 Solution

Use the uniform bound for harmonic sine polynomialsFor completeness, reduce to $0<t\le \pi$ and split at $K=\min (m,\lfloor 1/t\rfloor )$. The first part is bounded by $Kt\le 1$. Geometric-series summation bounds every interval sum of $\sin kt$ by $1/\sin (t/2)\le \pi /t$. Summation by parts bounds the remaining harmonic-weighted tail by $\pi /[t(K+1)]\le \pi$. Negative $t$ follows by oddness and $t=0$ is immediate.

Let $H_m={\sum }_{k=1}^{m}1/k$. Choose $m_j$ so large that $H_{m_j}\ge C_0{2}^{j+2}$, and positive integers $L_j>m_j$ such that the intervals $[L_j-m_j,L_j+m_j]$ are strictly separated and increase. DefineThe bound $\Vert Q_j{\Vert }_{\infty }\le 2^{-j-2}$ makes this a continuous function with $g(0)=0$.

Each Fourier coefficient of $P_m$ has modulus $1/(2k)$, so the sum of the absolute coefficients is $H_m$. Therefore every prefix of a normalized block $Q_j$ has norm at most one. At any Fourier cutoff, all earlier blocks are complete, at most one block is partial and all later blocks are absent. HenceAt zero, a completed block contributes zero, but its prefix through frequency $L_j$ consists of the negative-frequency sine coefficients and equals $i/2$. ThusBoth index sequences tend to infinity. The uniformly bounded partial sums fail to converge at the origin, even though the function is continuous and zero there.