I am trying to understand the linear speedup theorem:
Let $L$ be a decidable language. Let $X^L$ be the set of all deterministic Turing machines which decide $L$. For a word $x \in \Sigma^*$ define: $t^L(x) := \min_{M \in X^L} t_M(x)$ where $t_M(x)$ is the time of $M$ on input $x$. Now let's choose some machine $M$ which satisfies $t^L(x) = t_M(x)$. Then for all $A \in X^L$ we have:
$$t_M(x) \le t_A(x)\,.\qquad(**)$$
Now let's apply the linear speedup theorem for the machine $M$: We get a new machine $N$ such that:
$$t_N(x) \le t_M(x)/2 + |x| + 2\,.$$
But by (**) we have: $t_M(x) \le t_N(x)$ hence it follows, that $t_M(x) \le t_N(x) \le t_M(x)/2 + |x| + 2$. From this it follows that $t_M(x) \le 2 |x| + 4$.
But how can this be? It means that, for every word $x \in \Sigma^*$ there exists a deterministic Turing machine $M$ which decides $L$ such that the time of $M$ on input $x$ is at most $2|x| + 4$.
This seems very absurd. What am I doing wrong?