Denote the plain Kolmogorov complexity by $C(x)$.
Let $\phi(t,x)$ be a recursive function and $\lim_{t\to\infty} \phi(t,x) = C(x)$ for all $x$. For each $t$ define $\psi_t(x) := \phi(t,x)$ for all $x$. Then $C$ is the limit of the sequence of functions $\psi_1, \psi_2, \ldots$. Show that for each error $\epsilon$ and all $t$ there are infinitely many $x$ such that $\psi_t(x) - C(x) > \epsilon$.
As $C(x)$ is not recursive, we could not have for example for $\epsilon = \frac{1}{4}$ that $| \phi_t(x) - C(x) | < \varepsilon$ for all $x$, because then $\phi_t(x) = C(x)$ as the functions just attain integer values, but this is not possible if $\phi_t(x)$ is recursive. The above says a bit more, it says that there exists infinitely many such $x$ and that $\psi_t(x)$ is infinitely often by an $\epsilon$ larger then $C(x)$.
Do you have any ideas how to prove this?
