I'm referring to the proof outlined here (and wikipedia.org): https://proofwiki.org/wiki/Deterministic_Time_Hierarchy_Theorem
In my understanding, if I relaxed the conditions such that $K$ decides the input in time $O(f(m)^3)$, there should not be a contradiction in the final step because an appropriate TM $K$ should really exist.
Under this condition, $N$'s running time should be $O(f(2n+1)^3)$.
Finally, I arrive at the conclusion:
- If $N$ rejects $[N]$ (which we know it does in at most $f(2n+1)^3$ operations):
- By the definition of $N$, $K$ accepts $([N],[N])$.
- Therefore, by the definition of $K$, $([N],[N])\in H_f$.
- Therefore, by the definition of $H_f$, $N$ does accept $[N]$ in $f(n)$ steps.
To me it seems that 1. and 4. would still contradict. What did I do wrong?