= Solution
For a <function> on an interval, the <modulus of continuity> is $\omega(f,t)=\sup_{|x-y|\le t}|f(x)-f(y)|$, with the pair restricted to that interval. For a periodic <function> the equivalent definition is $\sup_{\theta,|h|\le t}|g(\theta+h)-g(\theta)|$. The <mean value theorem> and $|\sin\theta|\le1$ give
$$
|\cos(\theta+h)-\cos\theta|\le|h|.
$$
Both cosine values lie in $[-1,1]$. Hence for $|h|\le t$,
$$
|\widetilde f(\theta+h)-\widetilde f(\theta)|
=|f(\cos(\theta+h))-f(\cos\theta)|\le\omega(f,t).
$$
Taking the supremum proves
$$
\boxed{\omega(\widetilde f,t)\le\omega(f,t).}
$$
If the periodic <modulus of continuity> is defined using circular distance instead, choose lifts at that shortest distance and apply the same argument. The <cosine substitution for polynomial approximation> therefore does not enlarge this modulus.
Back to article page
