Is the multiplicative constant in the Big O notation are ignored because of Linear Speed-Up theorem?
Please note that the big O notation was invented before the proof of the linear speedup theorem, and even before Turing machines.
Also, the big O notation often gives an information independant of the multiplicative constant: &quot;if I multiply the input by $k$, then the total computing time will not be multiplied by more than …&quot;.
Finally, keep in mind that the linear speedup theorem gives a way to reduce the number of steps in the execution of a Turing Machine, but if you implement it on a real computer, it often also means that each step may be longer, so the real total time may not decrease.
The big O notation is a convenient way to compare asymptotic time complexity, but it is not always sufficient. For example, mergesort have an asymptotic complexity $\Theta(n\log n)$, but considering the multiplicative constant, it is often better to use insertion sort to sort small data, even if the asymptotic complexity of insertion sort is $\Theta(n^2)$ in average. Another example are Fibonacci heaps.