Do there exist x such that K(xx)<K(x), where K is Kolmogorov complextity. Let $K(x)$ denote the Kolmogorov complexity of a string $x$. Do there exist a string such that $K(xx)&lt;K(x)$. (Here $xx$ is the concatenation of $x$ with itself). A similar but different question was asked here, but the counterexample given in the answer to that question doesn't work for this one.
I am no expert on Kolmogorov complexity, but I think that such x can be constructed for every complexity function K as follows.  Since 1, 11, 1111, 11111111, …, 12n, … is an encoding of a natural number n, K(12n) cannot be o(log n).  However, when n=2m, obviously K(12n)=K(122m)=O(log m)=O(log log n).  Therefore the sequence K(1), K(11), K(1111), K(11111111), …, K(12n), … cannot be weakly monotonically increasing, which means that there exists a string x in the form 12n such that K(xx) &lt; K(x).