 Solution

The Weierstrass M-test gives a continuous periodic Weierstrass function, since $\sum a^{-k}<\infty$. For $5^m\le n<5^{m+1}$, the given best approximant leaves the tailIts supremum norm is at most the sum of its coefficients, and at $x=0$ all cosine terms equal one, attaining that bound. ConsequentlyPut $\alpha =\ln a/\ln 5$. Then $a^{-m}=(5^m{)}^{-\alpha }$ and $n/5^m\in [1,5)$, giving $E_n(g)\asymp n^{-\alpha }$, with constants depending only on $a$.

For $1<a<5$, part (a) gives $\omega (g,1/n)\le C_a{n}^{-\alpha }$. The first Jackson theorem for periodic approximation, $E_n(g)\le C\omega (g,1/n)$, supplies the matching lower bound. Hence the modulus of continuity has the precise orderAt $a=5$, the inverse theorem for trigonometric approximation gives the upper bound $C\delta \ln (1/\delta )$, and part (c) proves the matching lower bound. Thus the critical Weierstrass modulus isTo pass from scales $1/n$ to all small $\delta$, use monotonicity of the modulus of continuity and adjacent reciprocals, which are comparable to $\delta$.