# time complexity – is it true that if \$f(n)in O(g(n))\$ then \$f(h(n)) in O(g(h(n)))\$?

Formally it’s easy to bring counterexample: suppose $$f(1)=1$$, $$g(1)=0$$ and then, for other values of argument, $$f, g$$ are any pair of non-zero functions with property $$f(n)in O(g(n))$$. Now taking $$h(n)=1,forall n in mathbb{N}$$ makes impossible $$f(h(n)) in O(g(h(n)))$$.

On other hand, for example, if $$h$$ is strictly increasing function, then your claim will be true, because we obtain property for subsequence from sequence.