 Solution

For the Riemann-Lebesgue lemma, a half-period shift of the oscillation gives a direct proof. For a nonzero integer $r$, set $h=\pi /r$. Periodicity and a change of variable show that the Fourier coefficient of $t\mapsto f(t+h)$ at $r$ is $e^{irh}\hat{f}(r)=-\hat{f}(r)$. ThusThe right side tends to zero as $|r|\to \infty$ by uniform continuity on the circle. Consequently $\boxed{\hat{f}(r)\to 0\text{ as }|r|\to \infty }$.

There is nevertheless no prescribed rate shared by all continuous functions. Choose increasing positive integers $n_j$ with $k(n_j)\ge j{2}^j$, which is possible since $k(r)\to \infty$. DefineThe series converges absolutely and uniformly, so $g$ is continuous. Termwise integration is justified by uniform convergence, and the orthogonality of distinct exponential modes gives $\hat{g}(n_j)=2^{-j}$. HenceThis arbitrarily slow Fourier coefficient decay is compatible with the Riemann-Lebesgue lemma: for this particular function all Fourier coefficients still tend to zero, but along the selected subsequence they beat the proposed decay scale by an unbounded factor.