbol, possible writes some symbols while sitting still, and then moves right-or writes a blank symbol without moving.