A question on the linear speedup theorem 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?
I didn't check your working but your conclusion isn't at all absurd.  For every Turing machine that decides&nbsp;$L$, and for every $x\in\Sigma^*$, there is another Turing machine&nbsp;$M_x$ that decides&nbsp;$L$ and that runs for exactly $|x|+1$ steps on input&nbsp;$x$. Note that this is more than twice as fast as your conclusion of $2|x|+4$. This also means that, for all languages&nbsp;$L$ and all strings&nbsp;$x\in\Sigma^*$, $t^L(x)=|x|+1$.

$M_x$ operates as follows. First, it reads the first $|x|+1$ characters of the input.  If these are the string&nbsp;$x$ followed by a blank, it accepts if $x\in L$ and rejects if $x\notin L$; if they are not $x$ plus a blank, it returns to the left-hand end of the tape and does whatever $M$&nbsp;would do. Note that $M_x$&nbsp;does not need to contain any algorithm to decide whether $x\in L$. If $x\in L$, the machine is programmed "If my input is $x$, accept it; otherwise, rewind and do what $M$&nbsp;would do.", and if $x\notin L$, the machine is programmed "If my input is $x$, reject it; otherwise, rewind and do what $M$&nbsp;would do."