= 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 $e^{irh}\widehat f(r)=-\widehat f(r)$. Thus
$$
2\widehat f(r)=\frac1{2\pi}\int_{-\pi}^{\pi}(f(t)-f(t+h))e^{-irt}\,dt,\qquad |\widehat f(r)|\leq\frac12\sup_t|f(t)-f(t+\pi/r)|.
$$
The right side tends to zero as $|r|\to\infty$ by <uniform continuity> on the <circle>. Consequently $\boxed{\widehat f(r)\to0\text{ as }|r|\to\infty}$.
There is nevertheless no prescribed rate shared by all continuous functions. Choose increasing positive integers $n_j$ with $k(n_j)\geq j2^j$, which is possible since $k(r)\to\infty$. Define
$$
g(t)=\sum_{j=1}^{\infty}2^{-j}e^{in_jt}.
$$
The series converges absolutely and uniformly, so $g$ is continuous. Termwise integration is justified by <uniform convergence>, and the orthogonality of distinct exponential modes gives $\widehat g(n_j)=2^{-j}$. Hence
$$
k(n_j)|\widehat g(n_j)|\geq j,\qquad \boxed{\limsup_{r\to\infty}k(r)|\widehat g(r)|=\infty.}
$$
This <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.
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